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1,200 | ac636d2e-6ddd-11ea-a976-ccda262736ce | https://socratic.org/questions/what-volume-is-occupied-by-4-02-10-22-molecules-of-helium-gas-at-stp | 1.50 L | start physical_unit 10 11 volume l qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] helium gas [IN] L"}] | [{"type":"physical unit","value":"1.50 L"}] | [{"type":"physical unit","value":"Number [OF] helium gas molecules [=] \\pu{4.02 × 10^22}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What volume is occupied by #4.02 * 10^22# molecules of helium gas at STP? </h1> | null | 1.50 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In one mole of gas there are Avogadro's number of particles. There are <mathjax>#6.022#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^23#</mathjax> <mathjax>#"particles mol"^-1#</mathjax>.</p>
<p>So how many moles are there in <mathjax>#4.02xx10^22#</mathjax> helium particles?</p>
<p>It is simply the quotient: <mathjax>#(4.02xx10^22" helium particles")/(6.022xx10^23" helium particles mol"^-1 )#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#?? mol#</mathjax> </p>
<p>The volume of course is the prior quotient mulitplied by the molar volume of an ideal gas:</p>
<p><mathjax>#(4.02xx10^22" helium particles")/(6.022xx10^23" helium particles mol"^-1 )#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#22.4*L*mol^-1#</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>The molar volume of an ideal gas at STP is <mathjax>#22.4#</mathjax> <mathjax>#dm^3#</mathjax>, or <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In one mole of gas there are Avogadro's number of particles. There are <mathjax>#6.022#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^23#</mathjax> <mathjax>#"particles mol"^-1#</mathjax>.</p>
<p>So how many moles are there in <mathjax>#4.02xx10^22#</mathjax> helium particles?</p>
<p>It is simply the quotient: <mathjax>#(4.02xx10^22" helium particles")/(6.022xx10^23" helium particles mol"^-1 )#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#?? mol#</mathjax> </p>
<p>The volume of course is the prior quotient mulitplied by the molar volume of an ideal gas:</p>
<p><mathjax>#(4.02xx10^22" helium particles")/(6.022xx10^23" helium particles mol"^-1 )#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#22.4*L*mol^-1#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What volume is occupied by #4.02 * 10^22# molecules of helium gas at STP? </h1>
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<div class="markdown"><p>The molar volume of an ideal gas at STP is <mathjax>#22.4#</mathjax> <mathjax>#dm^3#</mathjax>, or <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In one mole of gas there are Avogadro's number of particles. There are <mathjax>#6.022#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#10^23#</mathjax> <mathjax>#"particles mol"^-1#</mathjax>.</p>
<p>So how many moles are there in <mathjax>#4.02xx10^22#</mathjax> helium particles?</p>
<p>It is simply the quotient: <mathjax>#(4.02xx10^22" helium particles")/(6.022xx10^23" helium particles mol"^-1 )#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#?? mol#</mathjax> </p>
<p>The volume of course is the prior quotient mulitplied by the molar volume of an ideal gas:</p>
<p><mathjax>#(4.02xx10^22" helium particles")/(6.022xx10^23" helium particles mol"^-1 )#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#22.4*L*mol^-1#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??L#</mathjax></p></div>
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</article> | What volume is occupied by #4.02 * 10^22# molecules of helium gas at STP? | null |
1,201 | a9c964ca-6ddd-11ea-9f47-ccda262736ce | https://socratic.org/questions/a-sample-of-helium-has-a-volume-of-325-ml-and-a-pressure-of-655-mm-hg-what-will- | 1703 mmHg | start physical_unit 1 3 pressure mmhg qc_end physical_unit 1 3 14 15 pressure qc_end physical_unit 1 3 8 9 volume qc_end physical_unit 1 3 27 28 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] helium sample [IN] mmHg"}] | [{"type":"physical unit","value":"1703 mmHg"}] | [{"type":"physical unit","value":"Pressure1 [OF] helium sample [=] \\pu{655 mmHg}"},{"type":"physical unit","value":"Volume1 [OF] helium sample [=] \\pu{325 mL}"},{"type":"physical unit","value":"Volume2 [OF] helium sample [=] \\pu{125 mL}"}] | <h1 class="questionTitle" itemprop="name">A sample of helium has a volume of 325 mL and a pressure of 655 mm Hg. What will be the pressure if the helium is compressed to 125 mL? </h1> | null | 1703 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a given amount of gas at constant temperature, varies inversely with the pressure. The equation is <mathjax>#P_1V_1=P_2V_2#</mathjax> </p>
<p><strong>Given/Known</strong><br/>
Pressure 1: <mathjax>#"655 mL Hg"#</mathjax><br/>
Volume 1: <mathjax>#"325 mL"#</mathjax><br/>
Volume 2: <mathjax>#125 mL"#</mathjax></p>
<p><strong>Unknown</strong><br/>
Pressure 2: <mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax> and solve.</p>
<p><mathjax>#P_2=(P_1V_1)/V_2#</mathjax></p>
<p><mathjax>#P_2=((655"mm Hg"xx325cancel"mL"))/(125cancel"mL")="1703 mmHg"#</mathjax></p></div>
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<div class="markdown"><p>The pressure at <mathjax>#"125 mL"#</mathjax> will be <mathjax>#"1703 mmHg"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>, which states that the volume of a given amount of gas at constant temperature, varies inversely with the pressure. The equation is <mathjax>#P_1V_1=P_2V_2#</mathjax> </p>
<p><strong>Given/Known</strong><br/>
Pressure 1: <mathjax>#"655 mL Hg"#</mathjax><br/>
Volume 1: <mathjax>#"325 mL"#</mathjax><br/>
Volume 2: <mathjax>#125 mL"#</mathjax></p>
<p><strong>Unknown</strong><br/>
Pressure 2: <mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax> and solve.</p>
<p><mathjax>#P_2=(P_1V_1)/V_2#</mathjax></p>
<p><mathjax>#P_2=((655"mm Hg"xx325cancel"mL"))/(125cancel"mL")="1703 mmHg"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A sample of helium has a volume of 325 mL and a pressure of 655 mm Hg. What will be the pressure if the helium is compressed to 125 mL? </h1>
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<div class="markdown"><p>The pressure at <mathjax>#"125 mL"#</mathjax> will be <mathjax>#"1703 mmHg"#</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="http://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 at constant temperature, varies inversely with the pressure. The equation is <mathjax>#P_1V_1=P_2V_2#</mathjax> </p>
<p><strong>Given/Known</strong><br/>
Pressure 1: <mathjax>#"655 mL Hg"#</mathjax><br/>
Volume 1: <mathjax>#"325 mL"#</mathjax><br/>
Volume 2: <mathjax>#125 mL"#</mathjax></p>
<p><strong>Unknown</strong><br/>
Pressure 2: <mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax> and solve.</p>
<p><mathjax>#P_2=(P_1V_1)/V_2#</mathjax></p>
<p><mathjax>#P_2=((655"mm Hg"xx325cancel"mL"))/(125cancel"mL")="1703 mmHg"#</mathjax></p></div>
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</article> | A sample of helium has a volume of 325 mL and a pressure of 655 mm Hg. What will be the pressure if the helium is compressed to 125 mL? | null |
1,202 | ac703eda-6ddd-11ea-872e-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-enthalpy-of-vaporization-for-1-mole-of-water | 40.68 kJ | start physical_unit 11 11 enthalpy_of_vaporization kj qc_end physical_unit 11 11 8 9 mole qc_end end | [{"type":"physical unit","value":"Enthalpy of vaporization [OF] water [IN] kJ"}] | [{"type":"physical unit","value":"40.68 kJ"}] | [{"type":"physical unit","value":"Mole [OF] water [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">How do you calculate enthalpy of vaporization for 1 mole of water?</h1> | null | 40.68 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Enthalpy or Heat of vaporization is given by;</p>
<p><mathjax>#Q = mH_v#</mathjax></p>
<p>Where;</p>
<p><mathjax>#H_v = 2260 Jg^-1 ("water at" 100^o C)#</mathjax></p>
<p><mathjax>#Q = "Enthalpy or Heat of vaporization"#</mathjax></p>
<p><mathjax>#m = "mass"#</mathjax></p>
<p>Now we are given <mathjax>#1#</mathjax> mole of water..</p>
<p>Recall;</p>
<p><mathjax>#"No of moles" = "mass"/"molar mass"#</mathjax></p>
<p><mathjax>#n = m/(Mm)#</mathjax></p>
<p><mathjax>#:. m = n xx Mm#</mathjax></p>
<p><mathjax>#n = 1"moles"#</mathjax></p>
<p><mathjax>#Mm color(white)x of color(white)x H_2O = (1 xx 2) + 16 = 2 + 16 = 16gmol^-1#</mathjax> </p>
<p>Hence;</p>
<p><mathjax>#m = 1cancel(mol) xx 18gcancel(mol^-1)#</mathjax></p>
<p><mathjax>#m = 18g#</mathjax></p>
<p>Now plugging it into the formula;</p>
<p><mathjax>#Q = 18g xx 2260 Jg^-1#</mathjax></p>
<p><mathjax>#Q = 18cancelg xx 2260 Jcancel(g^-1)#</mathjax></p>
<p><mathjax>#Q = 40,680J or 40.68KJ#</mathjax></p>
<p>Therefore the enthalpy of vaporization is <mathjax>#+40,680J or +40.68KJ#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization is <mathjax>#+40,680J or +40.68KJ#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Enthalpy or Heat of vaporization is given by;</p>
<p><mathjax>#Q = mH_v#</mathjax></p>
<p>Where;</p>
<p><mathjax>#H_v = 2260 Jg^-1 ("water at" 100^o C)#</mathjax></p>
<p><mathjax>#Q = "Enthalpy or Heat of vaporization"#</mathjax></p>
<p><mathjax>#m = "mass"#</mathjax></p>
<p>Now we are given <mathjax>#1#</mathjax> mole of water..</p>
<p>Recall;</p>
<p><mathjax>#"No of moles" = "mass"/"molar mass"#</mathjax></p>
<p><mathjax>#n = m/(Mm)#</mathjax></p>
<p><mathjax>#:. m = n xx Mm#</mathjax></p>
<p><mathjax>#n = 1"moles"#</mathjax></p>
<p><mathjax>#Mm color(white)x of color(white)x H_2O = (1 xx 2) + 16 = 2 + 16 = 16gmol^-1#</mathjax> </p>
<p>Hence;</p>
<p><mathjax>#m = 1cancel(mol) xx 18gcancel(mol^-1)#</mathjax></p>
<p><mathjax>#m = 18g#</mathjax></p>
<p>Now plugging it into the formula;</p>
<p><mathjax>#Q = 18g xx 2260 Jg^-1#</mathjax></p>
<p><mathjax>#Q = 18cancelg xx 2260 Jcancel(g^-1)#</mathjax></p>
<p><mathjax>#Q = 40,680J or 40.68KJ#</mathjax></p>
<p>Therefore the enthalpy of vaporization is <mathjax>#+40,680J or +40.68KJ#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How do you calculate enthalpy of vaporization for 1 mole of water?</h1>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> of vaporization is <mathjax>#+40,680J or +40.68KJ#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Enthalpy or Heat of vaporization is given by;</p>
<p><mathjax>#Q = mH_v#</mathjax></p>
<p>Where;</p>
<p><mathjax>#H_v = 2260 Jg^-1 ("water at" 100^o C)#</mathjax></p>
<p><mathjax>#Q = "Enthalpy or Heat of vaporization"#</mathjax></p>
<p><mathjax>#m = "mass"#</mathjax></p>
<p>Now we are given <mathjax>#1#</mathjax> mole of water..</p>
<p>Recall;</p>
<p><mathjax>#"No of moles" = "mass"/"molar mass"#</mathjax></p>
<p><mathjax>#n = m/(Mm)#</mathjax></p>
<p><mathjax>#:. m = n xx Mm#</mathjax></p>
<p><mathjax>#n = 1"moles"#</mathjax></p>
<p><mathjax>#Mm color(white)x of color(white)x H_2O = (1 xx 2) + 16 = 2 + 16 = 16gmol^-1#</mathjax> </p>
<p>Hence;</p>
<p><mathjax>#m = 1cancel(mol) xx 18gcancel(mol^-1)#</mathjax></p>
<p><mathjax>#m = 18g#</mathjax></p>
<p>Now plugging it into the formula;</p>
<p><mathjax>#Q = 18g xx 2260 Jg^-1#</mathjax></p>
<p><mathjax>#Q = 18cancelg xx 2260 Jcancel(g^-1)#</mathjax></p>
<p><mathjax>#Q = 40,680J or 40.68KJ#</mathjax></p>
<p>Therefore the enthalpy of vaporization is <mathjax>#+40,680J or +40.68KJ#</mathjax></p></div>
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</article> | How do you calculate enthalpy of vaporization for 1 mole of water? | null |
1,203 | abefe55d-6ddd-11ea-9d3d-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-of-sodium-ions-in-a-solution-obtained-by-dissolving-3- | 0.24 M | start physical_unit 5 6 concentration mol/l qc_end physical_unit 18 18 13 14 mass qc_end physical_unit 9 9 24 25 volume qc_end c_other OTHER qc_end physical_unit 18 18 31 32 molar_mass qc_end end | [{"type":"physical unit","value":"Concentration [OF] sodium ions [IN] M"}] | [{"type":"physical unit","value":"0.24 M"}] | [{"type":"physical unit","value":"Mass [OF] Na3PO4 [=] \\pu{3.25 g}"},{"type":"physical unit","value":"Valume [OF] solution [=] \\pu{250.0 mL}"},{"type":"other","value":"Enough water."},{"type":"physical unit","value":"Molar mass [OF] Na3PO4 [=] \\pu{163.9 g/mol}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of sodium ions in a solution obtained by dissolving #"3.25 g"# of sodium phosphate, #"Na"_3"PO"_4#, in enough water to make #"250.0 mL"# of solution?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Molar mass of <mathjax>#"Na"_3"PO"_4: "163.9 g/mol"#</mathjax></p></div>
</h2>
</div>
</div> | 0.24 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to use the fact that <strong>all the moles</strong> of sodium phosphate that you dissolve to make this solution will <strong>dissociate</strong> to produce sodium cations to calculate the concentration of the sodium cations. </p>
<blockquote>
<p><mathjax>#"Na"_ 3"PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO"_ (4(aq))^(3-)#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of sodium phosphate to calculate the number of moles of salt used to make this solution. </p>
<blockquote>
<p><mathjax>#3.25 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_3"PO"_4)/(163.9color(red)(cancel(color(black)("g")))) = "0.01983 moles Na"_3"PO"_4#</mathjax></p>
</blockquote>
<p>Now, notice that <strong>every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of sodium phosphate that you dissolve in water dissociates to produce <mathjax>#3#</mathjax> <strong>moles</strong> of sodium cations in aqueous solution. </p>
<p>This means that the number of moles of sodium cations present in this solution will be</p>
<blockquote>
<p><mathjax>#0.01983 color(red)(cancel(color(black)("moles Na"_3"PO"_4))) * "3 moles Na"^(+)/(1color(red)(cancel(color(black)("mole Na"_3"PO"_4)))) = "0.05949 moles Na"^(+)#</mathjax></p>
</blockquote>
<p>Finally, to figure out the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the sodium cations, you need to figure out the number of moles of sodium cations present in <mathjax>#10^3"mL" = "1 L"#</mathjax> of this solution. </p>
<p>To do that, you can use the fact that this sample has a volume of <mathjax>#"250.0 mL"#</mathjax>.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.05949 moles Na"^(+)/(250.0color(red)(cancel(color(black)("mL solution")))) = "0.23796 moles Na"^(+)#</mathjax></p>
</blockquote>
<p>You can thus say that the molarity of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 0.238 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of sodium phosphate. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.238 mol L"^(-1)#</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 fact that <strong>all the moles</strong> of sodium phosphate that you dissolve to make this solution will <strong>dissociate</strong> to produce sodium cations to calculate the concentration of the sodium cations. </p>
<blockquote>
<p><mathjax>#"Na"_ 3"PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO"_ (4(aq))^(3-)#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of sodium phosphate to calculate the number of moles of salt used to make this solution. </p>
<blockquote>
<p><mathjax>#3.25 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_3"PO"_4)/(163.9color(red)(cancel(color(black)("g")))) = "0.01983 moles Na"_3"PO"_4#</mathjax></p>
</blockquote>
<p>Now, notice that <strong>every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of sodium phosphate that you dissolve in water dissociates to produce <mathjax>#3#</mathjax> <strong>moles</strong> of sodium cations in aqueous solution. </p>
<p>This means that the number of moles of sodium cations present in this solution will be</p>
<blockquote>
<p><mathjax>#0.01983 color(red)(cancel(color(black)("moles Na"_3"PO"_4))) * "3 moles Na"^(+)/(1color(red)(cancel(color(black)("mole Na"_3"PO"_4)))) = "0.05949 moles Na"^(+)#</mathjax></p>
</blockquote>
<p>Finally, to figure out the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the sodium cations, you need to figure out the number of moles of sodium cations present in <mathjax>#10^3"mL" = "1 L"#</mathjax> of this solution. </p>
<p>To do that, you can use the fact that this sample has a volume of <mathjax>#"250.0 mL"#</mathjax>.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.05949 moles Na"^(+)/(250.0color(red)(cancel(color(black)("mL solution")))) = "0.23796 moles Na"^(+)#</mathjax></p>
</blockquote>
<p>You can thus say that the molarity of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 0.238 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of sodium phosphate. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the concentration of sodium ions in a solution obtained by dissolving #"3.25 g"# of sodium phosphate, #"Na"_3"PO"_4#, in enough water to make #"250.0 mL"# of solution?</h1>
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<div class="markdown"><p>Molar mass of <mathjax>#"Na"_3"PO"_4: "163.9 g/mol"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"0.238 mol L"^(-1)#</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 fact that <strong>all the moles</strong> of sodium phosphate that you dissolve to make this solution will <strong>dissociate</strong> to produce sodium cations to calculate the concentration of the sodium cations. </p>
<blockquote>
<p><mathjax>#"Na"_ 3"PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO"_ (4(aq))^(3-)#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of sodium phosphate to calculate the number of moles of salt used to make this solution. </p>
<blockquote>
<p><mathjax>#3.25 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_3"PO"_4)/(163.9color(red)(cancel(color(black)("g")))) = "0.01983 moles Na"_3"PO"_4#</mathjax></p>
</blockquote>
<p>Now, notice that <strong>every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of sodium phosphate that you dissolve in water dissociates to produce <mathjax>#3#</mathjax> <strong>moles</strong> of sodium cations in aqueous solution. </p>
<p>This means that the number of moles of sodium cations present in this solution will be</p>
<blockquote>
<p><mathjax>#0.01983 color(red)(cancel(color(black)("moles Na"_3"PO"_4))) * "3 moles Na"^(+)/(1color(red)(cancel(color(black)("mole Na"_3"PO"_4)))) = "0.05949 moles Na"^(+)#</mathjax></p>
</blockquote>
<p>Finally, to figure out the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the sodium cations, you need to figure out the number of moles of sodium cations present in <mathjax>#10^3"mL" = "1 L"#</mathjax> of this solution. </p>
<p>To do that, you can use the fact that this sample has a volume of <mathjax>#"250.0 mL"#</mathjax>.</p>
<blockquote>
<p><mathjax>#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.05949 moles Na"^(+)/(250.0color(red)(cancel(color(black)("mL solution")))) = "0.23796 moles Na"^(+)#</mathjax></p>
</blockquote>
<p>You can thus say that the molarity of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 0.238 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of sodium phosphate. </p></div>
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</article> | What is the concentration of sodium ions in a solution obtained by dissolving #"3.25 g"# of sodium phosphate, #"Na"_3"PO"_4#, in enough water to make #"250.0 mL"# of solution? |
Molar mass of #"Na"_3"PO"_4: "163.9 g/mol"#
|
1,204 | a95a6562-6ddd-11ea-a035-ccda262736ce | https://socratic.org/questions/how-many-moles-of-h2o-could-be-obtained-by-reacting-0-75-mole-of-h2o2-in-the-rea | 1.50 moles | start physical_unit 4 4 mole mol qc_end physical_unit 13 13 10 11 mole qc_end chemical_equation 17 24 qc_end end | [{"type":"physical unit","value":"Mole [OF] H2O [IN] moles"}] | [{"type":"physical unit","value":"1.50 moles"}] | [{"type":"physical unit","value":"Mole [OF] H2O2 [=] \\pu{0.75 moles}"},{"type":"chemical equation","value":"H2O2 + H2S -> 2 H2O + S"}] | <h1 class="questionTitle" itemprop="name">How many moles of H2O could be obtained by reacting 0.75 mole of H2O2 in the reaction H2O2+H2S→2H2O+S?</h1> | null | 1.50 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have the stoichiometric reaction:</p>
<p><mathjax>#H_2O_2(l) + H_2S(g) rarr 2H_2O(l) + S(s)darr#</mathjax></p>
<p>Is it balanced? For every reactant particle is there a corresponding product particle? There must be if the reaction represents chemical reality. It is balanced, and you have done the work not me. The reaction tells me UNEQUIVOCALLY that <mathjax>#34*g#</mathjax> of hydrogen peroxide reacts with <mathjax>#34*g#</mathjax> hydrogen sulfide to give <mathjax>#36*g#</mathjax> water and <mathjax>#32*g#</mathjax> sulfur. All of these masses correspond to molar equivalents. Charge and mass are balanced as required. From where did I get these masses? Did I just look them up?</p>
<p>Your starting conditions propose that <mathjax>#0.75*mol#</mathjax> hydrogen peroxide reacts, to give, THEREFORE, <mathjax>#27*g#</mathjax> <mathjax>#H_2O#</mathjax>, and <mathjax>#24*g#</mathjax> sulfur. Do you agree? This is an important principle to master, and if you don't from where we are coming, ask again. </p>
<p>And note that <mathjax>#0.75*mol#</mathjax> <mathjax>#H_2O_2#</mathjax> represents a mass of <mathjax>#0.75*molxx34*g*mol^-1-=25.5*g#</mathjax>........etc........</p></div>
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<div>
<div class="markdown"><p>What does the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> say..........? It says that <mathjax>#1*mol#</mathjax> of hydrogen peroxide and <mathjax>#1*mol#</mathjax> hydrogen sulfide gives <mathjax>#2 *mol#</mathjax> of water and <mathjax>#1*mol#</mathjax> of sulfur.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have the stoichiometric reaction:</p>
<p><mathjax>#H_2O_2(l) + H_2S(g) rarr 2H_2O(l) + S(s)darr#</mathjax></p>
<p>Is it balanced? For every reactant particle is there a corresponding product particle? There must be if the reaction represents chemical reality. It is balanced, and you have done the work not me. The reaction tells me UNEQUIVOCALLY that <mathjax>#34*g#</mathjax> of hydrogen peroxide reacts with <mathjax>#34*g#</mathjax> hydrogen sulfide to give <mathjax>#36*g#</mathjax> water and <mathjax>#32*g#</mathjax> sulfur. All of these masses correspond to molar equivalents. Charge and mass are balanced as required. From where did I get these masses? Did I just look them up?</p>
<p>Your starting conditions propose that <mathjax>#0.75*mol#</mathjax> hydrogen peroxide reacts, to give, THEREFORE, <mathjax>#27*g#</mathjax> <mathjax>#H_2O#</mathjax>, and <mathjax>#24*g#</mathjax> sulfur. Do you agree? This is an important principle to master, and if you don't from where we are coming, ask again. </p>
<p>And note that <mathjax>#0.75*mol#</mathjax> <mathjax>#H_2O_2#</mathjax> represents a mass of <mathjax>#0.75*molxx34*g*mol^-1-=25.5*g#</mathjax>........etc........</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of H2O could be obtained by reacting 0.75 mole of H2O2 in the reaction H2O2+H2S→2H2O+S?</h1>
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<div class="markdown"><p>What does the <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> say..........? It says that <mathjax>#1*mol#</mathjax> of hydrogen peroxide and <mathjax>#1*mol#</mathjax> hydrogen sulfide gives <mathjax>#2 *mol#</mathjax> of water and <mathjax>#1*mol#</mathjax> of sulfur.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You have the stoichiometric reaction:</p>
<p><mathjax>#H_2O_2(l) + H_2S(g) rarr 2H_2O(l) + S(s)darr#</mathjax></p>
<p>Is it balanced? For every reactant particle is there a corresponding product particle? There must be if the reaction represents chemical reality. It is balanced, and you have done the work not me. The reaction tells me UNEQUIVOCALLY that <mathjax>#34*g#</mathjax> of hydrogen peroxide reacts with <mathjax>#34*g#</mathjax> hydrogen sulfide to give <mathjax>#36*g#</mathjax> water and <mathjax>#32*g#</mathjax> sulfur. All of these masses correspond to molar equivalents. Charge and mass are balanced as required. From where did I get these masses? Did I just look them up?</p>
<p>Your starting conditions propose that <mathjax>#0.75*mol#</mathjax> hydrogen peroxide reacts, to give, THEREFORE, <mathjax>#27*g#</mathjax> <mathjax>#H_2O#</mathjax>, and <mathjax>#24*g#</mathjax> sulfur. Do you agree? This is an important principle to master, and if you don't from where we are coming, ask again. </p>
<p>And note that <mathjax>#0.75*mol#</mathjax> <mathjax>#H_2O_2#</mathjax> represents a mass of <mathjax>#0.75*molxx34*g*mol^-1-=25.5*g#</mathjax>........etc........</p></div>
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Peter M.
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<div class="markdown"><p>1.5 moles of water could be obtained.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we always want the balanced chemical equation as this tells us the proportions that the reactants will react in, and how much of the products are formed.</p>
<p><mathjax>#H_2O_2+H_2Srarr2H_2O+S#</mathjax></p>
<p>This equation tells us that for every mole of <mathjax>#H_2O_2#</mathjax>, two moles of water are produced. Therefore:</p>
<p><mathjax>#n(H_2O)=2*n(H_2O_2)=2*0.75=1.5" "mol#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"1.5 mol H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"H"_2"O"_2 + "H"_2"S"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2H"_2"O" + "S"#</mathjax></p>
<p>Multiply the given mol <mathjax>#"H"_2"O"#</mathjax> by <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"H"_2"O"#</mathjax> and <mathjax>#"H"_2"O"_2#</mathjax>.</p>
<p><mathjax>#0.75color(red)cancel(color(black)("mol H"_2"O"_2))xx(2"mol H"_2"O")/(1color(red)cancel(color(black)("mol H"_2"O"_2)))="1.5 mol H"_2"O"#</mathjax></p>
<p>You can reason out the answer without having to do the math. Notice by looking at the balanced equation, that for every one mole of <mathjax>#"H"_2"O"_2"#</mathjax> in the reactants, there are two moles <mathjax>#"H"_2"O"#</mathjax> in the products. So any number of moles of <mathjax>#"H"_2"O"_2"#</mathjax> will produce twice as many moles of <mathjax>#"H"_2"O"#</mathjax>.</p></div>
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</article> | How many moles of H2O could be obtained by reacting 0.75 mole of H2O2 in the reaction H2O2+H2S→2H2O+S? | null |
1,205 | aa69c97a-6ddd-11ea-8737-ccda262736ce | https://socratic.org/questions/what-mass-of-fe-oh-3-is-produced-when-35-ml-of-0-250-m-fe-no-3-3-solution-is-mix | 0.35 grams | start physical_unit 3 3 mass g qc_end physical_unit 13 13 7 8 volume qc_end physical_unit 3 3 10 11 molarity qc_end physical_unit 23 24 17 18 volume qc_end physical_unit 23 24 21 22 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] Fe(OH)3 [IN] grams"}] | [{"type":"physical unit","value":"0.35 grams"}] | [{"type":"physical unit","value":"Volume [OF] Fe(OH)3 solution [=] \\pu{35 mL}"},{"type":"physical unit","value":"Molarity [OF] Fe(OH)3 solution [=] \\pu{0.250 M}"},{"type":"physical unit","value":"Volume [OF] KOH solution [=] \\pu{55 mL}"},{"type":"physical unit","value":"Molarity [OF] KOH solution [=] \\pu{0.180 M}"}] | <h1 class="questionTitle" itemprop="name">What mass of #Fe(OH)_3# is produced when 35 mL of 0.250 M #Fe(NO_3)_3# solution is mixed with 55 mL of a 0.180 M #KOH# solution?</h1> | null | 0.35 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction between iron(III) nitrate, <mathjax>#"Fe"("NO"_3)_3#</mathjax>, and potassium hydroxide, <mathjax>#"KOH"#</mathjax>, will produce iron(III) hydroxide, <mathjax>#"Fe"("OH")_3#</mathjax>, an <strong>insoluble solid</strong> that precipitates out of solution, and aqueous potassium nitrate, <mathjax>#"KNO"_3#</mathjax></p>
<blockquote>
<p><mathjax>#"Fe"("NO"_ 3)_ (3(aq)) + color(red)(3)"KOH"_ ((aq)) -> "Fe"("OH")_ (3(s)) darr + 3"KNO"_ (3(aq))#</mathjax></p>
</blockquote>
<p>The first thin to do here is use the <em>molarities</em> and <em>volumes</em> of the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> to figure out how many <strong>moles</strong> of each reactant you're mixing</p>
<blockquote>
<p><mathjax>#35 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.250 moles Fe"("NO"_3)_3)/(1 color(red)(cancel(color(black)("L solution")))) = "0.00875 moles Fe"("NO"_3)_3#</mathjax></p>
<p><mathjax>#55 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.180 moles KOH")/(1 color(red)(cancel(color(black)("L solution")))) = "0.00990 moles KOH"#</mathjax></p>
</blockquote>
<p>Now, do you have <em>enough moles</em> of potassium hydroxide to allow for <strong>all the moles</strong> of iron(III) nitrate to react? Use the <mathjax>#1:color(red)(3)#</mathjax> <strong>mole ratio</strong> that exists between the two reactants to figure this out</p>
<blockquote>
<p><mathjax>#0.00875 color(red)(cancel(color(black)("moles Fe"("NO"_3)_3))) *(color(red)(3)color(white)(.)"moles KOH")/(1color(red)(cancel(color(black)("mole Fe"("NO"_3)_3)))) = "0.02625 moles KOH"#</mathjax></p>
</blockquote>
<p>As you can see, you'd need <mathjax>#0.02625#</mathjax> <strong>moles</strong> of potassium hydroxide in order for all the moles of iron(III) hydroxide to react. Since you have <strong>fewer</strong> moles of potassium hydroxide than you'd need, potassium hydroxide will act as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>.</p>
<p>This means that only </p>
<blockquote>
<p><mathjax>#0.00990 color(red)(cancel(color(black)("moles KOH"))) * ("1 mole Fe"("NO"_3)_3)/(color(red)(3)color(red)(cancel(color(black)("moles KOH")))) = "0.00330 moles Fe"("NO"_3)_3#</mathjax></p>
</blockquote>
<p>will take part in the reaction. Since iron(III) nitrate and iron(III) hydroxide are in a <mathjax>#1:1#</mathjax> mole ratio in the balanced chemical reaction, you can say that the reaction will produce <mathjax>#0.00330#</mathjax> <strong>moles</strong> of iron(III) hydroxide.</p>
<p>Use the compound's <strong>molar mass</strong> to convert this to <em>grams</em></p>
<blockquote>
<p><mathjax>#0.00330 color(red)(cancel(color(black)("moles Fe"("OH")_3))) * "106.87 g"/(1color(red)(cancel(color(black)("mole Fe"("OH")_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.35 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"0.35 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction between iron(III) nitrate, <mathjax>#"Fe"("NO"_3)_3#</mathjax>, and potassium hydroxide, <mathjax>#"KOH"#</mathjax>, will produce iron(III) hydroxide, <mathjax>#"Fe"("OH")_3#</mathjax>, an <strong>insoluble solid</strong> that precipitates out of solution, and aqueous potassium nitrate, <mathjax>#"KNO"_3#</mathjax></p>
<blockquote>
<p><mathjax>#"Fe"("NO"_ 3)_ (3(aq)) + color(red)(3)"KOH"_ ((aq)) -> "Fe"("OH")_ (3(s)) darr + 3"KNO"_ (3(aq))#</mathjax></p>
</blockquote>
<p>The first thin to do here is use the <em>molarities</em> and <em>volumes</em> of the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> to figure out how many <strong>moles</strong> of each reactant you're mixing</p>
<blockquote>
<p><mathjax>#35 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.250 moles Fe"("NO"_3)_3)/(1 color(red)(cancel(color(black)("L solution")))) = "0.00875 moles Fe"("NO"_3)_3#</mathjax></p>
<p><mathjax>#55 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.180 moles KOH")/(1 color(red)(cancel(color(black)("L solution")))) = "0.00990 moles KOH"#</mathjax></p>
</blockquote>
<p>Now, do you have <em>enough moles</em> of potassium hydroxide to allow for <strong>all the moles</strong> of iron(III) nitrate to react? Use the <mathjax>#1:color(red)(3)#</mathjax> <strong>mole ratio</strong> that exists between the two reactants to figure this out</p>
<blockquote>
<p><mathjax>#0.00875 color(red)(cancel(color(black)("moles Fe"("NO"_3)_3))) *(color(red)(3)color(white)(.)"moles KOH")/(1color(red)(cancel(color(black)("mole Fe"("NO"_3)_3)))) = "0.02625 moles KOH"#</mathjax></p>
</blockquote>
<p>As you can see, you'd need <mathjax>#0.02625#</mathjax> <strong>moles</strong> of potassium hydroxide in order for all the moles of iron(III) hydroxide to react. Since you have <strong>fewer</strong> moles of potassium hydroxide than you'd need, potassium hydroxide will act as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>.</p>
<p>This means that only </p>
<blockquote>
<p><mathjax>#0.00990 color(red)(cancel(color(black)("moles KOH"))) * ("1 mole Fe"("NO"_3)_3)/(color(red)(3)color(red)(cancel(color(black)("moles KOH")))) = "0.00330 moles Fe"("NO"_3)_3#</mathjax></p>
</blockquote>
<p>will take part in the reaction. Since iron(III) nitrate and iron(III) hydroxide are in a <mathjax>#1:1#</mathjax> mole ratio in the balanced chemical reaction, you can say that the reaction will produce <mathjax>#0.00330#</mathjax> <strong>moles</strong> of iron(III) hydroxide.</p>
<p>Use the compound's <strong>molar mass</strong> to convert this to <em>grams</em></p>
<blockquote>
<p><mathjax>#0.00330 color(red)(cancel(color(black)("moles Fe"("OH")_3))) * "106.87 g"/(1color(red)(cancel(color(black)("mole Fe"("OH")_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.35 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">What mass of #Fe(OH)_3# is produced when 35 mL of 0.250 M #Fe(NO_3)_3# solution is mixed with 55 mL of a 0.180 M #KOH# solution?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-08-07T00:34:49" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#"0.35 g"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction between iron(III) nitrate, <mathjax>#"Fe"("NO"_3)_3#</mathjax>, and potassium hydroxide, <mathjax>#"KOH"#</mathjax>, will produce iron(III) hydroxide, <mathjax>#"Fe"("OH")_3#</mathjax>, an <strong>insoluble solid</strong> that precipitates out of solution, and aqueous potassium nitrate, <mathjax>#"KNO"_3#</mathjax></p>
<blockquote>
<p><mathjax>#"Fe"("NO"_ 3)_ (3(aq)) + color(red)(3)"KOH"_ ((aq)) -> "Fe"("OH")_ (3(s)) darr + 3"KNO"_ (3(aq))#</mathjax></p>
</blockquote>
<p>The first thin to do here is use the <em>molarities</em> and <em>volumes</em> of the two <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> to figure out how many <strong>moles</strong> of each reactant you're mixing</p>
<blockquote>
<p><mathjax>#35 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.250 moles Fe"("NO"_3)_3)/(1 color(red)(cancel(color(black)("L solution")))) = "0.00875 moles Fe"("NO"_3)_3#</mathjax></p>
<p><mathjax>#55 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.180 moles KOH")/(1 color(red)(cancel(color(black)("L solution")))) = "0.00990 moles KOH"#</mathjax></p>
</blockquote>
<p>Now, do you have <em>enough moles</em> of potassium hydroxide to allow for <strong>all the moles</strong> of iron(III) nitrate to react? Use the <mathjax>#1:color(red)(3)#</mathjax> <strong>mole ratio</strong> that exists between the two reactants to figure this out</p>
<blockquote>
<p><mathjax>#0.00875 color(red)(cancel(color(black)("moles Fe"("NO"_3)_3))) *(color(red)(3)color(white)(.)"moles KOH")/(1color(red)(cancel(color(black)("mole Fe"("NO"_3)_3)))) = "0.02625 moles KOH"#</mathjax></p>
</blockquote>
<p>As you can see, you'd need <mathjax>#0.02625#</mathjax> <strong>moles</strong> of potassium hydroxide in order for all the moles of iron(III) hydroxide to react. Since you have <strong>fewer</strong> moles of potassium hydroxide than you'd need, potassium hydroxide will act as a <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong>.</p>
<p>This means that only </p>
<blockquote>
<p><mathjax>#0.00990 color(red)(cancel(color(black)("moles KOH"))) * ("1 mole Fe"("NO"_3)_3)/(color(red)(3)color(red)(cancel(color(black)("moles KOH")))) = "0.00330 moles Fe"("NO"_3)_3#</mathjax></p>
</blockquote>
<p>will take part in the reaction. Since iron(III) nitrate and iron(III) hydroxide are in a <mathjax>#1:1#</mathjax> mole ratio in the balanced chemical reaction, you can say that the reaction will produce <mathjax>#0.00330#</mathjax> <strong>moles</strong> of iron(III) hydroxide.</p>
<p>Use the compound's <strong>molar mass</strong> to convert this to <em>grams</em></p>
<blockquote>
<p><mathjax>#0.00330 color(red)(cancel(color(black)("moles Fe"("OH")_3))) * "106.87 g"/(1color(red)(cancel(color(black)("mole Fe"("OH")_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.35 g")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | What mass of #Fe(OH)_3# is produced when 35 mL of 0.250 M #Fe(NO_3)_3# solution is mixed with 55 mL of a 0.180 M #KOH# solution? | null |
1,206 | acb002fa-6ddd-11ea-9ecc-ccda262736ce | https://socratic.org/questions/when-2-moles-of-potassium-chlorate-crystals-decompose-to-potassium-chloride-crys | 2 KClO3(s) -> 2 KCl(s) + 3 O2(g) deltaHrxn = -44.7 kJ | start chemical_equation qc_end physical_unit 4 6 1 2 mole qc_end substance 9 11 qc_end substance 13 14 qc_end c_other ConstantTemperaturePressue qc_end end | [{"type":"other","value":"Chemical Equation [OF] the thermochemical equation"}] | [{"type":"chemical equation","value":"2 KClO3(s) -> 2 KCl(s) + 3 O2(g) deltaHrxn = -44.7 kJ"}] | [{"type":"physical unit","value":"Mole [OF] potassium chlorate crystals [=] \\pu{2 moles}"},{"type":"substance name","value":"Potassium chloride crystals"},{"type":"substance name","value":"Oxygen gas"},{"type":"other","value":"ConstantTemperaturePressue"},{"type":"physical unit","value":"Given off energy [OF] the reaction [=] \\pu{44.7 kJ}"}] | <h1 class="questionTitle" itemprop="name">When 2 moles of potassium chlorate crystals decompose to potassium chloride crystals and oxygen gas at constant temperature and pressure, 44.7 kJ are given off. What is the thermochemical equation?</h1> | null | 2 KClO3(s) -> 2 KCl(s) + 3 O2(g) deltaHrxn = -44.7 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A <strong>thermochemical equation</strong> is simply a chemical equation that contains information about the <em><a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"#</mathjax>.</p>
<p>The first thing to do when writing a thermochemical equation is to make sure that you get the balnced chemical equation right. </p>
<p>In your case, you know that <strong>2 moles</strong> of potassium chlorate, <mathjax>#"KClO"_3#</mathjax>, undergo <a href="http://socratic.org/chemistry/chemical-reactions/decomposition-reactions">decomposition</a> to form potassium chloride, <mathjax>#"KCl"#</mathjax>, and oxygen gas, <mathjax>#"O"_2#</mathjax>. </p>
<p>So you know that you have</p>
<blockquote>
<p><mathjax>#2"KClO"_text(3(s]) -> "KCl"_text((s]) + "O"_text(2(g])#</mathjax></p>
</blockquote>
<p>Balance this equation by multiplying the products by <mathjax>#2#</mathjax> and <mathjax>#3#</mathjax>, respectively</p>
<blockquote>
<p><mathjax>#2"KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O"_text(2(g])#</mathjax></p>
</blockquote>
<p>Now focus on the enthalpy change of reaction. The problem tells you that the reaction <strong>given off</strong> <mathjax>#"44.7 kJ"#</mathjax>, which means that the reaction is actually <a href="http://socratic.org/chemistry/thermochemistry/exothermic-processes">exothermic</a>.</p>
<p>As you know, energy changes are being represent <em>from the point of view of the system</em>. This implies that for an exothermic reaction the enthalpy change will carry a <strong>negative sign</strong>, since the system is <em>losing heat</em> to the surroundings. </p>
<p>Therefore, the enthalpy change of reaction will be </p>
<blockquote>
<p><mathjax>#DeltaH = -"44. 7 kJ"#</mathjax></p>
</blockquote>
<p>As a result, the <em>thermochemical equation</em> for the decomposition of potassium chlorate looks like this</p>
<blockquote>
<p><mathjax>#2"KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O"_text(2(g]). " "DeltaH_text(rxn) = -"44.7 kJ"#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2"KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O"_text(2(g]). " "DeltaH_text(rxn) = -"44.7 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A <strong>thermochemical equation</strong> is simply a chemical equation that contains information about the <em><a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"#</mathjax>.</p>
<p>The first thing to do when writing a thermochemical equation is to make sure that you get the balnced chemical equation right. </p>
<p>In your case, you know that <strong>2 moles</strong> of potassium chlorate, <mathjax>#"KClO"_3#</mathjax>, undergo <a href="http://socratic.org/chemistry/chemical-reactions/decomposition-reactions">decomposition</a> to form potassium chloride, <mathjax>#"KCl"#</mathjax>, and oxygen gas, <mathjax>#"O"_2#</mathjax>. </p>
<p>So you know that you have</p>
<blockquote>
<p><mathjax>#2"KClO"_text(3(s]) -> "KCl"_text((s]) + "O"_text(2(g])#</mathjax></p>
</blockquote>
<p>Balance this equation by multiplying the products by <mathjax>#2#</mathjax> and <mathjax>#3#</mathjax>, respectively</p>
<blockquote>
<p><mathjax>#2"KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O"_text(2(g])#</mathjax></p>
</blockquote>
<p>Now focus on the enthalpy change of reaction. The problem tells you that the reaction <strong>given off</strong> <mathjax>#"44.7 kJ"#</mathjax>, which means that the reaction is actually <a href="http://socratic.org/chemistry/thermochemistry/exothermic-processes">exothermic</a>.</p>
<p>As you know, energy changes are being represent <em>from the point of view of the system</em>. This implies that for an exothermic reaction the enthalpy change will carry a <strong>negative sign</strong>, since the system is <em>losing heat</em> to the surroundings. </p>
<p>Therefore, the enthalpy change of reaction will be </p>
<blockquote>
<p><mathjax>#DeltaH = -"44. 7 kJ"#</mathjax></p>
</blockquote>
<p>As a result, the <em>thermochemical equation</em> for the decomposition of potassium chlorate looks like this</p>
<blockquote>
<p><mathjax>#2"KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O"_text(2(g]). " "DeltaH_text(rxn) = -"44.7 kJ"#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">When 2 moles of potassium chlorate crystals decompose to potassium chloride crystals and oxygen gas at constant temperature and pressure, 44.7 kJ are given off. What is the thermochemical equation?</h1>
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Stefan V.
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Nov 1, 2015
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<div class="markdown"><p><mathjax>#2"KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O"_text(2(g]). " "DeltaH_text(rxn) = -"44.7 kJ"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A <strong>thermochemical equation</strong> is simply a chemical equation that contains information about the <em><a href="http://socratic.org/chemistry/thermochemistry/enthalpy">enthalpy</a> change of reaction</em>, <mathjax>#DeltaH_"rxn"#</mathjax>.</p>
<p>The first thing to do when writing a thermochemical equation is to make sure that you get the balnced chemical equation right. </p>
<p>In your case, you know that <strong>2 moles</strong> of potassium chlorate, <mathjax>#"KClO"_3#</mathjax>, undergo <a href="http://socratic.org/chemistry/chemical-reactions/decomposition-reactions">decomposition</a> to form potassium chloride, <mathjax>#"KCl"#</mathjax>, and oxygen gas, <mathjax>#"O"_2#</mathjax>. </p>
<p>So you know that you have</p>
<blockquote>
<p><mathjax>#2"KClO"_text(3(s]) -> "KCl"_text((s]) + "O"_text(2(g])#</mathjax></p>
</blockquote>
<p>Balance this equation by multiplying the products by <mathjax>#2#</mathjax> and <mathjax>#3#</mathjax>, respectively</p>
<blockquote>
<p><mathjax>#2"KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O"_text(2(g])#</mathjax></p>
</blockquote>
<p>Now focus on the enthalpy change of reaction. The problem tells you that the reaction <strong>given off</strong> <mathjax>#"44.7 kJ"#</mathjax>, which means that the reaction is actually <a href="http://socratic.org/chemistry/thermochemistry/exothermic-processes">exothermic</a>.</p>
<p>As you know, energy changes are being represent <em>from the point of view of the system</em>. This implies that for an exothermic reaction the enthalpy change will carry a <strong>negative sign</strong>, since the system is <em>losing heat</em> to the surroundings. </p>
<p>Therefore, the enthalpy change of reaction will be </p>
<blockquote>
<p><mathjax>#DeltaH = -"44. 7 kJ"#</mathjax></p>
</blockquote>
<p>As a result, the <em>thermochemical equation</em> for the decomposition of potassium chlorate looks like this</p>
<blockquote>
<p><mathjax>#2"KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O"_text(2(g]). " "DeltaH_text(rxn) = -"44.7 kJ"#</mathjax></p>
</blockquote></div>
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</article> | When 2 moles of potassium chlorate crystals decompose to potassium chloride crystals and oxygen gas at constant temperature and pressure, 44.7 kJ are given off. What is the thermochemical equation? | null |
1,207 | abe47e18-6ddd-11ea-9eba-ccda262736ce | https://socratic.org/questions/58ea66fe11ef6b7c0b75aaab | 1.20 × 10^24 | start physical_unit 2 2 number none qc_end physical_unit 9 10 6 7 mole qc_end end | [{"type":"physical unit","value":"Number [OF] atoms"}] | [{"type":"physical unit","value":"1.20 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] carbon monoxide [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">How many atoms are there in one mole of carbon monoxide?</h1> | null | 1.20 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The number of atoms in any mole of substance is always <mathjax>#6.022 xx 10^23#</mathjax> however the mass of one mole will differ between <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>.</p>
<p>In CO there is one mole of Carbon and one mole of Oxygen so the number of atoms will be for two mols. <br/>
<mathjax>#6.022 xx 10^23 xx 2 mols = 1.2044 xx 10^24 #</mathjax></p></div>
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<div class="markdown"><p><mathjax>#1.2044 xx 10^24 #</mathjax> atoms </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The number of atoms in any mole of substance is always <mathjax>#6.022 xx 10^23#</mathjax> however the mass of one mole will differ between <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>.</p>
<p>In CO there is one mole of Carbon and one mole of Oxygen so the number of atoms will be for two mols. <br/>
<mathjax>#6.022 xx 10^23 xx 2 mols = 1.2044 xx 10^24 #</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many atoms are there in one mole of carbon monoxide?</h1>
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<div class="markdown"><p><mathjax>#1.2044 xx 10^24 #</mathjax> atoms </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The number of atoms in any mole of substance is always <mathjax>#6.022 xx 10^23#</mathjax> however the mass of one mole will differ between <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>.</p>
<p>In CO there is one mole of Carbon and one mole of Oxygen so the number of atoms will be for two mols. <br/>
<mathjax>#6.022 xx 10^23 xx 2 mols = 1.2044 xx 10^24 #</mathjax></p></div>
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</article> | How many atoms are there in one mole of carbon monoxide? | null |
1,208 | aacb0e52-6ddd-11ea-b73a-ccda262736ce | https://socratic.org/questions/56403dfe11ef6b4bd4374e23 | NH3 | start chemical_formula qc_end physical_unit 8 8 6 7 mass qc_end physical_unit 12 12 10 11 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound sample [IN] empirical"}] | [{"type":"chemical equation","value":"NH3"}] | [{"type":"physical unit","value":"Mass [OF] H [=] \\pu{0.100 g}"},{"type":"physical unit","value":"Mass [OF] N [=] \\pu{4.20 g}"}] | <h1 class="questionTitle" itemprop="name">A sample of a compound contains #"0.100 g H"# and #"4.20 g N"#. What is its empirical formula?</h1> | null | NH3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The simplest formula for a compound is its empirical formula, which represents the lowest whole number ratio of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound.</p>
<p>We must first determine the number of moles of each element by dividing its mass by its molar mass, which is its 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>#0.100"g H"xx(1"mol H")/(1.00794"g H")="0.0992 mol H"#</mathjax></p>
<p><mathjax>#4.20"g N"xx(1"mol N")/(14.007"g N")="0.300 mol N"#</mathjax></p>
<p>Now we must determine the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a> between the elements by dividing the moles of each element by the least number of moles.</p>
<p><mathjax>#"H":#</mathjax><mathjax>#(0.0992cancel"mol")/(0.0992cancel"mol")="1.00"#</mathjax></p>
<p><mathjax>#"N":#</mathjax><mathjax>#(0.300cancel"mol")/(0.0992cancel"mol")="3.02~~3#</mathjax></p>
<p>The empirical formula is <mathjax>#"HN"_3#</mathjax>.</p></div>
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<div class="markdown"><p>The empirical formula is <mathjax>#"HN"_3#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The simplest formula for a compound is its empirical formula, which represents the lowest whole number ratio of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound.</p>
<p>We must first determine the number of moles of each element by dividing its mass by its molar mass, which is its 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>#0.100"g H"xx(1"mol H")/(1.00794"g H")="0.0992 mol H"#</mathjax></p>
<p><mathjax>#4.20"g N"xx(1"mol N")/(14.007"g N")="0.300 mol N"#</mathjax></p>
<p>Now we must determine the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a> between the elements by dividing the moles of each element by the least number of moles.</p>
<p><mathjax>#"H":#</mathjax><mathjax>#(0.0992cancel"mol")/(0.0992cancel"mol")="1.00"#</mathjax></p>
<p><mathjax>#"N":#</mathjax><mathjax>#(0.300cancel"mol")/(0.0992cancel"mol")="3.02~~3#</mathjax></p>
<p>The empirical formula is <mathjax>#"HN"_3#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">A sample of a compound contains #"0.100 g H"# and #"4.20 g N"#. What is its empirical formula?</h1>
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<div class="markdown"><p>The empirical formula is <mathjax>#"HN"_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The simplest formula for a compound is its empirical formula, which represents the lowest whole number ratio of <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> in the compound.</p>
<p>We must first determine the number of moles of each element by dividing its mass by its molar mass, which is its 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>#0.100"g H"xx(1"mol H")/(1.00794"g H")="0.0992 mol H"#</mathjax></p>
<p><mathjax>#4.20"g N"xx(1"mol N")/(14.007"g N")="0.300 mol N"#</mathjax></p>
<p>Now we must determine the <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a> between the elements by dividing the moles of each element by the least number of moles.</p>
<p><mathjax>#"H":#</mathjax><mathjax>#(0.0992cancel"mol")/(0.0992cancel"mol")="1.00"#</mathjax></p>
<p><mathjax>#"N":#</mathjax><mathjax>#(0.300cancel"mol")/(0.0992cancel"mol")="3.02~~3#</mathjax></p>
<p>The empirical formula is <mathjax>#"HN"_3#</mathjax>.</p></div>
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</article> | A sample of a compound contains #"0.100 g H"# and #"4.20 g N"#. What is its empirical formula? | null |
1,209 | aa3240b4-6ddd-11ea-b377-ccda262736ce | https://socratic.org/questions/you-need-to-produce-a-buffer-solution-that-has-a-ph-of-5-12-you-already-have-a-s | 24 millimoles | start physical_unit 30 30 mole mmol qc_end physical_unit 5 6 12 12 ph qc_end physical_unit 24 25 20 21 mole qc_end c_other OTHER qc_end physical_unit 24 25 51 51 pka qc_end end | [{"type":"physical unit","value":"Added mole [OF] acetate [IN] millimoles"}] | [{"type":"physical unit","value":"24 millimoles"}] | [{"type":"physical unit","value":"pH [OF] the buffer solution [=] \\pu{5.12}"},{"type":"physical unit","value":"Mole [OF] acetic acid [=] \\pu{10 mmol}"},{"type":"other","value":"Acetate is the conjugate base of acetic acid."},{"type":"physical unit","value":"pKa [OF] acetic acid [=] \\pu{4.74}"}] | <h1 class="questionTitle" itemprop="name">You need to produce a buffer solution that has a pH of 5.12. You already have a solution that contains 10 mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? </h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>The pKa of acetic acid is 4.74.</p></div>
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</div> | 24 millimoles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your <strong><a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a></strong> contains acetic acid, <mathjax>#"CH"_3"COOH"#</mathjax>, a <strong>weak acid</strong>, and the acetate anion, <mathjax>#"CH"_3"COO"^(-)#</mathjax>, its <strong>conjugate base</strong>.</p>
<p>You can find the buffer's <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> by using the <strong>Henderson - Hasselbalch equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))|))#</mathjax></p>
</blockquote>
<p>Notice that the pH of the solution depends on the ratio that exists between the <strong>concentration</strong> of the weak acid and the <strong>concentration</strong> of the conjugate base. </p>
<p>Now, a solution's <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> essentially tells you how many moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> you get in <strong>one liter of solution</strong>. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n_"solute"/V_"solution"#</mathjax></p>
</blockquote>
<p>Since the volume of the solution is <strong>the same for both chemical species</strong>, the ratio that exist between their concentrations will be <strong>equivalent</strong> to the ratio that exists between their number of moles.</p>
<p>If you take <mathjax>#V#</mathjax> to be the volume of the buffer solution, you can say that</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = n_text(acetic acid)/V" "#</mathjax> and <mathjax>#" "["CH"_3"COO"^(-)] = n_text(acetate)/V#</mathjax></p>
</blockquote>
<p>This means that you will have</p>
<blockquote>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = n_text(acetate)/color(red)(cancel(color(black)(V))) * color(red)(cancel(color(black)(V)))/n_"acetic acid" = n_"acetate"/n_"acetic acid"#</mathjax></p>
</blockquote>
<p>The H- H equation becomes</p>
<blockquote>
<p><mathjax>#"pH" = pK_a + log( n_"acetate"/n_"acetic acid")#</mathjax></p>
</blockquote>
<p>Now, you need the pH of the solution to be equal to <mathjax>#5.12#</mathjax>. Right from the start, you can look at the <mathjax>#pK_a#</mathjax> of the acid and say that since you need</p>
<blockquote>
<p><mathjax>#"pH" > pK_a#</mathjax></p>
</blockquote>
<p>the buffer will contain <strong>more moles of conjugate base</strong> than of <strong>weak acid</strong>. </p>
<p>More specifically, you will have</p>
<blockquote>
<p><mathjax>#5.12 = 4.74 + log( n_"acetate"/n_"acetic acid")#</mathjax></p>
<p><mathjax>#log(n_"acetate"/n_"acetic acid") = 0.38#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#10^log( n_"acetate"/n_"acetic acid") = 10^0.38#</mathjax></p>
</blockquote>
<p>which will give you </p>
<blockquote>
<p><mathjax>#n_"acetate"/n_"acetic acid" = 2.40#</mathjax></p>
</blockquote>
<p>This means that the solution must contain</p>
<blockquote>
<p><mathjax>#n_"acetate"= 2.40 * "10 mmol" = color(green)(|bar(ul(color(white)(a/a)"24 mmol"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="http://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>#"24 mmol CH"_3"COO"^(-)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your <strong><a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a></strong> contains acetic acid, <mathjax>#"CH"_3"COOH"#</mathjax>, a <strong>weak acid</strong>, and the acetate anion, <mathjax>#"CH"_3"COO"^(-)#</mathjax>, its <strong>conjugate base</strong>.</p>
<p>You can find the buffer's <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> by using the <strong>Henderson - Hasselbalch equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))|))#</mathjax></p>
</blockquote>
<p>Notice that the pH of the solution depends on the ratio that exists between the <strong>concentration</strong> of the weak acid and the <strong>concentration</strong> of the conjugate base. </p>
<p>Now, a solution's <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> essentially tells you how many moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> you get in <strong>one liter of solution</strong>. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n_"solute"/V_"solution"#</mathjax></p>
</blockquote>
<p>Since the volume of the solution is <strong>the same for both chemical species</strong>, the ratio that exist between their concentrations will be <strong>equivalent</strong> to the ratio that exists between their number of moles.</p>
<p>If you take <mathjax>#V#</mathjax> to be the volume of the buffer solution, you can say that</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = n_text(acetic acid)/V" "#</mathjax> and <mathjax>#" "["CH"_3"COO"^(-)] = n_text(acetate)/V#</mathjax></p>
</blockquote>
<p>This means that you will have</p>
<blockquote>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = n_text(acetate)/color(red)(cancel(color(black)(V))) * color(red)(cancel(color(black)(V)))/n_"acetic acid" = n_"acetate"/n_"acetic acid"#</mathjax></p>
</blockquote>
<p>The H- H equation becomes</p>
<blockquote>
<p><mathjax>#"pH" = pK_a + log( n_"acetate"/n_"acetic acid")#</mathjax></p>
</blockquote>
<p>Now, you need the pH of the solution to be equal to <mathjax>#5.12#</mathjax>. Right from the start, you can look at the <mathjax>#pK_a#</mathjax> of the acid and say that since you need</p>
<blockquote>
<p><mathjax>#"pH" > pK_a#</mathjax></p>
</blockquote>
<p>the buffer will contain <strong>more moles of conjugate base</strong> than of <strong>weak acid</strong>. </p>
<p>More specifically, you will have</p>
<blockquote>
<p><mathjax>#5.12 = 4.74 + log( n_"acetate"/n_"acetic acid")#</mathjax></p>
<p><mathjax>#log(n_"acetate"/n_"acetic acid") = 0.38#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#10^log( n_"acetate"/n_"acetic acid") = 10^0.38#</mathjax></p>
</blockquote>
<p>which will give you </p>
<blockquote>
<p><mathjax>#n_"acetate"/n_"acetic acid" = 2.40#</mathjax></p>
</blockquote>
<p>This means that the solution must contain</p>
<blockquote>
<p><mathjax>#n_"acetate"= 2.40 * "10 mmol" = color(green)(|bar(ul(color(white)(a/a)"24 mmol"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">You need to produce a buffer solution that has a pH of 5.12. You already have a solution that contains 10 mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? </h1>
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<div class="markdown"><p>The pKa of acetic acid is 4.74.</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2016-03-20T12:33:13" itemprop="dateCreated">
Mar 20, 2016
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<div class="markdown"><p><mathjax>#"24 mmol CH"_3"COO"^(-)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your <strong><a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a></strong> contains acetic acid, <mathjax>#"CH"_3"COOH"#</mathjax>, a <strong>weak acid</strong>, and the acetate anion, <mathjax>#"CH"_3"COO"^(-)#</mathjax>, its <strong>conjugate base</strong>.</p>
<p>You can find the buffer's <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> by using the <strong>Henderson - Hasselbalch equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))|))#</mathjax></p>
</blockquote>
<p>Notice that the pH of the solution depends on the ratio that exists between the <strong>concentration</strong> of the weak acid and the <strong>concentration</strong> of the conjugate base. </p>
<p>Now, a solution's <em><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> essentially tells you how many moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> you get in <strong>one liter of solution</strong>. </p>
<blockquote>
<p><mathjax>#color(blue)(c = n_"solute"/V_"solution"#</mathjax></p>
</blockquote>
<p>Since the volume of the solution is <strong>the same for both chemical species</strong>, the ratio that exist between their concentrations will be <strong>equivalent</strong> to the ratio that exists between their number of moles.</p>
<p>If you take <mathjax>#V#</mathjax> to be the volume of the buffer solution, you can say that</p>
<blockquote>
<p><mathjax>#["CH"_3"COOH"] = n_text(acetic acid)/V" "#</mathjax> and <mathjax>#" "["CH"_3"COO"^(-)] = n_text(acetate)/V#</mathjax></p>
</blockquote>
<p>This means that you will have</p>
<blockquote>
<p><mathjax>#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = n_text(acetate)/color(red)(cancel(color(black)(V))) * color(red)(cancel(color(black)(V)))/n_"acetic acid" = n_"acetate"/n_"acetic acid"#</mathjax></p>
</blockquote>
<p>The H- H equation becomes</p>
<blockquote>
<p><mathjax>#"pH" = pK_a + log( n_"acetate"/n_"acetic acid")#</mathjax></p>
</blockquote>
<p>Now, you need the pH of the solution to be equal to <mathjax>#5.12#</mathjax>. Right from the start, you can look at the <mathjax>#pK_a#</mathjax> of the acid and say that since you need</p>
<blockquote>
<p><mathjax>#"pH" > pK_a#</mathjax></p>
</blockquote>
<p>the buffer will contain <strong>more moles of conjugate base</strong> than of <strong>weak acid</strong>. </p>
<p>More specifically, you will have</p>
<blockquote>
<p><mathjax>#5.12 = 4.74 + log( n_"acetate"/n_"acetic acid")#</mathjax></p>
<p><mathjax>#log(n_"acetate"/n_"acetic acid") = 0.38#</mathjax></p>
</blockquote>
<p>This will be equivalent to </p>
<blockquote>
<p><mathjax>#10^log( n_"acetate"/n_"acetic acid") = 10^0.38#</mathjax></p>
</blockquote>
<p>which will give you </p>
<blockquote>
<p><mathjax>#n_"acetate"/n_"acetic acid" = 2.40#</mathjax></p>
</blockquote>
<p>This means that the solution must contain</p>
<blockquote>
<p><mathjax>#n_"acetate"= 2.40 * "10 mmol" = color(green)(|bar(ul(color(white)(a/a)"24 mmol"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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</article> | You need to produce a buffer solution that has a pH of 5.12. You already have a solution that contains 10 mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? |
The pKa of acetic acid is 4.74.
|
1,210 | a9bf13b0-6ddd-11ea-a1dc-ccda262736ce | https://socratic.org/questions/how-many-milliliters-of-a-solution-of-4-00m-ki-are-needed-to-prepare-250-0-ml-of | 47.50 milliliters | start physical_unit 5 5 volume ml qc_end physical_unit 9 9 7 8 molarity qc_end physical_unit 9 9 17 18 molarity qc_end physical_unit 5 5 14 15 volume qc_end end | [{"type":"physical unit","value":"Volume1 [OF] KI solution [IN] milliliters"}] | [{"type":"physical unit","value":"47.50 milliliters"}] | [{"type":"physical unit","value":"Molarity1 [OF] KI solution [=] \\pu{4.00 M}"},{"type":"physical unit","value":"Molarity2 [OF] KI solution [=] \\pu{0.760 M}"},{"type":"physical unit","value":"Volume2 [OF] KI solution [=] \\pu{250.0 mL}"}] | <h1 class="questionTitle" itemprop="name">How many milliliters of a solution of 4.00M #KI# are needed to prepare 250.0 mL of 0.760M #KI#?</h1> | null | 47.50 milliliters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>What you're essentially doing here is <strong>diluting</strong> a concentrated solution of potassium iodide, <mathjax>#"KI"#</mathjax>, to a target solution of <em>known volume</em>. </p>
<p>The most important thing to remember about <strong>dilutions</strong> is that the <em>number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> <strong>must remain constant</strong>, i.e. the <em>diluted solution</em> will have the same number of solute as the sample of <em>concentrated solution</em>. </p>
<p>As you know, the <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of a solution tells you how many moles of solute you get <strong>per liter</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liter of solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>This means that you can <strong>decrease</strong> the concentration of a solution by <strong>increasing</strong> its total volume <strong>while</strong> keeping the number of moles of solute constant. </p>
<p>To increase the total volume of the solution, you would <strong>add <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong>, which in this case is <em>pure water</em>. </p>
<p><img alt="http://ipkitten.blogspot.ro/2006/12/trade-mark-dilution-changes-and.html" src="https://useruploads.socratic.org/evVUosuTkS6esk6r0nkb_z23.gif"/> </p>
<p>The problem wants you to figure out what volume of concentrated solution will contain <strong>the same number of moles</strong> of potassium iodide as <mathjax>#"250.0 mL"#</mathjax> of the <mathjax>#"0.760-M"#</mathjax> target solution. </p>
<p>So, use the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the target solution to find the number of moles of solute it must contain</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to find - <strong>do not</strong> forget to convert the volume from <em>milliliters</em> to <em>liters</em>!</p>
<blockquote>
<p><mathjax>#n_(KI) = "0.760 mol" color(red)(cancel(color(black)("L"^(-1)))) * 250.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.190 moles KI"#</mathjax></p>
</blockquote>
<p>Now use the molarity of the stock solution to find what volume would contain this many moles</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies V_"solution" = n_"solute"/c)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#V_"stock" = (0.190 color(red)(cancel(color(black)("moles"))))/(4.00color(red)(cancel(color(black)("mol"))) "L"^(-1)) = "0.0475 L" = color(green)(|bar(ul(color(white)(a/a)"47.5 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, to prepare this solution, you would start with <mathjax>#"47.5 mL"#</mathjax> of the stock solution and add <em>enough water</em> until the total volume of the solution is equal to <mathjax>#"250.0 mL"#</mathjax>.</p>
<p>This is the concept behind the formula for <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a></strong> </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(black)("moles of solute in initial solution")) = overbrace(c_2 xx V_2)^(color(black)("moles of solute in diluted solution")))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the initial solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution </p>
<p>Plug in your values to get the exact same result</p>
<blockquote>
<p><mathjax>#c_1V_1 = c_2V_2 implies V_1 = c_2/c_1 * V_1#</mathjax></p>
<p><mathjax>#V_1 = (0.760 color(red)(cancel(color(black)("M"))))/(4.00color(red)(cancel(color(black)("M")))) * "250.0 mL" = color(green)(|bar(ul(color(white)(a/a)"47.5 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://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>#"47.5 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>What you're essentially doing here is <strong>diluting</strong> a concentrated solution of potassium iodide, <mathjax>#"KI"#</mathjax>, to a target solution of <em>known volume</em>. </p>
<p>The most important thing to remember about <strong>dilutions</strong> is that the <em>number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> <strong>must remain constant</strong>, i.e. the <em>diluted solution</em> will have the same number of solute as the sample of <em>concentrated solution</em>. </p>
<p>As you know, the <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of a solution tells you how many moles of solute you get <strong>per liter</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liter of solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>This means that you can <strong>decrease</strong> the concentration of a solution by <strong>increasing</strong> its total volume <strong>while</strong> keeping the number of moles of solute constant. </p>
<p>To increase the total volume of the solution, you would <strong>add <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong>, which in this case is <em>pure water</em>. </p>
<p><img alt="http://ipkitten.blogspot.ro/2006/12/trade-mark-dilution-changes-and.html" src="https://useruploads.socratic.org/evVUosuTkS6esk6r0nkb_z23.gif"/> </p>
<p>The problem wants you to figure out what volume of concentrated solution will contain <strong>the same number of moles</strong> of potassium iodide as <mathjax>#"250.0 mL"#</mathjax> of the <mathjax>#"0.760-M"#</mathjax> target solution. </p>
<p>So, use the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the target solution to find the number of moles of solute it must contain</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to find - <strong>do not</strong> forget to convert the volume from <em>milliliters</em> to <em>liters</em>!</p>
<blockquote>
<p><mathjax>#n_(KI) = "0.760 mol" color(red)(cancel(color(black)("L"^(-1)))) * 250.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.190 moles KI"#</mathjax></p>
</blockquote>
<p>Now use the molarity of the stock solution to find what volume would contain this many moles</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies V_"solution" = n_"solute"/c)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#V_"stock" = (0.190 color(red)(cancel(color(black)("moles"))))/(4.00color(red)(cancel(color(black)("mol"))) "L"^(-1)) = "0.0475 L" = color(green)(|bar(ul(color(white)(a/a)"47.5 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, to prepare this solution, you would start with <mathjax>#"47.5 mL"#</mathjax> of the stock solution and add <em>enough water</em> until the total volume of the solution is equal to <mathjax>#"250.0 mL"#</mathjax>.</p>
<p>This is the concept behind the formula for <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a></strong> </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(black)("moles of solute in initial solution")) = overbrace(c_2 xx V_2)^(color(black)("moles of solute in diluted solution")))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the initial solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution </p>
<p>Plug in your values to get the exact same result</p>
<blockquote>
<p><mathjax>#c_1V_1 = c_2V_2 implies V_1 = c_2/c_1 * V_1#</mathjax></p>
<p><mathjax>#V_1 = (0.760 color(red)(cancel(color(black)("M"))))/(4.00color(red)(cancel(color(black)("M")))) * "250.0 mL" = color(green)(|bar(ul(color(white)(a/a)"47.5 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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<h1 class="questionTitle" itemprop="name">How many milliliters of a solution of 4.00M #KI# are needed to prepare 250.0 mL of 0.760M #KI#?</h1>
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Stefan V.
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Mar 14, 2016
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<div class="markdown"><p><mathjax>#"47.5 mL"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>What you're essentially doing here is <strong>diluting</strong> a concentrated solution of potassium iodide, <mathjax>#"KI"#</mathjax>, to a target solution of <em>known volume</em>. </p>
<p>The most important thing to remember about <strong>dilutions</strong> is that the <em>number of moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em> <strong>must remain constant</strong>, i.e. the <em>diluted solution</em> will have the same number of solute as the sample of <em>concentrated solution</em>. </p>
<p>As you know, the <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of a solution tells you how many moles of solute you get <strong>per liter</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liter of solution"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>This means that you can <strong>decrease</strong> the concentration of a solution by <strong>increasing</strong> its total volume <strong>while</strong> keeping the number of moles of solute constant. </p>
<p>To increase the total volume of the solution, you would <strong>add <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong>, which in this case is <em>pure water</em>. </p>
<p><img alt="http://ipkitten.blogspot.ro/2006/12/trade-mark-dilution-changes-and.html" src="https://useruploads.socratic.org/evVUosuTkS6esk6r0nkb_z23.gif"/> </p>
<p>The problem wants you to figure out what volume of concentrated solution will contain <strong>the same number of moles</strong> of potassium iodide as <mathjax>#"250.0 mL"#</mathjax> of the <mathjax>#"0.760-M"#</mathjax> target solution. </p>
<p>So, use the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the target solution to find the number of moles of solute it must contain</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to find - <strong>do not</strong> forget to convert the volume from <em>milliliters</em> to <em>liters</em>!</p>
<blockquote>
<p><mathjax>#n_(KI) = "0.760 mol" color(red)(cancel(color(black)("L"^(-1)))) * 250.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.190 moles KI"#</mathjax></p>
</blockquote>
<p>Now use the molarity of the stock solution to find what volume would contain this many moles</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies V_"solution" = n_"solute"/c)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Plug in your values to get</p>
<blockquote>
<p><mathjax>#V_"stock" = (0.190 color(red)(cancel(color(black)("moles"))))/(4.00color(red)(cancel(color(black)("mol"))) "L"^(-1)) = "0.0475 L" = color(green)(|bar(ul(color(white)(a/a)"47.5 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>So, to prepare this solution, you would start with <mathjax>#"47.5 mL"#</mathjax> of the stock solution and add <em>enough water</em> until the total volume of the solution is equal to <mathjax>#"250.0 mL"#</mathjax>.</p>
<p>This is the concept behind the formula for <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/dilution-calculations">dilution calculations</a></strong> </p>
<blockquote>
<p><mathjax>#color(blue)(overbrace(c_1 xx V_1)^(color(black)("moles of solute in initial solution")) = overbrace(c_2 xx V_2)^(color(black)("moles of solute in diluted solution")))#</mathjax></p>
</blockquote>
<p>Here</p>
<p><mathjax>#c_1#</mathjax>, <mathjax>#V_1#</mathjax> - the molarity and volume of the initial solution<br/>
<mathjax>#c_2#</mathjax>, <mathjax>#V_2#</mathjax> - the molarity and volume of the diluted solution </p>
<p>Plug in your values to get the exact same result</p>
<blockquote>
<p><mathjax>#c_1V_1 = c_2V_2 implies V_1 = c_2/c_1 * V_1#</mathjax></p>
<p><mathjax>#V_1 = (0.760 color(red)(cancel(color(black)("M"))))/(4.00color(red)(cancel(color(black)("M")))) * "250.0 mL" = color(green)(|bar(ul(color(white)(a/a)"47.5 mL"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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</article> | How many milliliters of a solution of 4.00M #KI# are needed to prepare 250.0 mL of 0.760M #KI#? | null |
1,211 | ac66824b-6ddd-11ea-a522-ccda262736ce | https://socratic.org/questions/how-would-you-find-the-volume-in-liters-of-7-moles-of-ch-4-at-standard-temperatu | 156.80 liters | start physical_unit 12 12 volume l qc_end physical_unit 12 12 9 10 mole qc_end physical_unit 12 12 18 20 temperature qc_end physical_unit 12 12 22 23 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] CH4 [IN] liters"}] | [{"type":"physical unit","value":"156.80 liters"}] | [{"type":"physical unit","value":"Mole [OF] CH4 [=] \\pu{7 moles}"},{"type":"physical unit","value":"Temperature [OF] CH4 [=] \\pu{0 deg celsius}"},{"type":"physical unit","value":"Pressure [OF] CH4 [=] \\pu{1 atm}"}] | <h1 class="questionTitle" itemprop="name">How would you find the volume (in liters) of 7 moles of #CH_4# at standard temperature and pressure (0 deg celsius and 1atm)?</h1> | null | 156.80 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given this information, if there are <mathjax>#7#</mathjax> <mathjax>#"moles"#</mathjax> of methane, the volume is simply the product:<br/>
<mathjax>#7*molxx22.4*L*mol^(-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>At <mathjax>#"STP"#</mathjax>, <mathjax>#1#</mathjax> mole of Ideal gas occupies a <mathjax>#22.4*L#</mathjax> volume.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given this information, if there are <mathjax>#7#</mathjax> <mathjax>#"moles"#</mathjax> of methane, the volume is simply the product:<br/>
<mathjax>#7*molxx22.4*L*mol^(-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you find the volume (in liters) of 7 moles of #CH_4# at standard temperature and pressure (0 deg celsius and 1atm)?</h1>
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<div class="markdown"><p>At <mathjax>#"STP"#</mathjax>, <mathjax>#1#</mathjax> mole of Ideal gas occupies a <mathjax>#22.4*L#</mathjax> volume.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Given this information, if there are <mathjax>#7#</mathjax> <mathjax>#"moles"#</mathjax> of methane, the volume is simply the product:<br/>
<mathjax>#7*molxx22.4*L*mol^(-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??L#</mathjax>.</p></div>
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</article> | How would you find the volume (in liters) of 7 moles of #CH_4# at standard temperature and pressure (0 deg celsius and 1atm)? | null |
1,212 | ac59824d-6ddd-11ea-af0e-ccda262736ce | https://socratic.org/questions/if-a-balloon-containing-3000-l-of-gas-at-39-c-and-99-kpa-rises-to-an-altitude-wh | 5496.62 L | start physical_unit 33 34 volume l qc_end physical_unit 7 7 4 5 volume qc_end physical_unit 7 7 9 10 temperature qc_end physical_unit 7 7 12 13 pressure qc_end physical_unit 33 34 28 29 temperature qc_end physical_unit 33 34 22 23 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the balloon [IN] L"}] | [{"type":"physical unit","value":"5496.62 L"}] | [{"type":"physical unit","value":"Volume1 [OF] gas [=] \\pu{3000 L}"},{"type":"physical unit","value":"Temperature1 [OF] gas [=] \\pu{39 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] gas [=] \\pu{99 kPa}"},{"type":"physical unit","value":"Temperature2 [OF] the balloon [=] \\pu{16 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] the balloon [=] \\pu{45.5 kPa}"}] | <h1 class="questionTitle" itemprop="name">If a balloon containing 3000 L of gas at 39 C and 99 kPa rises to an altitude where the pressure is 45.5 kPa and the temperature is 16 C, the volume ofthe balloon under these new conditions would be calculated using what conversion factor ratios?</h1> | null | 5496.62 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This assumes constant amount of gas. And we solve for <mathjax>#V_2#</mathjax> to get:</p>
<p><mathjax>#V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)#</mathjax>, which equation, clearly, has units of volume. </p>
<p>Do you agree?</p>
<p>So <mathjax>#V_2=(99*kPaxx3000*Lxx289*K)/(312*Kxx45.5*kPa)=??L#</mathjax></p>
<p>So the volume has effectively doubled. Of course, this is probably a CLOSED weather balloon. I have never been in a hot-air balloon, but I suspect (and I may be wrong), that doubling the volume of a hot-air balloon might dangerously stretch the fabric. If there are any hot-air balloonists on these boards, their insight would be welcome. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>By the combined gas equation: <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This assumes constant amount of gas. And we solve for <mathjax>#V_2#</mathjax> to get:</p>
<p><mathjax>#V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)#</mathjax>, which equation, clearly, has units of volume. </p>
<p>Do you agree?</p>
<p>So <mathjax>#V_2=(99*kPaxx3000*Lxx289*K)/(312*Kxx45.5*kPa)=??L#</mathjax></p>
<p>So the volume has effectively doubled. Of course, this is probably a CLOSED weather balloon. I have never been in a hot-air balloon, but I suspect (and I may be wrong), that doubling the volume of a hot-air balloon might dangerously stretch the fabric. If there are any hot-air balloonists on these boards, their insight would be welcome. </p></div>
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<h1 class="questionTitle" itemprop="name">If a balloon containing 3000 L of gas at 39 C and 99 kPa rises to an altitude where the pressure is 45.5 kPa and the temperature is 16 C, the volume ofthe balloon under these new conditions would be calculated using what conversion factor ratios?</h1>
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anor277
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Feb 11, 2017
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<div class="markdown"><p>By the combined gas equation: <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This assumes constant amount of gas. And we solve for <mathjax>#V_2#</mathjax> to get:</p>
<p><mathjax>#V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)#</mathjax>, which equation, clearly, has units of volume. </p>
<p>Do you agree?</p>
<p>So <mathjax>#V_2=(99*kPaxx3000*Lxx289*K)/(312*Kxx45.5*kPa)=??L#</mathjax></p>
<p>So the volume has effectively doubled. Of course, this is probably a CLOSED weather balloon. I have never been in a hot-air balloon, but I suspect (and I may be wrong), that doubling the volume of a hot-air balloon might dangerously stretch the fabric. If there are any hot-air balloonists on these boards, their insight would be welcome. </p></div>
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</article> | If a balloon containing 3000 L of gas at 39 C and 99 kPa rises to an altitude where the pressure is 45.5 kPa and the temperature is 16 C, the volume ofthe balloon under these new conditions would be calculated using what conversion factor ratios? | null |
1,213 | a838df63-6ddd-11ea-89f0-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-with-0-2-moles-of-potassium-permanganate-kmno | 0.40 M | start physical_unit 6 6 molarity mol/l qc_end physical_unit 13 13 8 9 mole qc_end physical_unit 6 6 21 22 volume qc_end substance 17 17 qc_end end | [{"type":"physical unit","value":"Molarity [OF] solution [IN] M"}] | [{"type":"physical unit","value":"0.40 M"}] | [{"type":"physical unit","value":"Mole [OF] KMnO4 [=] \\pu{0.2 moles}"},{"type":"physical unit","value":"Volume [OF] KMnO4 solution [=] \\pu{500.0 mL}"},{"type":"substance name","value":"water"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution with 0.2 moles of potassium permanganate (#KMnO_4#) dissolved in enough water to make a 500.0 mL solution?</h1> | null | 0.40 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of a solution is found by: <mathjax>#C_M=n/V#</mathjax> where, </p>
<p><mathjax>#n#</mathjax> is the number of mole of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and <mathjax>#V#</mathjax> is the volume of the solution.</p>
<p>Therefore, <mathjax>#C_M=(0.2mol)/(0.5000L)=0.4M#</mathjax></p></div>
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<div class="markdown"><p>The solution is <mathjax>#0.4M#</mathjax>.</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of a solution is found by: <mathjax>#C_M=n/V#</mathjax> where, </p>
<p><mathjax>#n#</mathjax> is the number of mole of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and <mathjax>#V#</mathjax> is the volume of the solution.</p>
<p>Therefore, <mathjax>#C_M=(0.2mol)/(0.5000L)=0.4M#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the molarity of a solution with 0.2 moles of potassium permanganate (#KMnO_4#) dissolved in enough water to make a 500.0 mL solution?</h1>
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Dr. Hayek
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<div class="markdown"><p>The solution is <mathjax>#0.4M#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of a solution is found by: <mathjax>#C_M=n/V#</mathjax> where, </p>
<p><mathjax>#n#</mathjax> is the number of mole of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> and <mathjax>#V#</mathjax> is the volume of the solution.</p>
<p>Therefore, <mathjax>#C_M=(0.2mol)/(0.5000L)=0.4M#</mathjax></p></div>
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</article> | What is the molarity of a solution with 0.2 moles of potassium permanganate (#KMnO_4#) dissolved in enough water to make a 500.0 mL solution? | null |
1,214 | a9ac7e4c-6ddd-11ea-9546-ccda262736ce | https://socratic.org/questions/what-volume-is-occupied-by-0-695-mol-of-co-2-at-342-3-k-and-629-mmhg | 23.60 L | start physical_unit 8 8 volume l qc_end physical_unit 8 8 5 6 mole qc_end physical_unit 8 8 10 11 temperature qc_end physical_unit 8 8 13 14 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] CO2 [IN] L"}] | [{"type":"physical unit","value":"23.60 L"}] | [{"type":"physical unit","value":"Mole [OF] CO2 [=] \\pu{0.695 mol}"},{"type":"physical unit","value":"Temperature [OF] CO2 [=] \\pu{342.3 K}"},{"type":"physical unit","value":"Pressure [OF] CO2 [=] \\pu{629 mmHg}"}] | <h1 class="questionTitle" itemprop="name">What volume is occupied by 0.695 mol of #CO_2# at 342.3 K and 629 mmHg?</h1> | null | 23.60 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the relationship <mathjax>#"760 mm Hg "-=" 1 atm."#</mathjax> And then we apply the Ideal Gas Equation,</p>
<p><mathjax>#V=(nRT)/P#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.695*molxx0.0821*(L*atm)/(K*mol)xx342.3*K)/((629*mm*Hg)/(760*mm*Hg*atm^-1))#</mathjax></p>
<p><mathjax>#=??L#</mathjax></p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#V~=24*L#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the relationship <mathjax>#"760 mm Hg "-=" 1 atm."#</mathjax> And then we apply the Ideal Gas Equation,</p>
<p><mathjax>#V=(nRT)/P#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.695*molxx0.0821*(L*atm)/(K*mol)xx342.3*K)/((629*mm*Hg)/(760*mm*Hg*atm^-1))#</mathjax></p>
<p><mathjax>#=??L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What volume is occupied by 0.695 mol of #CO_2# at 342.3 K and 629 mmHg?</h1>
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<div class="markdown"><p><mathjax>#V~=24*L#</mathjax></p></div>
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<div class="markdown"><p>We use the relationship <mathjax>#"760 mm Hg "-=" 1 atm."#</mathjax> And then we apply the Ideal Gas Equation,</p>
<p><mathjax>#V=(nRT)/P#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.695*molxx0.0821*(L*atm)/(K*mol)xx342.3*K)/((629*mm*Hg)/(760*mm*Hg*atm^-1))#</mathjax></p>
<p><mathjax>#=??L#</mathjax></p></div>
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</article> | What volume is occupied by 0.695 mol of #CO_2# at 342.3 K and 629 mmHg? | null |
1,215 | ab4505b8-6ddd-11ea-9915-ccda262736ce | https://socratic.org/questions/592cf6f87c014979d246d3d3 | 4 | start physical_unit 6 7 ph none qc_end physical_unit 6 7 3 3 poh qc_end end | [{"type":"physical unit","value":"pH [OF] aqueous solution"}] | [{"type":"physical unit","value":"4"}] | [{"type":"physical unit","value":"pOH [OF] aqueous solution [=] \\pu{10}"}] | <h1 class="questionTitle" itemprop="name">If #pOH=10# for an aqueous solution, what is #pH#?</h1> | null | 4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH+pOH=14#</mathjax> in aqueous solution under standard conditions. This is something you will simply have to remember.</p>
<p>And so <mathjax>#pOH=14-pH=14-4#</mathjax>, because <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p>See <a href="https://socratic.org/questions/ph-value-definition">here for background.</a> The use of logarithmic scales derives from pre-calculator days, when chemists, physicists, and ACCOUNTANTS used log tables to calculate the products of very large or very small numbers.</p></div>
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<div class="markdown"><p><mathjax>#pOH=10#</mathjax>.....................what is solution <mathjax>#pH#</mathjax>?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH+pOH=14#</mathjax> in aqueous solution under standard conditions. This is something you will simply have to remember.</p>
<p>And so <mathjax>#pOH=14-pH=14-4#</mathjax>, because <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p>See <a href="https://socratic.org/questions/ph-value-definition">here for background.</a> The use of logarithmic scales derives from pre-calculator days, when chemists, physicists, and ACCOUNTANTS used log tables to calculate the products of very large or very small numbers.</p></div>
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<h1 class="questionTitle" itemprop="name">If #pOH=10# for an aqueous solution, what is #pH#?</h1>
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<div class="markdown"><p><mathjax>#pOH=10#</mathjax>.....................what is solution <mathjax>#pH#</mathjax>?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH+pOH=14#</mathjax> in aqueous solution under standard conditions. This is something you will simply have to remember.</p>
<p>And so <mathjax>#pOH=14-pH=14-4#</mathjax>, because <mathjax>#pH=-log_(10)[H_3O^+]#</mathjax></p>
<p>See <a href="https://socratic.org/questions/ph-value-definition">here for background.</a> The use of logarithmic scales derives from pre-calculator days, when chemists, physicists, and ACCOUNTANTS used log tables to calculate the products of very large or very small numbers.</p></div>
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</article> | If #pOH=10# for an aqueous solution, what is #pH#? | null |
1,216 | ab11bc00-6ddd-11ea-8e6f-ccda262736ce | https://socratic.org/questions/chemists-commonly-use-a-rule-of-thumb-that-an-increase-of-10-k-in-temperature-do | 52.4 kJ/mol | start physical_unit 28 29 activation_barrier kj/mol qc_end c_other OTHER qc_end physical_unit 28 29 38 39 temperature qc_end physical_unit 28 29 41 42 temperature qc_end end | [{"type":"physical unit","value":"Activation energy [OF] this statement [IN] kJ/mol"}] | [{"type":"physical unit","value":"52.4 kJ/mol"}] | [{"type":"other","value":"Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. "},{"type":"physical unit","value":"Temperature1 [OF] this statement [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] this statement [=] \\pu{35 ℃}"}] | <h1 class="questionTitle" itemprop="name">Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to 35°C?</h1> | null | 52.4 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The Arrhenius equation says that the dependence of the rate constant <mathjax>#k#</mathjax> on temperature reads</p>
<p><mathjax>#k=Ae^(-(E_a)/(RT))#</mathjax> </p>
<p>For the sake of this problem, let us say that <mathjax>#k_1#</mathjax> is the initial value of the constant at a temperature <mathjax>#T_1#</mathjax>, and <mathjax>#k_2#</mathjax> is the value at temperature <mathjax>#T_2#</mathjax></p>
<p>To double the reaction rate, we must double the value of k, i.e. the ratio <mathjax>#k_2/k_1#</mathjax> must equal 2.</p>
<p>This will require the right hand side of the Arrhenius equation to change to</p>
<p><mathjax>#Ae^(-(E_a)/(RT_2))#</mathjax> from <mathjax>#Ae^(-(E_a)/(RT_1))#</mathjax> </p>
<p>all of which can be written in ratio forms as</p>
<p><mathjax>#k_2/k_1=e^((-E_a/R)(1/T_2-1/T_1))#</mathjax> </p>
<p>Taking the natural log of each side:</p>
<p><mathjax>#ln2 = -(E_a/R)(1/T_2-1/T_1)#</mathjax></p>
<p>or</p>
<p><mathjax>#E_a= (ln2 (R))/(1/T_2-1/T_1)#</mathjax></p>
<p>Using 308K for <mathjax>#T_2#</mathjax> and 298K for <mathjax>#T_1#</mathjax></p>
<p><mathjax>#E_a= (0.693 (8.314))/(1/308-1/298)#</mathjax></p>
<p><mathjax>#E_a=52.4 kJ/(mol)#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>The Arrhenius equation will tell us that the activation energy in this case must be 52.4 kJ/mol.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The Arrhenius equation says that the dependence of the rate constant <mathjax>#k#</mathjax> on temperature reads</p>
<p><mathjax>#k=Ae^(-(E_a)/(RT))#</mathjax> </p>
<p>For the sake of this problem, let us say that <mathjax>#k_1#</mathjax> is the initial value of the constant at a temperature <mathjax>#T_1#</mathjax>, and <mathjax>#k_2#</mathjax> is the value at temperature <mathjax>#T_2#</mathjax></p>
<p>To double the reaction rate, we must double the value of k, i.e. the ratio <mathjax>#k_2/k_1#</mathjax> must equal 2.</p>
<p>This will require the right hand side of the Arrhenius equation to change to</p>
<p><mathjax>#Ae^(-(E_a)/(RT_2))#</mathjax> from <mathjax>#Ae^(-(E_a)/(RT_1))#</mathjax> </p>
<p>all of which can be written in ratio forms as</p>
<p><mathjax>#k_2/k_1=e^((-E_a/R)(1/T_2-1/T_1))#</mathjax> </p>
<p>Taking the natural log of each side:</p>
<p><mathjax>#ln2 = -(E_a/R)(1/T_2-1/T_1)#</mathjax></p>
<p>or</p>
<p><mathjax>#E_a= (ln2 (R))/(1/T_2-1/T_1)#</mathjax></p>
<p>Using 308K for <mathjax>#T_2#</mathjax> and 298K for <mathjax>#T_1#</mathjax></p>
<p><mathjax>#E_a= (0.693 (8.314))/(1/308-1/298)#</mathjax></p>
<p><mathjax>#E_a=52.4 kJ/(mol)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to 35°C?</h1>
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<div class="markdown"><p>The Arrhenius equation will tell us that the activation energy in this case must be 52.4 kJ/mol.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The Arrhenius equation says that the dependence of the rate constant <mathjax>#k#</mathjax> on temperature reads</p>
<p><mathjax>#k=Ae^(-(E_a)/(RT))#</mathjax> </p>
<p>For the sake of this problem, let us say that <mathjax>#k_1#</mathjax> is the initial value of the constant at a temperature <mathjax>#T_1#</mathjax>, and <mathjax>#k_2#</mathjax> is the value at temperature <mathjax>#T_2#</mathjax></p>
<p>To double the reaction rate, we must double the value of k, i.e. the ratio <mathjax>#k_2/k_1#</mathjax> must equal 2.</p>
<p>This will require the right hand side of the Arrhenius equation to change to</p>
<p><mathjax>#Ae^(-(E_a)/(RT_2))#</mathjax> from <mathjax>#Ae^(-(E_a)/(RT_1))#</mathjax> </p>
<p>all of which can be written in ratio forms as</p>
<p><mathjax>#k_2/k_1=e^((-E_a/R)(1/T_2-1/T_1))#</mathjax> </p>
<p>Taking the natural log of each side:</p>
<p><mathjax>#ln2 = -(E_a/R)(1/T_2-1/T_1)#</mathjax></p>
<p>or</p>
<p><mathjax>#E_a= (ln2 (R))/(1/T_2-1/T_1)#</mathjax></p>
<p>Using 308K for <mathjax>#T_2#</mathjax> and 298K for <mathjax>#T_1#</mathjax></p>
<p><mathjax>#E_a= (0.693 (8.314))/(1/308-1/298)#</mathjax></p>
<p><mathjax>#E_a=52.4 kJ/(mol)#</mathjax></p></div>
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</article> | Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to 35°C? | null |
1,217 | a8638f48-6ddd-11ea-9b04-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-chemical-equation-for-the-reaction-that-takes-place-between-1 | Br2 + 2 NaI -> I2 + 2 NaBr | start chemical_equation qc_end substance 13 13 qc_end substance 15 16 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"Br2 + 2 NaI -> I2 + 2 NaBr"}] | [{"type":"substance name","value":"Bromine"},{"type":"substance name","value":"Sodium iodide"}] | <h1 class="questionTitle" itemprop="name">What is the balanced chemical equation for the reaction that takes place between bromine and sodium iodide? </h1> | null | Br2 + 2 NaI -> I2 + 2 NaBr | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The activity series for the halogens follows the order of the halogens from top to bottom down the group on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, with fluorine the most reactive halogen and astatine the least reactive. Each halogen can replace the halogen below it in the activity series (or the periodic table).</p>
<p><img alt="http://www.online-sciences.com/the-matter/the-general-properties-of-halogens-in-the-modern-periodic-table/" src="https://useruploads.socratic.org/TFnkjsBRWKctHQPyc5rC_halogens-4-300x222.gif"/> </p></div>
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<div class="markdown"><p><mathjax>#"Br"_2+"2NaI"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"I"_2+"2NaBr"#</mathjax></p>
<p>This is a single replacement (single displacement) reaction involving the halogens (group 17/VIIA on the periodic table), rather than metals.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The activity series for the halogens follows the order of the halogens from top to bottom down the group on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, with fluorine the most reactive halogen and astatine the least reactive. Each halogen can replace the halogen below it in the activity series (or the periodic table).</p>
<p><img alt="http://www.online-sciences.com/the-matter/the-general-properties-of-halogens-in-the-modern-periodic-table/" src="https://useruploads.socratic.org/TFnkjsBRWKctHQPyc5rC_halogens-4-300x222.gif"/> </p></div>
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<h1 class="questionTitle" itemprop="name">What is the balanced chemical equation for the reaction that takes place between bromine and sodium iodide? </h1>
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<div class="markdown"><p><mathjax>#"Br"_2+"2NaI"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"I"_2+"2NaBr"#</mathjax></p>
<p>This is a single replacement (single displacement) reaction involving the halogens (group 17/VIIA on the periodic table), rather than metals.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The activity series for the halogens follows the order of the halogens from top to bottom down the group on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, with fluorine the most reactive halogen and astatine the least reactive. Each halogen can replace the halogen below it in the activity series (or the periodic table).</p>
<p><img alt="http://www.online-sciences.com/the-matter/the-general-properties-of-halogens-in-the-modern-periodic-table/" src="https://useruploads.socratic.org/TFnkjsBRWKctHQPyc5rC_halogens-4-300x222.gif"/> </p></div>
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</article> | What is the balanced chemical equation for the reaction that takes place between bromine and sodium iodide? | null |
1,218 | abafbe02-6ddd-11ea-90c7-ccda262736ce | https://socratic.org/questions/a-student-decides-to-react-hcl-with-naoh-in-50-g-of-h2o-the-reaction-produces-he | HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l) + deltaH | start chemical_equation qc_end chemical_equation 5 5 qc_end chemical_equation 7 7 qc_end physical_unit 12 12 9 10 mass qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] the balanced chemical equation"}] | [{"type":"chemical equation","value":"HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l) + deltaH"}] | [{"type":"chemical equation","value":"HCl"},{"type":"chemical equation","value":"NaOH"},{"type":"physical unit","value":"Mass [OF] H2O [=] \\pu{50 g}"},{"type":"other","value":"The reaction produces heat."}] | <h1 class="questionTitle" itemprop="name">A student decides to react HCl with NaOH in 50 g of H2O. The reaction produces heat.. What is the balanced chemical equation for the reaction?</h1> | null | HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l) + deltaH | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now not only does the given reaction observes <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> with respect to atoms and molecules, it also observes stoichiometry with respect to the heat of reaction. <mathjax>#DeltaH#</mathjax> is usually reported per mole of reaction as written, and can usually be precisely measured. </p></div>
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<div class="markdown"><p><mathjax>#HCl(aq) + NaOH(aq) rarr NaCl(aq) + H_2O(l) + Delta_"rxn"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Now not only does the given reaction observes <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> with respect to atoms and molecules, it also observes stoichiometry with respect to the heat of reaction. <mathjax>#DeltaH#</mathjax> is usually reported per mole of reaction as written, and can usually be precisely measured. </p></div>
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<h1 class="questionTitle" itemprop="name">A student decides to react HCl with NaOH in 50 g of H2O. The reaction produces heat.. What is the balanced chemical equation for the reaction?</h1>
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<div class="markdown"><p><mathjax>#HCl(aq) + NaOH(aq) rarr NaCl(aq) + H_2O(l) + Delta_"rxn"#</mathjax></p></div>
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<div class="markdown"><p>Now not only does the given reaction observes <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> with respect to atoms and molecules, it also observes stoichiometry with respect to the heat of reaction. <mathjax>#DeltaH#</mathjax> is usually reported per mole of reaction as written, and can usually be precisely measured. </p></div>
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</article> | A student decides to react HCl with NaOH in 50 g of H2O. The reaction produces heat.. What is the balanced chemical equation for the reaction? | null |
1,219 | ab6ae6c1-6ddd-11ea-8f1f-ccda262736ce | https://socratic.org/questions/how-many-grams-of-br-are-in-195-g-of-cabr-2 | 156 grams | start physical_unit 4 4 mass g qc_end physical_unit 10 10 7 8 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] Br [IN] grams"}] | [{"type":"physical unit","value":"156 grams"}] | [{"type":"physical unit","value":"Mass [OF] CaBr2 [=] \\pu{195 g}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #Br# are in 195 g of #CaBr_2#?</h1> | null | 156 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to figure out how many grams of bromine you get in that many grams of <em>calcium bromide</em>, <mathjax>#"CaBr"_2#</mathjax>, you must find the compound's <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong>. </p>
<p>To do that, use the fact that <strong>one mole</strong> of calcium bromide contains</p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of calcium cations,</em> <mathjax>#"Ca"^(2+)#</mathjax></li>
<li><em><strong>two moles</strong> of bromide anions,</em> <mathjax>#2 xx "Br"^(-)#</mathjax></li>
</ul>
</blockquote>
<p>You can thus us the <strong>molar mass</strong> of calcium bromide and the <strong>molar mass</strong> of bromine to determine how many grams of bromine you get <em>per</em> <mathjax>#"100 g"#</mathjax> of calcium bromide. </p>
<p>The two molar mass are</p>
<blockquote>
<p><mathjax>#"For CaBr"_2: " "M_M = "199.89 g mol"^(-1)#</mathjax></p>
<p><mathjax>#"For Br:" " " " " " " M_M = "79.904 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>So, two moles of bromide anions for every one mole of calcium bromide will give you a <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of </p>
<blockquote>
<p><mathjax>#(2 xx 79.904 color(red)(cancel(color(black)("g mol"^(-1)))))/(199.89color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = "79.95% Br"#</mathjax></p>
</blockquote>
<p>This means that <strong>every</strong> <mathjax>#"100 g"#</mathjax> of calcium bromide will contain <mathjax>#"79.95 g"#</mathjax> of elemental bromine in the form of bromide cations. </p>
<p>All you have to do now is use this percent composition as a <strong>conversion factor</strong> to determine how many grams of bromine you get in that <mathjax>#"195-g"#</mathjax> sample of calcium bromide</p>
<blockquote>
<p><mathjax>#195color(red)(cancel(color(black)("g CaBr"_2))) * overbrace("79.95 g Br"/(100color(red)(cancel(color(black)("g CaBr"_2)))))^(color(purple)("79.95% Br")) = color(green)(|bar(ul(color(white)(a/a)"156 g Br"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<div class="markdown"><p><mathjax>#"156 g Br"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to figure out how many grams of bromine you get in that many grams of <em>calcium bromide</em>, <mathjax>#"CaBr"_2#</mathjax>, you must find the compound's <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong>. </p>
<p>To do that, use the fact that <strong>one mole</strong> of calcium bromide contains</p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of calcium cations,</em> <mathjax>#"Ca"^(2+)#</mathjax></li>
<li><em><strong>two moles</strong> of bromide anions,</em> <mathjax>#2 xx "Br"^(-)#</mathjax></li>
</ul>
</blockquote>
<p>You can thus us the <strong>molar mass</strong> of calcium bromide and the <strong>molar mass</strong> of bromine to determine how many grams of bromine you get <em>per</em> <mathjax>#"100 g"#</mathjax> of calcium bromide. </p>
<p>The two molar mass are</p>
<blockquote>
<p><mathjax>#"For CaBr"_2: " "M_M = "199.89 g mol"^(-1)#</mathjax></p>
<p><mathjax>#"For Br:" " " " " " " M_M = "79.904 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>So, two moles of bromide anions for every one mole of calcium bromide will give you a <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of </p>
<blockquote>
<p><mathjax>#(2 xx 79.904 color(red)(cancel(color(black)("g mol"^(-1)))))/(199.89color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = "79.95% Br"#</mathjax></p>
</blockquote>
<p>This means that <strong>every</strong> <mathjax>#"100 g"#</mathjax> of calcium bromide will contain <mathjax>#"79.95 g"#</mathjax> of elemental bromine in the form of bromide cations. </p>
<p>All you have to do now is use this percent composition as a <strong>conversion factor</strong> to determine how many grams of bromine you get in that <mathjax>#"195-g"#</mathjax> sample of calcium bromide</p>
<blockquote>
<p><mathjax>#195color(red)(cancel(color(black)("g CaBr"_2))) * overbrace("79.95 g Br"/(100color(red)(cancel(color(black)("g CaBr"_2)))))^(color(purple)("79.95% Br")) = color(green)(|bar(ul(color(white)(a/a)"156 g Br"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of #Br# are in 195 g of #CaBr_2#?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"156 g Br"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to figure out how many grams of bromine you get in that many grams of <em>calcium bromide</em>, <mathjax>#"CaBr"_2#</mathjax>, you must find the compound's <strong><a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong>. </p>
<p>To do that, use the fact that <strong>one mole</strong> of calcium bromide contains</p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of calcium cations,</em> <mathjax>#"Ca"^(2+)#</mathjax></li>
<li><em><strong>two moles</strong> of bromide anions,</em> <mathjax>#2 xx "Br"^(-)#</mathjax></li>
</ul>
</blockquote>
<p>You can thus us the <strong>molar mass</strong> of calcium bromide and the <strong>molar mass</strong> of bromine to determine how many grams of bromine you get <em>per</em> <mathjax>#"100 g"#</mathjax> of calcium bromide. </p>
<p>The two molar mass are</p>
<blockquote>
<p><mathjax>#"For CaBr"_2: " "M_M = "199.89 g mol"^(-1)#</mathjax></p>
<p><mathjax>#"For Br:" " " " " " " M_M = "79.904 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>So, two moles of bromide anions for every one mole of calcium bromide will give you a <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of </p>
<blockquote>
<p><mathjax>#(2 xx 79.904 color(red)(cancel(color(black)("g mol"^(-1)))))/(199.89color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = "79.95% Br"#</mathjax></p>
</blockquote>
<p>This means that <strong>every</strong> <mathjax>#"100 g"#</mathjax> of calcium bromide will contain <mathjax>#"79.95 g"#</mathjax> of elemental bromine in the form of bromide cations. </p>
<p>All you have to do now is use this percent composition as a <strong>conversion factor</strong> to determine how many grams of bromine you get in that <mathjax>#"195-g"#</mathjax> sample of calcium bromide</p>
<blockquote>
<p><mathjax>#195color(red)(cancel(color(black)("g CaBr"_2))) * overbrace("79.95 g Br"/(100color(red)(cancel(color(black)("g CaBr"_2)))))^(color(purple)("79.95% Br")) = color(green)(|bar(ul(color(white)(a/a)"156 g Br"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | How many grams of #Br# are in 195 g of #CaBr_2#? | null |
1,220 | ab01f025-6ddd-11ea-bb30-ccda262736ce | https://socratic.org/questions/given-the-equation-2h-2o-2h-2-o-2-how-many-moles-of-h-2o-be-required-to-produce- | 5.00 moles | start physical_unit 4 4 mole mol qc_end chemical_equation 3 9 qc_end physical_unit 9 9 19 20 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] H2O [IN] moles"}] | [{"type":"physical unit","value":"5.00 moles"}] | [{"type":"chemical equation","value":"2 H2O -> 2 H2 + O2"},{"type":"physical unit","value":"Mole [OF] O2 [=] \\pu{2.5 moles}"}] | <h1 class="questionTitle" itemprop="name">Given the equation #2H_2O -> 2H_2 + O_2#, how many moles of #H_2O# be required to produce 2.5 moles of #O_2#?</h1> | null | 5.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So every 2 molecules of water gives 1 molecule of oxygen.<br/>
This ratio (<mathjax>#2:1#</mathjax>) stays the same if we are talking about moles, because a mole is a set number of molecules (<mathjax>#~~6.02xx10^23#</mathjax>). This goes for any substance.<br/>
So to produce 2.5 moles of oxygen, you will need the double number of moles of water, i.e. 5 moles.</p></div>
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<div class="markdown"><p>From the equation you'll see that <mathjax>#2H_2O#</mathjax> gives <mathjax>#1O_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So every 2 molecules of water gives 1 molecule of oxygen.<br/>
This ratio (<mathjax>#2:1#</mathjax>) stays the same if we are talking about moles, because a mole is a set number of molecules (<mathjax>#~~6.02xx10^23#</mathjax>). This goes for any substance.<br/>
So to produce 2.5 moles of oxygen, you will need the double number of moles of water, i.e. 5 moles.</p></div>
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<h1 class="questionTitle" itemprop="name">Given the equation #2H_2O -> 2H_2 + O_2#, how many moles of #H_2O# be required to produce 2.5 moles of #O_2#?</h1>
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<div class="markdown"><p>From the equation you'll see that <mathjax>#2H_2O#</mathjax> gives <mathjax>#1O_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So every 2 molecules of water gives 1 molecule of oxygen.<br/>
This ratio (<mathjax>#2:1#</mathjax>) stays the same if we are talking about moles, because a mole is a set number of molecules (<mathjax>#~~6.02xx10^23#</mathjax>). This goes for any substance.<br/>
So to produce 2.5 moles of oxygen, you will need the double number of moles of water, i.e. 5 moles.</p></div>
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</article> | Given the equation #2H_2O -> 2H_2 + O_2#, how many moles of #H_2O# be required to produce 2.5 moles of #O_2#? | null |
1,221 | a9451b2c-6ddd-11ea-acce-ccda262736ce | https://socratic.org/questions/how-many-moles-of-lead-are-in-1-50-10-12-atoms-of-lead | 2.49 × 10^(-12) moles | start physical_unit 4 4 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] lead [IN] moles"}] | [{"type":"physical unit","value":"2.49 × 10^(-12) moles"}] | [{"type":"physical unit","value":"Number [OF] lead atoms [=] \\pu{1.50 × 10^12}"}] | <h1 class="questionTitle" itemprop="name">How many moles of lead are in #1.50 * 10^12# atoms of lead?</h1> | null | 2.49 × 10^(-12) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before doing any calculations, it's worth noting that atoms do not <em>contain moles</em>, it's the other way around. </p>
<p>A <strong>mole</strong> is simply a collection of atoms. More specifically, you need to have exactly <mathjax>#6.022 * 10^(23)#</mathjax> atoms of an element in order to have <strong>one mole</strong> of that element - this is known as <em>Avogadro's number</em>. </p>
<p>In your case, you must determine how many moles of lead <em>would contain</em> <mathjax>#1.50 * 10^(12)#</mathjax> atoms of lead.</p>
<p>Well, if you know that one mole of lead must contain <mathjax>#6.022 * 10^(23)#</mathjax> atoms of lead, it follows that you get <mathjax>#1.50 * 10^12#</mathjax> atoms of lead in </p>
<blockquote>
<p><mathjax>#1.50 * 10^12 color(red)(cancel(color(black)("atoms of Pb"))) * "1 mole Pb"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms of Pb")))) = color(green)(2.49 * 10^(-12)"moles Pb")#</mathjax></p>
</blockquote>
<p>Here's an excellent video on <em>Avogadro's number</em> and <em><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a></em> featuring Professor Martyn Poliakoff from <a href="https://www.youtube.com/user/periodicvideos/about" rel="nofollow">Periodic Videos</a></p>
<p>
<iframe src="https://www.youtube.com/embed/2dzS_LXvYA0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#2.49 * 10^(-12)"moles Pb"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before doing any calculations, it's worth noting that atoms do not <em>contain moles</em>, it's the other way around. </p>
<p>A <strong>mole</strong> is simply a collection of atoms. More specifically, you need to have exactly <mathjax>#6.022 * 10^(23)#</mathjax> atoms of an element in order to have <strong>one mole</strong> of that element - this is known as <em>Avogadro's number</em>. </p>
<p>In your case, you must determine how many moles of lead <em>would contain</em> <mathjax>#1.50 * 10^(12)#</mathjax> atoms of lead.</p>
<p>Well, if you know that one mole of lead must contain <mathjax>#6.022 * 10^(23)#</mathjax> atoms of lead, it follows that you get <mathjax>#1.50 * 10^12#</mathjax> atoms of lead in </p>
<blockquote>
<p><mathjax>#1.50 * 10^12 color(red)(cancel(color(black)("atoms of Pb"))) * "1 mole Pb"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms of Pb")))) = color(green)(2.49 * 10^(-12)"moles Pb")#</mathjax></p>
</blockquote>
<p>Here's an excellent video on <em>Avogadro's number</em> and <em><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a></em> featuring Professor Martyn Poliakoff from <a href="https://www.youtube.com/user/periodicvideos/about" rel="nofollow">Periodic Videos</a></p>
<p>
<iframe src="https://www.youtube.com/embed/2dzS_LXvYA0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<h1 class="questionTitle" itemprop="name">How many moles of lead are in #1.50 * 10^12# atoms of lead?</h1>
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<div class="markdown"><p><mathjax>#2.49 * 10^(-12)"moles Pb"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Before doing any calculations, it's worth noting that atoms do not <em>contain moles</em>, it's the other way around. </p>
<p>A <strong>mole</strong> is simply a collection of atoms. More specifically, you need to have exactly <mathjax>#6.022 * 10^(23)#</mathjax> atoms of an element in order to have <strong>one mole</strong> of that element - this is known as <em>Avogadro's number</em>. </p>
<p>In your case, you must determine how many moles of lead <em>would contain</em> <mathjax>#1.50 * 10^(12)#</mathjax> atoms of lead.</p>
<p>Well, if you know that one mole of lead must contain <mathjax>#6.022 * 10^(23)#</mathjax> atoms of lead, it follows that you get <mathjax>#1.50 * 10^12#</mathjax> atoms of lead in </p>
<blockquote>
<p><mathjax>#1.50 * 10^12 color(red)(cancel(color(black)("atoms of Pb"))) * "1 mole Pb"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms of Pb")))) = color(green)(2.49 * 10^(-12)"moles Pb")#</mathjax></p>
</blockquote>
<p>Here's an excellent video on <em>Avogadro's number</em> and <em><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a></em> featuring Professor Martyn Poliakoff from <a href="https://www.youtube.com/user/periodicvideos/about" rel="nofollow">Periodic Videos</a></p>
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<iframe src="https://www.youtube.com/embed/2dzS_LXvYA0?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | How many moles of lead are in #1.50 * 10^12# atoms of lead? | null |
1,222 | abe62bc2-6ddd-11ea-9134-ccda262736ce | https://socratic.org/questions/what-volume-of-oxygen-is-needed-to-react-with-solid-sulfur-to-form-6-20-l-of-sul | 6.20 L | start physical_unit 3 3 volume l qc_end physical_unit 16 17 13 14 volume qc_end substance 9 10 qc_end end | [{"type":"physical unit","value":"Volume [OF] oxygen [IN] L"}] | [{"type":"physical unit","value":"6.20 L"}] | [{"type":"physical unit","value":"Volume [OF] sulfur dioxide [=] \\pu{6.20 L}"},{"type":"substance name","value":"Solid sulfur"}] | <h1 class="questionTitle" itemprop="name">What volume of oxygen is needed to react with solid sulfur to form 6.20 L of sulfur dioxide?</h1> | null | 6.20 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given that the reaction proceeds quantitatively, whatever the conditions of temperature and pressure, if there were <mathjax>#6.20*L#</mathjax> of <mathjax>#SO_2(g)#</mathjax>, there were <mathjax>#6.20*L#</mathjax> <mathjax>#"dioxygen"#</mathjax> required to make it. <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">Stoichiometry</a> demands this. </p>
<p>How much oxygen is required to form <mathjax>#6.20*L#</mathjax> of <mathjax>#SO_3(g)#</mathjax> from elemental sulfur?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#S(s) + O_2(g) rarr SO_2(g)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given that the reaction proceeds quantitatively, whatever the conditions of temperature and pressure, if there were <mathjax>#6.20*L#</mathjax> of <mathjax>#SO_2(g)#</mathjax>, there were <mathjax>#6.20*L#</mathjax> <mathjax>#"dioxygen"#</mathjax> required to make it. <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">Stoichiometry</a> demands this. </p>
<p>How much oxygen is required to form <mathjax>#6.20*L#</mathjax> of <mathjax>#SO_3(g)#</mathjax> from elemental sulfur?</p></div>
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<div class="markdown"><p><mathjax>#S(s) + O_2(g) rarr SO_2(g)#</mathjax></p></div>
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<div class="markdown"><p>Given that the reaction proceeds quantitatively, whatever the conditions of temperature and pressure, if there were <mathjax>#6.20*L#</mathjax> of <mathjax>#SO_2(g)#</mathjax>, there were <mathjax>#6.20*L#</mathjax> <mathjax>#"dioxygen"#</mathjax> required to make it. <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">Stoichiometry</a> demands this. </p>
<p>How much oxygen is required to form <mathjax>#6.20*L#</mathjax> of <mathjax>#SO_3(g)#</mathjax> from elemental sulfur?</p></div>
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</article> | What volume of oxygen is needed to react with solid sulfur to form 6.20 L of sulfur dioxide? | null |
1,223 | aa38f909-6ddd-11ea-9e06-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-solute-dissolved-in-10-0-g-of-a-5-00-sugar-solution | 0.50 grams | start physical_unit 5 5 mass g qc_end physical_unit 13 14 8 9 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] solute [IN] grams"}] | [{"type":"physical unit","value":"0.50 grams"}] | [{"type":"physical unit","value":"Mass [OF] sugar solution [=] \\pu{10.0 g}"},{"type":"physical unit","value":"Pecentage [OF] solute in sugar solution [=] \\pu{5.00%}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of solute dissolved in 10.0 g of a 5.00% sugar solution? </h1> | null | 0.50 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since 5 % means 5 gm sugar is in 100 gm<br/>
or 100 gm sugar solution contains 5 gm of sugar<br/>
so 10 gm of solution will have <strong>0.5 gm of sugar solute</strong>.</p></div>
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<div class="markdown"><p>0.5 gm</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since 5 % means 5 gm sugar is in 100 gm<br/>
or 100 gm sugar solution contains 5 gm of sugar<br/>
so 10 gm of solution will have <strong>0.5 gm of sugar solute</strong>.</p></div>
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<div class="markdown"><p>0.5 gm</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Since 5 % means 5 gm sugar is in 100 gm<br/>
or 100 gm sugar solution contains 5 gm of sugar<br/>
so 10 gm of solution will have <strong>0.5 gm of sugar solute</strong>.</p></div>
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</article> | What is the mass of solute dissolved in 10.0 g of a 5.00% sugar solution? | null |
1,224 | acf4bfe4-6ddd-11ea-aa2b-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-6-2-x-10-2-mol-l-solution-of-nitric-acid | 1.21 | start physical_unit 10 10 ph none qc_end physical_unit 12 13 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] nitric acid solution"}] | [{"type":"physical unit","value":"1.21"}] | [{"type":"physical unit","value":"Molarity [OF] nitric acid solution [=] \\pu{6.2 × 10^(-2) mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a 6.2 x 10-2 mol/L solution of nitric acid?</h1> | null | 1.21 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p>We know that nitric acid is a moderately strong acid for which protonolysis of the water <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> is relatively complete, i.e. the given equilibrium lies to the right...........</p>
<p><mathjax>#HNO_3(aq) + H_2O(l) rarr H_3O^+ + NO_3^-#</mathjax></p>
<p>That is the solution is stoichiometric in <mathjax>#H_3O^+#</mathjax>.</p>
<p>And we simply take <mathjax>#log_10#</mathjax>............</p>
<p><mathjax>#-log_10(6.2xx10^-2)=-(-1.207)=1.21#</mathjax>.</p>
<p>For more details, see this <a href="https://socratic.org/questions/ph-value-definition">old answer.</a> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]=1.21#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p>We know that nitric acid is a moderately strong acid for which protonolysis of the water <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> is relatively complete, i.e. the given equilibrium lies to the right...........</p>
<p><mathjax>#HNO_3(aq) + H_2O(l) rarr H_3O^+ + NO_3^-#</mathjax></p>
<p>That is the solution is stoichiometric in <mathjax>#H_3O^+#</mathjax>.</p>
<p>And we simply take <mathjax>#log_10#</mathjax>............</p>
<p><mathjax>#-log_10(6.2xx10^-2)=-(-1.207)=1.21#</mathjax>.</p>
<p>For more details, see this <a href="https://socratic.org/questions/ph-value-definition">old answer.</a> </p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a 6.2 x 10-2 mol/L solution of nitric acid?</h1>
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<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]=1.21#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax></p>
<p>We know that nitric acid is a moderately strong acid for which protonolysis of the water <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> is relatively complete, i.e. the given equilibrium lies to the right...........</p>
<p><mathjax>#HNO_3(aq) + H_2O(l) rarr H_3O^+ + NO_3^-#</mathjax></p>
<p>That is the solution is stoichiometric in <mathjax>#H_3O^+#</mathjax>.</p>
<p>And we simply take <mathjax>#log_10#</mathjax>............</p>
<p><mathjax>#-log_10(6.2xx10^-2)=-(-1.207)=1.21#</mathjax>.</p>
<p>For more details, see this <a href="https://socratic.org/questions/ph-value-definition">old answer.</a> </p></div>
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</article> | What is the pH of a 6.2 x 10-2 mol/L solution of nitric acid? | null |
1,225 | a8a68a62-6ddd-11ea-9a35-ccda262736ce | https://socratic.org/questions/a-12-0-l-sample-of-gas-is-at-stp-what-would-be-its-new-volume-if-its-pressure-wa | 31.72 L | start physical_unit 3 5 volume l qc_end physical_unit 3 5 1 2 volume qc_end physical_unit 3 5 21 22 pressure qc_end c_other STP qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas sample [IN] L"}] | [{"type":"physical unit","value":"31.72 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas sample [=] \\pu{12.0 L }"},{"type":"physical unit","value":"Pressure2 [OF] the gas sample [=] \\pu{575 mmHg}"},{"type":"other","value":"STP"},{"type":"other","value":"Its temperature was doubled."}] | <h1 class="questionTitle" itemprop="name"> A 12.0 L sample of gas is at STP. What would be its new volume if its pressure was decreased to 575 mmHg and its temperature was doubled?</h1> | null | 31.72 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let us first list the given data:</p>
<p><strong>Position 1</strong>:<br/>
STP means that <mathjax>#P=1atm=760mmHg#</mathjax> and <mathjax>#T=273K#</mathjax>.<br/>
The volume occupied by the gas is <mathjax>#V=12.0L#</mathjax>.</p>
<p><strong>Position 2</strong>:<br/>
The new conditions are:<br/>
<mathjax>#P=575mmHg#</mathjax> and <mathjax>#T=2xx273=546K#</mathjax>.<br/>
<mathjax>#V=?#</mathjax></p>
<p>To find the volume of the gas at position 2, we can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax> to be:</p>
<p><mathjax>#(PV)/T=k#</mathjax> where <mathjax>#k=nR#</mathjax>, which is constant.</p>
<p>Therefore, <mathjax>#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#</mathjax></p>
<p><mathjax>#=>V_2=(P_1V_1)/(T_1)xx(T_2)/(P_2)=(760cancel(mmHg)xx12.0L)/(273cancel(K))xx(546cancel(K))/(575cancel(mmHg))=31.7L#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#V_2=31.7L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let us first list the given data:</p>
<p><strong>Position 1</strong>:<br/>
STP means that <mathjax>#P=1atm=760mmHg#</mathjax> and <mathjax>#T=273K#</mathjax>.<br/>
The volume occupied by the gas is <mathjax>#V=12.0L#</mathjax>.</p>
<p><strong>Position 2</strong>:<br/>
The new conditions are:<br/>
<mathjax>#P=575mmHg#</mathjax> and <mathjax>#T=2xx273=546K#</mathjax>.<br/>
<mathjax>#V=?#</mathjax></p>
<p>To find the volume of the gas at position 2, we can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax> to be:</p>
<p><mathjax>#(PV)/T=k#</mathjax> where <mathjax>#k=nR#</mathjax>, which is constant.</p>
<p>Therefore, <mathjax>#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#</mathjax></p>
<p><mathjax>#=>V_2=(P_1V_1)/(T_1)xx(T_2)/(P_2)=(760cancel(mmHg)xx12.0L)/(273cancel(K))xx(546cancel(K))/(575cancel(mmHg))=31.7L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name"> A 12.0 L sample of gas is at STP. What would be its new volume if its pressure was decreased to 575 mmHg and its temperature was doubled?</h1>
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<div class="markdown"><p><mathjax>#V_2=31.7L#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let us first list the given data:</p>
<p><strong>Position 1</strong>:<br/>
STP means that <mathjax>#P=1atm=760mmHg#</mathjax> and <mathjax>#T=273K#</mathjax>.<br/>
The volume occupied by the gas is <mathjax>#V=12.0L#</mathjax>.</p>
<p><strong>Position 2</strong>:<br/>
The new conditions are:<br/>
<mathjax>#P=575mmHg#</mathjax> and <mathjax>#T=2xx273=546K#</mathjax>.<br/>
<mathjax>#V=?#</mathjax></p>
<p>To find the volume of the gas at position 2, we can rearrange the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax> to be:</p>
<p><mathjax>#(PV)/T=k#</mathjax> where <mathjax>#k=nR#</mathjax>, which is constant.</p>
<p>Therefore, <mathjax>#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#</mathjax></p>
<p><mathjax>#=>V_2=(P_1V_1)/(T_1)xx(T_2)/(P_2)=(760cancel(mmHg)xx12.0L)/(273cancel(K))xx(546cancel(K))/(575cancel(mmHg))=31.7L#</mathjax></p></div>
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</article> | A 12.0 L sample of gas is at STP. What would be its new volume if its pressure was decreased to 575 mmHg and its temperature was doubled? | null |
1,226 | a98d4436-6ddd-11ea-97fc-ccda262736ce | https://socratic.org/questions/5905fe82b72cff52318434ca | -7079.40 kJ/mol | start physical_unit 14 14 enthalpy_of_combustion kj/mol qc_end physical_unit 4 4 1 2 mass qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 4 4 18 19 temperature qc_end physical_unit 4 4 21 22 temperature qc_end end | [{"type":"physical unit","value":"Enthalpy of combustion [OF] sucrose [IN] kJ/mol"}] | [{"type":"physical unit","value":"-7079.40 kJ/mol"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{1500 g}"},{"type":"physical unit","value":"Mass [OF] sucrose [=] \\pu{1 g}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{25.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{28.3 ℃}"}] | <h1 class="questionTitle" itemprop="name">If #"1500 g"# of water was heated by the combustion of #"1 g"# of sucrose to go from #25.0^@ "C"# to #28.3^@ "C"#, what is the enthalpy of combustion in #"kJ/mol"#?</h1> | null | -7079.40 kJ/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Heat released </strong></p>
<p><mathjax>#"Increase of water temperature = 3.3 K = (28.3-25.0) K"#</mathjax></p>
<p><mathjax>#"1500 g water" = "1500 g"/"18.02 g/mol" = "83.24 mol"#</mathjax> of water</p>
<p><mathjax>#q="83.24 mol" * "75.4 J·mol"^"-1""K"^"-1" *3.3 "K" = "20700 J"#</mathjax></p>
<p>The amount of heat released by burning 1 g of sucrose is 20.7 kJ per gram of sucrose.</p>
<p>However, 1 mol of sucrose is 342 g.</p>
<p>Therefore <mathjax>#Delta H#</mathjax> can be computed:</p>
<p><mathjax>#DeltaH = "-20.7 kJ/g" * "342 g/mol" = "-7080 kJ/mol"#</mathjax> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>-7080 kJ /mol of sucrose</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Heat released </strong></p>
<p><mathjax>#"Increase of water temperature = 3.3 K = (28.3-25.0) K"#</mathjax></p>
<p><mathjax>#"1500 g water" = "1500 g"/"18.02 g/mol" = "83.24 mol"#</mathjax> of water</p>
<p><mathjax>#q="83.24 mol" * "75.4 J·mol"^"-1""K"^"-1" *3.3 "K" = "20700 J"#</mathjax></p>
<p>The amount of heat released by burning 1 g of sucrose is 20.7 kJ per gram of sucrose.</p>
<p>However, 1 mol of sucrose is 342 g.</p>
<p>Therefore <mathjax>#Delta H#</mathjax> can be computed:</p>
<p><mathjax>#DeltaH = "-20.7 kJ/g" * "342 g/mol" = "-7080 kJ/mol"#</mathjax> </p></div>
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<h1 class="questionTitle" itemprop="name">If #"1500 g"# of water was heated by the combustion of #"1 g"# of sucrose to go from #25.0^@ "C"# to #28.3^@ "C"#, what is the enthalpy of combustion in #"kJ/mol"#?</h1>
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<div class="markdown"><p>-7080 kJ /mol of sucrose</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p><strong>Heat released </strong></p>
<p><mathjax>#"Increase of water temperature = 3.3 K = (28.3-25.0) K"#</mathjax></p>
<p><mathjax>#"1500 g water" = "1500 g"/"18.02 g/mol" = "83.24 mol"#</mathjax> of water</p>
<p><mathjax>#q="83.24 mol" * "75.4 J·mol"^"-1""K"^"-1" *3.3 "K" = "20700 J"#</mathjax></p>
<p>The amount of heat released by burning 1 g of sucrose is 20.7 kJ per gram of sucrose.</p>
<p>However, 1 mol of sucrose is 342 g.</p>
<p>Therefore <mathjax>#Delta H#</mathjax> can be computed:</p>
<p><mathjax>#DeltaH = "-20.7 kJ/g" * "342 g/mol" = "-7080 kJ/mol"#</mathjax> </p></div>
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</article> | If #"1500 g"# of water was heated by the combustion of #"1 g"# of sucrose to go from #25.0^@ "C"# to #28.3^@ "C"#, what is the enthalpy of combustion in #"kJ/mol"#? | null |
1,227 | a894d286-6ddd-11ea-885e-ccda262736ce | https://socratic.org/questions/how-do-you-find-the-ph-of-a-solution-whose-h-3-44-times-10-12 | 11.5 | start physical_unit 8 8 ph none qc_end physical_unit 8 8 12 14 [h+] qc_end end | [{"type":"physical unit","value":"pH [OF] solution"}] | [{"type":"physical unit","value":"11.5"}] | [{"type":"physical unit","value":"[H+] [OF] solution [=] \\pu{3.44 × 10^(−12)}"}] | <h1 class="questionTitle" itemprop="name">How do you find the pH of a solution whose #[H^+] = 3.44 times 10^-12?#?</h1> | null | 11.5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So here <mathjax>#pH=-log_10(3.44xx10^-12).............#</mathjax></p>
<p><mathjax>#=-(-11.46)=11.5#</mathjax></p>
<p>For a few more details see <a href="https://socratic.org/questions/what-is-the-result-of-dissociation-of-water">this old answer.</a> </p></div>
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<div class="markdown"><p>Well, by definition <mathjax>#pH=-log_10[H_3O^+].............#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So here <mathjax>#pH=-log_10(3.44xx10^-12).............#</mathjax></p>
<p><mathjax>#=-(-11.46)=11.5#</mathjax></p>
<p>For a few more details see <a href="https://socratic.org/questions/what-is-the-result-of-dissociation-of-water">this old answer.</a> </p></div>
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<h1 class="questionTitle" itemprop="name">How do you find the pH of a solution whose #[H^+] = 3.44 times 10^-12?#?</h1>
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<div class="markdown"><p>Well, by definition <mathjax>#pH=-log_10[H_3O^+].............#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So here <mathjax>#pH=-log_10(3.44xx10^-12).............#</mathjax></p>
<p><mathjax>#=-(-11.46)=11.5#</mathjax></p>
<p>For a few more details see <a href="https://socratic.org/questions/what-is-the-result-of-dissociation-of-water">this old answer.</a> </p></div>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> = 11.5</p></div>
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<div class="markdown"><p><img alt="My own work (Mert METİN)" src="https://useruploads.socratic.org/82fX22inTL6sAJvvNYmo_qwdsa.png"/> </p></div>
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</article> | How do you find the pH of a solution whose #[H^+] = 3.44 times 10^-12?#? | null |
1,228 | ab7312e9-6ddd-11ea-b7cd-ccda262736ce | https://socratic.org/questions/a-350-ml-solution-contains-12-6-g-nacl-what-is-the-molarity-of-the-solution | 0.62 M | start physical_unit 7 7 molarity mol/l qc_end physical_unit 3 3 1 2 volume qc_end physical_unit 7 7 5 6 mass qc_end end | [{"type":"physical unit","value":"Molarity [OF] NaCl solution [IN] M"}] | [{"type":"physical unit","value":"0.62 M"}] | [{"type":"physical unit","value":"Volume [OF] NaCl solution [=] \\pu{350 mL}"},{"type":"physical unit","value":"Mass [OF] NaCl [=] \\pu{12.6 g}"}] | <h1 class="questionTitle" itemprop="name">A 350-mL solution contains 12.6 g #NaCl#. What is the molarity of the solution?</h1> | null | 0.62 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's use this equation:</p>
<p><img alt="www.buzzle.com" src="https://useruploads.socratic.org/Z37DL3yCSpawMhOJPGL3_molarity-formula.jpg"/> </p>
<p>Okay, so we have a bit of work to do! The volume is given in terms of mL and we have to convert that to L by using the following conversion factor: <mathjax>#1L = 1000mL#</mathjax></p>
<p><mathjax>#350cancel"mL"xx(1L)/(1000cancel"mL") = 0.350L#</mathjax></p>
<p>Now we have to express NaCl in terms of moles. We can do this by using the molar mass of NaCl (58.44 g/mol) as a conversion factor.</p>
<p><mathjax>#12.6 cancelgxx(1mol)/(58.44g) = 0.216mol#</mathjax></p>
<p>Now we can divide the moles by the volume to obtain the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> like this:</p>
<p><mathjax># Molarity = (0.216mol)/(0.350L)#</mathjax></p>
<p>M =<strong>0.617</strong></p></div>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution is <strong>0.617 M.</strong></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's use this equation:</p>
<p><img alt="www.buzzle.com" src="https://useruploads.socratic.org/Z37DL3yCSpawMhOJPGL3_molarity-formula.jpg"/> </p>
<p>Okay, so we have a bit of work to do! The volume is given in terms of mL and we have to convert that to L by using the following conversion factor: <mathjax>#1L = 1000mL#</mathjax></p>
<p><mathjax>#350cancel"mL"xx(1L)/(1000cancel"mL") = 0.350L#</mathjax></p>
<p>Now we have to express NaCl in terms of moles. We can do this by using the molar mass of NaCl (58.44 g/mol) as a conversion factor.</p>
<p><mathjax>#12.6 cancelgxx(1mol)/(58.44g) = 0.216mol#</mathjax></p>
<p>Now we can divide the moles by the volume to obtain the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> like this:</p>
<p><mathjax># Molarity = (0.216mol)/(0.350L)#</mathjax></p>
<p>M =<strong>0.617</strong></p></div>
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<h1 class="questionTitle" itemprop="name">A 350-mL solution contains 12.6 g #NaCl#. What is the molarity of the solution?</h1>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the solution is <strong>0.617 M.</strong></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's use this equation:</p>
<p><img alt="www.buzzle.com" src="https://useruploads.socratic.org/Z37DL3yCSpawMhOJPGL3_molarity-formula.jpg"/> </p>
<p>Okay, so we have a bit of work to do! The volume is given in terms of mL and we have to convert that to L by using the following conversion factor: <mathjax>#1L = 1000mL#</mathjax></p>
<p><mathjax>#350cancel"mL"xx(1L)/(1000cancel"mL") = 0.350L#</mathjax></p>
<p>Now we have to express NaCl in terms of moles. We can do this by using the molar mass of NaCl (58.44 g/mol) as a conversion factor.</p>
<p><mathjax>#12.6 cancelgxx(1mol)/(58.44g) = 0.216mol#</mathjax></p>
<p>Now we can divide the moles by the volume to obtain the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> like this:</p>
<p><mathjax># Molarity = (0.216mol)/(0.350L)#</mathjax></p>
<p>M =<strong>0.617</strong></p></div>
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</article> | A 350-mL solution contains 12.6 g #NaCl#. What is the molarity of the solution? | null |
1,229 | a8aa0712-6ddd-11ea-9408-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-kcl-solution-made-by-dissolving-24-7-g-of-kcl-in-a-tot | 0.66 mol/L | start physical_unit 6 7 molarity mol/l qc_end physical_unit 6 6 11 12 mass qc_end physical_unit 6 7 20 21 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] KCl solution [IN] mol/L"}] | [{"type":"physical unit","value":"0.66 mol/L"}] | [{"type":"physical unit","value":"Mass [OF] KCl [=] \\pu{24.7 g}"},{"type":"physical unit","value":"Total volume [OF] KCl solution [=] \\pu{500 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a #KCl# solution made by dissolving 24.7 g of #KCl# in a total volume of 500. mL?</h1> | null | 0.66 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to calculate the <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of a solution, you need to know two things</p>
<blockquote>
<ul>
<li><em>the <strong>number of moles of solute</strong> present in solution</em></li>
<li><em>the <strong>total volume</strong> of the solution</em></li>
</ul>
</blockquote>
<p>The problem provides you with a <mathjax>#"24.7-g"#</mathjax> sample of potassium chloride, <mathjax>#"KCl"#</mathjax>, and a <strong>total volume</strong> of a solution of <mathjax>#"500. mL"#</mathjax>.</p>
<p>In order to find the <em>number of moles</em> of potassium chloride, your <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, use the compound's <strong>molar mass</strong>, which as you know tells you the mass of <strong>one mole</strong> of potassium chloride</p>
<blockquote>
<p><mathjax>#24.7 color(red)(cancel(color(black)("g"))) * overbrace("1 mole KCl"/(74.55color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of KCl")) = "0.3313 moles KCl"#</mathjax></p>
</blockquote>
<p>Now, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is expressed <strong>per liter</strong> of solution. Since you dissolve <mathjax>#0.3313#</mathjax> moles of potassium chloride in <mathjax>#"500. mL"#</mathjax> of solution, you can say that <mathjax>#"1.0 L"#</mathjax> will contain</p>
<blockquote>
<p><mathjax>#1.0color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.3313 moles"/(500. color(red)(cancel(color(black)("mL")))) = "0.6626 moles KCl"#</mathjax></p>
</blockquote>
<p>This means that the molarity of the solution will be </p>
<blockquote>
<p><mathjax>#c = color(green)("0.663 mol L"^(-1))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.663 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to calculate the <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of a solution, you need to know two things</p>
<blockquote>
<ul>
<li><em>the <strong>number of moles of solute</strong> present in solution</em></li>
<li><em>the <strong>total volume</strong> of the solution</em></li>
</ul>
</blockquote>
<p>The problem provides you with a <mathjax>#"24.7-g"#</mathjax> sample of potassium chloride, <mathjax>#"KCl"#</mathjax>, and a <strong>total volume</strong> of a solution of <mathjax>#"500. mL"#</mathjax>.</p>
<p>In order to find the <em>number of moles</em> of potassium chloride, your <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, use the compound's <strong>molar mass</strong>, which as you know tells you the mass of <strong>one mole</strong> of potassium chloride</p>
<blockquote>
<p><mathjax>#24.7 color(red)(cancel(color(black)("g"))) * overbrace("1 mole KCl"/(74.55color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of KCl")) = "0.3313 moles KCl"#</mathjax></p>
</blockquote>
<p>Now, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is expressed <strong>per liter</strong> of solution. Since you dissolve <mathjax>#0.3313#</mathjax> moles of potassium chloride in <mathjax>#"500. mL"#</mathjax> of solution, you can say that <mathjax>#"1.0 L"#</mathjax> will contain</p>
<blockquote>
<p><mathjax>#1.0color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.3313 moles"/(500. color(red)(cancel(color(black)("mL")))) = "0.6626 moles KCl"#</mathjax></p>
</blockquote>
<p>This means that the molarity of the solution will be </p>
<blockquote>
<p><mathjax>#c = color(green)("0.663 mol L"^(-1))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://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 is the molarity of a #KCl# solution made by dissolving 24.7 g of #KCl# in a total volume of 500. mL?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"0.663 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to calculate the <strong><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of a solution, you need to know two things</p>
<blockquote>
<ul>
<li><em>the <strong>number of moles of solute</strong> present in solution</em></li>
<li><em>the <strong>total volume</strong> of the solution</em></li>
</ul>
</blockquote>
<p>The problem provides you with a <mathjax>#"24.7-g"#</mathjax> sample of potassium chloride, <mathjax>#"KCl"#</mathjax>, and a <strong>total volume</strong> of a solution of <mathjax>#"500. mL"#</mathjax>.</p>
<p>In order to find the <em>number of moles</em> of potassium chloride, your <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, use the compound's <strong>molar mass</strong>, which as you know tells you the mass of <strong>one mole</strong> of potassium chloride</p>
<blockquote>
<p><mathjax>#24.7 color(red)(cancel(color(black)("g"))) * overbrace("1 mole KCl"/(74.55color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of KCl")) = "0.3313 moles KCl"#</mathjax></p>
</blockquote>
<p>Now, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> is expressed <strong>per liter</strong> of solution. Since you dissolve <mathjax>#0.3313#</mathjax> moles of potassium chloride in <mathjax>#"500. mL"#</mathjax> of solution, you can say that <mathjax>#"1.0 L"#</mathjax> will contain</p>
<blockquote>
<p><mathjax>#1.0color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.3313 moles"/(500. color(red)(cancel(color(black)("mL")))) = "0.6626 moles KCl"#</mathjax></p>
</blockquote>
<p>This means that the molarity of the solution will be </p>
<blockquote>
<p><mathjax>#c = color(green)("0.663 mol L"^(-1))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | What is the molarity of a #KCl# solution made by dissolving 24.7 g of #KCl# in a total volume of 500. mL? | null |
1,230 | ac7fa62d-6ddd-11ea-a47c-ccda262736ce | https://socratic.org/questions/546d5d1d581e2a0a5d3bb0a2 | 394 mL | start physical_unit 19 20 volume ml qc_end physical_unit 19 20 6 7 volume qc_end physical_unit 19 20 9 10 temperature qc_end physical_unit 19 20 12 13 pressure qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"394 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{435 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{98.7 kPa}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A gas has a volume of 435 mL at #25^@"C"# and #"98.7 kPa"#. What would the volume of the gas be at STP? </h1> | null | 394 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This problem can be solved using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> with the equation:</p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>The temperature scale used in the <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> is the Kelvin scale, so the Celsius temperatures will need to be converted to Kelvins.</p>
<p><strong>STP</strong> for the gas laws are <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"100 kPa"#</mathjax>. </p>
<p><strong>Given/Known:</strong></p>
<p><mathjax>#V_1= "435mL"#</mathjax></p>
<p><mathjax>#T_1 = "25"^o"C" + "273.15" = "298K"#</mathjax></p>
<p><mathjax>#P_1 = "98.7 kPa"#</mathjax></p>
<p><mathjax>#T_2 = "273.15K"#</mathjax></p>
<p><mathjax>#P_2 = "100 kPa"#</mathjax></p>
<p><strong>Unknown:</strong></p>
<p><mathjax>#"V"_2#</mathjax></p>
<p><strong>Solution:</strong></p>
<p>Rearrange the formula to isolate <mathjax>#V_2#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#"V"_2 = (V"_1P_1T_2)/(P_2T_1)#</mathjax></p>
<p><mathjax>#V_2 = ((435"mL")xx(98.7"kPa")xx(273.15"K"))/((100"kPa")xx(298"K")) = "394mL"#</mathjax> (rounded to three significant figures)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new volume would be <mathjax>#"394 mL"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This problem can be solved using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> with the equation:</p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>The temperature scale used in the <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> is the Kelvin scale, so the Celsius temperatures will need to be converted to Kelvins.</p>
<p><strong>STP</strong> for the gas laws are <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"100 kPa"#</mathjax>. </p>
<p><strong>Given/Known:</strong></p>
<p><mathjax>#V_1= "435mL"#</mathjax></p>
<p><mathjax>#T_1 = "25"^o"C" + "273.15" = "298K"#</mathjax></p>
<p><mathjax>#P_1 = "98.7 kPa"#</mathjax></p>
<p><mathjax>#T_2 = "273.15K"#</mathjax></p>
<p><mathjax>#P_2 = "100 kPa"#</mathjax></p>
<p><strong>Unknown:</strong></p>
<p><mathjax>#"V"_2#</mathjax></p>
<p><strong>Solution:</strong></p>
<p>Rearrange the formula to isolate <mathjax>#V_2#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#"V"_2 = (V"_1P_1T_2)/(P_2T_1)#</mathjax></p>
<p><mathjax>#V_2 = ((435"mL")xx(98.7"kPa")xx(273.15"K"))/((100"kPa")xx(298"K")) = "394mL"#</mathjax> (rounded to three significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">A gas has a volume of 435 mL at #25^@"C"# and #"98.7 kPa"#. What would the volume of the gas be at STP? </h1>
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<div class="markdown"><p>The new volume would be <mathjax>#"394 mL"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This problem can be solved using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> with the equation:</p>
<p><mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax></p>
<p>The temperature scale used in the <a href="http://socratic.org/chemistry/the-behavior-of-gases/gas-laws">gas laws</a> is the Kelvin scale, so the Celsius temperatures will need to be converted to Kelvins.</p>
<p><strong>STP</strong> for the gas laws are <mathjax>#"273.15 K"#</mathjax> and <mathjax>#"100 kPa"#</mathjax>. </p>
<p><strong>Given/Known:</strong></p>
<p><mathjax>#V_1= "435mL"#</mathjax></p>
<p><mathjax>#T_1 = "25"^o"C" + "273.15" = "298K"#</mathjax></p>
<p><mathjax>#P_1 = "98.7 kPa"#</mathjax></p>
<p><mathjax>#T_2 = "273.15K"#</mathjax></p>
<p><mathjax>#P_2 = "100 kPa"#</mathjax></p>
<p><strong>Unknown:</strong></p>
<p><mathjax>#"V"_2#</mathjax></p>
<p><strong>Solution:</strong></p>
<p>Rearrange the formula to isolate <mathjax>#V_2#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#"V"_2 = (V"_1P_1T_2)/(P_2T_1)#</mathjax></p>
<p><mathjax>#V_2 = ((435"mL")xx(98.7"kPa")xx(273.15"K"))/((100"kPa")xx(298"K")) = "394mL"#</mathjax> (rounded to three significant figures)</p></div>
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</article> | A gas has a volume of 435 mL at #25^@"C"# and #"98.7 kPa"#. What would the volume of the gas be at STP? | null |
1,231 | a92f493a-6ddd-11ea-b440-ccda262736ce | https://socratic.org/questions/zn-s-2hcl-aq-h-2-g-zncl-2-aq-heat-energy-if-30-0-g-of-zn-react-how-many-grams-of | 1.85 grams | start physical_unit 5 5 mass g qc_end physical_unit 0 0 12 13 mass qc_end chemical_equation 0 10 qc_end end | [{"type":"physical unit","value":"Mass [OF] H2 [IN] grams"}] | [{"type":"physical unit","value":"1.85 grams"}] | [{"type":"physical unit","value":"Mass [OF] Zn [=] \\pu{30.0 g}"},{"type":"chemical equation","value":"Zn(s) + 2 HCl(aq) -> H2(g) + ZnCl2(aq) + heat energy"}] | <h1 class="questionTitle" itemprop="name">#Zn(s) + 2HCl(aq) -> H_2 (g) + ZnCl_2(aq) + #heat energy.
If 30.0 g of #Zn# react, how many grams of #H_2# will form?</h1> | null | 1.85 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>1. Write the balanced equation</strong>.</p>
<p><mathjax>#"Zn" + "2HCl" → "ZnCl"_2 + "H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Calculate the moles of <mathjax>#"Zn"#</mathjax></strong>.</p>
<p><mathjax>#"Moles of Zn" = 30.0 color(red)(cancel(color(black)("g Zn"))) × ("1 mol Zn")/(65.38 color(red)(cancel(color(black)("g Zn")))) = "0.4589 mol Zn"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. Calculate the moles of <mathjax>#"H"_2#</mathjax></strong>.</p>
<p><mathjax>#"Moles of H"_2 = 0.4589 color(red)(cancel(color(black)("mol Zn"))) × ("2 mol H"_2)/(1 color(red)(cancel(color(black)("mol Zn")))) = "0.9177 mol H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Calculate the mass of <mathjax>#"H"_2#</mathjax></strong>.</p>
<p><mathjax>#"Mass of H"_2 = 0.9177 color(red)(cancel(color(black)("mol H"_2))) × ("2.016 g H"_2)/(1 color(red)(cancel(color(black)("mol H"_2)))) = "1.85 g H"_2#</mathjax></p></div>
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<div class="markdown"><p>1.85 g of hydrogen will form.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>1. Write the balanced equation</strong>.</p>
<p><mathjax>#"Zn" + "2HCl" → "ZnCl"_2 + "H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Calculate the moles of <mathjax>#"Zn"#</mathjax></strong>.</p>
<p><mathjax>#"Moles of Zn" = 30.0 color(red)(cancel(color(black)("g Zn"))) × ("1 mol Zn")/(65.38 color(red)(cancel(color(black)("g Zn")))) = "0.4589 mol Zn"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. Calculate the moles of <mathjax>#"H"_2#</mathjax></strong>.</p>
<p><mathjax>#"Moles of H"_2 = 0.4589 color(red)(cancel(color(black)("mol Zn"))) × ("2 mol H"_2)/(1 color(red)(cancel(color(black)("mol Zn")))) = "0.9177 mol H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Calculate the mass of <mathjax>#"H"_2#</mathjax></strong>.</p>
<p><mathjax>#"Mass of H"_2 = 0.9177 color(red)(cancel(color(black)("mol H"_2))) × ("2.016 g H"_2)/(1 color(red)(cancel(color(black)("mol H"_2)))) = "1.85 g H"_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">#Zn(s) + 2HCl(aq) -> H_2 (g) + ZnCl_2(aq) + #heat energy.
If 30.0 g of #Zn# react, how many grams of #H_2# will form?</h1>
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<div class="markdown"><p>1.85 g of hydrogen will form.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>1. Write the balanced equation</strong>.</p>
<p><mathjax>#"Zn" + "2HCl" → "ZnCl"_2 + "H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>2. Calculate the moles of <mathjax>#"Zn"#</mathjax></strong>.</p>
<p><mathjax>#"Moles of Zn" = 30.0 color(red)(cancel(color(black)("g Zn"))) × ("1 mol Zn")/(65.38 color(red)(cancel(color(black)("g Zn")))) = "0.4589 mol Zn"#</mathjax></p>
<blockquote></blockquote>
<p><strong>3. Calculate the moles of <mathjax>#"H"_2#</mathjax></strong>.</p>
<p><mathjax>#"Moles of H"_2 = 0.4589 color(red)(cancel(color(black)("mol Zn"))) × ("2 mol H"_2)/(1 color(red)(cancel(color(black)("mol Zn")))) = "0.9177 mol H"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>4. Calculate the mass of <mathjax>#"H"_2#</mathjax></strong>.</p>
<p><mathjax>#"Mass of H"_2 = 0.9177 color(red)(cancel(color(black)("mol H"_2))) × ("2.016 g H"_2)/(1 color(red)(cancel(color(black)("mol H"_2)))) = "1.85 g H"_2#</mathjax></p></div>
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</article> | #Zn(s) + 2HCl(aq) -> H_2 (g) + ZnCl_2(aq) + #heat energy.
If 30.0 g of #Zn# react, how many grams of #H_2# will form? | null |
1,232 | a931935a-6ddd-11ea-ba0e-ccda262736ce | https://socratic.org/questions/a-4-0-g-sample-of-iron-was-heated-from-0-c-to-20-c-it-absorbed-35-2-j-of-energy- | 0.44 J/(g * ℃) | start physical_unit 28 31 specific_heat j/(°c_·_g) qc_end physical_unit 3 5 1 2 mass qc_end physical_unit 3 5 9 10 temperature qc_end physical_unit 3 5 12 13 temperature qc_end physical_unit 3 5 16 17 heat_energy qc_end end | [{"type":"physical unit","value":"Specific heat [OF] this piece of iron [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"0.44 J/(g * ℃)"}] | [{"type":"physical unit","value":"Mass [OF] iron sample [=] \\pu{4.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] iron sample [=] \\pu{0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] iron sample [=] \\pu{20 ℃}"},{"type":"physical unit","value":"Absorbed energy [OF] iron sample [=] \\pu{35.2 J}"}] | <h1 class="questionTitle" itemprop="name">A 4.0 g sample of iron was heated from 0°C to 20°C. It absorbed 35.2 J of energy as heat. What is the specific heat of this piece of iron? </h1> | null | 0.44 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for the heat absorbed by a substance is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) q = mcΔT color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#q#</mathjax> is the quantity of heat</p>
<p><mathjax>#m#</mathjax> is the mass of the substance</p>
<p><mathjax>#c#</mathjax> is the specific heat capacity of the material</p>
<p><mathjax>#ΔT#</mathjax> is the temperature change</p>
<blockquote></blockquote>
<p>You can rearrange the formula to calculate the specific heat capacity:</p>
<p><mathjax>#c = q/(mΔT)#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#ΔT = "(20 - 0) °C" = "20 °C"#</mathjax></p>
<p><mathjax>#c = "35.2 J"/"4.0 g × 20 °C" = "0.44 J·°C"^"-1""g"^"-1"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of the iron is <mathjax>#"0.44 J·°C"^"-1""g"^"-1"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for the heat absorbed by a substance is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) q = mcΔT color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#q#</mathjax> is the quantity of heat</p>
<p><mathjax>#m#</mathjax> is the mass of the substance</p>
<p><mathjax>#c#</mathjax> is the specific heat capacity of the material</p>
<p><mathjax>#ΔT#</mathjax> is the temperature change</p>
<blockquote></blockquote>
<p>You can rearrange the formula to calculate the specific heat capacity:</p>
<p><mathjax>#c = q/(mΔT)#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#ΔT = "(20 - 0) °C" = "20 °C"#</mathjax></p>
<p><mathjax>#c = "35.2 J"/"4.0 g × 20 °C" = "0.44 J·°C"^"-1""g"^"-1"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 4.0 g sample of iron was heated from 0°C to 20°C. It absorbed 35.2 J of energy as heat. What is the specific heat of this piece of iron? </h1>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity of the iron is <mathjax>#"0.44 J·°C"^"-1""g"^"-1"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The formula for the heat absorbed by a substance is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) q = mcΔT color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>where</p>
<p><mathjax>#q#</mathjax> is the quantity of heat</p>
<p><mathjax>#m#</mathjax> is the mass of the substance</p>
<p><mathjax>#c#</mathjax> is the specific heat capacity of the material</p>
<p><mathjax>#ΔT#</mathjax> is the temperature change</p>
<blockquote></blockquote>
<p>You can rearrange the formula to calculate the specific heat capacity:</p>
<p><mathjax>#c = q/(mΔT)#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#ΔT = "(20 - 0) °C" = "20 °C"#</mathjax></p>
<p><mathjax>#c = "35.2 J"/"4.0 g × 20 °C" = "0.44 J·°C"^"-1""g"^"-1"#</mathjax></p></div>
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</article> | A 4.0 g sample of iron was heated from 0°C to 20°C. It absorbed 35.2 J of energy as heat. What is the specific heat of this piece of iron? | null |
1,233 | aca5fd86-6ddd-11ea-8c88-ccda262736ce | https://socratic.org/questions/58d2fa117c01491a168abfca | 54.30 grams | start physical_unit 4 4 mass g qc_end physical_unit 11 11 9 10 mass qc_end physical_unit 19 20 14 14 mass_percent qc_end end | [{"type":"physical unit","value":"Mass [OF] solute [IN] grams"}] | [{"type":"physical unit","value":"54.30 grams"}] | [{"type":"physical unit","value":"Mass [OF] sample [=] \\pu{1500 g}"},{"type":"physical unit","value":"Percent by mass [OF] sodium hypochlorite in solution [=] \\pu{3.62%}"}] | <h1 class="questionTitle" itemprop="name">How many grams of solute are present in a #"1500-g"# sample of a #3.62%# by mass solution of sodium hypochlorite?</h1> | null | 54.30 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A solution's <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> <strong>by mass</strong> is a measure of the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>In your case, a solution that is <mathjax>#3.62%#</mathjax> <strong>by mass</strong> sodium hypochlorite will contain <mathjax>#"3.62 g"#</mathjax> of sodium hypochlorite, the solute, <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>As you know, <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>, which implies that they have the same composition throughout. This allows you to use the percent concentration by mass as a <strong>conversion factor</strong> to help you find the number of grams of solute present in <mathjax>#'1500 g"#</mathjax> of solution. </p>
<p>More specifically, you will have</p>
<blockquote>
<p><mathjax>#1500 color(red)(cancel(color(black)("g solution"))) * "3.62 g NaClO"/(100color(red)(cancel(color(black)("g solution")))) = color(darkgreen)(ul(color(black)("54 g NaClO")))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"54 g NaClO"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A solution's <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> <strong>by mass</strong> is a measure of the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>In your case, a solution that is <mathjax>#3.62%#</mathjax> <strong>by mass</strong> sodium hypochlorite will contain <mathjax>#"3.62 g"#</mathjax> of sodium hypochlorite, the solute, <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>As you know, <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>, which implies that they have the same composition throughout. This allows you to use the percent concentration by mass as a <strong>conversion factor</strong> to help you find the number of grams of solute present in <mathjax>#'1500 g"#</mathjax> of solution. </p>
<p>More specifically, you will have</p>
<blockquote>
<p><mathjax>#1500 color(red)(cancel(color(black)("g solution"))) * "3.62 g NaClO"/(100color(red)(cancel(color(black)("g solution")))) = color(darkgreen)(ul(color(black)("54 g NaClO")))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of the solution. </p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of solute are present in a #"1500-g"# sample of a #3.62%# by mass solution of sodium hypochlorite?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2017-03-23T00:02:41" itemprop="dateCreated">
Mar 23, 2017
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<div class="markdown"><p><mathjax>#"54 g NaClO"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>A solution's <a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> <strong>by mass</strong> is a measure of the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>In your case, a solution that is <mathjax>#3.62%#</mathjax> <strong>by mass</strong> sodium hypochlorite will contain <mathjax>#"3.62 g"#</mathjax> of sodium hypochlorite, the solute, <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>As you know, <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>, which implies that they have the same composition throughout. This allows you to use the percent concentration by mass as a <strong>conversion factor</strong> to help you find the number of grams of solute present in <mathjax>#'1500 g"#</mathjax> of solution. </p>
<p>More specifically, you will have</p>
<blockquote>
<p><mathjax>#1500 color(red)(cancel(color(black)("g solution"))) * "3.62 g NaClO"/(100color(red)(cancel(color(black)("g solution")))) = color(darkgreen)(ul(color(black)("54 g NaClO")))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the mass of the solution. </p></div>
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</article> | How many grams of solute are present in a #"1500-g"# sample of a #3.62%# by mass solution of sodium hypochlorite? | null |
1,234 | a96c230a-6ddd-11ea-9ab7-ccda262736ce | https://socratic.org/questions/what-is-the-weak-base-ionization-constant-k-b-for-hs-equal-to | 10^(-7.1) | start physical_unit 9 9 kb none qc_end chemical_equation 9 9 qc_end end | [{"type":"physical unit","value":"Kb [OF] HS-"}] | [{"type":"physical unit","value":"10^(-7.1)"}] | [{"type":"chemical equation","value":"HS-"}] | <h1 class="questionTitle" itemprop="name">What is the weak base ionization constant (#K_b#) for #HS^-# equal to?</h1> | null | 10^(-7.1) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hydrogen sulfide dissociates according to the reaction......</p>
<p><mathjax>#H_2S(aq) + H_2O rightleftharpoonsHS^(-) + H_3O^+#</mathjax></p>
<p>Now <mathjax>#K_a(H_2S)=10^(-6.9)=1.25xx10^-7#</mathjax>, and of course <mathjax>#pK_a=6.9#</mathjax>.</p>
<p>In aqueous solution, we know that <mathjax>#pK_a+pK_b=14#</mathjax></p>
<p>And thus <mathjax>#pK_b=14-pK_a=14-6.9=7.1#</mathjax></p>
<p><mathjax>#K_b=10^(-7.1)#</mathjax>...........<mathjax>#=??#</mathjax></p>
<p>Note that the question SHOULD have quoted at least <mathjax>#pK_(a1)#</mathjax> for <mathjax>#H_2S#</mathjax>. <mathjax>#K_a#</mathjax> and <mathjax>#K_b#</mathjax> are established by measurement; you could not possibly be expected to remember them. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well <mathjax>#pK_b=10^(14-pK_a)#</mathjax>............; and thus <mathjax>#K_b=1.25xx10^-7#</mathjax>............</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hydrogen sulfide dissociates according to the reaction......</p>
<p><mathjax>#H_2S(aq) + H_2O rightleftharpoonsHS^(-) + H_3O^+#</mathjax></p>
<p>Now <mathjax>#K_a(H_2S)=10^(-6.9)=1.25xx10^-7#</mathjax>, and of course <mathjax>#pK_a=6.9#</mathjax>.</p>
<p>In aqueous solution, we know that <mathjax>#pK_a+pK_b=14#</mathjax></p>
<p>And thus <mathjax>#pK_b=14-pK_a=14-6.9=7.1#</mathjax></p>
<p><mathjax>#K_b=10^(-7.1)#</mathjax>...........<mathjax>#=??#</mathjax></p>
<p>Note that the question SHOULD have quoted at least <mathjax>#pK_(a1)#</mathjax> for <mathjax>#H_2S#</mathjax>. <mathjax>#K_a#</mathjax> and <mathjax>#K_b#</mathjax> are established by measurement; you could not possibly be expected to remember them. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the weak base ionization constant (#K_b#) for #HS^-# equal to?</h1>
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anor277
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<div class="markdown"><p>Well <mathjax>#pK_b=10^(14-pK_a)#</mathjax>............; and thus <mathjax>#K_b=1.25xx10^-7#</mathjax>............</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hydrogen sulfide dissociates according to the reaction......</p>
<p><mathjax>#H_2S(aq) + H_2O rightleftharpoonsHS^(-) + H_3O^+#</mathjax></p>
<p>Now <mathjax>#K_a(H_2S)=10^(-6.9)=1.25xx10^-7#</mathjax>, and of course <mathjax>#pK_a=6.9#</mathjax>.</p>
<p>In aqueous solution, we know that <mathjax>#pK_a+pK_b=14#</mathjax></p>
<p>And thus <mathjax>#pK_b=14-pK_a=14-6.9=7.1#</mathjax></p>
<p><mathjax>#K_b=10^(-7.1)#</mathjax>...........<mathjax>#=??#</mathjax></p>
<p>Note that the question SHOULD have quoted at least <mathjax>#pK_(a1)#</mathjax> for <mathjax>#H_2S#</mathjax>. <mathjax>#K_a#</mathjax> and <mathjax>#K_b#</mathjax> are established by measurement; you could not possibly be expected to remember them. </p></div>
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</article> | What is the weak base ionization constant (#K_b#) for #HS^-# equal to? | null |
1,235 | a83e100c-6ddd-11ea-9f51-ccda262736ce | https://socratic.org/questions/580045eb11ef6b5b42acd730 | 73.05 g | start physical_unit 3 4 mass g qc_end physical_unit 15 15 9 10 volume qc_end physical_unit 15 15 13 14 molarity qc_end end | [{"type":"physical unit","value":"mass [OF] sodium chloride [IN] g"}] | [{"type":"physical unit","value":"73.05 g"}] | [{"type":"physical unit","value":"volume [OF] NaCl(aq) [=] \\pu{ 250 mL}"},{"type":"physical unit","value":"Molarity [OF] NaCl(aq) [=] \\pu{5.0 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What mass of sodium chloride is present in a #250*mL# volume of #5.0*mol*L^-1# #NaCl(aq)#?</h1> | null | 73.05 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the relationship:</p>
<p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p>We know <mathjax>#"Concentration"#</mathjax> and <mathjax>#"Volume"#</mathjax>, and thus </p>
<p><mathjax>#"Moles of solute"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Concentration"xx"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#5.0*mol*cancel(L^-1)xx250*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#1.25*mol#</mathjax></p>
<p>And a <mathjax>#1.25*mol#</mathjax> quantity of sodium chloride has a mass of <mathjax>#1.25*cancel(mol)xx58.44*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Mass of NaCl"~=75*g#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the relationship:</p>
<p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p>We know <mathjax>#"Concentration"#</mathjax> and <mathjax>#"Volume"#</mathjax>, and thus </p>
<p><mathjax>#"Moles of solute"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Concentration"xx"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#5.0*mol*cancel(L^-1)xx250*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#1.25*mol#</mathjax></p>
<p>And a <mathjax>#1.25*mol#</mathjax> quantity of sodium chloride has a mass of <mathjax>#1.25*cancel(mol)xx58.44*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of sodium chloride is present in a #250*mL# volume of #5.0*mol*L^-1# #NaCl(aq)#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#"Mass of NaCl"~=75*g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the relationship:</p>
<p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p>We know <mathjax>#"Concentration"#</mathjax> and <mathjax>#"Volume"#</mathjax>, and thus </p>
<p><mathjax>#"Moles of solute"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Concentration"xx"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#5.0*mol*cancel(L^-1)xx250*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#1.25*mol#</mathjax></p>
<p>And a <mathjax>#1.25*mol#</mathjax> quantity of sodium chloride has a mass of <mathjax>#1.25*cancel(mol)xx58.44*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax></p></div>
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</article> | What mass of sodium chloride is present in a #250*mL# volume of #5.0*mol*L^-1# #NaCl(aq)#? | null |
1,236 | aa2282d2-6ddd-11ea-bbc8-ccda262736ce | https://socratic.org/questions/a-balloon-is-filled-with-14-l-of-gas-at-302-k-what-is-its-temperature-in-kelvin- | 431.43 Kelvin | start physical_unit 8 8 temperature k qc_end physical_unit 8 8 5 6 volume qc_end physical_unit 8 8 10 11 temperature qc_end physical_unit 8 8 23 24 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] Kelvin"}] | [{"type":"physical unit","value":"431.43 Kelvin"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{14 L}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{302 K}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{20 L}"},{"type":"other","value":"Constant pressure."}] | <h1 class="questionTitle" itemprop="name">A balloon is filled with 14 L of gas at 302 K. What is its temperature in Kelvin when the volume expands to 20 L, assuming pressure remains constant?</h1> | null | 431.43 Kelvin | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember about <em>volume</em> and <em>temperature</em> is that they have a <strong>direct relationship</strong> when pressure and number of moles of gas <em><strong>remain constant</strong></em>. </p>
<p>This relationship is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a></strong> and states that when those conditions are met, <strong>increasing</strong> the temperature of the gas will result in an <strong>increase</strong> in volume.</p>
<p>Similarly, <strong>decreasing</strong> the temperature of the gas will result in a <strong>decrease</strong> in volume. </p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://www.bbc.co.uk/education/guides/zc4xsbk/revision/3" src="https://d2jmvrsizmvf4x.cloudfront.net/XBoTIe34T7Kfe4fZzx60_small"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Mathematically, this is expressed as</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)V_1/T_1 = V_2/T_2color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the volume and temperature of the gas at an initial state<br/>
<mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the volume and temperature of the gas at a final stat</p>
<p>In your case, the volume <strong>expanded</strong> from an initial value of <mathjax>#"14 L"#</mathjax> to a final value of <mathjax>#"20 L"#</mathjax>, which can only mean that the temperature of the gas <strong>increased</strong> as well, i.e. <mathjax>#T_2 > T_1#</mathjax>.</p>
<p>Rearrange the above equation to solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#T_2 = (20 color(red)(cancel(color(black)("L"))))/(14color(red)(cancel(color(black)("L")))) * "302 K" = "431.43 K"#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one sig fig for the final volume of the gas</p>
<blockquote>
<p><mathjax>#T_2 = color(green)(|bar(ul(color(white)(a/a)color(black)("430 K")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"430 K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember about <em>volume</em> and <em>temperature</em> is that they have a <strong>direct relationship</strong> when pressure and number of moles of gas <em><strong>remain constant</strong></em>. </p>
<p>This relationship is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a></strong> and states that when those conditions are met, <strong>increasing</strong> the temperature of the gas will result in an <strong>increase</strong> in volume.</p>
<p>Similarly, <strong>decreasing</strong> the temperature of the gas will result in a <strong>decrease</strong> in volume. </p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://www.bbc.co.uk/education/guides/zc4xsbk/revision/3" src="https://d2jmvrsizmvf4x.cloudfront.net/XBoTIe34T7Kfe4fZzx60_small"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Mathematically, this is expressed as</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)V_1/T_1 = V_2/T_2color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the volume and temperature of the gas at an initial state<br/>
<mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the volume and temperature of the gas at a final stat</p>
<p>In your case, the volume <strong>expanded</strong> from an initial value of <mathjax>#"14 L"#</mathjax> to a final value of <mathjax>#"20 L"#</mathjax>, which can only mean that the temperature of the gas <strong>increased</strong> as well, i.e. <mathjax>#T_2 > T_1#</mathjax>.</p>
<p>Rearrange the above equation to solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#T_2 = (20 color(red)(cancel(color(black)("L"))))/(14color(red)(cancel(color(black)("L")))) * "302 K" = "431.43 K"#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one sig fig for the final volume of the gas</p>
<blockquote>
<p><mathjax>#T_2 = color(green)(|bar(ul(color(white)(a/a)color(black)("430 K")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A balloon is filled with 14 L of gas at 302 K. What is its temperature in Kelvin when the volume expands to 20 L, assuming pressure remains constant?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-20T14:28:49" itemprop="dateCreated">
Jul 20, 2016
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<div class="markdown"><p><mathjax>#"430 K"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember about <em>volume</em> and <em>temperature</em> is that they have a <strong>direct relationship</strong> when pressure and number of moles of gas <em><strong>remain constant</strong></em>. </p>
<p>This relationship is known as <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a></strong> and states that when those conditions are met, <strong>increasing</strong> the temperature of the gas will result in an <strong>increase</strong> in volume.</p>
<p>Similarly, <strong>decreasing</strong> the temperature of the gas will result in a <strong>decrease</strong> in volume. </p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><img alt="http://www.bbc.co.uk/education/guides/zc4xsbk/revision/3" src="https://d2jmvrsizmvf4x.cloudfront.net/XBoTIe34T7Kfe4fZzx60_small"/> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>Mathematically, this is expressed as</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)V_1/T_1 = V_2/T_2color(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the volume and temperature of the gas at an initial state<br/>
<mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the volume and temperature of the gas at a final stat</p>
<p>In your case, the volume <strong>expanded</strong> from an initial value of <mathjax>#"14 L"#</mathjax> to a final value of <mathjax>#"20 L"#</mathjax>, which can only mean that the temperature of the gas <strong>increased</strong> as well, i.e. <mathjax>#T_2 > T_1#</mathjax>.</p>
<p>Rearrange the above equation to solve for <mathjax>#T_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#T_2 = (20 color(red)(cancel(color(black)("L"))))/(14color(red)(cancel(color(black)("L")))) * "302 K" = "431.43 K"#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one sig fig for the final volume of the gas</p>
<blockquote>
<p><mathjax>#T_2 = color(green)(|bar(ul(color(white)(a/a)color(black)("430 K")color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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1,237 | ab22e95c-6ddd-11ea-b597-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-salt-solution-made-by-dissolving-280-mg-of-nacl-in-2-0 | 2.40 M | start physical_unit 8 9 molarity mol/l qc_end physical_unit 16 16 13 14 mass qc_end physical_unit 9 9 18 19 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] salt solution [IN] M"}] | [{"type":"physical unit","value":"2.40 M"}] | [{"type":"physical unit","value":"Mass [OF] NaCl [=] \\pu{280 mg}"},{"type":"physical unit","value":"Volume [OF] solution [=] \\pu{2.00 mL}"}] | <h1 class="questionTitle" itemprop="name">How do you find the molarity of a salt solution made by dissolving 280 mg of NaCl in 2.00 mL of solution? </h1> | null | 2.40 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to determine the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of a solution, you basically need to know two things </p>
<ul>
<li><em>how many <strong>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> you have in your sample</em></li>
<li><em>the volume of this sample - expressed in <strong>liters</strong></em></li>
</ul>
<p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as </p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>Your solute is <em>sodium chloride</em>, <mathjax>#"NaCl"#</mathjax>. You know that you're dissolving a <mathjax>#"280-mg"#</mathjax> sample of sodium chloride in enough water to get a <mathjax>#"2.00-mL"#</mathjax> <em>volume of solution</em>. </p>
<p>Notice that the problem provides you with the volume of the solution, but that it is expressed in <em>milliliters</em>. This means that you will have to convert it to <strong>liters</strong> by using the conversion factor </p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>The volume of the solution will thus be</p>
<blockquote>
<p><mathjax>#2.00 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 2.00 * 10^(-3)"L"#</mathjax></p>
</blockquote>
<p>Now, in order to get the number of moles of solute, you need to use sodium chloride's <strong>molar mass</strong>, which tells you the mass of <strong>one mole</strong> of this compound.</p>
<p>Sodium chloride has a molar mass of <mathjax>#"58.44 g/mol"#</mathjax>, which tells you that <strong>one mole</strong> of sodium chloride has a mass of <mathjax>#"58.44 g"#</mathjax>. </p>
<p>Notice that the sample you have is expressed in <em>milligrams</em>. Once again, a <a href="http://socratic.org/chemistry/measurement-in-chemistry/unit-conversions">unit conversion</a> is in order.</p>
<blockquote>
<p><mathjax>#280 color(red)(cancel(color(black)("mg"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = 280 * 10^(-3)"g"#</mathjax></p>
</blockquote>
<p>This means that your sample will contain </p>
<blockquote>
<p><mathjax>#280 * 10^(-3)color(red)(cancel(color(black)("g NaCl"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g NaCl")))) = 4.79 * 10^(-3)"moles NaCl"#</mathjax></p>
</blockquote>
<p>Therefore, the molarity of this solution will be </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
<p><mathjax>#c = (4.79 * color(purple)(cancel(color(black)(10^3)))"moles")/(2.00 * color(purple)(cancel(color(black)(10^3)))"L") = "2.395 M"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the mass of sodium chloride, the answer will be </p>
<blockquote>
<p><mathjax>#c = color(green)("2.4 M")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>Because sodium chloride is soluble in aqueous solution, which means that it exists as sodium cations</em>, <mathjax>#"Na"^(+)#</mathjax>, <em>and chloride anions</em>, <mathjax>#"Cl"^(-)#</mathjax>, <em>you wouldn't actually say that you have a</em> <mathjax>#"2.4-M"#</mathjax> <em>sodium chloride solution</em>.</p>
<p><em>Instead, you would say that you have a solution that is</em> <mathjax>#"2.4 M"#</mathjax> <em>in</em> <mathjax>#"Na"^(+)#</mathjax> <em>and</em> <mathjax>#"2.4 M"#</mathjax> <em>in</em> <mathjax>#"Cl"^(-)#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2.4 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to determine the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of a solution, you basically need to know two things </p>
<ul>
<li><em>how many <strong>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> you have in your sample</em></li>
<li><em>the volume of this sample - expressed in <strong>liters</strong></em></li>
</ul>
<p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as </p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>Your solute is <em>sodium chloride</em>, <mathjax>#"NaCl"#</mathjax>. You know that you're dissolving a <mathjax>#"280-mg"#</mathjax> sample of sodium chloride in enough water to get a <mathjax>#"2.00-mL"#</mathjax> <em>volume of solution</em>. </p>
<p>Notice that the problem provides you with the volume of the solution, but that it is expressed in <em>milliliters</em>. This means that you will have to convert it to <strong>liters</strong> by using the conversion factor </p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>The volume of the solution will thus be</p>
<blockquote>
<p><mathjax>#2.00 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 2.00 * 10^(-3)"L"#</mathjax></p>
</blockquote>
<p>Now, in order to get the number of moles of solute, you need to use sodium chloride's <strong>molar mass</strong>, which tells you the mass of <strong>one mole</strong> of this compound.</p>
<p>Sodium chloride has a molar mass of <mathjax>#"58.44 g/mol"#</mathjax>, which tells you that <strong>one mole</strong> of sodium chloride has a mass of <mathjax>#"58.44 g"#</mathjax>. </p>
<p>Notice that the sample you have is expressed in <em>milligrams</em>. Once again, a <a href="http://socratic.org/chemistry/measurement-in-chemistry/unit-conversions">unit conversion</a> is in order.</p>
<blockquote>
<p><mathjax>#280 color(red)(cancel(color(black)("mg"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = 280 * 10^(-3)"g"#</mathjax></p>
</blockquote>
<p>This means that your sample will contain </p>
<blockquote>
<p><mathjax>#280 * 10^(-3)color(red)(cancel(color(black)("g NaCl"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g NaCl")))) = 4.79 * 10^(-3)"moles NaCl"#</mathjax></p>
</blockquote>
<p>Therefore, the molarity of this solution will be </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
<p><mathjax>#c = (4.79 * color(purple)(cancel(color(black)(10^3)))"moles")/(2.00 * color(purple)(cancel(color(black)(10^3)))"L") = "2.395 M"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the mass of sodium chloride, the answer will be </p>
<blockquote>
<p><mathjax>#c = color(green)("2.4 M")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>Because sodium chloride is soluble in aqueous solution, which means that it exists as sodium cations</em>, <mathjax>#"Na"^(+)#</mathjax>, <em>and chloride anions</em>, <mathjax>#"Cl"^(-)#</mathjax>, <em>you wouldn't actually say that you have a</em> <mathjax>#"2.4-M"#</mathjax> <em>sodium chloride solution</em>.</p>
<p><em>Instead, you would say that you have a solution that is</em> <mathjax>#"2.4 M"#</mathjax> <em>in</em> <mathjax>#"Na"^(+)#</mathjax> <em>and</em> <mathjax>#"2.4 M"#</mathjax> <em>in</em> <mathjax>#"Cl"^(-)#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you find the molarity of a salt solution made by dissolving 280 mg of NaCl in 2.00 mL of solution? </h1>
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Stefan V.
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Jan 6, 2016
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<div class="markdown"><p><mathjax>#"2.4 M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to determine the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of a solution, you basically need to know two things </p>
<ul>
<li><em>how many <strong>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> you have in your sample</em></li>
<li><em>the volume of this sample - expressed in <strong>liters</strong></em></li>
</ul>
<p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as </p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>Your solute is <em>sodium chloride</em>, <mathjax>#"NaCl"#</mathjax>. You know that you're dissolving a <mathjax>#"280-mg"#</mathjax> sample of sodium chloride in enough water to get a <mathjax>#"2.00-mL"#</mathjax> <em>volume of solution</em>. </p>
<p>Notice that the problem provides you with the volume of the solution, but that it is expressed in <em>milliliters</em>. This means that you will have to convert it to <strong>liters</strong> by using the conversion factor </p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>The volume of the solution will thus be</p>
<blockquote>
<p><mathjax>#2.00 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 2.00 * 10^(-3)"L"#</mathjax></p>
</blockquote>
<p>Now, in order to get the number of moles of solute, you need to use sodium chloride's <strong>molar mass</strong>, which tells you the mass of <strong>one mole</strong> of this compound.</p>
<p>Sodium chloride has a molar mass of <mathjax>#"58.44 g/mol"#</mathjax>, which tells you that <strong>one mole</strong> of sodium chloride has a mass of <mathjax>#"58.44 g"#</mathjax>. </p>
<p>Notice that the sample you have is expressed in <em>milligrams</em>. Once again, a <a href="http://socratic.org/chemistry/measurement-in-chemistry/unit-conversions">unit conversion</a> is in order.</p>
<blockquote>
<p><mathjax>#280 color(red)(cancel(color(black)("mg"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = 280 * 10^(-3)"g"#</mathjax></p>
</blockquote>
<p>This means that your sample will contain </p>
<blockquote>
<p><mathjax>#280 * 10^(-3)color(red)(cancel(color(black)("g NaCl"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g NaCl")))) = 4.79 * 10^(-3)"moles NaCl"#</mathjax></p>
</blockquote>
<p>Therefore, the molarity of this solution will be </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V)#</mathjax></p>
<p><mathjax>#c = (4.79 * color(purple)(cancel(color(black)(10^3)))"moles")/(2.00 * color(purple)(cancel(color(black)(10^3)))"L") = "2.395 M"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the mass of sodium chloride, the answer will be </p>
<blockquote>
<p><mathjax>#c = color(green)("2.4 M")#</mathjax></p>
</blockquote>
<p><strong>SIDE NOTE</strong> <em>Because sodium chloride is soluble in aqueous solution, which means that it exists as sodium cations</em>, <mathjax>#"Na"^(+)#</mathjax>, <em>and chloride anions</em>, <mathjax>#"Cl"^(-)#</mathjax>, <em>you wouldn't actually say that you have a</em> <mathjax>#"2.4-M"#</mathjax> <em>sodium chloride solution</em>.</p>
<p><em>Instead, you would say that you have a solution that is</em> <mathjax>#"2.4 M"#</mathjax> <em>in</em> <mathjax>#"Na"^(+)#</mathjax> <em>and</em> <mathjax>#"2.4 M"#</mathjax> <em>in</em> <mathjax>#"Cl"^(-)#</mathjax>.</p></div>
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</article> | How do you find the molarity of a salt solution made by dissolving 280 mg of NaCl in 2.00 mL of solution? | null |
1,238 | aaea499c-6ddd-11ea-ac7f-ccda262736ce | https://socratic.org/questions/5874f7e611ef6b34c6a6cacf | 1.78 × 10^(-11) M | start physical_unit 9 9 concentration mol/l qc_end physical_unit 11 12 3 3 poh qc_end end | [{"type":"physical unit","value":"Concentration [OF] [HO-] [IN] M"}] | [{"type":"physical unit","value":"1.78 × 10^(-11) M"}] | [{"type":"physical unit","value":"pOH [OF] the solution [=] \\pu{10.75}"}] | <h1 class="questionTitle" itemprop="name">If #pOH=10.75#, what is the concentration of #[HO^-]# in the solution?</h1> | null | 1.78 × 10^(-11) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p>If <mathjax>#pOH=10.75#</mathjax>, then <mathjax>#[HO^-]=10^(-10.75)=1.78xx10^-11*mol*L^-1.#</mathjax></p>
<p>What are <mathjax>#pH#</mathjax>, and <mathjax>#[H_3O^+]#</mathjax> in this solution?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#[HO^-]=1.78xx10^-11*mol*L^-1.#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p>If <mathjax>#pOH=10.75#</mathjax>, then <mathjax>#[HO^-]=10^(-10.75)=1.78xx10^-11*mol*L^-1.#</mathjax></p>
<p>What are <mathjax>#pH#</mathjax>, and <mathjax>#[H_3O^+]#</mathjax> in this solution?</p></div>
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<h1 class="questionTitle" itemprop="name">If #pOH=10.75#, what is the concentration of #[HO^-]# in the solution?</h1>
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<div class="markdown"><p><mathjax>#[HO^-]=1.78xx10^-11*mol*L^-1.#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pOH=-log_10[HO^-]#</mathjax></p>
<p>If <mathjax>#pOH=10.75#</mathjax>, then <mathjax>#[HO^-]=10^(-10.75)=1.78xx10^-11*mol*L^-1.#</mathjax></p>
<p>What are <mathjax>#pH#</mathjax>, and <mathjax>#[H_3O^+]#</mathjax> in this solution?</p></div>
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</article> | If #pOH=10.75#, what is the concentration of #[HO^-]# in the solution? | null |
1,239 | ac312a46-6ddd-11ea-9f05-ccda262736ce | https://socratic.org/questions/how-many-grams-of-sro-should-be-dissolved-in-sufficient-water-to-make-2-00-l-of- | 0.06 grams | start physical_unit 4 4 mass g qc_end c_other OTHER qc_end physical_unit 17 17 13 14 volume qc_end physical_unit 17 17 22 22 ph qc_end end | [{"type":"physical unit","value":"Mass [OF] SrO [IN] grams"}] | [{"type":"physical unit","value":"0.06 grams"}] | [{"type":"other","value":"Sufficient water."},{"type":"physical unit","value":"Volume [OF] SrO solution [=] \\pu{2.00 L}"},{"type":"physical unit","value":"pH [OF] SrO solution [=] \\pu{11.00}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #SrO# should be dissolved in sufficient water to make 2.00 L of a solution with a pH= 11.00?</h1> | null | 0.06 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Concentration of <mathjax>#["OH"]^"-"#</mathjax></strong></p>
<p><mathjax>#"pH" = 11.00#</mathjax></p>
<p>∴ <mathjax>#"pOH" = "14 00 - 11.00" = 3.00#</mathjax></p>
<p><mathjax>#["OH"^"-"] = 10^"-pOH"color(white)(l) "mol/L" = 10^"-3.00"color(white)(l) "mol/L" = 1.00 × 10^"-3"color(white)(l)"mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Mass of <mathjax>#"SrO"#</mathjax></strong></p>
<p><mathjax>#"Moles of OH"^"-" = 2.00 color(red)(cancel(color(black)("L"))) × (1.00 × 10^"-3"color(white)(l) "mol OH"^"-")/(1 color(red)(cancel(color(black)("L")))) = 2.00 × 10^"-3"color(white)(l) "mol OH"^"-"#</mathjax></p>
<p>The net ionic equation for the reaction is</p>
<blockquote>
<blockquote>
<p><mathjax>#"O"^"2-" + "H"_2"O" → "2OH"^"-"#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#"Moles of O"^"2-" = 2.00 × 10^"-3" color(red)(cancel(color(black)("mol OH"^"-"))) × ("1 mol O"_2^"-")/(2 color(red)(cancel(color(black)("mol OH"^"-")))) = 1.00 × 10^"-3"color(white)(l) "mol O"^"2-"#</mathjax></p>
<p><mathjax>#"Moles of SrO" = 1.00 × 10^"-3" color(red)(cancel(color(black)("mol O"^"2-"))) × "1 mol SrO"/(1 color(red)(cancel(color(black)("mol O"^"2-")))) = 1.00 × 10^"-3"color(white)(l) "mol SrO"#</mathjax></p>
<p><mathjax>#"Mass of SrO" = 1.00 × 10^"-3" color(red)(cancel(color(black)("mol SrO"))) × "64.06 g SrO"/(1 color(red)(cancel(color(black)("mol SrO")))) = "0.0641 g SrO"#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>You should use 0.0641 g of <mathjax>#"SrO"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Concentration of <mathjax>#["OH"]^"-"#</mathjax></strong></p>
<p><mathjax>#"pH" = 11.00#</mathjax></p>
<p>∴ <mathjax>#"pOH" = "14 00 - 11.00" = 3.00#</mathjax></p>
<p><mathjax>#["OH"^"-"] = 10^"-pOH"color(white)(l) "mol/L" = 10^"-3.00"color(white)(l) "mol/L" = 1.00 × 10^"-3"color(white)(l)"mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Mass of <mathjax>#"SrO"#</mathjax></strong></p>
<p><mathjax>#"Moles of OH"^"-" = 2.00 color(red)(cancel(color(black)("L"))) × (1.00 × 10^"-3"color(white)(l) "mol OH"^"-")/(1 color(red)(cancel(color(black)("L")))) = 2.00 × 10^"-3"color(white)(l) "mol OH"^"-"#</mathjax></p>
<p>The net ionic equation for the reaction is</p>
<blockquote>
<blockquote>
<p><mathjax>#"O"^"2-" + "H"_2"O" → "2OH"^"-"#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#"Moles of O"^"2-" = 2.00 × 10^"-3" color(red)(cancel(color(black)("mol OH"^"-"))) × ("1 mol O"_2^"-")/(2 color(red)(cancel(color(black)("mol OH"^"-")))) = 1.00 × 10^"-3"color(white)(l) "mol O"^"2-"#</mathjax></p>
<p><mathjax>#"Moles of SrO" = 1.00 × 10^"-3" color(red)(cancel(color(black)("mol O"^"2-"))) × "1 mol SrO"/(1 color(red)(cancel(color(black)("mol O"^"2-")))) = 1.00 × 10^"-3"color(white)(l) "mol SrO"#</mathjax></p>
<p><mathjax>#"Mass of SrO" = 1.00 × 10^"-3" color(red)(cancel(color(black)("mol SrO"))) × "64.06 g SrO"/(1 color(red)(cancel(color(black)("mol SrO")))) = "0.0641 g SrO"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of #SrO# should be dissolved in sufficient water to make 2.00 L of a solution with a pH= 11.00?</h1>
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<div class="markdown"><p>You should use 0.0641 g of <mathjax>#"SrO"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Concentration of <mathjax>#["OH"]^"-"#</mathjax></strong></p>
<p><mathjax>#"pH" = 11.00#</mathjax></p>
<p>∴ <mathjax>#"pOH" = "14 00 - 11.00" = 3.00#</mathjax></p>
<p><mathjax>#["OH"^"-"] = 10^"-pOH"color(white)(l) "mol/L" = 10^"-3.00"color(white)(l) "mol/L" = 1.00 × 10^"-3"color(white)(l)"mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Mass of <mathjax>#"SrO"#</mathjax></strong></p>
<p><mathjax>#"Moles of OH"^"-" = 2.00 color(red)(cancel(color(black)("L"))) × (1.00 × 10^"-3"color(white)(l) "mol OH"^"-")/(1 color(red)(cancel(color(black)("L")))) = 2.00 × 10^"-3"color(white)(l) "mol OH"^"-"#</mathjax></p>
<p>The net ionic equation for the reaction is</p>
<blockquote>
<blockquote>
<p><mathjax>#"O"^"2-" + "H"_2"O" → "2OH"^"-"#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#"Moles of O"^"2-" = 2.00 × 10^"-3" color(red)(cancel(color(black)("mol OH"^"-"))) × ("1 mol O"_2^"-")/(2 color(red)(cancel(color(black)("mol OH"^"-")))) = 1.00 × 10^"-3"color(white)(l) "mol O"^"2-"#</mathjax></p>
<p><mathjax>#"Moles of SrO" = 1.00 × 10^"-3" color(red)(cancel(color(black)("mol O"^"2-"))) × "1 mol SrO"/(1 color(red)(cancel(color(black)("mol O"^"2-")))) = 1.00 × 10^"-3"color(white)(l) "mol SrO"#</mathjax></p>
<p><mathjax>#"Mass of SrO" = 1.00 × 10^"-3" color(red)(cancel(color(black)("mol SrO"))) × "64.06 g SrO"/(1 color(red)(cancel(color(black)("mol SrO")))) = "0.0641 g SrO"#</mathjax></p></div>
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</article> | How many grams of #SrO# should be dissolved in sufficient water to make 2.00 L of a solution with a pH= 11.00? | null |
1,240 | ac0c1778-6ddd-11ea-84d4-ccda262736ce | https://socratic.org/questions/what-is-the-ionic-form-of-the-following-unbalanced-equation-mno-2-hno-2-mn-no-3- | MnO2(s) + HNO2(aq) + H+ -> Mn^2+ + H2O(l) + NO3- | start chemical_equation qc_end chemical_equation 10 16 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the ionic form"}] | [{"type":"chemical equation","value":"MnO2(s) + HNO2(aq) + H+ -> Mn^2+ + H2O(l) + NO3-"}] | [{"type":"chemical equation","value":"MnO2 + HNO2 -> Mn(NO3)2 + H2O"}] | <h1 class="questionTitle" itemprop="name">What is the ionic form of the following unbalanced equation #MnO_2 + HNO_2 -> Mn(NO_3)_2 + H_2O#?</h1> | null | MnO2(s) + HNO2(aq) + H+ -> Mn^2+ + H2O(l) + NO3- | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How do we get to this?</p>
<p>Well, <mathjax>#"manganese(IV) oxide"#</mathjax> is reduced to <mathjax>#Mn(II)#</mathjax>:</p>
<p><mathjax>#MnO_2(s) +4H^(+) + 2e^(-) rarr Mn^(2+) +2H_2O(l)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And, <mathjax>#"nitrous acid"#</mathjax> is oxidized to <mathjax>#"nitrate ion"#</mathjax>:</p>
<p><mathjax>#HONO(aq) +H_2O(l) rarr NO_3^(-)+ 3H^+ + 2e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And we just add <mathjax>#(i) + (ii)#</mathjax> together......</p>
<p><mathjax>#HONO(aq) +cancel(H_2O(l)) +MnO_2(s) +cancel4H^(+) + cancel(2e^(-))rarr Mn^(2+) +cancel2H_2O(l)+NO_3^(-)+ cancel(3H^+ + 2e^(-))#</mathjax></p>
<p>To give finally.....</p>
<p><mathjax>#HONO(aq) +MnO_2(s) +H^(+) rarr Mn^(2+) +H_2O(l)+NO_3^(-)#</mathjax></p></div>
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<div class="markdown"><p>You gots a redox reaction of the form....</p>
<p><mathjax>#HONO(aq) +MnO_2(s) +H^(+) rarr Mn^(2+) +H_2O(l)+NO_3^(-)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How do we get to this?</p>
<p>Well, <mathjax>#"manganese(IV) oxide"#</mathjax> is reduced to <mathjax>#Mn(II)#</mathjax>:</p>
<p><mathjax>#MnO_2(s) +4H^(+) + 2e^(-) rarr Mn^(2+) +2H_2O(l)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And, <mathjax>#"nitrous acid"#</mathjax> is oxidized to <mathjax>#"nitrate ion"#</mathjax>:</p>
<p><mathjax>#HONO(aq) +H_2O(l) rarr NO_3^(-)+ 3H^+ + 2e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And we just add <mathjax>#(i) + (ii)#</mathjax> together......</p>
<p><mathjax>#HONO(aq) +cancel(H_2O(l)) +MnO_2(s) +cancel4H^(+) + cancel(2e^(-))rarr Mn^(2+) +cancel2H_2O(l)+NO_3^(-)+ cancel(3H^+ + 2e^(-))#</mathjax></p>
<p>To give finally.....</p>
<p><mathjax>#HONO(aq) +MnO_2(s) +H^(+) rarr Mn^(2+) +H_2O(l)+NO_3^(-)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the ionic form of the following unbalanced equation #MnO_2 + HNO_2 -> Mn(NO_3)_2 + H_2O#?</h1>
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anor277
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<div class="markdown"><p>You gots a redox reaction of the form....</p>
<p><mathjax>#HONO(aq) +MnO_2(s) +H^(+) rarr Mn^(2+) +H_2O(l)+NO_3^(-)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How do we get to this?</p>
<p>Well, <mathjax>#"manganese(IV) oxide"#</mathjax> is reduced to <mathjax>#Mn(II)#</mathjax>:</p>
<p><mathjax>#MnO_2(s) +4H^(+) + 2e^(-) rarr Mn^(2+) +2H_2O(l)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And, <mathjax>#"nitrous acid"#</mathjax> is oxidized to <mathjax>#"nitrate ion"#</mathjax>:</p>
<p><mathjax>#HONO(aq) +H_2O(l) rarr NO_3^(-)+ 3H^+ + 2e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And we just add <mathjax>#(i) + (ii)#</mathjax> together......</p>
<p><mathjax>#HONO(aq) +cancel(H_2O(l)) +MnO_2(s) +cancel4H^(+) + cancel(2e^(-))rarr Mn^(2+) +cancel2H_2O(l)+NO_3^(-)+ cancel(3H^+ + 2e^(-))#</mathjax></p>
<p>To give finally.....</p>
<p><mathjax>#HONO(aq) +MnO_2(s) +H^(+) rarr Mn^(2+) +H_2O(l)+NO_3^(-)#</mathjax></p></div>
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</article> | What is the ionic form of the following unbalanced equation #MnO_2 + HNO_2 -> Mn(NO_3)_2 + H_2O#? | null |
1,241 | abfdbf7a-6ddd-11ea-816d-ccda262736ce | https://socratic.org/questions/a-helium-balloon-with-an-internal-pressure-of-1-00-atm-and-a-volume-of-4-50-l-at | 6.47 L | start physical_unit 1 2 volume l qc_end physical_unit 1 2 14 15 volume qc_end physical_unit 1 2 8 9 pressure qc_end physical_unit 1 2 17 18 temperature qc_end physical_unit 1 2 34 35 pressure qc_end physical_unit 1 2 40 41 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the helium balloon [IN] L"}] | [{"type":"physical unit","value":"6.47 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the helium balloon [=] \\pu{4.50 L}"},{"type":"physical unit","value":"Pressure1 [OF] the helium balloon [=] \\pu{1.00 atm}"},{"type":"physical unit","value":"Temperature1 [OF] the helium balloon [=] \\pu{20.0 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] the helium balloon [=] \\pu{0.600 atm}"},{"type":"physical unit","value":"Temperature2 [OF] the helium balloon [=] \\pu{-20.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">A helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 20.0°C is released. What volume will the balloon occupy at an altitude where the pressure is 0.600 atm and the temperature is -20.0° C?</h1> | null | 6.47 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can use the Ideal gas equation PV = nRT .....(a)</p>
<p><mathjax>#p_1#</mathjax> <mathjax>#v_1#</mathjax> / <mathjax>#T_1#</mathjax> = <mathjax>#p_2#</mathjax> <mathjax>#v_2#</mathjax> / <mathjax>#T_2#</mathjax> </p>
<p><mathjax>#p_1#</mathjax> = 1 atm, <mathjax>#v_1#</mathjax> = 4.50 L , <mathjax>#T_1#</mathjax> = 293 K</p>
<p><mathjax>#p_2#</mathjax> = 0.6 atm, <mathjax>#v_2#</mathjax> = ? , <mathjax>#T_2#</mathjax> = 253 K</p>
<p>Plugging in the values;</p>
<p>(1 atm x 4.50 L ) / 293 K = 0.6 atm x <mathjax>#v_2#</mathjax> / 253 K</p>
<p>(4.50 atm L x 253 K ) / ( 0.6 atm x 293 K) = <mathjax>#v_2#</mathjax> </p>
<p><mathjax>#v_2#</mathjax> = 1138.5 atm L K / 175.8 atm K </p>
<p><mathjax>#v_2#</mathjax> = 6.47 L approx </p></div>
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<div class="markdown"><p>6.47 L </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can use the Ideal gas equation PV = nRT .....(a)</p>
<p><mathjax>#p_1#</mathjax> <mathjax>#v_1#</mathjax> / <mathjax>#T_1#</mathjax> = <mathjax>#p_2#</mathjax> <mathjax>#v_2#</mathjax> / <mathjax>#T_2#</mathjax> </p>
<p><mathjax>#p_1#</mathjax> = 1 atm, <mathjax>#v_1#</mathjax> = 4.50 L , <mathjax>#T_1#</mathjax> = 293 K</p>
<p><mathjax>#p_2#</mathjax> = 0.6 atm, <mathjax>#v_2#</mathjax> = ? , <mathjax>#T_2#</mathjax> = 253 K</p>
<p>Plugging in the values;</p>
<p>(1 atm x 4.50 L ) / 293 K = 0.6 atm x <mathjax>#v_2#</mathjax> / 253 K</p>
<p>(4.50 atm L x 253 K ) / ( 0.6 atm x 293 K) = <mathjax>#v_2#</mathjax> </p>
<p><mathjax>#v_2#</mathjax> = 1138.5 atm L K / 175.8 atm K </p>
<p><mathjax>#v_2#</mathjax> = 6.47 L approx </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 20.0°C is released. What volume will the balloon occupy at an altitude where the pressure is 0.600 atm and the temperature is -20.0° C?</h1>
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<div class="markdown"><p>6.47 L </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can use the Ideal gas equation PV = nRT .....(a)</p>
<p><mathjax>#p_1#</mathjax> <mathjax>#v_1#</mathjax> / <mathjax>#T_1#</mathjax> = <mathjax>#p_2#</mathjax> <mathjax>#v_2#</mathjax> / <mathjax>#T_2#</mathjax> </p>
<p><mathjax>#p_1#</mathjax> = 1 atm, <mathjax>#v_1#</mathjax> = 4.50 L , <mathjax>#T_1#</mathjax> = 293 K</p>
<p><mathjax>#p_2#</mathjax> = 0.6 atm, <mathjax>#v_2#</mathjax> = ? , <mathjax>#T_2#</mathjax> = 253 K</p>
<p>Plugging in the values;</p>
<p>(1 atm x 4.50 L ) / 293 K = 0.6 atm x <mathjax>#v_2#</mathjax> / 253 K</p>
<p>(4.50 atm L x 253 K ) / ( 0.6 atm x 293 K) = <mathjax>#v_2#</mathjax> </p>
<p><mathjax>#v_2#</mathjax> = 1138.5 atm L K / 175.8 atm K </p>
<p><mathjax>#v_2#</mathjax> = 6.47 L approx </p></div>
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</article> | A helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 20.0°C is released. What volume will the balloon occupy at an altitude where the pressure is 0.600 atm and the temperature is -20.0° C? | null |
1,242 | ac11a162-6ddd-11ea-8b65-ccda262736ce | https://socratic.org/questions/na-3n-decomposes-to-form-sodium-and-nitrogen-gas-at-stp-if-13-7-l-of-nitrogen-is | 1.22 moles | start physical_unit 0 0 mole mol qc_end c_other STP qc_end physical_unit 6 6 11 12 volume qc_end c_other OTHER qc_end substance 4 4 qc_end end | [{"type":"physical unit","value":"Mole [OF] Na3N [IN] moles"}] | [{"type":"physical unit","value":"1.22 moles"}] | [{"type":"other","value":"STP"},{"type":"physical unit","value":"Volume [OF] nitrogen [=] \\pu{13.7 L}"},{"type":"other","value":"22.4 L = 1 mole of any gas"},{"type":"substance name","value":"Sodium"}] | <h1 class="questionTitle" itemprop="name">#Na_3N# decomposes to form sodium and nitrogen gas at STP. If 13.7 L of nitrogen is produced, how many moles of #Na_3N# was used (22.4 L = 1 mole of any gas) ?</h1> | null | 1.22 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Given:</strong> <mathjax>#Na_3N#</mathjax> decomposes to form sodium (Na) and Nitrogen (<mathjax>#N_2)#</mathjax> at STP (standard temperature and pressure). At STP <mathjax>#22.4 L = 1#</mathjax> mole of gas</p>
<p>First you need to write a skeleton equation and then balance it. You need to realize that nitrogen is a diatomic element <mathjax>#N_2#</mathjax>. </p>
<p>skeleton equation: <mathjax>#" "Na_3N -> Na + N_2#</mathjax></p>
<p>balanced: <mathjax>#" "2Na_3N -> 6Na + N_2#</mathjax></p>
<p>The balanced equation gives us <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio of sodium nitrate to nitrogen produced: <mathjax>#2/1#</mathjax>.</p>
<p><mathjax>#13.7 cancel(L N_2) xx (1 cancel(mol N_2))/(22.4 cancel(L N_2)) xx (2 mol Na_3N)/(1 cancel(mol N_2)) = 1.22" mol "Na_3N#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#1.22#</mathjax> mol <mathjax>#Na_3N#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Given:</strong> <mathjax>#Na_3N#</mathjax> decomposes to form sodium (Na) and Nitrogen (<mathjax>#N_2)#</mathjax> at STP (standard temperature and pressure). At STP <mathjax>#22.4 L = 1#</mathjax> mole of gas</p>
<p>First you need to write a skeleton equation and then balance it. You need to realize that nitrogen is a diatomic element <mathjax>#N_2#</mathjax>. </p>
<p>skeleton equation: <mathjax>#" "Na_3N -> Na + N_2#</mathjax></p>
<p>balanced: <mathjax>#" "2Na_3N -> 6Na + N_2#</mathjax></p>
<p>The balanced equation gives us <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio of sodium nitrate to nitrogen produced: <mathjax>#2/1#</mathjax>.</p>
<p><mathjax>#13.7 cancel(L N_2) xx (1 cancel(mol N_2))/(22.4 cancel(L N_2)) xx (2 mol Na_3N)/(1 cancel(mol N_2)) = 1.22" mol "Na_3N#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">#Na_3N# decomposes to form sodium and nitrogen gas at STP. If 13.7 L of nitrogen is produced, how many moles of #Na_3N# was used (22.4 L = 1 mole of any gas) ?</h1>
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<div class="markdown"><p><mathjax>#1.22#</mathjax> mol <mathjax>#Na_3N#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Given:</strong> <mathjax>#Na_3N#</mathjax> decomposes to form sodium (Na) and Nitrogen (<mathjax>#N_2)#</mathjax> at STP (standard temperature and pressure). At STP <mathjax>#22.4 L = 1#</mathjax> mole of gas</p>
<p>First you need to write a skeleton equation and then balance it. You need to realize that nitrogen is a diatomic element <mathjax>#N_2#</mathjax>. </p>
<p>skeleton equation: <mathjax>#" "Na_3N -> Na + N_2#</mathjax></p>
<p>balanced: <mathjax>#" "2Na_3N -> 6Na + N_2#</mathjax></p>
<p>The balanced equation gives us <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio of sodium nitrate to nitrogen produced: <mathjax>#2/1#</mathjax>.</p>
<p><mathjax>#13.7 cancel(L N_2) xx (1 cancel(mol N_2))/(22.4 cancel(L N_2)) xx (2 mol Na_3N)/(1 cancel(mol N_2)) = 1.22" mol "Na_3N#</mathjax></p></div>
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</article> | #Na_3N# decomposes to form sodium and nitrogen gas at STP. If 13.7 L of nitrogen is produced, how many moles of #Na_3N# was used (22.4 L = 1 mole of any gas) ? | null |
1,243 | ac11a165-6ddd-11ea-810e-ccda262736ce | https://socratic.org/questions/what-is-the-volume-of-1-2-moles-of-oxygen-gas-at-stp | 26.90 L | start physical_unit 8 9 volume l qc_end physical_unit 8 9 5 6 mole qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] oxygen gas [IN] L"}] | [{"type":"physical unit","value":"26.90 L"}] | [{"type":"physical unit","value":"Mole [OF] oxygen gas [=] \\pu{1.2 moles}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume of 1.2 moles of oxygen gas at STP?</h1> | null | 26.90 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since we are at STP and are given only one set of conditions, we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>:</p>
<p><img alt="chemistryhungergames.com" src="https://useruploads.socratic.org/Ccuv15kkTrmcWZwIc4Ye_ideal-gas-law.png"/> </p>
<p>I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.</p>
<p>List your known and unknown variables. Our only unknown is the volume of <mathjax>#O_2(g)#</mathjax>. Our known variables are P,n,R, and T. </p>
<p>At STP, the temperature is 273K and the pressure is 1 atm.</p>
<p>Now we have to rearrange the equation to solve for V:</p>
<p><mathjax># (nxxRxxT)/P#</mathjax></p>
<p><mathjax>#V = (1.2 cancel"mol"xx0.0821Lxxcancel(atm)/cancel(molxxK)xx273cancelK)/(1cancel(atm)#</mathjax></p>
<p><mathjax>#V = 27 L#</mathjax></p></div>
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<div class="markdown"><p>At STP, 1.2 moles of <mathjax>#O_2#</mathjax> has a volume of <strong>27 L.</strong></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since we are at STP and are given only one set of conditions, we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>:</p>
<p><img alt="chemistryhungergames.com" src="https://useruploads.socratic.org/Ccuv15kkTrmcWZwIc4Ye_ideal-gas-law.png"/> </p>
<p>I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.</p>
<p>List your known and unknown variables. Our only unknown is the volume of <mathjax>#O_2(g)#</mathjax>. Our known variables are P,n,R, and T. </p>
<p>At STP, the temperature is 273K and the pressure is 1 atm.</p>
<p>Now we have to rearrange the equation to solve for V:</p>
<p><mathjax># (nxxRxxT)/P#</mathjax></p>
<p><mathjax>#V = (1.2 cancel"mol"xx0.0821Lxxcancel(atm)/cancel(molxxK)xx273cancelK)/(1cancel(atm)#</mathjax></p>
<p><mathjax>#V = 27 L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume of 1.2 moles of oxygen gas at STP?</h1>
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<div class="markdown"><p>At STP, 1.2 moles of <mathjax>#O_2#</mathjax> has a volume of <strong>27 L.</strong></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since we are at STP and are given only one set of conditions, we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>:</p>
<p><img alt="chemistryhungergames.com" src="https://useruploads.socratic.org/Ccuv15kkTrmcWZwIc4Ye_ideal-gas-law.png"/> </p>
<p>I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.</p>
<p>List your known and unknown variables. Our only unknown is the volume of <mathjax>#O_2(g)#</mathjax>. Our known variables are P,n,R, and T. </p>
<p>At STP, the temperature is 273K and the pressure is 1 atm.</p>
<p>Now we have to rearrange the equation to solve for V:</p>
<p><mathjax># (nxxRxxT)/P#</mathjax></p>
<p><mathjax>#V = (1.2 cancel"mol"xx0.0821Lxxcancel(atm)/cancel(molxxK)xx273cancelK)/(1cancel(atm)#</mathjax></p>
<p><mathjax>#V = 27 L#</mathjax></p></div>
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</article> | What is the volume of 1.2 moles of oxygen gas at STP? | null |
1,244 | a91249a8-6ddd-11ea-b15a-ccda262736ce | https://socratic.org/questions/what-volume-of-1-25-m-caf-2-solution-is-needed-to-provide-4-00-mol-caf-2 | 3.2 L | start physical_unit 5 6 volume l qc_end physical_unit 5 5 11 12 mole qc_end physical_unit 5 6 3 4 molarity qc_end end | [{"type":"physical unit","value":"Volume [OF] CaF2 solution [IN] L"}] | [{"type":"physical unit","value":"3.2 L"}] | [{"type":"physical unit","value":"Mole [OF] CaF2 [=] \\pu{4.00 mol}"},{"type":"physical unit","value":"Molarity [OF] CaF2 solution [=] \\pu{1.25 M}"}] | <h1 class="questionTitle" itemprop="name">What volume of 1.25 M #CaF_2# solution is needed to provide 4.00 mol #CaF_2#? </h1> | null | 3.2 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this problem, you will need to know the formula for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>.</p>
<p><mathjax>#Molarity = (mol) / (volume(L))#</mathjax></p>
<p>or</p>
<p><mathjax>#M= n/V#</mathjax></p>
<p>Given:</p>
<p>M = 1.25<br/>
n = 4.00 mol</p>
<p>Plug in your given data:</p>
<p><mathjax>#1.25 = (4.00 mol)/V#</mathjax></p>
<p><strong><mathjax>#V#</mathjax> = 3.2 L</strong></p></div>
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<div class="markdown"><p><strong>3.2 L</strong> of 1.25M CaF2</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this problem, you will need to know the formula for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>.</p>
<p><mathjax>#Molarity = (mol) / (volume(L))#</mathjax></p>
<p>or</p>
<p><mathjax>#M= n/V#</mathjax></p>
<p>Given:</p>
<p>M = 1.25<br/>
n = 4.00 mol</p>
<p>Plug in your given data:</p>
<p><mathjax>#1.25 = (4.00 mol)/V#</mathjax></p>
<p><strong><mathjax>#V#</mathjax> = 3.2 L</strong></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume of 1.25 M #CaF_2# solution is needed to provide 4.00 mol #CaF_2#? </h1>
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<div class="markdown"><p><strong>3.2 L</strong> of 1.25M CaF2</p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this problem, you will need to know the formula for <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>.</p>
<p><mathjax>#Molarity = (mol) / (volume(L))#</mathjax></p>
<p>or</p>
<p><mathjax>#M= n/V#</mathjax></p>
<p>Given:</p>
<p>M = 1.25<br/>
n = 4.00 mol</p>
<p>Plug in your given data:</p>
<p><mathjax>#1.25 = (4.00 mol)/V#</mathjax></p>
<p><strong><mathjax>#V#</mathjax> = 3.2 L</strong></p></div>
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</article> | What volume of 1.25 M #CaF_2# solution is needed to provide 4.00 mol #CaF_2#? | null |
1,245 | aa797e27-6ddd-11ea-b96a-ccda262736ce | https://socratic.org/questions/how-many-oxygen-atoms-are-present-on-the-reactant-side-of-the-chemical-equation- | 6 | start physical_unit 2 3 number none qc_end chemical_equation 14 21 qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"6"}] | [{"type":"chemical equation","value":"4 Fe + 3 O2 -> 2 Fe2O"}] | <h1 class="questionTitle" itemprop="name">How many oxygen atoms are present on the reactant side of the chemical equation #4Fe + 3O_2 -> 2Fe_2O#?</h1> | null | 6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Clearly, there are <mathjax>#6#</mathjax> oxygen atoms on the left hand side of the equation; and, equally clearly, there are <mathjax>#6#</mathjax> oxygen atoms on the right hand side of the equation, as indeed there must be in a stoichiometrically balanced equation. </p>
<p>The oxidation of iron in structures is of special concern; see <a href="https://socratic.org/questions/how-are-redox-reactions-involved-in-corrosion?source=search">here.</a> </p></div>
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<div class="markdown"><p><mathjax>#4Fe(s)+3O_2(g)rarr2Fe_2O_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Clearly, there are <mathjax>#6#</mathjax> oxygen atoms on the left hand side of the equation; and, equally clearly, there are <mathjax>#6#</mathjax> oxygen atoms on the right hand side of the equation, as indeed there must be in a stoichiometrically balanced equation. </p>
<p>The oxidation of iron in structures is of special concern; see <a href="https://socratic.org/questions/how-are-redox-reactions-involved-in-corrosion?source=search">here.</a> </p></div>
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<h1 class="questionTitle" itemprop="name">How many oxygen atoms are present on the reactant side of the chemical equation #4Fe + 3O_2 -> 2Fe_2O#?</h1>
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<div class="markdown"><p><mathjax>#4Fe(s)+3O_2(g)rarr2Fe_2O_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Clearly, there are <mathjax>#6#</mathjax> oxygen atoms on the left hand side of the equation; and, equally clearly, there are <mathjax>#6#</mathjax> oxygen atoms on the right hand side of the equation, as indeed there must be in a stoichiometrically balanced equation. </p>
<p>The oxidation of iron in structures is of special concern; see <a href="https://socratic.org/questions/how-are-redox-reactions-involved-in-corrosion?source=search">here.</a> </p></div>
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</article> | How many oxygen atoms are present on the reactant side of the chemical equation #4Fe + 3O_2 -> 2Fe_2O#? | null |
1,246 | a96144d2-6ddd-11ea-a888-ccda262736ce | https://socratic.org/questions/an-aqueous-solution-has-a-hydronium-ion-concentration-of-6-10-9-mole-per-liter-a | 8 | start physical_unit 1 2 ph none qc_end physical_unit 5 6 9 14 concentration qc_end physical_unit 1 2 16 17 temperature qc_end end | [{"type":"physical unit","value":"pH [OF] aqueous solution"}] | [{"type":"physical unit","value":"8"}] | [{"type":"physical unit","value":"Concentration [OF] hydronium ion [=] \\pu{6 × 10^(−9) mole per liter}"},{"type":"physical unit","value":"Temperature [OF] aqueous solution [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">An aqueous solution has a hydronium ion concentration of #6 * 10^-9# mole per liter at 25°C. What is its pH?</h1> | null | 8 | <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> of a solution could be calculated using the concentration of hydronium <mathjax>#H_3O^+#</mathjax> by: <mathjax>#pH=-log[H_3O^+]#</mathjax></p>
<p>Therefore, <mathjax>#pH=-log(6xx10^(-9))=8.22 =8 ("1 significant figures")#</mathjax></p>
<p><strong>Acids & Bases | Nature, Strength & pH Scale.</strong><br/>
<iframe src="https://www.youtube.com/embed/3rPRpozi64Q?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div class="markdown"><p><mathjax>#pH=8#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of a solution could be calculated using the concentration of hydronium <mathjax>#H_3O^+#</mathjax> by: <mathjax>#pH=-log[H_3O^+]#</mathjax></p>
<p>Therefore, <mathjax>#pH=-log(6xx10^(-9))=8.22 =8 ("1 significant figures")#</mathjax></p>
<p><strong>Acids & Bases | Nature, Strength & pH Scale.</strong><br/>
<iframe src="https://www.youtube.com/embed/3rPRpozi64Q?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<h1 class="questionTitle" itemprop="name">An aqueous solution has a hydronium ion concentration of #6 * 10^-9# mole per liter at 25°C. What is its pH?</h1>
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Dr. Hayek
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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> of a solution could be calculated using the concentration of hydronium <mathjax>#H_3O^+#</mathjax> by: <mathjax>#pH=-log[H_3O^+]#</mathjax></p>
<p>Therefore, <mathjax>#pH=-log(6xx10^(-9))=8.22 =8 ("1 significant figures")#</mathjax></p>
<p><strong>Acids & Bases | Nature, Strength & pH Scale.</strong><br/>
<iframe src="https://www.youtube.com/embed/3rPRpozi64Q?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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anor277
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May 9, 2016
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<div class="markdown"><p><mathjax>#pH =8.22#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So what is <mathjax>#pH#</mathjax>?</p>
<p>Here it is simply the function <mathjax>#pH=-log_10[H_3O^+]#</mathjax>, <mathjax>#pH#</mathjax> derives from the French <mathjax>#"pouvoir hydrogene"#</mathjax>, <mathjax>#"the power of hydrogen"#</mathjax>.</p>
<p>We have <mathjax>#[H_3O^+]#</mathjax> and we have a calculator, so we press the buttons. </p>
<p>It is worthwhile to review the logarithmic function, and see why the use of logarithms (and slide rules) was absolutely widespread among scientists and engineers in a pre-calculator world.</p></div>
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</article> | An aqueous solution has a hydronium ion concentration of #6 * 10^-9# mole per liter at 25°C. What is its pH? | null |
1,247 | ac617165-6ddd-11ea-838f-ccda262736ce | https://socratic.org/questions/how-many-moles-of-solute-are-needed-to-make-a-0-5-l-solution-of-2-5-m-hcl | 1.25 moles | start physical_unit 4 4 mole mol qc_end physical_unit 12 12 10 11 volume qc_end physical_unit 16 16 14 15 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] solute [IN] moles"}] | [{"type":"physical unit","value":"1.25 moles"}] | [{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{0.5 L}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{2.5 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of solute are needed to make a 0.5 L solution of 2.5 M #HCl#?</h1> | null | 1.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as the number of moles of solute per liter of solution. <br/>
Molarity = moles of solute / liter of solution</p>
<p>We are given the molarity and volume, both of which have the correct units. All we have to do is rearrange the equation to find the number of moles. You can do this by multiplying both sides of the equation by the volume to cancel it out on the right hand side. Afterwards, you should end up having the volume multiplied by the molarity equaling the number of moles of solute like so:</p>
<p>Moles of solute = Molarity * Volume </p>
<p>2.5M HCl * 0.5 L = 1.25 moles of HCl</p>
<p>I hope this made sense.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>1.25 moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as the number of moles of solute per liter of solution. <br/>
Molarity = moles of solute / liter of solution</p>
<p>We are given the molarity and volume, both of which have the correct units. All we have to do is rearrange the equation to find the number of moles. You can do this by multiplying both sides of the equation by the volume to cancel it out on the right hand side. Afterwards, you should end up having the volume multiplied by the molarity equaling the number of moles of solute like so:</p>
<p>Moles of solute = Molarity * Volume </p>
<p>2.5M HCl * 0.5 L = 1.25 moles of HCl</p>
<p>I hope this made sense.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How many moles of solute are needed to make a 0.5 L solution of 2.5 M #HCl#?</h1>
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<div class="markdown"><p>1.25 moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as the number of moles of solute per liter of solution. <br/>
Molarity = moles of solute / liter of solution</p>
<p>We are given the molarity and volume, both of which have the correct units. All we have to do is rearrange the equation to find the number of moles. You can do this by multiplying both sides of the equation by the volume to cancel it out on the right hand side. Afterwards, you should end up having the volume multiplied by the molarity equaling the number of moles of solute like so:</p>
<p>Moles of solute = Molarity * Volume </p>
<p>2.5M HCl * 0.5 L = 1.25 moles of HCl</p>
<p>I hope this made sense.</p></div>
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</article> | How many moles of solute are needed to make a 0.5 L solution of 2.5 M #HCl#? | null |
1,248 | a85cb0c2-6ddd-11ea-a063-ccda262736ce | https://socratic.org/questions/iron-pyrite-fool-s-gold-is-iron-li-sulfide-what-is-its-formula | FeS2 | start chemical_formula qc_end substance 5 7 qc_end end | [{"type":"other","value":"Chemical Formula [OF] iron (lI) sulfide [IN] default"}] | [{"type":"chemical equation","value":"FeS2"}] | [{"type":"substance name","value":"Iron (lI) sulfide"}] | <h1 class="questionTitle" itemprop="name">Iron pyrite (fool's gold) is iron (lI) sulfide. What is its formula?</h1> | null | FeS2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The sulfur binds as a <mathjax>#"persulfide"#</mathjax>, i.e. <mathjax>#""^(-)S-S^-#</mathjax>; think <mathjax>#"peroxide"#</mathjax>, <mathjax>#O_2^(2-)#</mathjax>. The formal oxidation state of <mathjax>#S#</mathjax> here is <mathjax>#-I#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Iron pyrites is <mathjax>#"iron persulfide"#</mathjax>, <mathjax>#FeS_2#</mathjax>, with a formal <mathjax>#Fe(II+)#</mathjax> oxidation state.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The sulfur binds as a <mathjax>#"persulfide"#</mathjax>, i.e. <mathjax>#""^(-)S-S^-#</mathjax>; think <mathjax>#"peroxide"#</mathjax>, <mathjax>#O_2^(2-)#</mathjax>. The formal oxidation state of <mathjax>#S#</mathjax> here is <mathjax>#-I#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">Iron pyrite (fool's gold) is iron (lI) sulfide. What is its formula?</h1>
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<div class="markdown"><p>Iron pyrites is <mathjax>#"iron persulfide"#</mathjax>, <mathjax>#FeS_2#</mathjax>, with a formal <mathjax>#Fe(II+)#</mathjax> oxidation state.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The sulfur binds as a <mathjax>#"persulfide"#</mathjax>, i.e. <mathjax>#""^(-)S-S^-#</mathjax>; think <mathjax>#"peroxide"#</mathjax>, <mathjax>#O_2^(2-)#</mathjax>. The formal oxidation state of <mathjax>#S#</mathjax> here is <mathjax>#-I#</mathjax>.</p></div>
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</article> | Iron pyrite (fool's gold) is iron (lI) sulfide. What is its formula? | null |
1,249 | a8ca8a1b-6ddd-11ea-9af2-ccda262736ce | https://socratic.org/questions/at-40-c-a-sample-of-gas-has-a-volume-of-233-ml-on-heating-the-volume-expands-to- | 40 ℃ | start physical_unit 4 6 temperature °c qc_end physical_unit 4 6 1 2 temperature qc_end physical_unit 4 6 11 12 volume qc_end physical_unit 4 6 19 20 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Celsius temperature2 [OF] gas sample [IN] ℃"}] | [{"type":"physical unit","value":"40 ℃"}] | [{"type":"physical unit","value":"Celsius temperature1 [OF] gas sample [=] \\pu{-40 ℃}"},{"type":"physical unit","value":"Volume1 [OF] gas sample [=] \\pu{233 mL}"},{"type":"physical unit","value":"Volume2 [OF] gas sample [=] \\pu{313 mL}"},{"type":"other","value":"Constant pressure."}] | <h1 class="questionTitle" itemprop="name">At -40°C, a sample of gas has a volume of 233 mL. On heating, the volume expands to 313 mL. Assuming constant pressure, what is the new Celsius temperature?</h1> | null | 40 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This isthe an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a> which states the volume of a gas that is held at a constant pressure varies directly with the Kelvin temperature. The formula used to solve this problem is <mathjax>#V_1/T_1=V_2/T_2#</mathjax>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V_1="233 mL"#</mathjax><br/>
<mathjax>#T_1=-40^"o""C"+273.15="233 K"#</mathjax><br/>
<mathjax>#V_2="313 mL"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax> in degrees C</p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax> and solve.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(313cancel"mL"xx233"K")/(233cancel"mL")="313 K"#</mathjax></p>
<p>Convert temperature in Kelvins to degrees Celsius.</p>
<p><mathjax>#""^"o""C"="K-273.15"#</mathjax></p>
<p><mathjax>#""^"o""C"=313 "K"-273.15=40""^"o""C"#</mathjax> (rounded to a whole number)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new temperature in degrees Celsius is 40 degrees C.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This isthe an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a> which states the volume of a gas that is held at a constant pressure varies directly with the Kelvin temperature. The formula used to solve this problem is <mathjax>#V_1/T_1=V_2/T_2#</mathjax>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V_1="233 mL"#</mathjax><br/>
<mathjax>#T_1=-40^"o""C"+273.15="233 K"#</mathjax><br/>
<mathjax>#V_2="313 mL"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax> in degrees C</p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax> and solve.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(313cancel"mL"xx233"K")/(233cancel"mL")="313 K"#</mathjax></p>
<p>Convert temperature in Kelvins to degrees Celsius.</p>
<p><mathjax>#""^"o""C"="K-273.15"#</mathjax></p>
<p><mathjax>#""^"o""C"=313 "K"-273.15=40""^"o""C"#</mathjax> (rounded to a whole number)</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">At -40°C, a sample of gas has a volume of 233 mL. On heating, the volume expands to 313 mL. Assuming constant pressure, what is the new Celsius temperature?</h1>
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<div class="markdown"><p>The new temperature in degrees Celsius is 40 degrees C.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This isthe an example of <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' law</a> which states the volume of a gas that is held at a constant pressure varies directly with the Kelvin temperature. The formula used to solve this problem is <mathjax>#V_1/T_1=V_2/T_2#</mathjax>.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#V_1="233 mL"#</mathjax><br/>
<mathjax>#T_1=-40^"o""C"+273.15="233 K"#</mathjax><br/>
<mathjax>#V_2="313 mL"#</mathjax></p>
<p><strong>Unknown</strong><br/>
<mathjax>#T_2#</mathjax> in degrees C</p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#T_2#</mathjax> and solve.</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p><mathjax>#T_2=(V_2T_1)/V_1#</mathjax></p>
<p><mathjax>#T_2=(313cancel"mL"xx233"K")/(233cancel"mL")="313 K"#</mathjax></p>
<p>Convert temperature in Kelvins to degrees Celsius.</p>
<p><mathjax>#""^"o""C"="K-273.15"#</mathjax></p>
<p><mathjax>#""^"o""C"=313 "K"-273.15=40""^"o""C"#</mathjax> (rounded to a whole number)</p></div>
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</article> | At -40°C, a sample of gas has a volume of 233 mL. On heating, the volume expands to 313 mL. Assuming constant pressure, what is the new Celsius temperature? | null |
1,250 | aa11e61a-6ddd-11ea-8fc9-ccda262736ce | https://socratic.org/questions/a-compound-is-54-53-c-9-15-h-and-36-32-o-by-mass-what-is-its-empirical-formula | C2H4O | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"C2H4O"}] | [{"type":"physical unit","value":"Percentage by mass [OF] C in the compound [=] \\pu{54.53%}"},{"type":"physical unit","value":"Percentage by mass [OF] H in the compound [=] \\pu{9.15%}"},{"type":"physical unit","value":"Percentage by mass [OF] O in the compound [=] \\pu{36.32%}"}] | <h1 class="questionTitle" itemprop="name">A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula?</h1> | null | C2H4O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The ratio of no of atoms of C,H&O=<mathjax>#(54.53/12):(9.15/1):(36.32/16)#</mathjax></p>
<p>=<mathjax>#4.54:9.15:2.27#</mathjax></p>
<p>=<mathjax>#4.54/2.27:9.15/2.27:2.27/2.27#</mathjax></p>
<p>=<mathjax>#2:4:1#</mathjax></p>
<p>Empirical formula <mathjax>#C_2H_4O#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#C_2H_4O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The ratio of no of atoms of C,H&O=<mathjax>#(54.53/12):(9.15/1):(36.32/16)#</mathjax></p>
<p>=<mathjax>#4.54:9.15:2.27#</mathjax></p>
<p>=<mathjax>#4.54/2.27:9.15/2.27:2.27/2.27#</mathjax></p>
<p>=<mathjax>#2:4:1#</mathjax></p>
<p>Empirical formula <mathjax>#C_2H_4O#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula?</h1>
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<span class="dateCreated" datetime="2016-02-26T08:20:20" itemprop="dateCreated">
Feb 26, 2016
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<div class="markdown"><p><mathjax>#C_2H_4O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The ratio of no of atoms of C,H&O=<mathjax>#(54.53/12):(9.15/1):(36.32/16)#</mathjax></p>
<p>=<mathjax>#4.54:9.15:2.27#</mathjax></p>
<p>=<mathjax>#4.54/2.27:9.15/2.27:2.27/2.27#</mathjax></p>
<p>=<mathjax>#2:4:1#</mathjax></p>
<p>Empirical formula <mathjax>#C_2H_4O#</mathjax></p></div>
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</article> | A compound is 54.53% C, 9.15% H, and 36.32% O by mass. What is its empirical formula? | null |
1,251 | ac1347d2-6ddd-11ea-9056-ccda262736ce | https://socratic.org/questions/a-60-g-piece-of-metal-at-an-initial-temperature-of-92-c-is-transferred-into-70-g | -1.45 J/(g * ℃) | start physical_unit 5 5 specific_heat j/(°c_·_g) qc_end physical_unit 5 5 1 2 mass qc_end physical_unit 5 5 11 12 temperature qc_end physical_unit 19 19 16 17 mass qc_end physical_unit 19 19 22 23 temperature qc_end physical_unit 28 32 34 35 temperature qc_end end | [{"type":"physical unit","value":"Specific heat [OF] metal [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"-1.45 J/(g * ℃)"}] | [{"type":"physical unit","value":"Mass [OF] metal [=] \\pu{60 g}"},{"type":"physical unit","value":"Temperature1 [OF] metal [=] \\pu{92 ℃}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{70 g}"},{"type":"physical unit","value":"Tem[erature1 [OF] water [=] \\pu{24.5 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the water and the metal [=] \\pu{40 ℃}"}] | <h1 class="questionTitle" itemprop="name">A 60 g piece of metal at an initial temperature of 92°C is transferred into 70 g of water initially at 24.5°C. The final temperature of the water and the metal is 40°C. How do you determine the specific heat of the metal?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>Assume perfect heat transfer.</p></div>
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</div> | -1.45 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#q_"gain" = q_"lost"#</mathjax></p>
<p><mathjax>#(60g)(92^o - 40^oC)(x) = (70g)(24.5^oC - 40^oC)(4.18)"#</mathjax></p>
<p>4.18 is specific of water. Solve for x</p>
<p><mathjax>#3120x=-4535.3#</mathjax></p>
<p><mathjax>#x = 4535.3/3120 or 45353/31200#</mathjax></p>
<p><mathjax>#x = -1.45362#</mathjax></p>
<p>This is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></p></div>
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<div class="markdown"><p><mathjax>#(-1.45362J)/(gram C^o)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#q_"gain" = q_"lost"#</mathjax></p>
<p><mathjax>#(60g)(92^o - 40^oC)(x) = (70g)(24.5^oC - 40^oC)(4.18)"#</mathjax></p>
<p>4.18 is specific of water. Solve for x</p>
<p><mathjax>#3120x=-4535.3#</mathjax></p>
<p><mathjax>#x = 4535.3/3120 or 45353/31200#</mathjax></p>
<p><mathjax>#x = -1.45362#</mathjax></p>
<p>This is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></p></div>
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<h1 class="questionTitle" itemprop="name">A 60 g piece of metal at an initial temperature of 92°C is transferred into 70 g of water initially at 24.5°C. The final temperature of the water and the metal is 40°C. How do you determine the specific heat of the metal?</h1>
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<div class="markdown"><p>Assume perfect heat transfer.</p></div>
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Mar 16, 2017
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<div class="markdown"><p><mathjax>#(-1.45362J)/(gram C^o)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#q_"gain" = q_"lost"#</mathjax></p>
<p><mathjax>#(60g)(92^o - 40^oC)(x) = (70g)(24.5^oC - 40^oC)(4.18)"#</mathjax></p>
<p>4.18 is specific of water. Solve for x</p>
<p><mathjax>#3120x=-4535.3#</mathjax></p>
<p><mathjax>#x = 4535.3/3120 or 45353/31200#</mathjax></p>
<p><mathjax>#x = -1.45362#</mathjax></p>
<p>This is the <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></p></div>
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</article> | A 60 g piece of metal at an initial temperature of 92°C is transferred into 70 g of water initially at 24.5°C. The final temperature of the water and the metal is 40°C. How do you determine the specific heat of the metal? |
Assume perfect heat transfer.
|
1,252 | acd3a6c6-6ddd-11ea-9f5a-ccda262736ce | https://socratic.org/questions/a-0-642-g-sample-of-an-unknown-gas-was-collected-over-water-at-25-0-degrees-c-an | 103 g/mol | start physical_unit 40 42 molar_mass g/mol qc_end physical_unit 6 7 1 2 mass qc_end physical_unit 6 7 13 15 temperature qc_end physical_unit 6 7 17 18 pressure qc_end physical_unit 6 7 23 24 volume qc_end end | [{"type":"physical unit","value":"Molar mass [OF] the unknown gas [IN] g/mol"}] | [{"type":"physical unit","value":"103 g/mol"}] | [{"type":"physical unit","value":"Mass [OF] unknown gas sample [=] \\pu{0.642 g}"},{"type":"physical unit","value":"Temperature [OF] unknown gas sample [=] \\pu{25.0 degrees C}"},{"type":"physical unit","value":"Pressure [OF] unknown gas sample [=] \\pu{1.04 atm}"},{"type":"physical unit","value":"Volume [OF] unknown gas sample [=] \\pu{151.3 mL}"}] | <h1 class="questionTitle" itemprop="name">A 0.642 g sample of an unknown gas was collected over water at 25.0 degrees C and 1.04 atm. The collection cylinder contained 151.3 mL of gas after the sample was released. How do you find the molar mass of the unknown gas?</h1> | null | 103 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = m/M#</mathjax>, we can rearrange this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#PV = (m/M)RT#</mathjax></p>
</blockquote>
</blockquote>
<p>And we can solve this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (mRT)/(PV)#</mathjax></p>
</blockquote>
</blockquote>
<p>Since you are collecting the gas over water,</p>
<p><mathjax>#P_"atm" = P_"gas" + P_"water"#</mathjax></p>
<p>At 25.0 °C, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of water is 0.0313 atm</p>
<p>∴ <mathjax># P_"gas" = P_"atm" - P_"water" = "1.04 atm - 0.0313 atm" = "1.009 atm"#</mathjax></p>
<blockquote></blockquote>
<p>Thus, in your problem,</p>
<p><mathjax>#m = "0.642 g"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "25.0 °C" = "298.15 K"#</mathjax><br/>
<mathjax>#P = "1.009 atm"#</mathjax><br/>
<mathjax>#V = "151.3 mL" = "0.1513 L"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = (0.642 color(red)(cancel(color(black)("g"))) × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 298.15 color(red)(cancel(color(black)("K"))))/(1.009 color(red)(cancel(color(black)("atm"))) × 0.1513 color(red)(cancel(color(black)("L")))) = "103 g/mol"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The molar mass of the gas is 103 g/mol.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = m/M#</mathjax>, we can rearrange this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#PV = (m/M)RT#</mathjax></p>
</blockquote>
</blockquote>
<p>And we can solve this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (mRT)/(PV)#</mathjax></p>
</blockquote>
</blockquote>
<p>Since you are collecting the gas over water,</p>
<p><mathjax>#P_"atm" = P_"gas" + P_"water"#</mathjax></p>
<p>At 25.0 °C, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of water is 0.0313 atm</p>
<p>∴ <mathjax># P_"gas" = P_"atm" - P_"water" = "1.04 atm - 0.0313 atm" = "1.009 atm"#</mathjax></p>
<blockquote></blockquote>
<p>Thus, in your problem,</p>
<p><mathjax>#m = "0.642 g"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "25.0 °C" = "298.15 K"#</mathjax><br/>
<mathjax>#P = "1.009 atm"#</mathjax><br/>
<mathjax>#V = "151.3 mL" = "0.1513 L"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = (0.642 color(red)(cancel(color(black)("g"))) × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 298.15 color(red)(cancel(color(black)("K"))))/(1.009 color(red)(cancel(color(black)("atm"))) × 0.1513 color(red)(cancel(color(black)("L")))) = "103 g/mol"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 0.642 g sample of an unknown gas was collected over water at 25.0 degrees C and 1.04 atm. The collection cylinder contained 151.3 mL of gas after the sample was released. How do you find the molar mass of the unknown gas?</h1>
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Ernest Z.
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<span class="dateCreated" datetime="2016-12-29T02:27:20" itemprop="dateCreated">
Dec 29, 2016
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<div class="markdown"><p>The molar mass of the gas is 103 g/mol.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We can use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> to solve this problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = m/M#</mathjax>, we can rearrange this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#PV = (m/M)RT#</mathjax></p>
</blockquote>
</blockquote>
<p>And we can solve this equation to get</p>
<blockquote>
<blockquote>
<p><mathjax>#M = (mRT)/(PV)#</mathjax></p>
</blockquote>
</blockquote>
<p>Since you are collecting the gas over water,</p>
<p><mathjax>#P_"atm" = P_"gas" + P_"water"#</mathjax></p>
<p>At 25.0 °C, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of water is 0.0313 atm</p>
<p>∴ <mathjax># P_"gas" = P_"atm" - P_"water" = "1.04 atm - 0.0313 atm" = "1.009 atm"#</mathjax></p>
<blockquote></blockquote>
<p>Thus, in your problem,</p>
<p><mathjax>#m = "0.642 g"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "25.0 °C" = "298.15 K"#</mathjax><br/>
<mathjax>#P = "1.009 atm"#</mathjax><br/>
<mathjax>#V = "151.3 mL" = "0.1513 L"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = (0.642 color(red)(cancel(color(black)("g"))) × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 298.15 color(red)(cancel(color(black)("K"))))/(1.009 color(red)(cancel(color(black)("atm"))) × 0.1513 color(red)(cancel(color(black)("L")))) = "103 g/mol"#</mathjax></p></div>
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</article> | A 0.642 g sample of an unknown gas was collected over water at 25.0 degrees C and 1.04 atm. The collection cylinder contained 151.3 mL of gas after the sample was released. How do you find the molar mass of the unknown gas? | null |
1,253 | ab0105e8-6ddd-11ea-a2d1-ccda262736ce | https://socratic.org/questions/58f94b5f7c01493da0d8272d | 0.33 mol/L | start physical_unit 8 8 concentration mol/l qc_end physical_unit 8 8 6 7 concentration qc_end physical_unit 8 8 17 18 concentration qc_end end | [{"type":"physical unit","value":"Concentration3 [OF] HCl(aq) solution [IN] mol/L"}] | [{"type":"physical unit","value":"0.33 mol/L"}] | [{"type":"physical unit","value":"Concentration1 [OF] HCl(aq) solution [=] \\pu{0.130 mol/L}"},{"type":"physical unit","value":"Volume1 [OF] HCl(aq) solution [=] \\pu{16.0 mL}"},{"type":"physical unit","value":"Concentration2 [OF] HCl(aq) solution [=] \\pu{0.600 mol/L}"},{"type":"physical unit","value":"Volume2 [OF] HCl(aq) solution [=] \\pu{12.0 mL}"}] | <h1 class="questionTitle" itemprop="name">
A #=16.0*mL# volume of #0.130*mol*L^-1# #HCl(aq)# was mixed with a #12.0*mL# volume of #0.600*mol*L^-1# #HCl(aq)#. What is the resultant concentration?
</h1> | null | 0.33 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the final concentration of the <mathjax>#"HCl"#</mathjax> solution after two separate <mathjax>#"HCl"#</mathjax> <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed.</p>
<p>To do this, we can use the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> equation</strong>:</p>
<blockquote>
<p><mathjax>#"molarity" = "mol HCl"/"L solution"#</mathjax></p>
</blockquote>
<p>Since we're combining two separate solutions, this can also be represented as</p>
<blockquote>
<p><mathjax>#"molarity" = ("mol"_1 + "mol"_2)/("L"_1 + "L"_2)#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li>
<p><mathjax>#"mol"_1#</mathjax> and <mathjax>#"mol"_2#</mathjax> are the number of <strong>moles</strong> of <mathjax>#"HCl"#</mathjax> in solutions <mathjax>#1#</mathjax> and <mathjax>#2#</mathjax></p>
</li>
<li>
<p><mathjax>#"L"_1#</mathjax> and <mathjax>#"L"_2#</mathjax> are the volumes, in <strong>liters</strong>, of the two <mathjax>#"HCl"#</mathjax> solutions</p>
<blockquote></blockquote>
</li>
</ul>
<p>We're actually going to use the molarity equation to find the number of <strong>moles</strong> of <mathjax>#"HCl"#</mathjax> in each solution:</p>
<p><strong>Solution 1</strong>:</p>
<blockquote>
<p><mathjax>#"mol"_1 = ("molarity"_1)("L"_1) = (0.130"mol"/(cancel("L")))overbrace((0.0160cancel("L")))^"converted to liters" = color(red)(ul(0.00208color(white)(l)"mol HCl"#</mathjax></p>
</blockquote>
<p><strong>Solution 2</strong>:</p>
<blockquote>
<p><mathjax>#"mol"_2 = ("molarity"_2)("L"_2) = (0.600"mol"/(cancel("L")))overbrace((0.0120cancel("L")))^"converted to liters" = color(green)(ul(0.00720color(white)(l)"mol HCl"#</mathjax></p>
</blockquote>
<p>The molarity of the final solution is thus</p>
<blockquote>
<p><mathjax>#"molarity" = ("mol"_1 + "mol"_2)/("L"_1 + "L"_2)#</mathjax></p>
<p><mathjax>#"molarity" = (color(red)(0.00208color(white)(l)"mol") + color(green)(0.00720color(white)(l)"mol"))/(0.0160color(white)(l)"L" + 0.0120color(white)(l)"L") = color(blue)(ulbar(|stackrel(" ")(" "0.331color(white)(l)"mol/L"" ")|)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"molarity" = 0.331#</mathjax> <mathjax>#"mol/L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the final concentration of the <mathjax>#"HCl"#</mathjax> solution after two separate <mathjax>#"HCl"#</mathjax> <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed.</p>
<p>To do this, we can use the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> equation</strong>:</p>
<blockquote>
<p><mathjax>#"molarity" = "mol HCl"/"L solution"#</mathjax></p>
</blockquote>
<p>Since we're combining two separate solutions, this can also be represented as</p>
<blockquote>
<p><mathjax>#"molarity" = ("mol"_1 + "mol"_2)/("L"_1 + "L"_2)#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li>
<p><mathjax>#"mol"_1#</mathjax> and <mathjax>#"mol"_2#</mathjax> are the number of <strong>moles</strong> of <mathjax>#"HCl"#</mathjax> in solutions <mathjax>#1#</mathjax> and <mathjax>#2#</mathjax></p>
</li>
<li>
<p><mathjax>#"L"_1#</mathjax> and <mathjax>#"L"_2#</mathjax> are the volumes, in <strong>liters</strong>, of the two <mathjax>#"HCl"#</mathjax> solutions</p>
<blockquote></blockquote>
</li>
</ul>
<p>We're actually going to use the molarity equation to find the number of <strong>moles</strong> of <mathjax>#"HCl"#</mathjax> in each solution:</p>
<p><strong>Solution 1</strong>:</p>
<blockquote>
<p><mathjax>#"mol"_1 = ("molarity"_1)("L"_1) = (0.130"mol"/(cancel("L")))overbrace((0.0160cancel("L")))^"converted to liters" = color(red)(ul(0.00208color(white)(l)"mol HCl"#</mathjax></p>
</blockquote>
<p><strong>Solution 2</strong>:</p>
<blockquote>
<p><mathjax>#"mol"_2 = ("molarity"_2)("L"_2) = (0.600"mol"/(cancel("L")))overbrace((0.0120cancel("L")))^"converted to liters" = color(green)(ul(0.00720color(white)(l)"mol HCl"#</mathjax></p>
</blockquote>
<p>The molarity of the final solution is thus</p>
<blockquote>
<p><mathjax>#"molarity" = ("mol"_1 + "mol"_2)/("L"_1 + "L"_2)#</mathjax></p>
<p><mathjax>#"molarity" = (color(red)(0.00208color(white)(l)"mol") + color(green)(0.00720color(white)(l)"mol"))/(0.0160color(white)(l)"L" + 0.0120color(white)(l)"L") = color(blue)(ulbar(|stackrel(" ")(" "0.331color(white)(l)"mol/L"" ")|)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">
A #=16.0*mL# volume of #0.130*mol*L^-1# #HCl(aq)# was mixed with a #12.0*mL# volume of #0.600*mol*L^-1# #HCl(aq)#. What is the resultant concentration?
</h1>
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<div class="markdown"><p><mathjax>#"molarity" = 0.331#</mathjax> <mathjax>#"mol/L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the final concentration of the <mathjax>#"HCl"#</mathjax> solution after two separate <mathjax>#"HCl"#</mathjax> <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are mixed.</p>
<p>To do this, we can use the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> equation</strong>:</p>
<blockquote>
<p><mathjax>#"molarity" = "mol HCl"/"L solution"#</mathjax></p>
</blockquote>
<p>Since we're combining two separate solutions, this can also be represented as</p>
<blockquote>
<p><mathjax>#"molarity" = ("mol"_1 + "mol"_2)/("L"_1 + "L"_2)#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li>
<p><mathjax>#"mol"_1#</mathjax> and <mathjax>#"mol"_2#</mathjax> are the number of <strong>moles</strong> of <mathjax>#"HCl"#</mathjax> in solutions <mathjax>#1#</mathjax> and <mathjax>#2#</mathjax></p>
</li>
<li>
<p><mathjax>#"L"_1#</mathjax> and <mathjax>#"L"_2#</mathjax> are the volumes, in <strong>liters</strong>, of the two <mathjax>#"HCl"#</mathjax> solutions</p>
<blockquote></blockquote>
</li>
</ul>
<p>We're actually going to use the molarity equation to find the number of <strong>moles</strong> of <mathjax>#"HCl"#</mathjax> in each solution:</p>
<p><strong>Solution 1</strong>:</p>
<blockquote>
<p><mathjax>#"mol"_1 = ("molarity"_1)("L"_1) = (0.130"mol"/(cancel("L")))overbrace((0.0160cancel("L")))^"converted to liters" = color(red)(ul(0.00208color(white)(l)"mol HCl"#</mathjax></p>
</blockquote>
<p><strong>Solution 2</strong>:</p>
<blockquote>
<p><mathjax>#"mol"_2 = ("molarity"_2)("L"_2) = (0.600"mol"/(cancel("L")))overbrace((0.0120cancel("L")))^"converted to liters" = color(green)(ul(0.00720color(white)(l)"mol HCl"#</mathjax></p>
</blockquote>
<p>The molarity of the final solution is thus</p>
<blockquote>
<p><mathjax>#"molarity" = ("mol"_1 + "mol"_2)/("L"_1 + "L"_2)#</mathjax></p>
<p><mathjax>#"molarity" = (color(red)(0.00208color(white)(l)"mol") + color(green)(0.00720color(white)(l)"mol"))/(0.0160color(white)(l)"L" + 0.0120color(white)(l)"L") = color(blue)(ulbar(|stackrel(" ")(" "0.331color(white)(l)"mol/L"" ")|)#</mathjax></p>
</blockquote></div>
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anor277
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<div class="markdown"><p><mathjax>#[HCl]~~0.3*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the relationship, <mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>We assume (reasonably) that the volumes are additive, and we work out the entire number of moles of <mathjax>#HCl#</mathjax>.........</p>
<p><mathjax>#"Solution 1: Moles of HCl"#</mathjax> <mathjax>#=16.0*mLxx10^-3*L*mL^-1xx0.130*mol*L^-1=2.08xx10^-3*mol#</mathjax></p>
<p><mathjax>#"Solution 2: Moles of HCl"#</mathjax> <mathjax>#=12.0*mLxx10^-3*L*mL^-1xx0.600*mol*L^-1=7.20xx10^-3*mol#</mathjax></p>
<p>And thus the final concentration is given by the quotient......</p>
<p><mathjax>#(2.08xx10^-3*mol+7.20xx10^-3*mol)/(16.0xx10^-3*L+12.0xx10^-3*L)=0.331*mol*L^-1#</mathjax>.</p></div>
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A #=16.0*mL# volume of #0.130*mol*L^-1# #HCl(aq)# was mixed with a #12.0*mL# volume of #0.600*mol*L^-1# #HCl(aq)#. What is the resultant concentration?
| null |
1,254 | abdd2cba-6ddd-11ea-8351-ccda262736ce | https://socratic.org/questions/what-is-the-molar-concentration-of-sulfate-ions-in-a-0-150-m-na-2so-4-solution | 0.15 M | start physical_unit 6 7 molarity mol/l qc_end physical_unit 12 13 10 11 molarity qc_end end | [{"type":"physical unit","value":"Molar concentration [OF] sulfate ions [IN] M"}] | [{"type":"physical unit","value":"0.15 M"}] | [{"type":"physical unit","value":"Molar concentration [OF] Na2SO4 solution [=] \\pu{0.150 M}"}] | <h1 class="questionTitle" itemprop="name">What is the molar concentration of sulfate ions in a 0.150 M #Na_2SO_4# solution?</h1> | null | 0.15 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In solution, sodium sulfate dissociates according to the following equation:</p>
<p><mathjax>#Na_2SO_4(s)->2Na^(+)(aq)+SO_4^(2-)(aq)#</mathjax></p>
<p>Therefore, <mathjax>#1mol#</mathjax> of <mathjax>#Na_2SO_4#</mathjax> will give <mathjax>#1mol#</mathjax> of <mathjax>#SO_4^(2-)#</mathjax> and thus, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of sulfate ions will be equal to that of the sodium sulfate.</p>
<p><mathjax>#M_(SO_4^(2-))=0.150M#</mathjax></p>
<p>However, the molarity of sodium ions is the double of that of sodium sulfate since <mathjax>#1mol#</mathjax> of <mathjax>#Na_2SO_4#</mathjax> will give <mathjax>#2mol#</mathjax> of <mathjax>#Na^(+)#</mathjax>:</p>
<p><mathjax>#M_(Na^(+))=0.300M#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#M_(SO_4^(2-))=0.150M#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In solution, sodium sulfate dissociates according to the following equation:</p>
<p><mathjax>#Na_2SO_4(s)->2Na^(+)(aq)+SO_4^(2-)(aq)#</mathjax></p>
<p>Therefore, <mathjax>#1mol#</mathjax> of <mathjax>#Na_2SO_4#</mathjax> will give <mathjax>#1mol#</mathjax> of <mathjax>#SO_4^(2-)#</mathjax> and thus, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of sulfate ions will be equal to that of the sodium sulfate.</p>
<p><mathjax>#M_(SO_4^(2-))=0.150M#</mathjax></p>
<p>However, the molarity of sodium ions is the double of that of sodium sulfate since <mathjax>#1mol#</mathjax> of <mathjax>#Na_2SO_4#</mathjax> will give <mathjax>#2mol#</mathjax> of <mathjax>#Na^(+)#</mathjax>:</p>
<p><mathjax>#M_(Na^(+))=0.300M#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molar concentration of sulfate ions in a 0.150 M #Na_2SO_4# solution?</h1>
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<div class="markdown"><p><mathjax>#M_(SO_4^(2-))=0.150M#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In solution, sodium sulfate dissociates according to the following equation:</p>
<p><mathjax>#Na_2SO_4(s)->2Na^(+)(aq)+SO_4^(2-)(aq)#</mathjax></p>
<p>Therefore, <mathjax>#1mol#</mathjax> of <mathjax>#Na_2SO_4#</mathjax> will give <mathjax>#1mol#</mathjax> of <mathjax>#SO_4^(2-)#</mathjax> and thus, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of sulfate ions will be equal to that of the sodium sulfate.</p>
<p><mathjax>#M_(SO_4^(2-))=0.150M#</mathjax></p>
<p>However, the molarity of sodium ions is the double of that of sodium sulfate since <mathjax>#1mol#</mathjax> of <mathjax>#Na_2SO_4#</mathjax> will give <mathjax>#2mol#</mathjax> of <mathjax>#Na^(+)#</mathjax>:</p>
<p><mathjax>#M_(Na^(+))=0.300M#</mathjax></p></div>
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</article> | What is the molar concentration of sulfate ions in a 0.150 M #Na_2SO_4# solution? | null |
1,255 | aa0dc15c-6ddd-11ea-a4ef-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-1-00-x-10-4-m-solution-of-lithium-hydroxide-solution | 10.00 | start physical_unit 12 14 ph none qc_end physical_unit 12 14 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] lithium hydroxide solution"}] | [{"type":"physical unit","value":"10.00"}] | [{"type":"physical unit","value":"Molarity [OF] lithium hydroxide solution [=] \\pu{1.00 × 10^(-4) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a 1.00 x 10^-4 M solution of lithium hydroxide solution?</h1> | null | 10.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax>, <mathjax>#"pouvoir hydrogene"#</mathjax>, <mathjax>#=-log_10[H_3O^+]#</mathjax></p>
<p>Now it is a fact that in water the following autoprotolysis takes place:</p>
<p><mathjax>#2H_2O(l) rightleftharpoonsHO^(-) + H_3O^+#</mathjax></p>
<p>The ionic product at <mathjax>#298*K#</mathjax> <mathjax>#=#</mathjax> <mathjax>#K_w=[H_3O^+][HO^-]=10^-14#</mathjax>.</p>
<p>If we take <mathjax>#-log_10#</mathjax> of both sides we get:</p>
<p><mathjax>#pK_w =14=-log_10[H_3O^+]-log_10[HO^-]#</mathjax></p>
<p>But by definition, <mathjax>#pH =-log_10[H_3O^+]#</mathjax> and <mathjax>#pOH =-log_10[HO^-]#</mathjax></p>
<p>Thus <mathjax>#pK_w =14=pH+pOH#</mathjax></p>
<p>And so <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(10^-4)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4#</mathjax></p>
<p>And <mathjax>#pH=10#</mathjax>.</p>
<p>Capisce?</p></div>
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<div class="markdown"><p><mathjax>#pH=10#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax>, <mathjax>#"pouvoir hydrogene"#</mathjax>, <mathjax>#=-log_10[H_3O^+]#</mathjax></p>
<p>Now it is a fact that in water the following autoprotolysis takes place:</p>
<p><mathjax>#2H_2O(l) rightleftharpoonsHO^(-) + H_3O^+#</mathjax></p>
<p>The ionic product at <mathjax>#298*K#</mathjax> <mathjax>#=#</mathjax> <mathjax>#K_w=[H_3O^+][HO^-]=10^-14#</mathjax>.</p>
<p>If we take <mathjax>#-log_10#</mathjax> of both sides we get:</p>
<p><mathjax>#pK_w =14=-log_10[H_3O^+]-log_10[HO^-]#</mathjax></p>
<p>But by definition, <mathjax>#pH =-log_10[H_3O^+]#</mathjax> and <mathjax>#pOH =-log_10[HO^-]#</mathjax></p>
<p>Thus <mathjax>#pK_w =14=pH+pOH#</mathjax></p>
<p>And so <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(10^-4)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4#</mathjax></p>
<p>And <mathjax>#pH=10#</mathjax>.</p>
<p>Capisce?</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a 1.00 x 10^-4 M solution of lithium hydroxide solution?</h1>
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anor277
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<div class="markdown"><p><mathjax>#pH=10#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH#</mathjax>, <mathjax>#"pouvoir hydrogene"#</mathjax>, <mathjax>#=-log_10[H_3O^+]#</mathjax></p>
<p>Now it is a fact that in water the following autoprotolysis takes place:</p>
<p><mathjax>#2H_2O(l) rightleftharpoonsHO^(-) + H_3O^+#</mathjax></p>
<p>The ionic product at <mathjax>#298*K#</mathjax> <mathjax>#=#</mathjax> <mathjax>#K_w=[H_3O^+][HO^-]=10^-14#</mathjax>.</p>
<p>If we take <mathjax>#-log_10#</mathjax> of both sides we get:</p>
<p><mathjax>#pK_w =14=-log_10[H_3O^+]-log_10[HO^-]#</mathjax></p>
<p>But by definition, <mathjax>#pH =-log_10[H_3O^+]#</mathjax> and <mathjax>#pOH =-log_10[HO^-]#</mathjax></p>
<p>Thus <mathjax>#pK_w =14=pH+pOH#</mathjax></p>
<p>And so <mathjax>#pOH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_10(10^-4)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#4#</mathjax></p>
<p>And <mathjax>#pH=10#</mathjax>.</p>
<p>Capisce?</p></div>
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</article> | What is the pH of a 1.00 x 10^-4 M solution of lithium hydroxide solution? | null |
1,256 | aa6b4c8c-6ddd-11ea-8fc1-ccda262736ce | https://socratic.org/questions/what-is-the-number-of-grams-of-xenon-in-3-958-g-of-the-compound-xenon-tetrafluor | 2.51 grams | start physical_unit 7 7 mass g qc_end physical_unit 12 15 9 10 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] xenon [IN] grams"}] | [{"type":"physical unit","value":"2.51 grams"}] | [{"type":"physical unit","value":"Mass [OF] the compound xenon tetrafluoride [=] \\pu{3.958 g}"}] | <h1 class="questionTitle" itemprop="name">What is the number of grams of xenon in 3.958 g of the compound xenon tetrafluoride?</h1> | null | 2.51 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Moles of "XeF_4#</mathjax> <mathjax>#~=1/50*mol#</mathjax></p>
<p>How many moles of <mathjax>#F#</mathjax> are present here?</p></div>
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<div class="markdown"><p><mathjax>#"Moles of "XeF_4#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(3.958*g)/(207.28*g*mol^-1)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Moles of "XeF_4#</mathjax> <mathjax>#~=1/50*mol#</mathjax></p>
<p>How many moles of <mathjax>#F#</mathjax> are present here?</p></div>
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<div class="markdown"><p><mathjax>#"Moles of "XeF_4#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(3.958*g)/(207.28*g*mol^-1)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Moles of "XeF_4#</mathjax> <mathjax>#~=1/50*mol#</mathjax></p>
<p>How many moles of <mathjax>#F#</mathjax> are present here?</p></div>
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</article> | What is the number of grams of xenon in 3.958 g of the compound xenon tetrafluoride? | null |
1,257 | a8acc58c-6ddd-11ea-a841-ccda262736ce | https://socratic.org/questions/what-is-the-net-ionic-equation-for-the-reaction-between-hf-a-weak-acid-and-naoh | HF + OH- -> F- + H2O | start chemical_equation qc_end chemical_equation 15 15 qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"HF + OH- -> F- + H2O"}] | [{"type":"chemical equation","value":"NaOH"},{"type":"other","value":"HF is a weak acid."}] | <h1 class="questionTitle" itemprop="name">What is the net ionic equation for the reaction between #HF# (a weak acid) and #NaOH#?</h1> | null | HF + OH- -> F- + H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Hydrofluoric acid</em>, <mathjax>#"HF"#</mathjax>, a <strong>weak acid</strong>, will react with <em>sodium hydroxide</em>, <mathjax>#"NaOH"#</mathjax>, a <strong>strong base</strong>, to produce <em>aqueous</em> sodium fluoride, <mathjax>#"NaF"#</mathjax>, and water. </p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) + "NaOH"_ ((aq)) -> "NaF"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Now, the important thing to remember here is that hydrofluoric acid is a <strong>weak acid</strong>, which implies that it <em>does not</em> dissociate <strong>completely</strong> in aqueous solution to form hydrogen cations, <mathjax>#"H"^(+)#</mathjax>, usually referred to as <em>hydronium cations</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, and fluoride anions, <mathjax>#"F"^(-)#</mathjax>. </p>
<p>Instead, an equilibrium will be established between the undissociated acid and the dissociated ions</p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "F"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Sodium hydroxide, on the other hand, <em>will</em> dissociate <strong>completely</strong> in aqueous solution to produce sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Here's what happens at this point. The hydroxide anions and the hydrogen cations will <strong>neutralize</strong> each other to produce water. </p>
<blockquote>
<p><mathjax>#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>This will <strong>consume</strong> hydrogen cations and cause the dissociation equilibrium of hydrofluoric acid to <strong>shift to the right</strong> <mathjax>#->#</mathjax> more of the molecules of acid will dissociate. </p>
<p>This will go on until <strong>all the molecules of acid</strong> have dissociated and all the hydrogen cations have reacted with the hydroxide anions <mathjax>#->#</mathjax> a <strong>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong> took place. </p>
<p>Now, the <strong>complete ionic equation</strong> looks like this -- keep in mind that sodium fluoride is <strong>soluble</strong> in aqueous solution</p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "Na"_ ((aq))^(+) + "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Notice that because hydrofluoric acid does not dissociate completely, you must represent it in <em>molecular form</em>. </p>
<p>Eliminate the <em>spectator ions</em>, which in this case are the sodium cations</p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>to get the <strong>net ionic equation</strong> that describes this neutralization reaction</p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("HF"_ ((aq)) + "OH"_ ((aq))^(-) -> "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l)))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"HF"_ ((aq)) + "OH"_ ((aq))^(-) -> "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Hydrofluoric acid</em>, <mathjax>#"HF"#</mathjax>, a <strong>weak acid</strong>, will react with <em>sodium hydroxide</em>, <mathjax>#"NaOH"#</mathjax>, a <strong>strong base</strong>, to produce <em>aqueous</em> sodium fluoride, <mathjax>#"NaF"#</mathjax>, and water. </p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) + "NaOH"_ ((aq)) -> "NaF"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Now, the important thing to remember here is that hydrofluoric acid is a <strong>weak acid</strong>, which implies that it <em>does not</em> dissociate <strong>completely</strong> in aqueous solution to form hydrogen cations, <mathjax>#"H"^(+)#</mathjax>, usually referred to as <em>hydronium cations</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, and fluoride anions, <mathjax>#"F"^(-)#</mathjax>. </p>
<p>Instead, an equilibrium will be established between the undissociated acid and the dissociated ions</p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "F"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Sodium hydroxide, on the other hand, <em>will</em> dissociate <strong>completely</strong> in aqueous solution to produce sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Here's what happens at this point. The hydroxide anions and the hydrogen cations will <strong>neutralize</strong> each other to produce water. </p>
<blockquote>
<p><mathjax>#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>This will <strong>consume</strong> hydrogen cations and cause the dissociation equilibrium of hydrofluoric acid to <strong>shift to the right</strong> <mathjax>#->#</mathjax> more of the molecules of acid will dissociate. </p>
<p>This will go on until <strong>all the molecules of acid</strong> have dissociated and all the hydrogen cations have reacted with the hydroxide anions <mathjax>#->#</mathjax> a <strong>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong> took place. </p>
<p>Now, the <strong>complete ionic equation</strong> looks like this -- keep in mind that sodium fluoride is <strong>soluble</strong> in aqueous solution</p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "Na"_ ((aq))^(+) + "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Notice that because hydrofluoric acid does not dissociate completely, you must represent it in <em>molecular form</em>. </p>
<p>Eliminate the <em>spectator ions</em>, which in this case are the sodium cations</p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>to get the <strong>net ionic equation</strong> that describes this neutralization reaction</p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("HF"_ ((aq)) + "OH"_ ((aq))^(-) -> "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l)))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the net ionic equation for the reaction between #HF# (a weak acid) and #NaOH#?</h1>
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<div class="markdown"><p><mathjax>#"HF"_ ((aq)) + "OH"_ ((aq))^(-) -> "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Hydrofluoric acid</em>, <mathjax>#"HF"#</mathjax>, a <strong>weak acid</strong>, will react with <em>sodium hydroxide</em>, <mathjax>#"NaOH"#</mathjax>, a <strong>strong base</strong>, to produce <em>aqueous</em> sodium fluoride, <mathjax>#"NaF"#</mathjax>, and water. </p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) + "NaOH"_ ((aq)) -> "NaF"_ ((aq)) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Now, the important thing to remember here is that hydrofluoric acid is a <strong>weak acid</strong>, which implies that it <em>does not</em> dissociate <strong>completely</strong> in aqueous solution to form hydrogen cations, <mathjax>#"H"^(+)#</mathjax>, usually referred to as <em>hydronium cations</em>, <mathjax>#"H"_3"O"^(+)#</mathjax>, and fluoride anions, <mathjax>#"F"^(-)#</mathjax>. </p>
<p>Instead, an equilibrium will be established between the undissociated acid and the dissociated ions</p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "F"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Sodium hydroxide, on the other hand, <em>will</em> dissociate <strong>completely</strong> in aqueous solution to produce sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and hydroxide anions, <mathjax>#"OH"^(-)#</mathjax></p>
<blockquote>
<p><mathjax>#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Here's what happens at this point. The hydroxide anions and the hydrogen cations will <strong>neutralize</strong> each other to produce water. </p>
<blockquote>
<p><mathjax>#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>This will <strong>consume</strong> hydrogen cations and cause the dissociation equilibrium of hydrofluoric acid to <strong>shift to the right</strong> <mathjax>#->#</mathjax> more of the molecules of acid will dissociate. </p>
<p>This will go on until <strong>all the molecules of acid</strong> have dissociated and all the hydrogen cations have reacted with the hydroxide anions <mathjax>#->#</mathjax> a <strong>complete <a href="https://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a></strong> took place. </p>
<p>Now, the <strong>complete ionic equation</strong> looks like this -- keep in mind that sodium fluoride is <strong>soluble</strong> in aqueous solution</p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "Na"_ ((aq))^(+) + "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>Notice that because hydrofluoric acid does not dissociate completely, you must represent it in <em>molecular form</em>. </p>
<p>Eliminate the <em>spectator ions</em>, which in this case are the sodium cations</p>
<blockquote>
<p><mathjax>#"HF"_ ((aq)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#</mathjax></p>
</blockquote>
<p>to get the <strong>net ionic equation</strong> that describes this neutralization reaction</p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("HF"_ ((aq)) + "OH"_ ((aq))^(-) -> "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l)))color(white)(a/a)|)))#</mathjax></p>
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</article> | What is the net ionic equation for the reaction between #HF# (a weak acid) and #NaOH#? | null |
1,258 | ad05a576-6ddd-11ea-a875-ccda262736ce | https://socratic.org/questions/at-10-c-the-gas-in-a-cylinder-has-a-volume-of-0-250-l-the-gas-is-allowed-to-expa | 49.62 ℃ | start physical_unit 3 4 temperature °c qc_end physical_unit 3 4 1 2 temperature qc_end physical_unit 3 4 12 13 volume qc_end physical_unit 3 4 21 22 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] ℃"}] | [{"type":"physical unit","value":"49.62 ℃"}] | [{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{10 ℃}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{0.250 L}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{0.285 L}"},{"type":"other","value":"The pressure to remain constant."}] | <h1 class="questionTitle" itemprop="name">At 10°C, the gas in a cylinder has a volume of 0.250 L. The gas is allowed to expand to 0.285 L. What must the final temperature be for the pressure to remain constant? </h1> | null | 49.62 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the formula: <mathjax>#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#</mathjax></p>
<p>We know that we want <mathjax>#P_1=P_2#</mathjax> so the pressure will remain constant, so we can say that: <mathjax>#(color(blue)(P_1)V_1)/(T_1)=(color(blue)(P_1)V_2)/(T_2)#</mathjax></p>
<p>Plug in the values we know: <mathjax>#(P_1*0.250"L")/(283"K")=(P_1*0.285"L")/(T_2)#</mathjax><br/>
(Remember that temperature must be done in Kelvin.)</p>
<p>Cross multiply: <mathjax>#T_2*P_1*0.250"L"=283"K"*P_1*0.285"L"#</mathjax></p>
<p>Divide both sides by <mathjax>#P_1#</mathjax>.</p>
<p><mathjax>#T_2*0.250"L"=283"K"*0.285"L"#</mathjax></p>
<p><mathjax>#T_2=(283"K"*0.285"L")/(0.250"L")#</mathjax></p>
<p><mathjax>#T_2=323"K"#</mathjax></p>
<p>(This is also <mathjax>#50˚"C"#</mathjax>.)</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#50˚"C"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the formula: <mathjax>#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#</mathjax></p>
<p>We know that we want <mathjax>#P_1=P_2#</mathjax> so the pressure will remain constant, so we can say that: <mathjax>#(color(blue)(P_1)V_1)/(T_1)=(color(blue)(P_1)V_2)/(T_2)#</mathjax></p>
<p>Plug in the values we know: <mathjax>#(P_1*0.250"L")/(283"K")=(P_1*0.285"L")/(T_2)#</mathjax><br/>
(Remember that temperature must be done in Kelvin.)</p>
<p>Cross multiply: <mathjax>#T_2*P_1*0.250"L"=283"K"*P_1*0.285"L"#</mathjax></p>
<p>Divide both sides by <mathjax>#P_1#</mathjax>.</p>
<p><mathjax>#T_2*0.250"L"=283"K"*0.285"L"#</mathjax></p>
<p><mathjax>#T_2=(283"K"*0.285"L")/(0.250"L")#</mathjax></p>
<p><mathjax>#T_2=323"K"#</mathjax></p>
<p>(This is also <mathjax>#50˚"C"#</mathjax>.)</p></div>
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<h1 class="questionTitle" itemprop="name">At 10°C, the gas in a cylinder has a volume of 0.250 L. The gas is allowed to expand to 0.285 L. What must the final temperature be for the pressure to remain constant? </h1>
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<span class="dateCreated" datetime="2015-12-05T16:34:11" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#50˚"C"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the formula: <mathjax>#(P_1V_1)/(T_1)=(P_2V_2)/(T_2)#</mathjax></p>
<p>We know that we want <mathjax>#P_1=P_2#</mathjax> so the pressure will remain constant, so we can say that: <mathjax>#(color(blue)(P_1)V_1)/(T_1)=(color(blue)(P_1)V_2)/(T_2)#</mathjax></p>
<p>Plug in the values we know: <mathjax>#(P_1*0.250"L")/(283"K")=(P_1*0.285"L")/(T_2)#</mathjax><br/>
(Remember that temperature must be done in Kelvin.)</p>
<p>Cross multiply: <mathjax>#T_2*P_1*0.250"L"=283"K"*P_1*0.285"L"#</mathjax></p>
<p>Divide both sides by <mathjax>#P_1#</mathjax>.</p>
<p><mathjax>#T_2*0.250"L"=283"K"*0.285"L"#</mathjax></p>
<p><mathjax>#T_2=(283"K"*0.285"L")/(0.250"L")#</mathjax></p>
<p><mathjax>#T_2=323"K"#</mathjax></p>
<p>(This is also <mathjax>#50˚"C"#</mathjax>.)</p></div>
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</article> | At 10°C, the gas in a cylinder has a volume of 0.250 L. The gas is allowed to expand to 0.285 L. What must the final temperature be for the pressure to remain constant? | null |
1,259 | aa92f9cc-6ddd-11ea-92db-ccda262736ce | https://socratic.org/questions/a-sample-of-nitrogen-gas-at-a-pressure-of-3-atm-inside-a-rigid-metal-container-a-1 | 3.31 atm | start physical_unit 1 4 pressure atm qc_end physical_unit 1 4 9 10 pressure qc_end physical_unit 1 4 17 18 temperature qc_end physical_unit 1 4 27 28 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] nitrogen gas sample [IN] atm"}] | [{"type":"physical unit","value":"3.31 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] nitrogen gas sample [=] \\pu{3 atm}"},{"type":"physical unit","value":"Temperature1 [OF] nitrogen gas sample [=] \\pu{20.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] nitrogen gas sample [=] \\pu{50.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">A sample of nitrogen gas at a pressure of 3 atm inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. What is the pressure of the nitrogen after its temperature is increased? Answer
atm
</h1> | null | 3.31 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Gay-Lussac's Law states <mathjax>#P_1/T_1 = P_2/T_2#</mathjax></p>
<p>So let's take the <mathjax>#"N"_"2(g)"#</mathjax> and calculate the <mathjax>#P_1#</mathjax> divided by <mathjax>#T_1#</mathjax>, which is <mathjax>#"3 atm"#</mathjax> divided by <mathjax>#"293 K"#</mathjax>. Then we take <mathjax>#P_2#</mathjax> divided by <mathjax>#T_2#</mathjax> which is <mathjax>#P_2#</mathjax> divided by <mathjax>#"323 Kelvin"#</mathjax>.</p>
<p><mathjax>#"3 atm"/(293 cancel"Kelvin") = (P_2)/(323 cancel"Kelvin")#</mathjax></p>
<p><mathjax>#"0.010239 atm" = (P_2)/323#</mathjax></p>
<p>Then cross multiply, imagine the denominator of <mathjax>#0.010239#</mathjax> is <mathjax>#1#</mathjax>, since <mathjax>#0.010239#</mathjax> divided by <mathjax>#1#</mathjax> is <mathjax>#0.010239#</mathjax>.</p>
<p><mathjax>#"0.010239 atm" xx 323 = P_2 xx 1#</mathjax></p>
<p><mathjax>#"3.31 atm" = P_2#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"3.31 atm"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Gay-Lussac's Law states <mathjax>#P_1/T_1 = P_2/T_2#</mathjax></p>
<p>So let's take the <mathjax>#"N"_"2(g)"#</mathjax> and calculate the <mathjax>#P_1#</mathjax> divided by <mathjax>#T_1#</mathjax>, which is <mathjax>#"3 atm"#</mathjax> divided by <mathjax>#"293 K"#</mathjax>. Then we take <mathjax>#P_2#</mathjax> divided by <mathjax>#T_2#</mathjax> which is <mathjax>#P_2#</mathjax> divided by <mathjax>#"323 Kelvin"#</mathjax>.</p>
<p><mathjax>#"3 atm"/(293 cancel"Kelvin") = (P_2)/(323 cancel"Kelvin")#</mathjax></p>
<p><mathjax>#"0.010239 atm" = (P_2)/323#</mathjax></p>
<p>Then cross multiply, imagine the denominator of <mathjax>#0.010239#</mathjax> is <mathjax>#1#</mathjax>, since <mathjax>#0.010239#</mathjax> divided by <mathjax>#1#</mathjax> is <mathjax>#0.010239#</mathjax>.</p>
<p><mathjax>#"0.010239 atm" xx 323 = P_2 xx 1#</mathjax></p>
<p><mathjax>#"3.31 atm" = P_2#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A sample of nitrogen gas at a pressure of 3 atm inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. What is the pressure of the nitrogen after its temperature is increased? Answer
atm
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Stefan V.
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<div class="markdown"><p><mathjax>#"3.31 atm"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Gay-Lussac's Law states <mathjax>#P_1/T_1 = P_2/T_2#</mathjax></p>
<p>So let's take the <mathjax>#"N"_"2(g)"#</mathjax> and calculate the <mathjax>#P_1#</mathjax> divided by <mathjax>#T_1#</mathjax>, which is <mathjax>#"3 atm"#</mathjax> divided by <mathjax>#"293 K"#</mathjax>. Then we take <mathjax>#P_2#</mathjax> divided by <mathjax>#T_2#</mathjax> which is <mathjax>#P_2#</mathjax> divided by <mathjax>#"323 Kelvin"#</mathjax>.</p>
<p><mathjax>#"3 atm"/(293 cancel"Kelvin") = (P_2)/(323 cancel"Kelvin")#</mathjax></p>
<p><mathjax>#"0.010239 atm" = (P_2)/323#</mathjax></p>
<p>Then cross multiply, imagine the denominator of <mathjax>#0.010239#</mathjax> is <mathjax>#1#</mathjax>, since <mathjax>#0.010239#</mathjax> divided by <mathjax>#1#</mathjax> is <mathjax>#0.010239#</mathjax>.</p>
<p><mathjax>#"0.010239 atm" xx 323 = P_2 xx 1#</mathjax></p>
<p><mathjax>#"3.31 atm" = P_2#</mathjax></p></div>
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<div class="markdown"><p>See Below</p></div>
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<div class="markdown"><p>Rigid means the volume doesn't change. As long as it is sealed, the moles don't change, either.</p>
<p><mathjax>#n_1 = n_2......n="PV"/"RT"#</mathjax></p>
<p><mathjax>#(P_1V_1)/(RT_1) = (P_2V_2)/(RT_2)#</mathjax></p>
<p>R and R are the same, so they cancel. V1 and V2 are the same (rigid container), so they cancel. You are left with</p>
<p><mathjax>#(P_1)/(T_1) = (P_2)/(T_2)#</mathjax></p>
<p>Change 20C to 293K (T1) and 50C to 323K (T2) and solve for P2</p>
<p><mathjax>#P_2 = (P_1)/(T_1)xxT_2#</mathjax><br/>
<mathjax>#P_2 = (3atm)/(293)xx323#</mathjax></p>
<p><mathjax>#P_2 = 3.31 atm#</mathjax></p></div>
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</article> | A sample of nitrogen gas at a pressure of 3 atm inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. What is the pressure of the nitrogen after its temperature is increased? Answer
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1,260 | abe604c0-6ddd-11ea-a1b2-ccda262736ce | https://socratic.org/questions/if-iron-ii-oxide-gas-decomposes-into-iron-liquid-and-oxygen-gas-and-6-moles-of-i | 67.2 liters | start physical_unit 10 11 volume l qc_end physical_unit 1 4 13 14 mole qc_end substance 7 8 qc_end end | [{"type":"physical unit","value":"Volume [OF] oxygen gas [IN] liters"}] | [{"type":"physical unit","value":"67.2 liters"}] | [{"type":"physical unit","value":"Mole [OF] iron (II) oxide gas [=] \\pu{6 moles}"},{"type":"substance name","value":"Iron liquid"}] | <h1 class="questionTitle" itemprop="name">If iron (II) oxide gas decomposes into iron liquid and oxygen gas and 6 moles of iron (II) oxide gas are used, how of liters of oxygen gas are produced?</h1> | null | 67.2 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This means that for <mathjax>#6mol#</mathjax> of <mathjax>#FeO#</mathjax>, only <mathjax>#3mol O_2#</mathjax> is produced (half the moles).<br/>
The volume of the oxygen is dependant on temperature and pressure. </p>
<p><strong>E.g.</strong> : If you cooled it down to <mathjax>#0^oC#</mathjax> and get it to standard atmospheric pressure the volume would be <mathjax>#3*22.4L=67.2L#</mathjax> <br/>
The 22.4 L being the molar volume for ideal gases under those conditions. There are tables for other temperatures.</p></div>
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<div class="markdown"><p>The reaction equation is: <mathjax>#2FeO->2Fe+O_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This means that for <mathjax>#6mol#</mathjax> of <mathjax>#FeO#</mathjax>, only <mathjax>#3mol O_2#</mathjax> is produced (half the moles).<br/>
The volume of the oxygen is dependant on temperature and pressure. </p>
<p><strong>E.g.</strong> : If you cooled it down to <mathjax>#0^oC#</mathjax> and get it to standard atmospheric pressure the volume would be <mathjax>#3*22.4L=67.2L#</mathjax> <br/>
The 22.4 L being the molar volume for ideal gases under those conditions. There are tables for other temperatures.</p></div>
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<h1 class="questionTitle" itemprop="name">If iron (II) oxide gas decomposes into iron liquid and oxygen gas and 6 moles of iron (II) oxide gas are used, how of liters of oxygen gas are produced?</h1>
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<div class="markdown"><p>The reaction equation is: <mathjax>#2FeO->2Fe+O_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This means that for <mathjax>#6mol#</mathjax> of <mathjax>#FeO#</mathjax>, only <mathjax>#3mol O_2#</mathjax> is produced (half the moles).<br/>
The volume of the oxygen is dependant on temperature and pressure. </p>
<p><strong>E.g.</strong> : If you cooled it down to <mathjax>#0^oC#</mathjax> and get it to standard atmospheric pressure the volume would be <mathjax>#3*22.4L=67.2L#</mathjax> <br/>
The 22.4 L being the molar volume for ideal gases under those conditions. There are tables for other temperatures.</p></div>
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</article> | If iron (II) oxide gas decomposes into iron liquid and oxygen gas and 6 moles of iron (II) oxide gas are used, how of liters of oxygen gas are produced? | null |
1,261 | a92f705c-6ddd-11ea-afee-ccda262736ce | https://socratic.org/questions/5810d355b72cff3c33df94d2 | 4.78 | start physical_unit 11 12 ph none qc_end physical_unit 11 12 3 6 [oh-] qc_end end | [{"type":"physical unit","value":"pH [OF] this solution"}] | [{"type":"physical unit","value":"4.78"}] | [{"type":"physical unit","value":"[HO-] [OF] this solution [=] \\pu{6.0 × 10^(-10) mol/L}"}] | <h1 class="questionTitle" itemprop="name">If #[HO^-]=6.0xx10^-10*mol*L^-1#, what is #pH# for this solution?</h1> | null | 4.78 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqueous solution at <mathjax>#298*K#</mathjax>, water undergoes the following equilibrium reaction:</p>
<p><mathjax>#H_2O(l) rightleftharpoons H^(+) + HO^(-)#</mathjax></p>
<p>Alternatively, and perhaps now more commonly we would write:</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>This equilibrium reaction has been carefully and quantitatively measured, and under standard conditions:</p>
<p><mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax></p>
<p>If we take negative logarithms to the base 10 of both sides we get:</p>
<p><mathjax>#pK_w=14=pOH+pH#</mathjax>, where the <mathjax>#pH#</mathjax> functions means <mathjax>#-log_10[H_3O^+]#</mathjax>, etc. </p>
<p>We are given <mathjax>#[HO^-]=6.0xx10^-10#</mathjax>, so <mathjax>#pOH=-log_10(6.0xx10^-10)=-(-9.22)=9.22#</mathjax>, and thus <mathjax>#pH=14-9.22=4.78#</mathjax>.</p>
<p>Is this solution acidic or basic?</p>
<p>The given reaction occurs at <mathjax>#298*K#</mathjax>. Given that this is a bond-breaking reaction, how do you think <mathjax>#pK_w#</mathjax> would evolve at <mathjax>#323*K#</mathjax> or <mathjax>#373*K#</mathjax>?</p></div>
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<div class="markdown"><p><mathjax>#[H^+] ("or "[H_3O^+])=10^(-4.78)*mol*L^-1#</mathjax>, <mathjax>#pH=4.78#</mathjax>, <mathjax>#pOH=9.22#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In aqueous solution at <mathjax>#298*K#</mathjax>, water undergoes the following equilibrium reaction:</p>
<p><mathjax>#H_2O(l) rightleftharpoons H^(+) + HO^(-)#</mathjax></p>
<p>Alternatively, and perhaps now more commonly we would write:</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>This equilibrium reaction has been carefully and quantitatively measured, and under standard conditions:</p>
<p><mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax></p>
<p>If we take negative logarithms to the base 10 of both sides we get:</p>
<p><mathjax>#pK_w=14=pOH+pH#</mathjax>, where the <mathjax>#pH#</mathjax> functions means <mathjax>#-log_10[H_3O^+]#</mathjax>, etc. </p>
<p>We are given <mathjax>#[HO^-]=6.0xx10^-10#</mathjax>, so <mathjax>#pOH=-log_10(6.0xx10^-10)=-(-9.22)=9.22#</mathjax>, and thus <mathjax>#pH=14-9.22=4.78#</mathjax>.</p>
<p>Is this solution acidic or basic?</p>
<p>The given reaction occurs at <mathjax>#298*K#</mathjax>. Given that this is a bond-breaking reaction, how do you think <mathjax>#pK_w#</mathjax> would evolve at <mathjax>#323*K#</mathjax> or <mathjax>#373*K#</mathjax>?</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">If #[HO^-]=6.0xx10^-10*mol*L^-1#, what is #pH# for this solution?</h1>
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anor277
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<div class="markdown"><p><mathjax>#[H^+] ("or "[H_3O^+])=10^(-4.78)*mol*L^-1#</mathjax>, <mathjax>#pH=4.78#</mathjax>, <mathjax>#pOH=9.22#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In aqueous solution at <mathjax>#298*K#</mathjax>, water undergoes the following equilibrium reaction:</p>
<p><mathjax>#H_2O(l) rightleftharpoons H^(+) + HO^(-)#</mathjax></p>
<p>Alternatively, and perhaps now more commonly we would write:</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>This equilibrium reaction has been carefully and quantitatively measured, and under standard conditions:</p>
<p><mathjax>#[H_3O^+][HO^-]=10^-14#</mathjax></p>
<p>If we take negative logarithms to the base 10 of both sides we get:</p>
<p><mathjax>#pK_w=14=pOH+pH#</mathjax>, where the <mathjax>#pH#</mathjax> functions means <mathjax>#-log_10[H_3O^+]#</mathjax>, etc. </p>
<p>We are given <mathjax>#[HO^-]=6.0xx10^-10#</mathjax>, so <mathjax>#pOH=-log_10(6.0xx10^-10)=-(-9.22)=9.22#</mathjax>, and thus <mathjax>#pH=14-9.22=4.78#</mathjax>.</p>
<p>Is this solution acidic or basic?</p>
<p>The given reaction occurs at <mathjax>#298*K#</mathjax>. Given that this is a bond-breaking reaction, how do you think <mathjax>#pK_w#</mathjax> would evolve at <mathjax>#323*K#</mathjax> or <mathjax>#373*K#</mathjax>?</p></div>
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</article> | If #[HO^-]=6.0xx10^-10*mol*L^-1#, what is #pH# for this solution? | null |
1,262 | ab1b1f92-6ddd-11ea-bfbc-ccda262736ce | https://socratic.org/questions/the-same-amount-of-heat-which-will-change-the-temperature-of-50-0-g-of-water-by- | 0.23 J/(g * ℃) | start physical_unit 26 26 specific_heat j/(°c_·_g) qc_end c_other OTHER qc_end physical_unit 14 14 11 12 mass qc_end physical_unit 14 14 16 17 temperature qc_end physical_unit 26 26 23 24 mass qc_end physical_unit 26 26 28 29 temperature qc_end physical_unit 26 26 31 32 temperature qc_end end | [{"type":"physical unit","value":"Specific heat [OF] tin [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"0.23 J/(g * ℃)"}] | [{"type":"other","value":"The same amount of heat."},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{50.0 g}"},{"type":"physical unit","value":"Changed temperature [OF] water [=] \\pu{4.5 ℃}"},{"type":"physical unit","value":"Mass [OF] tin [=] \\pu{110.0 g}"},{"type":"physical unit","value":"Temperature1 [OF] tin [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] tin [=] \\pu{62.7 ℃}"}] | <h1 class="questionTitle" itemprop="name">The same amount of heat which will change the temperature of 50.0 g of water by 4.5° C will raise the temperature of 110.0 g of tin from 25° C to 62.7° C. What is the specific heat of tin?</h1> | null | 0.23 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that if the heat <strong>given off</strong> by the water is equal to the heat <strong>absorbed</strong> by the metal, then you can say that </p>
<blockquote>
<p><mathjax>#color(blue)(q_"tin" = -q_"water")#</mathjax></p>
</blockquote>
<p>The minus sign is used to because <em>heat lost</em> carries a negative sign. </p>
<p>Now, the equation that establishes a relationship between heat lost/gained and change in temperature looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/lost<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>The <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of water is equal to <mathjax>#4.18"J"/("g" ""^@"C")#</mathjax>.</p>
<p>For water, this equation will take the form </p>
<blockquote>
<p><mathjax>#q_"water" = 50.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (-4.5)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"water" = -"940.5 J"#</mathjax></p>
</blockquote>
<p>For the tin sample, the equation will take the form </p>
<blockquote>
<p><mathjax>#q_"tin" = "110.0 g" * c_"tin" * (62.7 - 25)^@"C"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#q_"tin" = -q_"water"#</mathjax></p>
<p><mathjax>#q_"tin" = - (-"940.5 J") = +"940.5 J"#</mathjax></p>
</blockquote>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#"940.5 J" = "110.0 g" * c_"tin" * 37.7^@"C"#</mathjax></p>
<p><mathjax>#c_"tin" = "940.5 J"/("110.0 g" * 37.7^@"C") = 0.2267"J"/("g" ""^@"C")#</mathjax></p>
</blockquote>
<p>You need to round this off to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the change in temperature of the water</p>
<blockquote>
<p><mathjax>#c_"tin" = color(green)(0.23"J"/("g" ""^@"C"))#</mathjax></p>
</blockquote>
<p>The accepted value for tin's specific heat is <mathjax>#0.21"J"/("g" ""^@"C")#</mathjax>, so your result is fairly accurate. </p>
<p><a href="http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html" rel="nofollow" target="_blank">http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html</a></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#0.23"J"/("g" ""^@"C")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that if the heat <strong>given off</strong> by the water is equal to the heat <strong>absorbed</strong> by the metal, then you can say that </p>
<blockquote>
<p><mathjax>#color(blue)(q_"tin" = -q_"water")#</mathjax></p>
</blockquote>
<p>The minus sign is used to because <em>heat lost</em> carries a negative sign. </p>
<p>Now, the equation that establishes a relationship between heat lost/gained and change in temperature looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/lost<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>The <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of water is equal to <mathjax>#4.18"J"/("g" ""^@"C")#</mathjax>.</p>
<p>For water, this equation will take the form </p>
<blockquote>
<p><mathjax>#q_"water" = 50.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (-4.5)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"water" = -"940.5 J"#</mathjax></p>
</blockquote>
<p>For the tin sample, the equation will take the form </p>
<blockquote>
<p><mathjax>#q_"tin" = "110.0 g" * c_"tin" * (62.7 - 25)^@"C"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#q_"tin" = -q_"water"#</mathjax></p>
<p><mathjax>#q_"tin" = - (-"940.5 J") = +"940.5 J"#</mathjax></p>
</blockquote>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#"940.5 J" = "110.0 g" * c_"tin" * 37.7^@"C"#</mathjax></p>
<p><mathjax>#c_"tin" = "940.5 J"/("110.0 g" * 37.7^@"C") = 0.2267"J"/("g" ""^@"C")#</mathjax></p>
</blockquote>
<p>You need to round this off to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the change in temperature of the water</p>
<blockquote>
<p><mathjax>#c_"tin" = color(green)(0.23"J"/("g" ""^@"C"))#</mathjax></p>
</blockquote>
<p>The accepted value for tin's specific heat is <mathjax>#0.21"J"/("g" ""^@"C")#</mathjax>, so your result is fairly accurate. </p>
<p><a href="http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html" rel="nofollow" target="_blank">http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html</a></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">The same amount of heat which will change the temperature of 50.0 g of water by 4.5° C will raise the temperature of 110.0 g of tin from 25° C to 62.7° C. What is the specific heat of tin?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-11-29T22:29:46" itemprop="dateCreated">
Nov 29, 2015
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<div class="markdown"><p><mathjax>#0.23"J"/("g" ""^@"C")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that if the heat <strong>given off</strong> by the water is equal to the heat <strong>absorbed</strong> by the metal, then you can say that </p>
<blockquote>
<p><mathjax>#color(blue)(q_"tin" = -q_"water")#</mathjax></p>
</blockquote>
<p>The minus sign is used to because <em>heat lost</em> carries a negative sign. </p>
<p>Now, the equation that establishes a relationship between heat lost/gained and change in temperature looks like this </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - heat absorbed/lost<br/>
<mathjax>#m#</mathjax> - the mass of the sample<br/>
<mathjax>#c#</mathjax> - the <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of the substance<br/>
<mathjax>#DeltaT#</mathjax> - the change in temperature, defined as <em>final temperature</em> minus <em>initial temperature</em></p>
<p>The <a href="http://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> of water is equal to <mathjax>#4.18"J"/("g" ""^@"C")#</mathjax>.</p>
<p>For water, this equation will take the form </p>
<blockquote>
<p><mathjax>#q_"water" = 50.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (-4.5)color(red)(cancel(color(black)(""^@"C")))#</mathjax></p>
<p><mathjax>#q_"water" = -"940.5 J"#</mathjax></p>
</blockquote>
<p>For the tin sample, the equation will take the form </p>
<blockquote>
<p><mathjax>#q_"tin" = "110.0 g" * c_"tin" * (62.7 - 25)^@"C"#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#q_"tin" = -q_"water"#</mathjax></p>
<p><mathjax>#q_"tin" = - (-"940.5 J") = +"940.5 J"#</mathjax></p>
</blockquote>
<p>Therefore, </p>
<blockquote>
<p><mathjax>#"940.5 J" = "110.0 g" * c_"tin" * 37.7^@"C"#</mathjax></p>
<p><mathjax>#c_"tin" = "940.5 J"/("110.0 g" * 37.7^@"C") = 0.2267"J"/("g" ""^@"C")#</mathjax></p>
</blockquote>
<p>You need to round this off to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the change in temperature of the water</p>
<blockquote>
<p><mathjax>#c_"tin" = color(green)(0.23"J"/("g" ""^@"C"))#</mathjax></p>
</blockquote>
<p>The accepted value for tin's specific heat is <mathjax>#0.21"J"/("g" ""^@"C")#</mathjax>, so your result is fairly accurate. </p>
<p><a href="http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html" rel="nofollow" target="_blank">http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html</a></p></div>
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</article> | The same amount of heat which will change the temperature of 50.0 g of water by 4.5° C will raise the temperature of 110.0 g of tin from 25° C to 62.7° C. What is the specific heat of tin? | null |
1,263 | a8d7f856-6ddd-11ea-8619-ccda262736ce | https://socratic.org/questions/how-do-you-balance-c-3h-8-o-2-co-2-h-2o | C3H8 + 5 O2 -> 3 CO2 + 4 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":"C3H8 + 5 O2 -> 3 CO2 + 4 H2O"}] | [{"type":"chemical equation","value":"C3H8 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #C_3H_8 + O_2 -> CO_2 + H_2O#?</h1> | null | C3H8 + 5 O2 -> 3 CO2 + 4 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Garbage in must equal garbage out. Complete combustion specifies that all of the carbon reactant is oxidized to <mathjax>#CO_2#</mathjax>. </p>
<p>To make sure that it is balanced look at the left hand sides and right sides: LHS: <mathjax>#3xxC+8xxH+10xxO#</mathjax>; RHS: <mathjax>#3xxC+8xxH+10xxO#</mathjax>. Reactants and products are stoichiometrically balanced as they must be.</p>
<p>Can you represent the complete combustions of pentane, <mathjax>#C_5H_12#</mathjax>, and then hexane, <mathjax>#C_6H_14#</mathjax>?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>If you mean complete combustion of propane:</p>
<p><mathjax>#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Garbage in must equal garbage out. Complete combustion specifies that all of the carbon reactant is oxidized to <mathjax>#CO_2#</mathjax>. </p>
<p>To make sure that it is balanced look at the left hand sides and right sides: LHS: <mathjax>#3xxC+8xxH+10xxO#</mathjax>; RHS: <mathjax>#3xxC+8xxH+10xxO#</mathjax>. Reactants and products are stoichiometrically balanced as they must be.</p>
<p>Can you represent the complete combustions of pentane, <mathjax>#C_5H_12#</mathjax>, and then hexane, <mathjax>#C_6H_14#</mathjax>?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #C_3H_8 + O_2 -> CO_2 + H_2O#?</h1>
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anor277
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Mar 1, 2016
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<div class="markdown"><p>If you mean complete combustion of propane:</p>
<p><mathjax>#C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Garbage in must equal garbage out. Complete combustion specifies that all of the carbon reactant is oxidized to <mathjax>#CO_2#</mathjax>. </p>
<p>To make sure that it is balanced look at the left hand sides and right sides: LHS: <mathjax>#3xxC+8xxH+10xxO#</mathjax>; RHS: <mathjax>#3xxC+8xxH+10xxO#</mathjax>. Reactants and products are stoichiometrically balanced as they must be.</p>
<p>Can you represent the complete combustions of pentane, <mathjax>#C_5H_12#</mathjax>, and then hexane, <mathjax>#C_6H_14#</mathjax>?</p></div>
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</article> | How do you balance #C_3H_8 + O_2 -> CO_2 + H_2O#? | null |
1,264 | aa97463e-6ddd-11ea-b02f-ccda262736ce | https://socratic.org/questions/an-aqueous-stock-solution-is-36-0-hcl-by-mass-and-its-density-is-1-18g-ml-what-v | 209 millilitres | start physical_unit 18 19 volume ml qc_end physical_unit 3 3 24 25 volume qc_end physical_unit 29 29 27 28 molarity qc_end end | [{"type":"physical unit","value":"Volume2 [OF] this solution [IN] millilitres"}] | [{"type":"physical unit","value":"209 millilitres"}] | [{"type":"physical unit","value":"Percent by mass [OF] HCl in the aqueous stock solution [=] \\pu{36.0%}"},{"type":"physical unit","value":"Density [OF] HCl aqueous stock solution [=] \\pu{1.18 g/mL}"},{"type":"physical unit","value":"Volume1 [OF] HCl(aq) solution [=] \\pu{1.00 L}"},{"type":"physical unit","value":"Molarity1 [OF] HCl(aq) solution [=] \\pu{1.75 mol/L}"}] | <h1 class="questionTitle" itemprop="name">An aqueous stock solution is #36.0%# HCl by mass and its density is #"1.18 g/mL"#. What volume of this solution is required to make #"1.00 L"# of #"1.75 mol/L"# #"HCl"_((aq))#? Give your answer in millilitres, accurate to three significant figures.</h1> | null | 209 millilitres | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, a <mathjax>#"1.75-mol L"^(-1)#</mathjax> hydrochloric acid solution will contain <mathjax>#1.75#</mathjax> <strong>moles</strong> of hydrochloric acid, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, <strong>for every</strong> <mathjax>#"1 L"#</mathjax> of solution.</p>
<p>So right from the start, you know that your solution must contain <mathjax>#1.75#</mathjax> <strong>moles</strong> of hydrochloric acid. </p>
<p>Use the compound's <strong>molar mass</strong> to convert this to <em>grams</em></p>
<blockquote>
<p><mathjax>#1.75 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = "63.805 g"#</mathjax></p>
</blockquote>
<p>Now, you know that your stock solution is <mathjax>#36.0%#</mathjax> hydrochloric acid <strong>by mass</strong>, which implies that in order to have <mathjax>#"36.0 g"#</mathjax> of hydrochloric acid, you need <mathjax>#"100 g"#</mathjax> of this solution. </p>
<p>You can thus say that in order for your target solution to contain <mathjax>#"63.805 g"#</mathjax> of hydrochloric acid, the sample you take from the stock solution must have a <em>mass</em> of </p>
<blockquote>
<p><mathjax>#63.805 color(red)(cancel(color(black)("g HCl"))) * "100 g solution"/(36.0color(red)(cancel(color(black)("g HCl")))) = "177.24 g"#</mathjax></p>
</blockquote>
<p>Use the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of the stock solution to figure out the <em>volume</em> of the sample</p>
<blockquote>
<p><mathjax>#177.24 color(red)(cancel(color(black)("g solution"))) * "1 mL"/(1.18color(red)(cancel(color(black)("g solution")))) = color(darkgreen)(ul(color(black)("209 mL")))#</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>
<p>So what you would do is take <mathjax>#"209 mL"#</mathjax> of the stock solution and <strong>add it</strong> to enough water to get the total volume of the resulting solution to <mathjax>#"1.00 L"#</mathjax>. </p>
<p>Since you're working with concentrated hydrochloric acid, you should use an ice bath to cool the water before adding the stock solution to it. </p>
<p>You should also add the stock solution in <em>multiple steps</em> and stir the solution properly between each step. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"209 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, a <mathjax>#"1.75-mol L"^(-1)#</mathjax> hydrochloric acid solution will contain <mathjax>#1.75#</mathjax> <strong>moles</strong> of hydrochloric acid, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, <strong>for every</strong> <mathjax>#"1 L"#</mathjax> of solution.</p>
<p>So right from the start, you know that your solution must contain <mathjax>#1.75#</mathjax> <strong>moles</strong> of hydrochloric acid. </p>
<p>Use the compound's <strong>molar mass</strong> to convert this to <em>grams</em></p>
<blockquote>
<p><mathjax>#1.75 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = "63.805 g"#</mathjax></p>
</blockquote>
<p>Now, you know that your stock solution is <mathjax>#36.0%#</mathjax> hydrochloric acid <strong>by mass</strong>, which implies that in order to have <mathjax>#"36.0 g"#</mathjax> of hydrochloric acid, you need <mathjax>#"100 g"#</mathjax> of this solution. </p>
<p>You can thus say that in order for your target solution to contain <mathjax>#"63.805 g"#</mathjax> of hydrochloric acid, the sample you take from the stock solution must have a <em>mass</em> of </p>
<blockquote>
<p><mathjax>#63.805 color(red)(cancel(color(black)("g HCl"))) * "100 g solution"/(36.0color(red)(cancel(color(black)("g HCl")))) = "177.24 g"#</mathjax></p>
</blockquote>
<p>Use the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of the stock solution to figure out the <em>volume</em> of the sample</p>
<blockquote>
<p><mathjax>#177.24 color(red)(cancel(color(black)("g solution"))) * "1 mL"/(1.18color(red)(cancel(color(black)("g solution")))) = color(darkgreen)(ul(color(black)("209 mL")))#</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>
<p>So what you would do is take <mathjax>#"209 mL"#</mathjax> of the stock solution and <strong>add it</strong> to enough water to get the total volume of the resulting solution to <mathjax>#"1.00 L"#</mathjax>. </p>
<p>Since you're working with concentrated hydrochloric acid, you should use an ice bath to cool the water before adding the stock solution to it. </p>
<p>You should also add the stock solution in <em>multiple steps</em> and stir the solution properly between each step. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">An aqueous stock solution is #36.0%# HCl by mass and its density is #"1.18 g/mL"#. What volume of this solution is required to make #"1.00 L"# of #"1.75 mol/L"# #"HCl"_((aq))#? Give your answer in millilitres, accurate to three significant figures.</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"209 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, a <mathjax>#"1.75-mol L"^(-1)#</mathjax> hydrochloric acid solution will contain <mathjax>#1.75#</mathjax> <strong>moles</strong> of hydrochloric acid, the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, <strong>for every</strong> <mathjax>#"1 L"#</mathjax> of solution.</p>
<p>So right from the start, you know that your solution must contain <mathjax>#1.75#</mathjax> <strong>moles</strong> of hydrochloric acid. </p>
<p>Use the compound's <strong>molar mass</strong> to convert this to <em>grams</em></p>
<blockquote>
<p><mathjax>#1.75 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = "63.805 g"#</mathjax></p>
</blockquote>
<p>Now, you know that your stock solution is <mathjax>#36.0%#</mathjax> hydrochloric acid <strong>by mass</strong>, which implies that in order to have <mathjax>#"36.0 g"#</mathjax> of hydrochloric acid, you need <mathjax>#"100 g"#</mathjax> of this solution. </p>
<p>You can thus say that in order for your target solution to contain <mathjax>#"63.805 g"#</mathjax> of hydrochloric acid, the sample you take from the stock solution must have a <em>mass</em> of </p>
<blockquote>
<p><mathjax>#63.805 color(red)(cancel(color(black)("g HCl"))) * "100 g solution"/(36.0color(red)(cancel(color(black)("g HCl")))) = "177.24 g"#</mathjax></p>
</blockquote>
<p>Use the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of the stock solution to figure out the <em>volume</em> of the sample</p>
<blockquote>
<p><mathjax>#177.24 color(red)(cancel(color(black)("g solution"))) * "1 mL"/(1.18color(red)(cancel(color(black)("g solution")))) = color(darkgreen)(ul(color(black)("209 mL")))#</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>
<p>So what you would do is take <mathjax>#"209 mL"#</mathjax> of the stock solution and <strong>add it</strong> to enough water to get the total volume of the resulting solution to <mathjax>#"1.00 L"#</mathjax>. </p>
<p>Since you're working with concentrated hydrochloric acid, you should use an ice bath to cool the water before adding the stock solution to it. </p>
<p>You should also add the stock solution in <em>multiple steps</em> and stir the solution properly between each step. </p></div>
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</article> | An aqueous stock solution is #36.0%# HCl by mass and its density is #"1.18 g/mL"#. What volume of this solution is required to make #"1.00 L"# of #"1.75 mol/L"# #"HCl"_((aq))#? Give your answer in millilitres, accurate to three significant figures. | null |
1,265 | ab67ae9e-6ddd-11ea-998a-ccda262736ce | https://socratic.org/questions/calculate-the-mass-of-ethyl-acetate-formed-at-25-c-from-1-66-moles-of-acetic-aci | 110 g | start physical_unit 4 5 mass g qc_end physical_unit 14 15 11 12 mole qc_end physical_unit 20 21 17 18 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] ethyl acetate [IN] g"}] | [{"type":"physical unit","value":"110 g"}] | [{"type":"physical unit","value":"Temperature [OF] the reaction [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Mole [OF] acetic acid [=] \\pu{1.66 moles}"},{"type":"physical unit","value":"Mole [OF] ethyl alcohol [=] \\pu{2.17 moles}"}] | <h1 class="questionTitle" itemprop="name">Calculate the mass of ethyl acetate formed at 25 C from 1.66 moles of acetic acid and 2.17 moles of ethyl alcohol ? . Kc = 4 </h1> | null | 110 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We need the volumes of the reactants, so we can calculate the molarities and do the equilibrium calculation.</p>
<blockquote></blockquote>
<p><strong>Volumes</strong></p>
<p><strong>(a) Acetic acid</strong></p>
<p><mathjax>#V_a = 1.66 color(red)(cancel(color(black)("mol"))) × (60.05 color(red)(cancel(color(black)("g"))))/(1 color(red)(cancel(color(black)("mol")))) × "1 mL"/(1.0497 color(red)(cancel(color(black)("g")))) = "95.0 mL"#</mathjax></p>
<p><strong>(b) Ethanol</strong></p>
<p><mathjax>#V_e = 2.17 color(red)(cancel(color(black)("mol"))) × (46.07 color(red)(cancel(color(black)("g"))))/(1 color(red)(cancel(color(black)("mol")))) × "1 mL"/(0.789 color(red)(cancel(color(black)("g")))) = "126.7 mL"#</mathjax></p>
<p><strong>(c) Total volume</strong></p>
<p><mathjax>#V_"tot" = V_a + V_e = "95.0 mL + 126.7 mL" = "221.7mL"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Initial concentrations</strong></p>
<p><mathjax>#["Acetic acid"] = "1.66 mol"/"0.2217 L" = "7.49 mol/L"#</mathjax></p>
<p><mathjax>#["Ethanol"] = "2.17 mol"/"0.2217 L" = "9.79 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>The calculations</strong></p>
<p><mathjax>#color(white)(mmmmmm)"acetic acid" + "ethanol" ⇌ "ethyl acetate" + "water"#</mathjax><br/>
<mathjax>#color(white)(mmmmmmmm)"A"color(white)(mml) +color(white)(ml) "B"color(white)(mll) ⇌color(white)(mml) "C"color(white)(mmm) + color(white)(m)"D"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1": color(white)(mm)7.49color(white)(mmmm) 9.79color(white)(mmmmmll) 0color(white)(mmmmmll) 0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mml)"-"x color(white)(mmmmm)"-"x color(white)(mmmmm)+xcolor(white)(mmmm) +x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(mm) 7.49-xcolor(white)(m) 9.79-xcolor(white)(mmmm) x color(white)(mmmmmll)x#</mathjax></p>
<p><mathjax>#K_"c" = (["C"]["D"])/(["A"]["B"]) = (x·x)/((7.49-x)(9.79-x)) = 4#</mathjax></p>
<p><mathjax>#x^2 = 4(7.49-x)(9.79-x) = 4(73.33 - 17.28x + x^2) = 293.3 - 69.12x + 4x^2#</mathjax></p>
<p><mathjax>#3x^2 - 69.12x +293.3 = 0#</mathjax></p>
<p><mathjax>#x = 5.61#</mathjax></p>
<p>The equilibrium concentration of ethyl acetate is 5.61 mol/L.</p>
<p>If the volume change is minimal, the mass of ethyl acetate formed is</p>
<p><mathjax>#0.2217 color(red)(cancel(color(black)("L"))) × (5.61 color(red)(cancel(color(black)("mol"))))/(1 color(red)(cancel(color(black)("L")))) × "88.10 g"/(1 color(red)(cancel(color(black)("mol")))) = "110 g"#</mathjax></p>
<p>The theoretical yield of ethyl acetate is 110 g.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The theoretical yield is 110 g of ethyl acetate.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We need the volumes of the reactants, so we can calculate the molarities and do the equilibrium calculation.</p>
<blockquote></blockquote>
<p><strong>Volumes</strong></p>
<p><strong>(a) Acetic acid</strong></p>
<p><mathjax>#V_a = 1.66 color(red)(cancel(color(black)("mol"))) × (60.05 color(red)(cancel(color(black)("g"))))/(1 color(red)(cancel(color(black)("mol")))) × "1 mL"/(1.0497 color(red)(cancel(color(black)("g")))) = "95.0 mL"#</mathjax></p>
<p><strong>(b) Ethanol</strong></p>
<p><mathjax>#V_e = 2.17 color(red)(cancel(color(black)("mol"))) × (46.07 color(red)(cancel(color(black)("g"))))/(1 color(red)(cancel(color(black)("mol")))) × "1 mL"/(0.789 color(red)(cancel(color(black)("g")))) = "126.7 mL"#</mathjax></p>
<p><strong>(c) Total volume</strong></p>
<p><mathjax>#V_"tot" = V_a + V_e = "95.0 mL + 126.7 mL" = "221.7mL"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Initial concentrations</strong></p>
<p><mathjax>#["Acetic acid"] = "1.66 mol"/"0.2217 L" = "7.49 mol/L"#</mathjax></p>
<p><mathjax>#["Ethanol"] = "2.17 mol"/"0.2217 L" = "9.79 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>The calculations</strong></p>
<p><mathjax>#color(white)(mmmmmm)"acetic acid" + "ethanol" ⇌ "ethyl acetate" + "water"#</mathjax><br/>
<mathjax>#color(white)(mmmmmmmm)"A"color(white)(mml) +color(white)(ml) "B"color(white)(mll) ⇌color(white)(mml) "C"color(white)(mmm) + color(white)(m)"D"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1": color(white)(mm)7.49color(white)(mmmm) 9.79color(white)(mmmmmll) 0color(white)(mmmmmll) 0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mml)"-"x color(white)(mmmmm)"-"x color(white)(mmmmm)+xcolor(white)(mmmm) +x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(mm) 7.49-xcolor(white)(m) 9.79-xcolor(white)(mmmm) x color(white)(mmmmmll)x#</mathjax></p>
<p><mathjax>#K_"c" = (["C"]["D"])/(["A"]["B"]) = (x·x)/((7.49-x)(9.79-x)) = 4#</mathjax></p>
<p><mathjax>#x^2 = 4(7.49-x)(9.79-x) = 4(73.33 - 17.28x + x^2) = 293.3 - 69.12x + 4x^2#</mathjax></p>
<p><mathjax>#3x^2 - 69.12x +293.3 = 0#</mathjax></p>
<p><mathjax>#x = 5.61#</mathjax></p>
<p>The equilibrium concentration of ethyl acetate is 5.61 mol/L.</p>
<p>If the volume change is minimal, the mass of ethyl acetate formed is</p>
<p><mathjax>#0.2217 color(red)(cancel(color(black)("L"))) × (5.61 color(red)(cancel(color(black)("mol"))))/(1 color(red)(cancel(color(black)("L")))) × "88.10 g"/(1 color(red)(cancel(color(black)("mol")))) = "110 g"#</mathjax></p>
<p>The theoretical yield of ethyl acetate is 110 g.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">Calculate the mass of ethyl acetate formed at 25 C from 1.66 moles of acetic acid and 2.17 moles of ethyl alcohol ? . Kc = 4 </h1>
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<div class="markdown"><p>The theoretical yield is 110 g of ethyl acetate.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>We need the volumes of the reactants, so we can calculate the molarities and do the equilibrium calculation.</p>
<blockquote></blockquote>
<p><strong>Volumes</strong></p>
<p><strong>(a) Acetic acid</strong></p>
<p><mathjax>#V_a = 1.66 color(red)(cancel(color(black)("mol"))) × (60.05 color(red)(cancel(color(black)("g"))))/(1 color(red)(cancel(color(black)("mol")))) × "1 mL"/(1.0497 color(red)(cancel(color(black)("g")))) = "95.0 mL"#</mathjax></p>
<p><strong>(b) Ethanol</strong></p>
<p><mathjax>#V_e = 2.17 color(red)(cancel(color(black)("mol"))) × (46.07 color(red)(cancel(color(black)("g"))))/(1 color(red)(cancel(color(black)("mol")))) × "1 mL"/(0.789 color(red)(cancel(color(black)("g")))) = "126.7 mL"#</mathjax></p>
<p><strong>(c) Total volume</strong></p>
<p><mathjax>#V_"tot" = V_a + V_e = "95.0 mL + 126.7 mL" = "221.7mL"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Initial concentrations</strong></p>
<p><mathjax>#["Acetic acid"] = "1.66 mol"/"0.2217 L" = "7.49 mol/L"#</mathjax></p>
<p><mathjax>#["Ethanol"] = "2.17 mol"/"0.2217 L" = "9.79 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>The calculations</strong></p>
<p><mathjax>#color(white)(mmmmmm)"acetic acid" + "ethanol" ⇌ "ethyl acetate" + "water"#</mathjax><br/>
<mathjax>#color(white)(mmmmmmmm)"A"color(white)(mml) +color(white)(ml) "B"color(white)(mll) ⇌color(white)(mml) "C"color(white)(mmm) + color(white)(m)"D"#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1": color(white)(mm)7.49color(white)(mmmm) 9.79color(white)(mmmmmll) 0color(white)(mmmmmll) 0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mml)"-"x color(white)(mmmmm)"-"x color(white)(mmmmm)+xcolor(white)(mmmm) +x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1":color(white)(mm) 7.49-xcolor(white)(m) 9.79-xcolor(white)(mmmm) x color(white)(mmmmmll)x#</mathjax></p>
<p><mathjax>#K_"c" = (["C"]["D"])/(["A"]["B"]) = (x·x)/((7.49-x)(9.79-x)) = 4#</mathjax></p>
<p><mathjax>#x^2 = 4(7.49-x)(9.79-x) = 4(73.33 - 17.28x + x^2) = 293.3 - 69.12x + 4x^2#</mathjax></p>
<p><mathjax>#3x^2 - 69.12x +293.3 = 0#</mathjax></p>
<p><mathjax>#x = 5.61#</mathjax></p>
<p>The equilibrium concentration of ethyl acetate is 5.61 mol/L.</p>
<p>If the volume change is minimal, the mass of ethyl acetate formed is</p>
<p><mathjax>#0.2217 color(red)(cancel(color(black)("L"))) × (5.61 color(red)(cancel(color(black)("mol"))))/(1 color(red)(cancel(color(black)("L")))) × "88.10 g"/(1 color(red)(cancel(color(black)("mol")))) = "110 g"#</mathjax></p>
<p>The theoretical yield of ethyl acetate is 110 g.</p></div>
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</article> | Calculate the mass of ethyl acetate formed at 25 C from 1.66 moles of acetic acid and 2.17 moles of ethyl alcohol ? . Kc = 4 | null |
1,266 | a99ecea8-6ddd-11ea-acb0-ccda262736ce | https://socratic.org/questions/how-many-electrons-must-be-shown-in-the-lewis-structure-of-the-hydroxide-ion-oh | 8 | start physical_unit 2 2 number none qc_end chemical_equation 14 14 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Number [OF] electrons"}] | [{"type":"physical unit","value":"8"}] | [{"type":"chemical equation","value":"OH−"},{"type":"other","value":"in the Lewis structure"}] | <h1 class="questionTitle" itemprop="name">How many electrons must be shown in the Lewis structure of the hydroxide ion, #OH^-#?</h1> | null | 8 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Only <a href="http://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> are shown on a Lewis structure - these are the electrons in the outermost shell of an atom. The oxygen atom has 6 <a href="http://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>, while the hydrogen atom has just 1.</p>
<p>However, the hydroxide ion carries a negative charge, which means 1 extra electron has been introduced. We must add that to the valence electrons already present.</p>
<p>This gives a total of <mathjax>#6+1+1 = 8#</mathjax> electrons.</p></div>
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<div class="markdown"><p>8 electrons.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Only <a href="http://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> are shown on a Lewis structure - these are the electrons in the outermost shell of an atom. The oxygen atom has 6 <a href="http://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>, while the hydrogen atom has just 1.</p>
<p>However, the hydroxide ion carries a negative charge, which means 1 extra electron has been introduced. We must add that to the valence electrons already present.</p>
<p>This gives a total of <mathjax>#6+1+1 = 8#</mathjax> electrons.</p></div>
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<h1 class="questionTitle" itemprop="name">How many electrons must be shown in the Lewis structure of the hydroxide ion, #OH^-#?</h1>
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<div class="markdown"><p>8 electrons.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Only <a href="http://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table">valence electrons</a> are shown on a Lewis structure - these are the electrons in the outermost shell of an atom. The oxygen atom has 6 <a href="http://socratic.org/chemistry/the-electron-configuration-of-atoms/valence-electrons">valence electrons</a>, while the hydrogen atom has just 1.</p>
<p>However, the hydroxide ion carries a negative charge, which means 1 extra electron has been introduced. We must add that to the valence electrons already present.</p>
<p>This gives a total of <mathjax>#6+1+1 = 8#</mathjax> electrons.</p></div>
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</article> | How many electrons must be shown in the Lewis structure of the hydroxide ion, #OH^-#? | null |
1,267 | a8e1895c-6ddd-11ea-8697-ccda262736ce | https://socratic.org/questions/a-container-of-gas-with-a-pressure-of-450-kpa-contains-three-different-gasses-hy | 115 kPa | start physical_unit 17 17 partial_pressure kpa qc_end physical_unit 0 1 8 9 pressure qc_end physical_unit 14 14 26 27 partial_pressure qc_end physical_unit 15 15 36 37 partial_pressure qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] nitrogen [IN] kPa"}] | [{"type":"physical unit","value":"115 kPa"}] | [{"type":"physical unit","value":"Pressure [OF] a container [=] \\pu{450 kPa}"},{"type":"physical unit","value":"Partial pressure [OF] hydrogen [=] \\pu{210 kPa}"},{"type":"physical unit","value":"Partial pressure [OF] oxygen [=] \\pu{125 kPa}"}] | <h1 class="questionTitle" itemprop="name">A container of gas with a pressure of 450 kPa contains three different gasses -hydrogen, oxygen, and nitrogen. If the partial pressure of the hydrogen is 210 kPa and the partial pressure of the oxygen is 125 kPa, what is the partial pressure of nitrogen?</h1> | null | 115 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>According to <strong>Dalton's law of partial pressures</strong>, the total pressure is equivalent to the sum of the pressures of the individual gases (if they were present alone). </p>
<p>Therefore, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of <mathjax>#"N"_2#</mathjax> is simply</p>
<p><mathjax>#overbrace(450color(white)(l)"kPa")^(P_"total")#</mathjax> <mathjax>#- overbrace(210color(white)(l)"kPa")^(P_("H"_2))#</mathjax> <mathjax>#- overbrace(125color(white)(l)"kPa")^(P_(""O"_2))#</mathjax> <mathjax>#= color(red)(115#</mathjax> <mathjax>#color(red)("kPa"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#P_ ("N"_2) = 115#</mathjax> <mathjax>#"kPa"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>According to <strong>Dalton's law of partial pressures</strong>, the total pressure is equivalent to the sum of the pressures of the individual gases (if they were present alone). </p>
<p>Therefore, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of <mathjax>#"N"_2#</mathjax> is simply</p>
<p><mathjax>#overbrace(450color(white)(l)"kPa")^(P_"total")#</mathjax> <mathjax>#- overbrace(210color(white)(l)"kPa")^(P_("H"_2))#</mathjax> <mathjax>#- overbrace(125color(white)(l)"kPa")^(P_(""O"_2))#</mathjax> <mathjax>#= color(red)(115#</mathjax> <mathjax>#color(red)("kPa"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A container of gas with a pressure of 450 kPa contains three different gasses -hydrogen, oxygen, and nitrogen. If the partial pressure of the hydrogen is 210 kPa and the partial pressure of the oxygen is 125 kPa, what is the partial pressure of nitrogen?</h1>
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Nathan L.
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<div class="markdown"><p><mathjax>#P_ ("N"_2) = 115#</mathjax> <mathjax>#"kPa"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>According to <strong>Dalton's law of partial pressures</strong>, the total pressure is equivalent to the sum of the pressures of the individual gases (if they were present alone). </p>
<p>Therefore, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> of <mathjax>#"N"_2#</mathjax> is simply</p>
<p><mathjax>#overbrace(450color(white)(l)"kPa")^(P_"total")#</mathjax> <mathjax>#- overbrace(210color(white)(l)"kPa")^(P_("H"_2))#</mathjax> <mathjax>#- overbrace(125color(white)(l)"kPa")^(P_(""O"_2))#</mathjax> <mathjax>#= color(red)(115#</mathjax> <mathjax>#color(red)("kPa"#</mathjax></p></div>
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</article> | A container of gas with a pressure of 450 kPa contains three different gasses -hydrogen, oxygen, and nitrogen. If the partial pressure of the hydrogen is 210 kPa and the partial pressure of the oxygen is 125 kPa, what is the partial pressure of nitrogen? | null |
1,268 | ac2bbcd9-6ddd-11ea-9bbd-ccda262736ce | https://socratic.org/questions/what-is-the-chemical-equation-for-iron-iii-chloride-sodium-hydroxide | FeCl3(aq) + 3 NaOH(aq) -> Fe(OH)3(s) + 3 NaCl(aq) | start chemical_equation qc_end substance 6 7 qc_end substance 9 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the chemical equation"}] | [{"type":"chemical equation","value":"FeCl3(aq) + 3 NaOH(aq) -> Fe(OH)3(s) + 3 NaCl(aq)"}] | [{"type":"substance name","value":"Iron(III) Chloride"},{"type":"substance name","value":"Sodium Hydroxide"}] | <h1 class="questionTitle" itemprop="name">What is the chemical equation for Iron(III)Chloride + Sodium Hydroxide?</h1> | null | FeCl3(aq) + 3 NaOH(aq) -> Fe(OH)3(s) + 3 NaCl(aq) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Most hydroxides are insoluble. The iron(III)hydroxide is probably an hydrous oxide, that is very poorly characterized. </p></div>
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<div class="markdown"><p><mathjax>#FeCl_3(aq) + 3NaOH(aq) rarr Fe(OH)_3(s)darr + 3NaCl(aq)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Most hydroxides are insoluble. The iron(III)hydroxide is probably an hydrous oxide, that is very poorly characterized. </p></div>
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<div class="markdown"><p><mathjax>#FeCl_3(aq) + 3NaOH(aq) rarr Fe(OH)_3(s)darr + 3NaCl(aq)#</mathjax></p></div>
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<div class="markdown"><p>Most hydroxides are insoluble. The iron(III)hydroxide is probably an hydrous oxide, that is very poorly characterized. </p></div>
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</article> | What is the chemical equation for Iron(III)Chloride + Sodium Hydroxide? | null |
1,269 | a92e5f94-6ddd-11ea-87a9-ccda262736ce | https://socratic.org/questions/a-balloon-has-a-volume-of-20-0-l-at-atmospheric-pressure-and-20-0-c-one-wants-to | 79 degrees Celsius | start physical_unit 1 1 temperature °c qc_end physical_unit 1 1 6 7 volume qc_end physical_unit 1 1 12 13 temperature qc_end c_other OTHER qc_end physical_unit 1 1 21 22 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the balloon [IN] degrees Celsius"}] | [{"type":"physical unit","value":"79 degrees Celsius"}] | [{"type":"physical unit","value":"Volume1 [OF] the balloon [=] \\pu{20.0 L}"},{"type":"physical unit","value":"Temperature1 [OF] the balloon [=] \\pu{20.0 ℃}"},{"type":"other","value":"Atmospheric pressure."},{"type":"physical unit","value":"Volume2 [OF] the balloon [=] \\pu{24.0 L}"},{"type":"other","value":"Constant pressure."}] | <h1 class="questionTitle" itemprop="name">A balloon has a volume of 20.0 L at atmospheric pressure and 20.0 C. One wants to increase the volume to 24.0 L at constant pressure. To what temperature in degrees Celsius does one need to increase the temperature to?</h1> | null | 79 degrees Celsius | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As the initial volume of 20 L is less than the final volume of 24 L, the temperature must increase.</p>
<p>This is based on Charles' Law, which states that there is a direct relationship between temperature and volume.</p>
<p>If your answer is less than 20 L, the answer is wrong.</p>
<p>It is important to remember that Charles' Law is based on <a href="https://socratic.org/chemistry/the-behavior-of-gases/kelvin-temperatures-and-absolute-zero">Kelvin temperatures</a>. </p>
<p><strong>Step one</strong> is to change Celsius degrees into kelvins.</p>
<p><mathjax>#"0 °C" = "273 K"#</mathjax></p>
<p>∴ <mathjax>#T_1 = "(20 + 273) K" = "293 K"#</mathjax></p>
<p><strong>Step two</strong> is to insert the values into the Charles' Law expression.</p>
<p><mathjax># T_1/V_1 = T_2/V_2#</mathjax></p>
<p><mathjax>#"293 K"/"20 L" = T_2/"24 L"#</mathjax> </p>
<p><mathjax># 24 × "293 K"/20 = T_2#</mathjax></p>
<p><mathjax>#"352 K" = T_2#</mathjax></p>
<p><strong>Step three</strong> is to convert the Kelvin temperature to Celsius.</p>
<p>To change this answer back to Celsius degrees, subtract 273.</p>
<p><mathjax>#"(352-273) °C"color(white)(l) =color(white)(l) "79 °C"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>The answer is 79 °C.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As the initial volume of 20 L is less than the final volume of 24 L, the temperature must increase.</p>
<p>This is based on Charles' Law, which states that there is a direct relationship between temperature and volume.</p>
<p>If your answer is less than 20 L, the answer is wrong.</p>
<p>It is important to remember that Charles' Law is based on <a href="https://socratic.org/chemistry/the-behavior-of-gases/kelvin-temperatures-and-absolute-zero">Kelvin temperatures</a>. </p>
<p><strong>Step one</strong> is to change Celsius degrees into kelvins.</p>
<p><mathjax>#"0 °C" = "273 K"#</mathjax></p>
<p>∴ <mathjax>#T_1 = "(20 + 273) K" = "293 K"#</mathjax></p>
<p><strong>Step two</strong> is to insert the values into the Charles' Law expression.</p>
<p><mathjax># T_1/V_1 = T_2/V_2#</mathjax></p>
<p><mathjax>#"293 K"/"20 L" = T_2/"24 L"#</mathjax> </p>
<p><mathjax># 24 × "293 K"/20 = T_2#</mathjax></p>
<p><mathjax>#"352 K" = T_2#</mathjax></p>
<p><strong>Step three</strong> is to convert the Kelvin temperature to Celsius.</p>
<p>To change this answer back to Celsius degrees, subtract 273.</p>
<p><mathjax>#"(352-273) °C"color(white)(l) =color(white)(l) "79 °C"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A balloon has a volume of 20.0 L at atmospheric pressure and 20.0 C. One wants to increase the volume to 24.0 L at constant pressure. To what temperature in degrees Celsius does one need to increase the temperature to?</h1>
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<div class="markdown"><p>The answer is 79 °C.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As the initial volume of 20 L is less than the final volume of 24 L, the temperature must increase.</p>
<p>This is based on Charles' Law, which states that there is a direct relationship between temperature and volume.</p>
<p>If your answer is less than 20 L, the answer is wrong.</p>
<p>It is important to remember that Charles' Law is based on <a href="https://socratic.org/chemistry/the-behavior-of-gases/kelvin-temperatures-and-absolute-zero">Kelvin temperatures</a>. </p>
<p><strong>Step one</strong> is to change Celsius degrees into kelvins.</p>
<p><mathjax>#"0 °C" = "273 K"#</mathjax></p>
<p>∴ <mathjax>#T_1 = "(20 + 273) K" = "293 K"#</mathjax></p>
<p><strong>Step two</strong> is to insert the values into the Charles' Law expression.</p>
<p><mathjax># T_1/V_1 = T_2/V_2#</mathjax></p>
<p><mathjax>#"293 K"/"20 L" = T_2/"24 L"#</mathjax> </p>
<p><mathjax># 24 × "293 K"/20 = T_2#</mathjax></p>
<p><mathjax>#"352 K" = T_2#</mathjax></p>
<p><strong>Step three</strong> is to convert the Kelvin temperature to Celsius.</p>
<p>To change this answer back to Celsius degrees, subtract 273.</p>
<p><mathjax>#"(352-273) °C"color(white)(l) =color(white)(l) "79 °C"#</mathjax></p></div>
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</article> | A balloon has a volume of 20.0 L at atmospheric pressure and 20.0 C. One wants to increase the volume to 24.0 L at constant pressure. To what temperature in degrees Celsius does one need to increase the temperature to? | null |
1,270 | a98308ae-6ddd-11ea-a425-ccda262736ce | https://socratic.org/questions/what-is-the-partial-pressure-of-co-2 | 40 mmHg | start physical_unit 6 6 partial_pressure mmhg qc_end chemical_equation 6 6 qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] CO2 [IN] mmHg"}] | [{"type":"physical unit","value":"40 mmHg"}] | [{"type":"chemical equation","value":"CO2"}] | <h1 class="questionTitle" itemprop="name">What is the partial pressure of #CO_2#?</h1> | null | 40 mmHg | <div class="answerDescription">
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<div class="markdown"><p>Is <mathjax>#P_"carbon dioxide"#</mathjax> historically increasing or decreasing? Why?</p></div>
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<div class="markdown"><p>In air? At sea level, <mathjax>#P_"carbon dioxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#40#</mathjax> <mathjax>#mm*Hg#</mathjax></p></div>
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<div class="markdown"><p>Is <mathjax>#P_"carbon dioxide"#</mathjax> historically increasing or decreasing? Why?</p></div>
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<div class="markdown"><p>In air? At sea level, <mathjax>#P_"carbon dioxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#40#</mathjax> <mathjax>#mm*Hg#</mathjax></p></div>
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<div class="markdown"><p>Is <mathjax>#P_"carbon dioxide"#</mathjax> historically increasing or decreasing? Why?</p></div>
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</article> | What is the partial pressure of #CO_2#? | null |
1,271 | aac9b3ae-6ddd-11ea-a75f-ccda262736ce | https://socratic.org/questions/the-label-on-an-ocean-spray-cran-raspberry-drink-lists-30-g-of-sugar-in-240-ml-o | 11.95% | start physical_unit 38 41 percent_composition none qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 40 41 14 15 volume qc_end physical_unit 40 41 30 31 mass qc_end end | [{"type":"physical unit","value":"Percent composition [OF] sugar in the drink"}] | [{"type":"physical unit","value":"11.95%"}] | [{"type":"physical unit","value":"Mass [OF] sugar [=] \\pu{30 g}"},{"type":"physical unit","value":"Volume [OF] the drink [=] \\pu{240 mL}"},{"type":"physical unit","value":"Mass [OF] the drink [=] \\pu{251 g}"}] | <h1 class="questionTitle" itemprop="name"> The label on an Ocean Spray Cran-Raspberry drink lists 30 g of sugar in 240 mL of drink. I weighed 240 mL of drink and found its mass to be 251 g. What is the percent composition of sugar in the drink?</h1> | null | 11.95% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> as a measure of how many grams of a given substance are present in exactly <mathjax>#"100 g"#</mathjax> of a compound or a mixture. </p>
<p>In this case, the percent composition of sugar in the beverage will tell you how many grams of sugar you get <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of beverage</strong>. </p>
<p>Now, you know that you have <mathjax>#"30 g"#</mathjax> of sugar in <mathjax>#"251 g"#</mathjax> of beverage. This is the known composition of the drink. You can use this composition to determine the mass of sugar present in <mathjax>#"100 g"#</mathjax> of beverage. </p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g drink"))) * "30 g sugar"/(251color(red)(cancel(color(black)("g drink")))) = "11.95 g sugar"#</mathjax></p>
</blockquote>
<p>This means that the percent composition of sugar in the drink will be </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% sugar = 12%")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one significant figure for the mass of sugar. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"12% sugar"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> as a measure of how many grams of a given substance are present in exactly <mathjax>#"100 g"#</mathjax> of a compound or a mixture. </p>
<p>In this case, the percent composition of sugar in the beverage will tell you how many grams of sugar you get <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of beverage</strong>. </p>
<p>Now, you know that you have <mathjax>#"30 g"#</mathjax> of sugar in <mathjax>#"251 g"#</mathjax> of beverage. This is the known composition of the drink. You can use this composition to determine the mass of sugar present in <mathjax>#"100 g"#</mathjax> of beverage. </p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g drink"))) * "30 g sugar"/(251color(red)(cancel(color(black)("g drink")))) = "11.95 g sugar"#</mathjax></p>
</blockquote>
<p>This means that the percent composition of sugar in the drink will be </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% sugar = 12%")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one significant figure for the mass of sugar. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> The label on an Ocean Spray Cran-Raspberry drink lists 30 g of sugar in 240 mL of drink. I weighed 240 mL of drink and found its mass to be 251 g. What is the percent composition of sugar in the drink?</h1>
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Stefan V.
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Mar 3, 2017
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<div class="markdown"><p><mathjax>#"12% sugar"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> as a measure of how many grams of a given substance are present in exactly <mathjax>#"100 g"#</mathjax> of a compound or a mixture. </p>
<p>In this case, the percent composition of sugar in the beverage will tell you how many grams of sugar you get <strong>for every</strong> <mathjax>#"100 g"#</mathjax> <strong>of beverage</strong>. </p>
<p>Now, you know that you have <mathjax>#"30 g"#</mathjax> of sugar in <mathjax>#"251 g"#</mathjax> of beverage. This is the known composition of the drink. You can use this composition to determine the mass of sugar present in <mathjax>#"100 g"#</mathjax> of beverage. </p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g drink"))) * "30 g sugar"/(251color(red)(cancel(color(black)("g drink")))) = "11.95 g sugar"#</mathjax></p>
</blockquote>
<p>This means that the percent composition of sugar in the drink will be </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% sugar = 12%")))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, but keep in mind that you only have one significant figure for the mass of sugar. </p></div>
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</article> | The label on an Ocean Spray Cran-Raspberry drink lists 30 g of sugar in 240 mL of drink. I weighed 240 mL of drink and found its mass to be 251 g. What is the percent composition of sugar in the drink? | null |
1,272 | ab00a263-6ddd-11ea-a2ba-ccda262736ce | https://socratic.org/questions/what-is-the-molar-mass-of-a-gas-that-has-a-density-of-1-02-g-l-at-0-990-atm-pres | 26.3 g/mol | start physical_unit 7 7 molar_mass g/mol qc_end physical_unit 7 7 20 21 temperature qc_end physical_unit 7 7 13 14 density qc_end physical_unit 7 7 16 17 pressure qc_end end | [{"type":"physical unit","value":"Molar mass [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"26.3 g/mol"}] | [{"type":"physical unit","value":"Temperature [OF] the gas [=] \\pu{37 ℃}"},{"type":"physical unit","value":"Density [OF] the gas [=] \\pu{1.02 g/L}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{0.990 atm}"}] | <h1 class="questionTitle" itemprop="name">What is the molar mass of a gas that has a density of 1.02 g/L at 0.990 atm pressure and 37°C?</h1> | null | 26.3 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your starting point here will be the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
<p>Now, you will have to manipulate this equation in order to find a relationship between the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of the gas, <mathjax>#rho#</mathjax>, under those conditions for pressure and temperature, and its <strong>molar mass</strong>, <mathjax>#M_M#</mathjax>. </p>
<p>You know that the molar mass of a substance tells you the mass of exactly <strong>one mole</strong> of that substance. This means that for a given mass <mathjax>#m#</mathjax> of this gas, you can express its molar mass as the ratio between <mathjax>#m#</mathjax> and <mathjax>#n#</mathjax>, the <em>number of moles</em> it contains</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)color(black)(M_M = m/n)color(white)(a/a)|)))" " " "color(orange)((1))#</mathjax></p>
</blockquote>
<p>Similarly, the density of the substance tells you the mass of exactly <strong>one unit of volume</strong> of that substance. </p>
<p>This means that for the mass <mathjax>#m#</mathjax> of this gas, you can express its density as the ratio between <mathjax>#m#</mathjax> and the <em>volume</em> it occupies</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)color(black)(rho = m/V)color(white)(a/a)|)))" " " "color(orange)((2))#</mathjax></p>
</blockquote>
<p>Plug equation <mathjax>#color(orange)((1))#</mathjax> into the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to get </p>
<blockquote>
<p><mathjax>#PV = m/M_M * RT#</mathjax></p>
</blockquote>
<p>Rearrange to get </p>
<blockquote>
<p><mathjax>#PV * M_M = m * RT#</mathjax></p>
<p><mathjax>#P * M_M = m/V * RT#</mathjax></p>
<p><mathjax>#M_M = m/V * (RT)/P#</mathjax></p>
</blockquote>
<p>Finally, use equation <mathjax>#color(orange)((2))#</mathjax> to write</p>
<blockquote>
<p><mathjax>#M_M = rho * (RT)/P#</mathjax></p>
</blockquote>
<p>Convert the temperature of the gas from <em>degrees Celsius</em> to <em>Kelvin</em> then plug in your values to find</p>
<blockquote>
<p><mathjax>#M_M = 1.02 "g"/color(red)(cancel(color(black)("L"))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 37)color(red)(cancel(color(black)("K"))))/(0.990color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_M = color(green)(|bar(ul(color(white)(a/a)color(black)("26.3 g mol"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"26.3 g mol"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your starting point here will be the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
<p>Now, you will have to manipulate this equation in order to find a relationship between the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of the gas, <mathjax>#rho#</mathjax>, under those conditions for pressure and temperature, and its <strong>molar mass</strong>, <mathjax>#M_M#</mathjax>. </p>
<p>You know that the molar mass of a substance tells you the mass of exactly <strong>one mole</strong> of that substance. This means that for a given mass <mathjax>#m#</mathjax> of this gas, you can express its molar mass as the ratio between <mathjax>#m#</mathjax> and <mathjax>#n#</mathjax>, the <em>number of moles</em> it contains</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)color(black)(M_M = m/n)color(white)(a/a)|)))" " " "color(orange)((1))#</mathjax></p>
</blockquote>
<p>Similarly, the density of the substance tells you the mass of exactly <strong>one unit of volume</strong> of that substance. </p>
<p>This means that for the mass <mathjax>#m#</mathjax> of this gas, you can express its density as the ratio between <mathjax>#m#</mathjax> and the <em>volume</em> it occupies</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)color(black)(rho = m/V)color(white)(a/a)|)))" " " "color(orange)((2))#</mathjax></p>
</blockquote>
<p>Plug equation <mathjax>#color(orange)((1))#</mathjax> into the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to get </p>
<blockquote>
<p><mathjax>#PV = m/M_M * RT#</mathjax></p>
</blockquote>
<p>Rearrange to get </p>
<blockquote>
<p><mathjax>#PV * M_M = m * RT#</mathjax></p>
<p><mathjax>#P * M_M = m/V * RT#</mathjax></p>
<p><mathjax>#M_M = m/V * (RT)/P#</mathjax></p>
</blockquote>
<p>Finally, use equation <mathjax>#color(orange)((2))#</mathjax> to write</p>
<blockquote>
<p><mathjax>#M_M = rho * (RT)/P#</mathjax></p>
</blockquote>
<p>Convert the temperature of the gas from <em>degrees Celsius</em> to <em>Kelvin</em> then plug in your values to find</p>
<blockquote>
<p><mathjax>#M_M = 1.02 "g"/color(red)(cancel(color(black)("L"))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 37)color(red)(cancel(color(black)("K"))))/(0.990color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_M = color(green)(|bar(ul(color(white)(a/a)color(black)("26.3 g mol"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the molar mass of a gas that has a density of 1.02 g/L at 0.990 atm pressure and 37°C?</h1>
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Stefan V.
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May 30, 2016
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<div class="markdown"><p><mathjax>#"26.3 g mol"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your starting point here will be the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a></strong> equation</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P#</mathjax> - the pressure of the gas<br/>
<mathjax>#V#</mathjax> - the volume it occupies<br/>
<mathjax>#n#</mathjax> - the number of moles of gas<br/>
<mathjax>#R#</mathjax> - the <em>universal gas constant</em>, usually given as <mathjax>#0.0821("atm" * "L")/("mol" * "K")#</mathjax><br/>
<mathjax>#T#</mathjax> - the <strong>absolute temperature</strong> of the gas</p>
<p>Now, you will have to manipulate this equation in order to find a relationship between the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of the gas, <mathjax>#rho#</mathjax>, under those conditions for pressure and temperature, and its <strong>molar mass</strong>, <mathjax>#M_M#</mathjax>. </p>
<p>You know that the molar mass of a substance tells you the mass of exactly <strong>one mole</strong> of that substance. This means that for a given mass <mathjax>#m#</mathjax> of this gas, you can express its molar mass as the ratio between <mathjax>#m#</mathjax> and <mathjax>#n#</mathjax>, the <em>number of moles</em> it contains</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)color(black)(M_M = m/n)color(white)(a/a)|)))" " " "color(orange)((1))#</mathjax></p>
</blockquote>
<p>Similarly, the density of the substance tells you the mass of exactly <strong>one unit of volume</strong> of that substance. </p>
<p>This means that for the mass <mathjax>#m#</mathjax> of this gas, you can express its density as the ratio between <mathjax>#m#</mathjax> and the <em>volume</em> it occupies</p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)color(black)(rho = m/V)color(white)(a/a)|)))" " " "color(orange)((2))#</mathjax></p>
</blockquote>
<p>Plug equation <mathjax>#color(orange)((1))#</mathjax> into the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation to get </p>
<blockquote>
<p><mathjax>#PV = m/M_M * RT#</mathjax></p>
</blockquote>
<p>Rearrange to get </p>
<blockquote>
<p><mathjax>#PV * M_M = m * RT#</mathjax></p>
<p><mathjax>#P * M_M = m/V * RT#</mathjax></p>
<p><mathjax>#M_M = m/V * (RT)/P#</mathjax></p>
</blockquote>
<p>Finally, use equation <mathjax>#color(orange)((2))#</mathjax> to write</p>
<blockquote>
<p><mathjax>#M_M = rho * (RT)/P#</mathjax></p>
</blockquote>
<p>Convert the temperature of the gas from <em>degrees Celsius</em> to <em>Kelvin</em> then plug in your values to find</p>
<blockquote>
<p><mathjax>#M_M = 1.02 "g"/color(red)(cancel(color(black)("L"))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 37)color(red)(cancel(color(black)("K"))))/(0.990color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#M_M = color(green)(|bar(ul(color(white)(a/a)color(black)("26.3 g mol"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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</article> | What is the molar mass of a gas that has a density of 1.02 g/L at 0.990 atm pressure and 37°C? | null |
1,273 | ab26a25c-6ddd-11ea-9d46-ccda262736ce | https://socratic.org/questions/what-is-the-mole-ratio-of-co-g-to-co-2-g-in-the-reaction-2co-g-o-2-g-2co-2-g | 1:1 | start physical_unit 6 8 mole_fraction none qc_end chemical_equation 12 18 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">What is the mole ratio of #CO(g)# to #CO_2(g)# in the reaction #2CO(g) + O_2(g) -> 2CO_2(g)#?</h1> | null | 1:1 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>mole ratio</strong> that exists between two chemical species that take part in a chemical reaction is simply the ratio that exists between the <strong>stoichiometric coefficients</strong> added in front of said species. </p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#color(blue)(2)"CO"_ ((g)) + "O"_ (2(g)) -> color(red)(2)"CO"_ (2(g))#</mathjax></p>
</blockquote>
<p>According to the balanced chemical equation that describes this reaction, the reaction consumes <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon monoxide and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas and produces <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of carbon dioxide. </p>
<p>So, carbon monoxide has a coefficient of <mathjax>#color(blue)(2)#</mathjax>. The same can be said about carbon dioxide, which has a coefficient of <mathjax>#color(red)(2)#</mathjax> as well. </p>
<p>This means that the <strong>mole ratio</strong> that exists between carbon monoxide and carbon dioxide will be</p>
<blockquote>
<p><mathjax>#(color(blue)(2)color(white)(.)"moles CO")/(color(red)(2)color(white)(.)"moles CO"_2) = color(blue)(2)/color(red)(2) = 1/1#</mathjax></p>
</blockquote>
<p>Because the two coefficients are equal, we say that the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> have a <mathjax>#1:1#</mathjax> mole ratio. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1:1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>mole ratio</strong> that exists between two chemical species that take part in a chemical reaction is simply the ratio that exists between the <strong>stoichiometric coefficients</strong> added in front of said species. </p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#color(blue)(2)"CO"_ ((g)) + "O"_ (2(g)) -> color(red)(2)"CO"_ (2(g))#</mathjax></p>
</blockquote>
<p>According to the balanced chemical equation that describes this reaction, the reaction consumes <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon monoxide and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas and produces <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of carbon dioxide. </p>
<p>So, carbon monoxide has a coefficient of <mathjax>#color(blue)(2)#</mathjax>. The same can be said about carbon dioxide, which has a coefficient of <mathjax>#color(red)(2)#</mathjax> as well. </p>
<p>This means that the <strong>mole ratio</strong> that exists between carbon monoxide and carbon dioxide will be</p>
<blockquote>
<p><mathjax>#(color(blue)(2)color(white)(.)"moles CO")/(color(red)(2)color(white)(.)"moles CO"_2) = color(blue)(2)/color(red)(2) = 1/1#</mathjax></p>
</blockquote>
<p>Because the two coefficients are equal, we say that the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> have a <mathjax>#1:1#</mathjax> mole ratio. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the mole ratio of #CO(g)# to #CO_2(g)# in the reaction #2CO(g) + O_2(g) -> 2CO_2(g)#?</h1>
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Stefan V.
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Mar 12, 2017
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<div class="markdown"><p><mathjax>#1:1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <strong>mole ratio</strong> that exists between two chemical species that take part in a chemical reaction is simply the ratio that exists between the <strong>stoichiometric coefficients</strong> added in front of said species. </p>
<p>In your case, you have</p>
<blockquote>
<p><mathjax>#color(blue)(2)"CO"_ ((g)) + "O"_ (2(g)) -> color(red)(2)"CO"_ (2(g))#</mathjax></p>
</blockquote>
<p>According to the balanced chemical equation that describes this reaction, the reaction consumes <mathjax>#color(blue)(2)#</mathjax> <strong>moles</strong> of carbon monoxide and <mathjax>#1#</mathjax> <strong>mole</strong> of oxygen gas and produces <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of carbon dioxide. </p>
<p>So, carbon monoxide has a coefficient of <mathjax>#color(blue)(2)#</mathjax>. The same can be said about carbon dioxide, which has a coefficient of <mathjax>#color(red)(2)#</mathjax> as well. </p>
<p>This means that the <strong>mole ratio</strong> that exists between carbon monoxide and carbon dioxide will be</p>
<blockquote>
<p><mathjax>#(color(blue)(2)color(white)(.)"moles CO")/(color(red)(2)color(white)(.)"moles CO"_2) = color(blue)(2)/color(red)(2) = 1/1#</mathjax></p>
</blockquote>
<p>Because the two coefficients are equal, we say that the two <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> have a <mathjax>#1:1#</mathjax> mole ratio. </p></div>
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</article> | What is the mole ratio of #CO(g)# to #CO_2(g)# in the reaction #2CO(g) + O_2(g) -> 2CO_2(g)#? | null |
1,274 | aa0d4c2e-6ddd-11ea-85e3-ccda262736ce | https://socratic.org/questions/a-gas-with-a-volume-of-3-00-x-10-2-ml-at-150-0-degrees-c-is-heated-until-its-vol | 846.3 K | start physical_unit 30 31 temperature k qc_end physical_unit 30 31 11 13 temperature qc_end physical_unit 30 31 6 9 volume qc_end physical_unit 30 31 20 23 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas [IN] K"}] | [{"type":"physical unit","value":"846.3 K"}] | [{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{150.0 degrees C}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{3.00 × 10^2 mL}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{6.00 × 10^2 mL}"},{"type":"other","value":"The pressure remains constant during the heating process."}] | <h1 class="questionTitle" itemprop="name">A gas with a volume of #3.00 x 10^2# mL at 150.0 degrees C is heated until its volume is #6.00 X 10^2# mL. What is the new temperature of the gas if the pressure remains constant during the heating process?</h1> | null | 846.3 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use old Charles' Law, which states that at constant pressure, and a given quantity of gas, volume varies with temperature.........i.e. <mathjax>#VpropT#</mathjax>.</p>
<p>And thus <mathjax>#V=kT#</mathjax>. And if we solve for <mathjax>#k#</mathjax>, we gets the relationship....</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax> for a given quantity of gas.......</p>
<p><mathjax>#T_2=(V_2xxT_1)/V_1#</mathjax>, <mathjax>#"absolute temperature"#</mathjax> is used......</p>
<p><mathjax>#=(600*mLxx423.15*K)/(300*mL)=846.3*K#</mathjax>. Can you convert this temperature to <mathjax>#""^@C#</mathjax>?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use old <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a>, <mathjax>#VpropT#</mathjax>; units of <mathjax>#"absolute temperature"#</mathjax> must be used, <mathjax>#0#</mathjax> <mathjax>#""^@C=273.15*K#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use old Charles' Law, which states that at constant pressure, and a given quantity of gas, volume varies with temperature.........i.e. <mathjax>#VpropT#</mathjax>.</p>
<p>And thus <mathjax>#V=kT#</mathjax>. And if we solve for <mathjax>#k#</mathjax>, we gets the relationship....</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax> for a given quantity of gas.......</p>
<p><mathjax>#T_2=(V_2xxT_1)/V_1#</mathjax>, <mathjax>#"absolute temperature"#</mathjax> is used......</p>
<p><mathjax>#=(600*mLxx423.15*K)/(300*mL)=846.3*K#</mathjax>. Can you convert this temperature to <mathjax>#""^@C#</mathjax>?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas with a volume of #3.00 x 10^2# mL at 150.0 degrees C is heated until its volume is #6.00 X 10^2# mL. What is the new temperature of the gas if the pressure remains constant during the heating process?</h1>
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<div class="markdown"><p>We use old <a href="https://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a>, <mathjax>#VpropT#</mathjax>; units of <mathjax>#"absolute temperature"#</mathjax> must be used, <mathjax>#0#</mathjax> <mathjax>#""^@C=273.15*K#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use old Charles' Law, which states that at constant pressure, and a given quantity of gas, volume varies with temperature.........i.e. <mathjax>#VpropT#</mathjax>.</p>
<p>And thus <mathjax>#V=kT#</mathjax>. And if we solve for <mathjax>#k#</mathjax>, we gets the relationship....</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax> for a given quantity of gas.......</p>
<p><mathjax>#T_2=(V_2xxT_1)/V_1#</mathjax>, <mathjax>#"absolute temperature"#</mathjax> is used......</p>
<p><mathjax>#=(600*mLxx423.15*K)/(300*mL)=846.3*K#</mathjax>. Can you convert this temperature to <mathjax>#""^@C#</mathjax>?</p></div>
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</article> | A gas with a volume of #3.00 x 10^2# mL at 150.0 degrees C is heated until its volume is #6.00 X 10^2# mL. What is the new temperature of the gas if the pressure remains constant during the heating process? | null |
1,275 | a8ce5ad2-6ddd-11ea-ab15-ccda262736ce | https://socratic.org/questions/29-4ml-of-an-ch-3cooh-solution-were-titrated-with-18-5ml-of-a-0-175m-lioh-soluti | 0.11 M | start physical_unit 4 5 molarity mol/l qc_end physical_unit 4 5 0 1 volume qc_end physical_unit 15 16 9 10 volume qc_end physical_unit 15 16 13 14 molarity qc_end end | [{"type":"physical unit","value":"Molarity [OF] CH3COOH solution [IN] M"}] | [{"type":"physical unit","value":"0.11 M"}] | [{"type":"physical unit","value":"Volume [OF] CH3COOH solution [=] \\pu{29.4 mL}"},{"type":"physical unit","value":"Volume [OF] LiOH solution [=] \\pu{18.5 mL}"},{"type":"physical unit","value":"Molarity [OF] LiOH solution [=] \\pu{0.175 M}"}] | <h1 class="questionTitle" itemprop="name">29.4mL of an #CH_3COOH# solution were titrated with 18.5mL of a 0.175M #LiOH# solution to reach the equivalence point What is the molarity of the #CH_3COOH# solution?</h1> | null | 0.11 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the equation:</p>
<p><mathjax>#sf(CH_3COOH_((aq))+LiOH_((aq))rarrCH_3COO^(-)Li_((aq))^(+)+H_2O_((l))#</mathjax></p>
<p>Concentration = amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> / volume of solution</p>
<p><mathjax>#sf(c=n/v)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(n=cxxv)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(nLiOH=0.175xx18.5/1000=0.003238)#</mathjax></p>
<p>Where <mathjax>#sf(n)#</mathjax> refers to the number of moles.</p>
<p>From the 1:1 molar ratio as shown by the equation, we can say that the number of moles of ethanoic acid must be the same.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf([CH_3COOH_((aq))]=n/v=0.003238/(29.4/1000)=0.11color(white)(x)"mol/l")#</mathjax></p>
<p>Note that I have converted <mathjax>#sf(ml)#</mathjax> to <mathjax>#sf(litre)#</mathjax> by dividing by 1000.</p></div>
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<div class="markdown"><p><mathjax>#sf(0.11color(white)(x)"mol/l")#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the equation:</p>
<p><mathjax>#sf(CH_3COOH_((aq))+LiOH_((aq))rarrCH_3COO^(-)Li_((aq))^(+)+H_2O_((l))#</mathjax></p>
<p>Concentration = amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> / volume of solution</p>
<p><mathjax>#sf(c=n/v)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(n=cxxv)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(nLiOH=0.175xx18.5/1000=0.003238)#</mathjax></p>
<p>Where <mathjax>#sf(n)#</mathjax> refers to the number of moles.</p>
<p>From the 1:1 molar ratio as shown by the equation, we can say that the number of moles of ethanoic acid must be the same.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf([CH_3COOH_((aq))]=n/v=0.003238/(29.4/1000)=0.11color(white)(x)"mol/l")#</mathjax></p>
<p>Note that I have converted <mathjax>#sf(ml)#</mathjax> to <mathjax>#sf(litre)#</mathjax> by dividing by 1000.</p></div>
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<h1 class="questionTitle" itemprop="name">29.4mL of an #CH_3COOH# solution were titrated with 18.5mL of a 0.175M #LiOH# solution to reach the equivalence point What is the molarity of the #CH_3COOH# solution?</h1>
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Michael
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<div class="markdown"><p><mathjax>#sf(0.11color(white)(x)"mol/l")#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the equation:</p>
<p><mathjax>#sf(CH_3COOH_((aq))+LiOH_((aq))rarrCH_3COO^(-)Li_((aq))^(+)+H_2O_((l))#</mathjax></p>
<p>Concentration = amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> / volume of solution</p>
<p><mathjax>#sf(c=n/v)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(n=cxxv)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(nLiOH=0.175xx18.5/1000=0.003238)#</mathjax></p>
<p>Where <mathjax>#sf(n)#</mathjax> refers to the number of moles.</p>
<p>From the 1:1 molar ratio as shown by the equation, we can say that the number of moles of ethanoic acid must be the same.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf([CH_3COOH_((aq))]=n/v=0.003238/(29.4/1000)=0.11color(white)(x)"mol/l")#</mathjax></p>
<p>Note that I have converted <mathjax>#sf(ml)#</mathjax> to <mathjax>#sf(litre)#</mathjax> by dividing by 1000.</p></div>
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</article> | 29.4mL of an #CH_3COOH# solution were titrated with 18.5mL of a 0.175M #LiOH# solution to reach the equivalence point What is the molarity of the #CH_3COOH# solution? | null |
1,276 | ab807055-6ddd-11ea-800a-ccda262736ce | https://socratic.org/questions/57ad6c8d11ef6b1193487623 | 1.52 M | start physical_unit 6 8 molarity mol/l qc_end physical_unit 6 7 11 12 mole qc_end physical_unit 6 8 17 18 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] sodium chloride solution [IN] M"}] | [{"type":"physical unit","value":"1.52 M"}] | [{"type":"physical unit","value":"Mole [OF] sodium chloride [=] \\pu{3.8 moles}"},{"type":"physical unit","value":"Volume [OF] sodium chloride solution [=] \\pu{2.5 L}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a sodium chloride solution that contains #"3.8 moles"# of sodium chloride in #"2.5 L"# of the solution?</h1> | null | 1.52 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem provides you with the composition of a sodium chloride solution in terms of <strong>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> and <strong>total volume of solution</strong>.</p>
<p>In order to find the solution's <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em>, all you have to do is figure out how many moles of solute would be present in <mathjax>#"1 L"#</mathjax> of solution. This is the <em>definition</em> of molarity -- the number of moles of solute, which in your case is sodium chloride, present in <strong>One liter of solution</strong>. </p>
<p>You already know that <mathjax>#3.8#</mathjax> <strong>moles</strong> are present in <mathjax>#"2.5 L"#</mathjax>, so use this known composition to "scale down" the solution to a volume of <mathjax>#"1 L"#</mathjax>. You can do that because, by definition, the particles of solute are <strong>evenly mixed</strong> with the particles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>. </p>
<p>You can thus say that you have</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * overbrace("3.8 moles NaCl"/(2.5color(red)(cancel(color(black)("L solution")))))^(color(blue)("known composition")) = "1.52 moles NaCl"#</mathjax></p>
</blockquote>
<p>Since <em>one liter</em> of solution contains <mathjax>#1.52#</mathjax> <strong>moles</strong> of solute, you can say that the moalrity of the solution, <mathjax>#c#</mathjax>, will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(c = "1.5 mol L"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<div class="markdown"><p><mathjax>#"1.5 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem provides you with the composition of a sodium chloride solution in terms of <strong>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> and <strong>total volume of solution</strong>.</p>
<p>In order to find the solution's <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em>, all you have to do is figure out how many moles of solute would be present in <mathjax>#"1 L"#</mathjax> of solution. This is the <em>definition</em> of molarity -- the number of moles of solute, which in your case is sodium chloride, present in <strong>One liter of solution</strong>. </p>
<p>You already know that <mathjax>#3.8#</mathjax> <strong>moles</strong> are present in <mathjax>#"2.5 L"#</mathjax>, so use this known composition to "scale down" the solution to a volume of <mathjax>#"1 L"#</mathjax>. You can do that because, by definition, the particles of solute are <strong>evenly mixed</strong> with the particles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>. </p>
<p>You can thus say that you have</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * overbrace("3.8 moles NaCl"/(2.5color(red)(cancel(color(black)("L solution")))))^(color(blue)("known composition")) = "1.52 moles NaCl"#</mathjax></p>
</blockquote>
<p>Since <em>one liter</em> of solution contains <mathjax>#1.52#</mathjax> <strong>moles</strong> of solute, you can say that the moalrity of the solution, <mathjax>#c#</mathjax>, will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(c = "1.5 mol L"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a sodium chloride solution that contains #"3.8 moles"# of sodium chloride in #"2.5 L"# of the solution?</h1>
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<div class="markdown"><p><mathjax>#"1.5 mol L"^(-1)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The problem provides you with the composition of a sodium chloride solution in terms of <strong>moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></strong> and <strong>total volume of solution</strong>.</p>
<p>In order to find the solution's <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em>, all you have to do is figure out how many moles of solute would be present in <mathjax>#"1 L"#</mathjax> of solution. This is the <em>definition</em> of molarity -- the number of moles of solute, which in your case is sodium chloride, present in <strong>One liter of solution</strong>. </p>
<p>You already know that <mathjax>#3.8#</mathjax> <strong>moles</strong> are present in <mathjax>#"2.5 L"#</mathjax>, so use this known composition to "scale down" the solution to a volume of <mathjax>#"1 L"#</mathjax>. You can do that because, by definition, the particles of solute are <strong>evenly mixed</strong> with the particles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>. </p>
<p>You can thus say that you have</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("L solution"))) * overbrace("3.8 moles NaCl"/(2.5color(red)(cancel(color(black)("L solution")))))^(color(blue)("known composition")) = "1.52 moles NaCl"#</mathjax></p>
</blockquote>
<p>Since <em>one liter</em> of solution contains <mathjax>#1.52#</mathjax> <strong>moles</strong> of solute, you can say that the moalrity of the solution, <mathjax>#c#</mathjax>, will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(c = "1.5 mol L"^(-1))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | What is the molarity of a sodium chloride solution that contains #"3.8 moles"# of sodium chloride in #"2.5 L"# of the solution? | null |
1,277 | aa8ae388-6ddd-11ea-8b3c-ccda262736ce | https://socratic.org/questions/57e073a611ef6b0db0b642ee | 0.64 moles | start physical_unit 4 5 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] chlorine gas [IN] moles"}] | [{"type":"physical unit","value":"0.64 moles"}] | [{"type":"physical unit","value":"Mass [OF] chlorine element [=] \\pu{45 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles of chlorine gas are associated with a #45*g# mass of the element?</h1> | null | 0.64 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All (well most) elemental gases are binuclear: i.e. <mathjax>#O_2#</mathjax>, <mathjax>#N_2#</mathjax>, <mathjax>#H_2#</mathjax>, <mathjax>#F_2#</mathjax>. The exceptions are the Noble Gases, which are unimolecular. In any case all the halogens (gas, liquid, and solid) are bimolecular, i.e. <mathjax>#X_2#</mathjax>.</p>
<p><mathjax>#"Moles of stuff"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass of stuff"/"Molar mass of stuff"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(45.12*cancelg)/(70.9*cancelg*mol^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#>0.6*mol#</mathjax> <mathjax>#Cl_2#</mathjax> gas.</p>
<p>Note that you simply have to know that chlorine, and fluorine gases, and liquid bromine and solid iodine are BIMOLECULAR. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>A bit over a <mathjax>#1/2#</mathjax> mole of <mathjax>#Cl_2#</mathjax> gas. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All (well most) elemental gases are binuclear: i.e. <mathjax>#O_2#</mathjax>, <mathjax>#N_2#</mathjax>, <mathjax>#H_2#</mathjax>, <mathjax>#F_2#</mathjax>. The exceptions are the Noble Gases, which are unimolecular. In any case all the halogens (gas, liquid, and solid) are bimolecular, i.e. <mathjax>#X_2#</mathjax>.</p>
<p><mathjax>#"Moles of stuff"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass of stuff"/"Molar mass of stuff"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(45.12*cancelg)/(70.9*cancelg*mol^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#>0.6*mol#</mathjax> <mathjax>#Cl_2#</mathjax> gas.</p>
<p>Note that you simply have to know that chlorine, and fluorine gases, and liquid bromine and solid iodine are BIMOLECULAR. </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of chlorine gas are associated with a #45*g# mass of the element?</h1>
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anor277
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Sep 20, 2016
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<div class="markdown"><p>A bit over a <mathjax>#1/2#</mathjax> mole of <mathjax>#Cl_2#</mathjax> gas. </p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All (well most) elemental gases are binuclear: i.e. <mathjax>#O_2#</mathjax>, <mathjax>#N_2#</mathjax>, <mathjax>#H_2#</mathjax>, <mathjax>#F_2#</mathjax>. The exceptions are the Noble Gases, which are unimolecular. In any case all the halogens (gas, liquid, and solid) are bimolecular, i.e. <mathjax>#X_2#</mathjax>.</p>
<p><mathjax>#"Moles of stuff"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass of stuff"/"Molar mass of stuff"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(45.12*cancelg)/(70.9*cancelg*mol^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#>0.6*mol#</mathjax> <mathjax>#Cl_2#</mathjax> gas.</p>
<p>Note that you simply have to know that chlorine, and fluorine gases, and liquid bromine and solid iodine are BIMOLECULAR. </p></div>
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</article> | How many moles of chlorine gas are associated with a #45*g# mass of the element? | null |
1,278 | ab422415-6ddd-11ea-a7e1-ccda262736ce | https://socratic.org/questions/if-2-0-mol-of-propane-are-burned-reacted-with-oxygen-how-many-moles-of-carbon-di | 6.00 moles | start physical_unit 14 15 mole mol qc_end physical_unit 4 4 1 2 mole qc_end chemical_equation 46 55 qc_end end | [{"type":"physical unit","value":"Mole [OF] carbon dioxide [IN] moles"}] | [{"type":"physical unit","value":"6.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] propane [=] \\pu{2.0 mol}"},{"type":"chemical equation","value":"C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(g)"}] | <h1 class="questionTitle" itemprop="name">If 2.0 mol of propane are burned (reacted with oxygen), how many moles of carbon dioxide will be produced? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Propane, C3H8, a fuel used in many outdoor grills, burns in oxygen to produce carbon dioxide gas, CO2, and water vapor, H2O, as shown in the reaction</p>
<p><mathjax>#"C"_3"H"_8 (g) + 5"O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O" (g)#</mathjax></p></div>
</h2>
</div>
</div> | 6.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember about <em>balanced</em> <a href="https://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a> is that the <strong>stoichiometric coefficients</strong> added to each chemical species that takes part in the reaction are equivalent to <strong>moles</strong>.</p>
<p>In your case, you know that the chemical equation that describes the combustion of propane looks like this</p>
<blockquote>
<p><mathjax>#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> color(red)(3)"CO"_ (2(g)) + 4"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>Notice that you don't have a coefficient for propane, which implies that the coefficient is actually equal to <mathjax>#1#</mathjax>, and that you have a <mathjax>#color(red)(3)#</mathjax> for carbon dioxide. </p>
<p>This tells you that when <mathjax>#1#</mathjax> <strong>mole</strong> of propane undergoes combustion, <mathjax>#color(red)(3)#</mathjax> <strong>moles</strong> of carbon dioxide are produced. </p>
<p>You can thus say that when <mathjax>#2.0#</mathjax> <strong>moles</strong> of propane react, you will get</p>
<blockquote>
<p><mathjax>#2.0 color(red)(cancel(color(black)("moles C"_3"H"_8))) * (color(red)(3)color(white)(.)"moles CO"_2)/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = color(darkgreen)(ul(color(black)("6.0 moles CO"_2)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"6.0 moles CO"_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember about <em>balanced</em> <a href="https://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a> is that the <strong>stoichiometric coefficients</strong> added to each chemical species that takes part in the reaction are equivalent to <strong>moles</strong>.</p>
<p>In your case, you know that the chemical equation that describes the combustion of propane looks like this</p>
<blockquote>
<p><mathjax>#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> color(red)(3)"CO"_ (2(g)) + 4"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>Notice that you don't have a coefficient for propane, which implies that the coefficient is actually equal to <mathjax>#1#</mathjax>, and that you have a <mathjax>#color(red)(3)#</mathjax> for carbon dioxide. </p>
<p>This tells you that when <mathjax>#1#</mathjax> <strong>mole</strong> of propane undergoes combustion, <mathjax>#color(red)(3)#</mathjax> <strong>moles</strong> of carbon dioxide are produced. </p>
<p>You can thus say that when <mathjax>#2.0#</mathjax> <strong>moles</strong> of propane react, you will get</p>
<blockquote>
<p><mathjax>#2.0 color(red)(cancel(color(black)("moles C"_3"H"_8))) * (color(red)(3)color(white)(.)"moles CO"_2)/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = color(darkgreen)(ul(color(black)("6.0 moles CO"_2)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">If 2.0 mol of propane are burned (reacted with oxygen), how many moles of carbon dioxide will be produced? </h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Propane, C3H8, a fuel used in many outdoor grills, burns in oxygen to produce carbon dioxide gas, CO2, and water vapor, H2O, as shown in the reaction</p>
<p><mathjax>#"C"_3"H"_8 (g) + 5"O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O" (g)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"6.0 moles CO"_2#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The thing to remember about <em>balanced</em> <a href="https://socratic.org/chemistry/chemical-reactions/chemical-equations">chemical equations</a> is that the <strong>stoichiometric coefficients</strong> added to each chemical species that takes part in the reaction are equivalent to <strong>moles</strong>.</p>
<p>In your case, you know that the chemical equation that describes the combustion of propane looks like this</p>
<blockquote>
<p><mathjax>#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> color(red)(3)"CO"_ (2(g)) + 4"H"_ 2"O"_ ((g))#</mathjax></p>
</blockquote>
<p>Notice that you don't have a coefficient for propane, which implies that the coefficient is actually equal to <mathjax>#1#</mathjax>, and that you have a <mathjax>#color(red)(3)#</mathjax> for carbon dioxide. </p>
<p>This tells you that when <mathjax>#1#</mathjax> <strong>mole</strong> of propane undergoes combustion, <mathjax>#color(red)(3)#</mathjax> <strong>moles</strong> of carbon dioxide are produced. </p>
<p>You can thus say that when <mathjax>#2.0#</mathjax> <strong>moles</strong> of propane react, you will get</p>
<blockquote>
<p><mathjax>#2.0 color(red)(cancel(color(black)("moles C"_3"H"_8))) * (color(red)(3)color(white)(.)"moles CO"_2)/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = color(darkgreen)(ul(color(black)("6.0 moles CO"_2)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | If 2.0 mol of propane are burned (reacted with oxygen), how many moles of carbon dioxide will be produced? |
Propane, C3H8, a fuel used in many outdoor grills, burns in oxygen to produce carbon dioxide gas, CO2, and water vapor, H2O, as shown in the reaction
#"C"_3"H"_8 (g) + 5"O"_2(g) -> 3"CO"_2(g) + 4"H"_2"O" (g)#
|
1,279 | aae7d89f-6ddd-11ea-9bbe-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-5-l-of-h-2-at-stp | 0.22 moles | start physical_unit 8 8 mole mol qc_end physical_unit 8 8 5 6 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Mole [OF] H2 [IN] moles"}] | [{"type":"physical unit","value":"0.22 moles"}] | [{"type":"physical unit","value":"Volume [OF] H2 [=] \\pu{5 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">How many moles are in 5 L of #H_2# at STP?</h1> | null | 0.22 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this type of problem you would use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, PV = nRT. Where P represents pressure (must have units of atm), V represents volume (must have units of liters), n represents the number of moles, R is the proportionality constant<br/>
(has units of L<em> atm/ mol </em>K), and T represents the temperature, which must be in Kelvins. </p>
<p>Now what you want to do is list your known and unknown variables. Our only unknown is the number of moles of hydrogen. Our known variables are P,V,R, and T. Since we are at STP, the temperature is 273K and the pressure is 1 atm. We are given volume and the proportionality constant, R, is equal to 0.0821 L<em> atm/ mol </em>K.</p>
<p>Now all we have to do is rearrange the equation and solve for n like so: <br/>
n = PV/RT</p>
<p>n = (1 atm) (5 L) / (0.0821 L<em> atm/ mol </em>K) (273K)<br/>
Which gives you 0.223 moles of hydrogen gas. </p>
<p>I hope this makes sense, and <a href="http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Physical_Properties_of_Matter/States_of_Matter/Gases/Gas_Laws/The_Ideal_Gas_Law" rel="nofollow">here is a link for additional help!</a> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>0.223 moles. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this type of problem you would use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, PV = nRT. Where P represents pressure (must have units of atm), V represents volume (must have units of liters), n represents the number of moles, R is the proportionality constant<br/>
(has units of L<em> atm/ mol </em>K), and T represents the temperature, which must be in Kelvins. </p>
<p>Now what you want to do is list your known and unknown variables. Our only unknown is the number of moles of hydrogen. Our known variables are P,V,R, and T. Since we are at STP, the temperature is 273K and the pressure is 1 atm. We are given volume and the proportionality constant, R, is equal to 0.0821 L<em> atm/ mol </em>K.</p>
<p>Now all we have to do is rearrange the equation and solve for n like so: <br/>
n = PV/RT</p>
<p>n = (1 atm) (5 L) / (0.0821 L<em> atm/ mol </em>K) (273K)<br/>
Which gives you 0.223 moles of hydrogen gas. </p>
<p>I hope this makes sense, and <a href="http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Physical_Properties_of_Matter/States_of_Matter/Gases/Gas_Laws/The_Ideal_Gas_Law" rel="nofollow">here is a link for additional help!</a> </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are in 5 L of #H_2# at STP?</h1>
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<div class="markdown"><p>0.223 moles. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this type of problem you would use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation, PV = nRT. Where P represents pressure (must have units of atm), V represents volume (must have units of liters), n represents the number of moles, R is the proportionality constant<br/>
(has units of L<em> atm/ mol </em>K), and T represents the temperature, which must be in Kelvins. </p>
<p>Now what you want to do is list your known and unknown variables. Our only unknown is the number of moles of hydrogen. Our known variables are P,V,R, and T. Since we are at STP, the temperature is 273K and the pressure is 1 atm. We are given volume and the proportionality constant, R, is equal to 0.0821 L<em> atm/ mol </em>K.</p>
<p>Now all we have to do is rearrange the equation and solve for n like so: <br/>
n = PV/RT</p>
<p>n = (1 atm) (5 L) / (0.0821 L<em> atm/ mol </em>K) (273K)<br/>
Which gives you 0.223 moles of hydrogen gas. </p>
<p>I hope this makes sense, and <a href="http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Physical_Properties_of_Matter/States_of_Matter/Gases/Gas_Laws/The_Ideal_Gas_Law" rel="nofollow">here is a link for additional help!</a> </p></div>
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</article> | How many moles are in 5 L of #H_2# at STP? | null |
1,280 | ad0cd10c-6ddd-11ea-94c8-ccda262736ce | https://socratic.org/questions/how-do-you-balance-al-cl-2-alcl-3 | 2 Al + 3 Cl2 -> 2 AlCl3 | start chemical_equation qc_end chemical_equation 4 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 Al + 3 Cl2 -> 2 AlCl3"}] | [{"type":"chemical equation","value":"Al + Cl2 -> AlCl3"}] | <h1 class="questionTitle" itemprop="name">How do you balance #Al+Cl_2->AlCl_3#?</h1> | null | 2 Al + 3 Cl2 -> 2 AlCl3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's balance this equation by using <em><a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a></em>. </p>
<p>Start by assigning oxidation numbers to the atoms that take part in the reaction</p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Al") _ ((s)) + stackrel(color(blue)(0))("Cl")_ (2(g)) -> stackrel(color(blue)(+3))("Al") stackrel(color(blue)(-1))("Cl")_ (3(s))#</mathjax></p>
</blockquote>
<p>Notice that the oxidation number of aluminium <strong>increases</strong> from <mathjax>#color(blue)(0)#</mathjax> on the reactants' side to <mathjax>#color(blue)(+3)#</mathjax> on the products' side, which means that it's being <strong>oxidized</strong>. </p>
<p>On the other hand, the oxidation number of chlorine <strong>decreases</strong> from <mathjax>#color(blue)(0)#</mathjax> on the reactants' side to <mathjax>#-1#</mathjax> on the products' side, which means that chlorine is being <strong>reduced</strong>. </p>
<p>The <strong>oxidation half-reaction</strong> looks like this </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Al") -> stackrel(color(blue)(+3))("Al")""^(3+) + 3"e"^(-)#</mathjax></p>
</blockquote>
<p>The <strong>reduction half-reaction</strong> looks like this</p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Cl")_ 2 + "e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-)#</mathjax></p>
</blockquote>
<p>Balance the chlorine atoms by multiplying the product by <mathjax>#2#</mathjax>. Since <strong>one atom</strong> of chlorine gains <mathjax>#1#</mathjax> <strong>electron</strong>, it follows that <strong>two atoms</strong> will gain <mathjax>#2#</mathjax> <strong>electrons</strong></p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Cl")_ 2 + 2"e"^(-) -> 2stackrel(color(blue)(-1))("Cl")""^(-)#</mathjax></p>
</blockquote>
<p>Now, in any redox reaction, the number of electrons <strong>lost</strong> in the oxidation half-reaction <em><strong>must be equal to</strong></em> the number of electrons <strong>gained</strong> in the reduction half-reaction.</p>
<p>To get them to balance out, multiply the oxidation half-reaction by <mathjax>#2#</mathjax> and the reduction half-reaction by <mathjax>#3#</mathjax>., then add the two half-reactions to get</p>
<blockquote>
<p><mathjax>#{(color(white)(aaaaaa)stackrel(color(blue)(0))("Al") -> stackrel(color(blue)(+3))("Al")""^(3+) + 3"e"^(-)" "| xx 2), (stackrel(color(blue)(0))("Cl")_ 2 + 2"e"^(-) -> 2stackrel(color(blue)(-1))("Cl")""^(-)" "color(white)(aaaaa)| xx 3) :}#</mathjax><br/>
<mathjax>#color(white)(a)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#</mathjax></p>
<p><mathjax>#2"Al" + 3"Cl"_ 2 + color(red)(cancel(color(black)(6"e"^(-)))) -> 2"Al"^(3+) + 6"Cl"^(-) + color(red)(cancel(color(black)(6"e"^(-))))#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#2"Al" + 3"Cl"_ 2 -> 2 xx ["Al"^(3+) + 3"Cl"^(-)]#</mathjax></p>
</blockquote>
<p>which of course gets you </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(2"Al"_ ((s)) + 3"Cl"_ (2(g)) -> 2"AlCl"_ (3(s)))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2"Al"_ ((s)) + 3"Cl"_ (2(g)) -> 2"AlCl"_ (3(s))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's balance this equation by using <em><a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a></em>. </p>
<p>Start by assigning oxidation numbers to the atoms that take part in the reaction</p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Al") _ ((s)) + stackrel(color(blue)(0))("Cl")_ (2(g)) -> stackrel(color(blue)(+3))("Al") stackrel(color(blue)(-1))("Cl")_ (3(s))#</mathjax></p>
</blockquote>
<p>Notice that the oxidation number of aluminium <strong>increases</strong> from <mathjax>#color(blue)(0)#</mathjax> on the reactants' side to <mathjax>#color(blue)(+3)#</mathjax> on the products' side, which means that it's being <strong>oxidized</strong>. </p>
<p>On the other hand, the oxidation number of chlorine <strong>decreases</strong> from <mathjax>#color(blue)(0)#</mathjax> on the reactants' side to <mathjax>#-1#</mathjax> on the products' side, which means that chlorine is being <strong>reduced</strong>. </p>
<p>The <strong>oxidation half-reaction</strong> looks like this </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Al") -> stackrel(color(blue)(+3))("Al")""^(3+) + 3"e"^(-)#</mathjax></p>
</blockquote>
<p>The <strong>reduction half-reaction</strong> looks like this</p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Cl")_ 2 + "e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-)#</mathjax></p>
</blockquote>
<p>Balance the chlorine atoms by multiplying the product by <mathjax>#2#</mathjax>. Since <strong>one atom</strong> of chlorine gains <mathjax>#1#</mathjax> <strong>electron</strong>, it follows that <strong>two atoms</strong> will gain <mathjax>#2#</mathjax> <strong>electrons</strong></p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Cl")_ 2 + 2"e"^(-) -> 2stackrel(color(blue)(-1))("Cl")""^(-)#</mathjax></p>
</blockquote>
<p>Now, in any redox reaction, the number of electrons <strong>lost</strong> in the oxidation half-reaction <em><strong>must be equal to</strong></em> the number of electrons <strong>gained</strong> in the reduction half-reaction.</p>
<p>To get them to balance out, multiply the oxidation half-reaction by <mathjax>#2#</mathjax> and the reduction half-reaction by <mathjax>#3#</mathjax>., then add the two half-reactions to get</p>
<blockquote>
<p><mathjax>#{(color(white)(aaaaaa)stackrel(color(blue)(0))("Al") -> stackrel(color(blue)(+3))("Al")""^(3+) + 3"e"^(-)" "| xx 2), (stackrel(color(blue)(0))("Cl")_ 2 + 2"e"^(-) -> 2stackrel(color(blue)(-1))("Cl")""^(-)" "color(white)(aaaaa)| xx 3) :}#</mathjax><br/>
<mathjax>#color(white)(a)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#</mathjax></p>
<p><mathjax>#2"Al" + 3"Cl"_ 2 + color(red)(cancel(color(black)(6"e"^(-)))) -> 2"Al"^(3+) + 6"Cl"^(-) + color(red)(cancel(color(black)(6"e"^(-))))#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#2"Al" + 3"Cl"_ 2 -> 2 xx ["Al"^(3+) + 3"Cl"^(-)]#</mathjax></p>
</blockquote>
<p>which of course gets you </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(2"Al"_ ((s)) + 3"Cl"_ (2(g)) -> 2"AlCl"_ (3(s)))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you balance #Al+Cl_2->AlCl_3#?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#2"Al"_ ((s)) + 3"Cl"_ (2(g)) -> 2"AlCl"_ (3(s))#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let's balance this equation by using <em><a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a></em>. </p>
<p>Start by assigning oxidation numbers to the atoms that take part in the reaction</p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Al") _ ((s)) + stackrel(color(blue)(0))("Cl")_ (2(g)) -> stackrel(color(blue)(+3))("Al") stackrel(color(blue)(-1))("Cl")_ (3(s))#</mathjax></p>
</blockquote>
<p>Notice that the oxidation number of aluminium <strong>increases</strong> from <mathjax>#color(blue)(0)#</mathjax> on the reactants' side to <mathjax>#color(blue)(+3)#</mathjax> on the products' side, which means that it's being <strong>oxidized</strong>. </p>
<p>On the other hand, the oxidation number of chlorine <strong>decreases</strong> from <mathjax>#color(blue)(0)#</mathjax> on the reactants' side to <mathjax>#-1#</mathjax> on the products' side, which means that chlorine is being <strong>reduced</strong>. </p>
<p>The <strong>oxidation half-reaction</strong> looks like this </p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Al") -> stackrel(color(blue)(+3))("Al")""^(3+) + 3"e"^(-)#</mathjax></p>
</blockquote>
<p>The <strong>reduction half-reaction</strong> looks like this</p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Cl")_ 2 + "e"^(-) -> stackrel(color(blue)(-1))("Cl")""^(-)#</mathjax></p>
</blockquote>
<p>Balance the chlorine atoms by multiplying the product by <mathjax>#2#</mathjax>. Since <strong>one atom</strong> of chlorine gains <mathjax>#1#</mathjax> <strong>electron</strong>, it follows that <strong>two atoms</strong> will gain <mathjax>#2#</mathjax> <strong>electrons</strong></p>
<blockquote>
<p><mathjax>#stackrel(color(blue)(0))("Cl")_ 2 + 2"e"^(-) -> 2stackrel(color(blue)(-1))("Cl")""^(-)#</mathjax></p>
</blockquote>
<p>Now, in any redox reaction, the number of electrons <strong>lost</strong> in the oxidation half-reaction <em><strong>must be equal to</strong></em> the number of electrons <strong>gained</strong> in the reduction half-reaction.</p>
<p>To get them to balance out, multiply the oxidation half-reaction by <mathjax>#2#</mathjax> and the reduction half-reaction by <mathjax>#3#</mathjax>., then add the two half-reactions to get</p>
<blockquote>
<p><mathjax>#{(color(white)(aaaaaa)stackrel(color(blue)(0))("Al") -> stackrel(color(blue)(+3))("Al")""^(3+) + 3"e"^(-)" "| xx 2), (stackrel(color(blue)(0))("Cl")_ 2 + 2"e"^(-) -> 2stackrel(color(blue)(-1))("Cl")""^(-)" "color(white)(aaaaa)| xx 3) :}#</mathjax><br/>
<mathjax>#color(white)(a)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#</mathjax></p>
<p><mathjax>#2"Al" + 3"Cl"_ 2 + color(red)(cancel(color(black)(6"e"^(-)))) -> 2"Al"^(3+) + 6"Cl"^(-) + color(red)(cancel(color(black)(6"e"^(-))))#</mathjax></p>
</blockquote>
<p>This will be equivalent to</p>
<blockquote>
<p><mathjax>#2"Al" + 3"Cl"_ 2 -> 2 xx ["Al"^(3+) + 3"Cl"^(-)]#</mathjax></p>
</blockquote>
<p>which of course gets you </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)(2"Al"_ ((s)) + 3"Cl"_ (2(g)) -> 2"AlCl"_ (3(s)))color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | How do you balance #Al+Cl_2->AlCl_3#? | null |
1,281 | a90e3186-6ddd-11ea-b1bf-ccda262736ce | https://socratic.org/questions/what-amount-of-zn-will-be-required-to-produce-h2-by-its-action-on-dilute-h2so4-w | 42.56 g | start physical_unit 3 3 mass g qc_end physical_unit 29 29 26 27 mass qc_end chemical_equation 9 9 qc_end chemical_equation 15 15 qc_end substance 22 22 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] Zn [IN] g"}] | [{"type":"physical unit","value":"42.56 g"}] | [{"type":"physical unit","value":"Mass [OF] KClO3 [=] \\pu{30 g}"},{"type":"chemical equation","value":"H2"},{"type":"chemical equation","value":"H2SO4"},{"type":"substance name","value":"Oxygen"},{"type":"other","value":"Completely react."}] | <h1 class="questionTitle" itemprop="name">What amount of #Zn# will be required to produce #H_2# by its action on dilute #H_2SO_4# which will completely react with the oxygen, produced by heating #30g# of #KClO_3#?</h1> | null | 42.56 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the only known quantity in this question</p>
<p><mathjax>#m("KClO"_3) = 30 color(white)(l) g#</mathjax></p>
<p><mathjax>#n("KClO"_3) = 30 color(white)(l) g * (1 color(white)(l) "mol")/(138.55 color(white)(l) "g")=0.217 color(white)(l) "mol"#</mathjax></p>
<p>The balanced equation of the decomposition of potassium perchlorate</p>
<p><mathjax>#color(navy)(2) color(white)(l) "KClO"_3 (s) stackrel(Delta)(to) 2 color(white)(l) "KCl" (s) + color(purple)(3) color(white)(l) "O"_2 (g)#</mathjax></p>
<p>suggests the stoichiometric relationship </p>
<ul>
<li><mathjax>#color(navy)(2) color(white)(l) "mol" color(white)(l) "KClO"_3#</mathjax> decomposes to produce <mathjax>#color(purple)(3) color(white)(l) "mol" color(white)(l) "O"_2 (g)#</mathjax></li>
</ul>
<p>Hence the amount of oxygen required for the complete combustion of the unknown amount of <mathjax>#"H"_2 (g)#</mathjax> produced would be</p>
<p><mathjax>#n("O"_2) = 0.217 color(white)(l) "mol" color(white)(l) "KClO"_3 * (color(purple)(3) color(white)(l) "mol" color(white)(l) "O"_2 )/(color(navy)(2) color(white)(l) "mol" color(white)(l) "KClO"_3)=0.326 color(white)(l) "mol" color(white)(l) "O"_2 #</mathjax></p>
<p>Oxygen reacts with hydrogen by the equation</p>
<p><mathjax># color(purple)(1) color(white)(l) "O"_2 (g) + color(navy)(2) color(white)(l) "H"_2 (g)stackrel("*")(to) 2 color(white)(l) "H"_2"O" (g)#</mathjax></p>
<p>at a <mathjax>#color(purple)(1): color(navy)(2)#</mathjax> ratio, meaning that the combustion would consume </p>
<p><mathjax>#n("H"_2) = 0.326 color(white)(l) "mol" color(white)(l) "O"_2 * (color(navy)(2) color(white)(l) "H"_2 )/(color(purple)(1) color(white)(l) "O"_2)=0.651 color(white)(l) "mol" color(white)(l) "H"_2#</mathjax></p>
<p>of hydrogen. All these <mathjax>#"H"_2#</mathjax> came from the reaction between <mathjax>#"Zn"#</mathjax> and dilute sulfuric acid <mathjax>#"H"_2 "SO"_4#</mathjax> as seen in the following equation</p>
<p><mathjax>#color(navy)(1) color(white)(l) "Zn"(s) + "H"_2"SO"_4 (aq) to color(navy)(1) color(white)(l) "H"_2 (g) + "ZnSO"_4 (aq)#</mathjax></p>
<p>where for each mole of <mathjax>#"H"_2 (g)#</mathjax> produced, <mathjax>#color(navy)(1) color(white)(l) "mol"#</mathjax> of <mathjax>#"Zn"#</mathjax> is consumed. That is:</p>
<p><mathjax>#n("Zn") = n("H"_2) = 0.651 color(white)(l) "mol"#</mathjax></p>
<p>Hence the mass of <mathjax>#"Zn"#</mathjax></p>
<p><mathjax>#m("Zn") = n("Zn") * M("Zn") = 42.56 color(white)(l) g#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><ul>
<li><mathjax>#n("Zn")=0.651 color(white)(l) "mol"#</mathjax></li>
<li><mathjax>#m("Zn") = n("Zn") * M("Zn") = 42.56 color(white)(l) g#</mathjax></li>
</ul></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the only known quantity in this question</p>
<p><mathjax>#m("KClO"_3) = 30 color(white)(l) g#</mathjax></p>
<p><mathjax>#n("KClO"_3) = 30 color(white)(l) g * (1 color(white)(l) "mol")/(138.55 color(white)(l) "g")=0.217 color(white)(l) "mol"#</mathjax></p>
<p>The balanced equation of the decomposition of potassium perchlorate</p>
<p><mathjax>#color(navy)(2) color(white)(l) "KClO"_3 (s) stackrel(Delta)(to) 2 color(white)(l) "KCl" (s) + color(purple)(3) color(white)(l) "O"_2 (g)#</mathjax></p>
<p>suggests the stoichiometric relationship </p>
<ul>
<li><mathjax>#color(navy)(2) color(white)(l) "mol" color(white)(l) "KClO"_3#</mathjax> decomposes to produce <mathjax>#color(purple)(3) color(white)(l) "mol" color(white)(l) "O"_2 (g)#</mathjax></li>
</ul>
<p>Hence the amount of oxygen required for the complete combustion of the unknown amount of <mathjax>#"H"_2 (g)#</mathjax> produced would be</p>
<p><mathjax>#n("O"_2) = 0.217 color(white)(l) "mol" color(white)(l) "KClO"_3 * (color(purple)(3) color(white)(l) "mol" color(white)(l) "O"_2 )/(color(navy)(2) color(white)(l) "mol" color(white)(l) "KClO"_3)=0.326 color(white)(l) "mol" color(white)(l) "O"_2 #</mathjax></p>
<p>Oxygen reacts with hydrogen by the equation</p>
<p><mathjax># color(purple)(1) color(white)(l) "O"_2 (g) + color(navy)(2) color(white)(l) "H"_2 (g)stackrel("*")(to) 2 color(white)(l) "H"_2"O" (g)#</mathjax></p>
<p>at a <mathjax>#color(purple)(1): color(navy)(2)#</mathjax> ratio, meaning that the combustion would consume </p>
<p><mathjax>#n("H"_2) = 0.326 color(white)(l) "mol" color(white)(l) "O"_2 * (color(navy)(2) color(white)(l) "H"_2 )/(color(purple)(1) color(white)(l) "O"_2)=0.651 color(white)(l) "mol" color(white)(l) "H"_2#</mathjax></p>
<p>of hydrogen. All these <mathjax>#"H"_2#</mathjax> came from the reaction between <mathjax>#"Zn"#</mathjax> and dilute sulfuric acid <mathjax>#"H"_2 "SO"_4#</mathjax> as seen in the following equation</p>
<p><mathjax>#color(navy)(1) color(white)(l) "Zn"(s) + "H"_2"SO"_4 (aq) to color(navy)(1) color(white)(l) "H"_2 (g) + "ZnSO"_4 (aq)#</mathjax></p>
<p>where for each mole of <mathjax>#"H"_2 (g)#</mathjax> produced, <mathjax>#color(navy)(1) color(white)(l) "mol"#</mathjax> of <mathjax>#"Zn"#</mathjax> is consumed. That is:</p>
<p><mathjax>#n("Zn") = n("H"_2) = 0.651 color(white)(l) "mol"#</mathjax></p>
<p>Hence the mass of <mathjax>#"Zn"#</mathjax></p>
<p><mathjax>#m("Zn") = n("Zn") * M("Zn") = 42.56 color(white)(l) g#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What amount of #Zn# will be required to produce #H_2# by its action on dilute #H_2SO_4# which will completely react with the oxygen, produced by heating #30g# of #KClO_3#?</h1>
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Jacob T.
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<div class="markdown"><ul>
<li><mathjax>#n("Zn")=0.651 color(white)(l) "mol"#</mathjax></li>
<li><mathjax>#m("Zn") = n("Zn") * M("Zn") = 42.56 color(white)(l) g#</mathjax></li>
</ul></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start with the only known quantity in this question</p>
<p><mathjax>#m("KClO"_3) = 30 color(white)(l) g#</mathjax></p>
<p><mathjax>#n("KClO"_3) = 30 color(white)(l) g * (1 color(white)(l) "mol")/(138.55 color(white)(l) "g")=0.217 color(white)(l) "mol"#</mathjax></p>
<p>The balanced equation of the decomposition of potassium perchlorate</p>
<p><mathjax>#color(navy)(2) color(white)(l) "KClO"_3 (s) stackrel(Delta)(to) 2 color(white)(l) "KCl" (s) + color(purple)(3) color(white)(l) "O"_2 (g)#</mathjax></p>
<p>suggests the stoichiometric relationship </p>
<ul>
<li><mathjax>#color(navy)(2) color(white)(l) "mol" color(white)(l) "KClO"_3#</mathjax> decomposes to produce <mathjax>#color(purple)(3) color(white)(l) "mol" color(white)(l) "O"_2 (g)#</mathjax></li>
</ul>
<p>Hence the amount of oxygen required for the complete combustion of the unknown amount of <mathjax>#"H"_2 (g)#</mathjax> produced would be</p>
<p><mathjax>#n("O"_2) = 0.217 color(white)(l) "mol" color(white)(l) "KClO"_3 * (color(purple)(3) color(white)(l) "mol" color(white)(l) "O"_2 )/(color(navy)(2) color(white)(l) "mol" color(white)(l) "KClO"_3)=0.326 color(white)(l) "mol" color(white)(l) "O"_2 #</mathjax></p>
<p>Oxygen reacts with hydrogen by the equation</p>
<p><mathjax># color(purple)(1) color(white)(l) "O"_2 (g) + color(navy)(2) color(white)(l) "H"_2 (g)stackrel("*")(to) 2 color(white)(l) "H"_2"O" (g)#</mathjax></p>
<p>at a <mathjax>#color(purple)(1): color(navy)(2)#</mathjax> ratio, meaning that the combustion would consume </p>
<p><mathjax>#n("H"_2) = 0.326 color(white)(l) "mol" color(white)(l) "O"_2 * (color(navy)(2) color(white)(l) "H"_2 )/(color(purple)(1) color(white)(l) "O"_2)=0.651 color(white)(l) "mol" color(white)(l) "H"_2#</mathjax></p>
<p>of hydrogen. All these <mathjax>#"H"_2#</mathjax> came from the reaction between <mathjax>#"Zn"#</mathjax> and dilute sulfuric acid <mathjax>#"H"_2 "SO"_4#</mathjax> as seen in the following equation</p>
<p><mathjax>#color(navy)(1) color(white)(l) "Zn"(s) + "H"_2"SO"_4 (aq) to color(navy)(1) color(white)(l) "H"_2 (g) + "ZnSO"_4 (aq)#</mathjax></p>
<p>where for each mole of <mathjax>#"H"_2 (g)#</mathjax> produced, <mathjax>#color(navy)(1) color(white)(l) "mol"#</mathjax> of <mathjax>#"Zn"#</mathjax> is consumed. That is:</p>
<p><mathjax>#n("Zn") = n("H"_2) = 0.651 color(white)(l) "mol"#</mathjax></p>
<p>Hence the mass of <mathjax>#"Zn"#</mathjax></p>
<p><mathjax>#m("Zn") = n("Zn") * M("Zn") = 42.56 color(white)(l) g#</mathjax></p></div>
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</article> | What amount of #Zn# will be required to produce #H_2# by its action on dilute #H_2SO_4# which will completely react with the oxygen, produced by heating #30g# of #KClO_3#? | null |
1,282 | ab59d80b-6ddd-11ea-8799-ccda262736ce | https://socratic.org/questions/how-much-mass-is-in-a-3-25-mole-sample-of-nh-4oh | 114 g | start physical_unit 8 10 mass g qc_end physical_unit 8 10 6 7 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] NH4OH sample [IN] g"}] | [{"type":"physical unit","value":"114 g"}] | [{"type":"physical unit","value":"Mole [OF] NH4OH sample [=] \\pu{3.25 mole}"}] | <h1 class="questionTitle" itemprop="name">How much mass is in a 3.25-mole sample of #NH_4OH#?</h1> | null | 114 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we'll need to find out the mass of <mathjax>#1#</mathjax> mole of <mathjax>#NH_4OH#</mathjax>. <br/>
To do this, we need to add up the masses of one mole, or molar masses, of its constituent atoms:</p>
<p><mathjax>#"1 mole" = N+Hxx4+O+H = "14.01 g/mol" + "1.008 g/mol" xx 4 + "16.00 g/mol" + "1.008 g/mol" = "35.05 g/mol"#</mathjax></p>
<p>Now, to find the mass of <mathjax>#3.25#</mathjax> moles of <mathjax>#NH_4OH#</mathjax>, we'll just need to multiply its molar mass by <mathjax>#3.25#</mathjax>:</p>
<p><mathjax>#"mass" = "number of moles" xx "mass of 1 mole"#</mathjax><br/>
<mathjax>#= "number of moles" xx "molar mass"#</mathjax><br/>
<mathjax>#= "3.25 mol" xx "35.05 g/mol" = "114 g"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"114 g"#</mathjax>. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we'll need to find out the mass of <mathjax>#1#</mathjax> mole of <mathjax>#NH_4OH#</mathjax>. <br/>
To do this, we need to add up the masses of one mole, or molar masses, of its constituent atoms:</p>
<p><mathjax>#"1 mole" = N+Hxx4+O+H = "14.01 g/mol" + "1.008 g/mol" xx 4 + "16.00 g/mol" + "1.008 g/mol" = "35.05 g/mol"#</mathjax></p>
<p>Now, to find the mass of <mathjax>#3.25#</mathjax> moles of <mathjax>#NH_4OH#</mathjax>, we'll just need to multiply its molar mass by <mathjax>#3.25#</mathjax>:</p>
<p><mathjax>#"mass" = "number of moles" xx "mass of 1 mole"#</mathjax><br/>
<mathjax>#= "number of moles" xx "molar mass"#</mathjax><br/>
<mathjax>#= "3.25 mol" xx "35.05 g/mol" = "114 g"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How much mass is in a 3.25-mole sample of #NH_4OH#?</h1>
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<div class="markdown"><p><mathjax>#"114 g"#</mathjax>. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>First, we'll need to find out the mass of <mathjax>#1#</mathjax> mole of <mathjax>#NH_4OH#</mathjax>. <br/>
To do this, we need to add up the masses of one mole, or molar masses, of its constituent atoms:</p>
<p><mathjax>#"1 mole" = N+Hxx4+O+H = "14.01 g/mol" + "1.008 g/mol" xx 4 + "16.00 g/mol" + "1.008 g/mol" = "35.05 g/mol"#</mathjax></p>
<p>Now, to find the mass of <mathjax>#3.25#</mathjax> moles of <mathjax>#NH_4OH#</mathjax>, we'll just need to multiply its molar mass by <mathjax>#3.25#</mathjax>:</p>
<p><mathjax>#"mass" = "number of moles" xx "mass of 1 mole"#</mathjax><br/>
<mathjax>#= "number of moles" xx "molar mass"#</mathjax><br/>
<mathjax>#= "3.25 mol" xx "35.05 g/mol" = "114 g"#</mathjax></p></div>
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</article> | How much mass is in a 3.25-mole sample of #NH_4OH#? | null |
1,283 | a8df62b9-6ddd-11ea-b321-ccda262736ce | https://socratic.org/questions/56538ab7581e2a6511a9bd01 | SiO2(s) + 6 HF(aq) -> SiF6^2− + 2 H3O+ | start chemical_equation qc_end chemical_equation 3 3 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"SiO2(s) + 6 HF(aq) -> SiF6^2− + 2 H3O+"}] | [{"type":"chemical equation","value":"SiO2"}] | <h1 class="questionTitle" itemprop="name">How react is #SiO_2#?</h1> | null | SiO2(s) + 6 HF(aq) -> SiF6^2− + 2 H3O+ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are few materials as unreactive as silicon dioxide, quartz (or sand). Unlike carbon dioxide, silicon dioxide is a non-molecular substance that features <mathjax>#Si-O#</mathjax> bonds that extend along the lattice. <mathjax>#SiO_2#</mathjax> is VERY inert (i.e. UNREACTIVE). It will react with <mathjax>#HF#</mathjax>, but that's another story. </p>
<p><mathjax>#SiO_2(s) + 6HF(aq) rarr SiF_6^(2-) + 2H_3O^+#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#SiO_2#</mathjax> is fairly unreactive stuff. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There are few materials as unreactive as silicon dioxide, quartz (or sand). Unlike carbon dioxide, silicon dioxide is a non-molecular substance that features <mathjax>#Si-O#</mathjax> bonds that extend along the lattice. <mathjax>#SiO_2#</mathjax> is VERY inert (i.e. UNREACTIVE). It will react with <mathjax>#HF#</mathjax>, but that's another story. </p>
<p><mathjax>#SiO_2(s) + 6HF(aq) rarr SiF_6^(2-) + 2H_3O^+#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#SiO_2#</mathjax> is fairly unreactive stuff. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>There are few materials as unreactive as silicon dioxide, quartz (or sand). Unlike carbon dioxide, silicon dioxide is a non-molecular substance that features <mathjax>#Si-O#</mathjax> bonds that extend along the lattice. <mathjax>#SiO_2#</mathjax> is VERY inert (i.e. UNREACTIVE). It will react with <mathjax>#HF#</mathjax>, but that's another story. </p>
<p><mathjax>#SiO_2(s) + 6HF(aq) rarr SiF_6^(2-) + 2H_3O^+#</mathjax></p></div>
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</article> | How react is #SiO_2#? | null |
1,284 | ac0846ec-6ddd-11ea-ab71-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-calcium-carbonate | CaCO3 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] Calcium Carbonate [IN] default"}] | [{"type":"chemical equation","value":"CaCO3"}] | [{"type":"substance name","value":"Calcium Carbonate"}] | <h1 class="questionTitle" itemprop="name">What is the formula for Calcium Carbonate? </h1> | null | CaCO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Calcium carbonate"#</mathjax> is reasonably insoluble. When you bubble carbon dioxide gas thru lime water (<mathjax>#Ca(OH)_2(aq)#</mathjax>) you see a white precipitate of the carbonate:</p>
<p><mathjax>#Ca(OH)_2(aq) + CO_2(g) rarr CaCO_3(s)darr +H_2O(aq)#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#CaCO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Calcium carbonate"#</mathjax> is reasonably insoluble. When you bubble carbon dioxide gas thru lime water (<mathjax>#Ca(OH)_2(aq)#</mathjax>) you see a white precipitate of the carbonate:</p>
<p><mathjax>#Ca(OH)_2(aq) + CO_2(g) rarr CaCO_3(s)darr +H_2O(aq)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for Calcium Carbonate? </h1>
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<div class="markdown"><p><mathjax>#CaCO_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Calcium carbonate"#</mathjax> is reasonably insoluble. When you bubble carbon dioxide gas thru lime water (<mathjax>#Ca(OH)_2(aq)#</mathjax>) you see a white precipitate of the carbonate:</p>
<p><mathjax>#Ca(OH)_2(aq) + CO_2(g) rarr CaCO_3(s)darr +H_2O(aq)#</mathjax></p></div>
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</article> | What is the formula for Calcium Carbonate? | null |
1,285 | a98be912-6ddd-11ea-a6fe-ccda262736ce | https://socratic.org/questions/720-ml-of-h-2-gas-at-0-degree-c-and-126-6-kpa-is-changed-to-s-t-p-what-will-be-i | 907 mL | start physical_unit 3 4 volume ml qc_end physical_unit 3 4 0 1 volume qc_end physical_unit 3 4 6 8 temperature qc_end physical_unit 3 4 10 11 pressure qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] H2 gas [IN] mL"}] | [{"type":"physical unit","value":"907 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] H2 gas [=] \\pu{720 mL}"},{"type":"physical unit","value":"Temperature1 [OF] H2 gas [=] \\pu{0 degree C}"},{"type":"physical unit","value":"Pressure1 [OF] H2 gas [=] \\pu{126.6 kPa}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">720 mL of #H_2# gas at 0 degree C and 126.6 kPa is changed to S .T. P. What will be its new volume?</h1> | null | 907 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>STP is defined as 0 °C and 1 bar (100 kPa).</p>
<p>Since only the pressure changes, we can use <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> to solve the problem.</p>
<p>Boyle's Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange Boyle's Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_2 = V_1 × P_1/P_2#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#P_1 = "126 kPa"; V_1 = "720 mL"#</mathjax><br/>
<mathjax>#P_2 ="100 kPa"; V_2 = "?"#</mathjax></p>
<p>∴ <mathjax>#V_2 = "720 mL" × (126 color(red)(cancel(color(black)("kPa"))))/(100 color(red)(cancel(color(black)("kPa")))) = "907 mL"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The new volume will be 907 mL.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>STP is defined as 0 °C and 1 bar (100 kPa).</p>
<p>Since only the pressure changes, we can use <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> to solve the problem.</p>
<p>Boyle's Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange Boyle's Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_2 = V_1 × P_1/P_2#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#P_1 = "126 kPa"; V_1 = "720 mL"#</mathjax><br/>
<mathjax>#P_2 ="100 kPa"; V_2 = "?"#</mathjax></p>
<p>∴ <mathjax>#V_2 = "720 mL" × (126 color(red)(cancel(color(black)("kPa"))))/(100 color(red)(cancel(color(black)("kPa")))) = "907 mL"#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">720 mL of #H_2# gas at 0 degree C and 126.6 kPa is changed to S .T. P. What will be its new volume?</h1>
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Ernest Z.
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Jan 10, 2017
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<div class="markdown"><p>The new volume will be 907 mL.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>STP is defined as 0 °C and 1 bar (100 kPa).</p>
<p>Since only the pressure changes, we can use <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a></strong> to solve the problem.</p>
<p>Boyle's Law is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange Boyle's Law to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_2 = V_1 × P_1/P_2#</mathjax></p>
</blockquote>
</blockquote>
<p>In this problem,</p>
<p><mathjax>#P_1 = "126 kPa"; V_1 = "720 mL"#</mathjax><br/>
<mathjax>#P_2 ="100 kPa"; V_2 = "?"#</mathjax></p>
<p>∴ <mathjax>#V_2 = "720 mL" × (126 color(red)(cancel(color(black)("kPa"))))/(100 color(red)(cancel(color(black)("kPa")))) = "907 mL"#</mathjax></p></div>
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</article> | 720 mL of #H_2# gas at 0 degree C and 126.6 kPa is changed to S .T. P. What will be its new volume? | null |
1,286 | aa92d4d3-6ddd-11ea-b796-ccda262736ce | https://socratic.org/questions/how-many-grams-are-in-12-moles-of-helium | 48.04 grams | start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] helium [IN] grams"}] | [{"type":"physical unit","value":"48.04 grams"}] | [{"type":"physical unit","value":"Mole [OF] helium [=] \\pu{12 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams are in 12 moles of helium? </h1> | null | 48.04 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#He#</mathjax> has a molar mass of <mathjax>#4.003*g*mol^-1#</mathjax>.</p>
<p>Given your molar quantity:</p>
<p><mathjax>#"Mass" = 4.003*g*cancel(mol^-1)xx12*cancel(mol)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>Just under <mathjax>#50*g#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#He#</mathjax> has a molar mass of <mathjax>#4.003*g*mol^-1#</mathjax>.</p>
<p>Given your molar quantity:</p>
<p><mathjax>#"Mass" = 4.003*g*cancel(mol^-1)xx12*cancel(mol)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams are in 12 moles of helium? </h1>
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anor277
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<div class="markdown"><p>Just under <mathjax>#50*g#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#He#</mathjax> has a molar mass of <mathjax>#4.003*g*mol^-1#</mathjax>.</p>
<p>Given your molar quantity:</p>
<p><mathjax>#"Mass" = 4.003*g*cancel(mol^-1)xx12*cancel(mol)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??*g#</mathjax></p></div>
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</article> | How many grams are in 12 moles of helium? | null |
1,287 | abb1b494-6ddd-11ea-8ada-ccda262736ce | https://socratic.org/questions/58e8b5627c01490bf211d4fd | 2 Cr^3+ + 2 H2O + 3 H2O2 -> 2 CrO4^2- + 10 H+ | start chemical_equation qc_end chemical_equation 7 7 qc_end chemical_equation 10 10 qc_end substance 12 13 qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] the oxidation"}] | [{"type":"chemical equation","value":"2 Cr^3+ + 2 H2O + 3 H2O2 -> 2 CrO4^2- + 10 H+"}] | [{"type":"chemical equation","value":"Cr^3+"},{"type":"chemical equation","value":"CrO4^2-"},{"type":"substance name","value":"Hydrogen peroxide"},{"type":"other","value":"Using the method of half-equations."}] | <h1 class="questionTitle" itemprop="name">How do we represent the oxidation of #Cr^(3+)# ion to #CrO_4^(2-)# by hydrogen peroxide, using the method of half-equations?</h1> | null | 2 Cr^3+ + 2 H2O + 3 H2O2 -> 2 CrO4^2- + 10 H+ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Chromic chloride"#</mathjax>, <mathjax>#CrCl_3#</mathjax>, is oxidized from <mathjax>#Cr(+III)#</mathjax> to <mathjax>#Cr(+VI)#</mathjax> in chromate:</p>
<p><mathjax>#Cr^(3+) + 4H_2O rarr stackrel(VI+)(Cr)O_4^(2-) +8H^(+)+ 3e^(-) #</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Is this balanced with respect to mass and charge? It is your problem not mine.</p>
<p>Hydrogen peroxide, <mathjax>#H_2O_2#</mathjax>, is reduced from <mathjax>#O^(-I)#</mathjax> to <mathjax>#O^(-II)#</mathjax>. </p>
<p><mathjax>#H_2O_2 +2H^(+) + 2e^(-) rarr 2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Again is the thing balanced?</p>
<p>To give the overall redox equation, we cross-multiply the individual redox equations: <mathjax>#2xx(i) + 3xx(ii)#</mathjax>:</p>
<p><mathjax>#2Cr^(3+) + cancel(8)2H_2O +3H_2O_2 +cancel(6H^(+)) + cancel(6e^(-)) rarr cancel(6H_2O)+2CrO_4^(2-) +cancel(16)10H^(+)+ cancel(6e^(-)) #</mathjax></p>
<p>And then we SUBTRACT the superfluous reagents from each side:</p>
<p><mathjax>#2Cr^(3+) + 2H_2O +3H_2O_2 rarr 2CrO_4^(2-) +10H^(+) #</mathjax></p>
<p>But you wanted the reaction done under basic conditions. And so we add <mathjax>#10xxHO^(-)#</mathjax> to BOTH SIDES OF THE REACTION:</p>
<p><mathjax>#2Cr^(3+) +3H_2O_2 +10HO^(-) rarr 2CrO_4^(2-) +8H_2O #</mathjax></p>
<p>Is this balanced with respect to mass and charge?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>By the method of half-equations............</p>
<p><mathjax>#2Cr^(3+) + 2H_2O +3H_2O_2 rarr 2CrO_4^(2-) +10H^(+) #</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Chromic chloride"#</mathjax>, <mathjax>#CrCl_3#</mathjax>, is oxidized from <mathjax>#Cr(+III)#</mathjax> to <mathjax>#Cr(+VI)#</mathjax> in chromate:</p>
<p><mathjax>#Cr^(3+) + 4H_2O rarr stackrel(VI+)(Cr)O_4^(2-) +8H^(+)+ 3e^(-) #</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Is this balanced with respect to mass and charge? It is your problem not mine.</p>
<p>Hydrogen peroxide, <mathjax>#H_2O_2#</mathjax>, is reduced from <mathjax>#O^(-I)#</mathjax> to <mathjax>#O^(-II)#</mathjax>. </p>
<p><mathjax>#H_2O_2 +2H^(+) + 2e^(-) rarr 2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Again is the thing balanced?</p>
<p>To give the overall redox equation, we cross-multiply the individual redox equations: <mathjax>#2xx(i) + 3xx(ii)#</mathjax>:</p>
<p><mathjax>#2Cr^(3+) + cancel(8)2H_2O +3H_2O_2 +cancel(6H^(+)) + cancel(6e^(-)) rarr cancel(6H_2O)+2CrO_4^(2-) +cancel(16)10H^(+)+ cancel(6e^(-)) #</mathjax></p>
<p>And then we SUBTRACT the superfluous reagents from each side:</p>
<p><mathjax>#2Cr^(3+) + 2H_2O +3H_2O_2 rarr 2CrO_4^(2-) +10H^(+) #</mathjax></p>
<p>But you wanted the reaction done under basic conditions. And so we add <mathjax>#10xxHO^(-)#</mathjax> to BOTH SIDES OF THE REACTION:</p>
<p><mathjax>#2Cr^(3+) +3H_2O_2 +10HO^(-) rarr 2CrO_4^(2-) +8H_2O #</mathjax></p>
<p>Is this balanced with respect to mass and charge?</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do we represent the oxidation of #Cr^(3+)# ion to #CrO_4^(2-)# by hydrogen peroxide, using the method of half-equations?</h1>
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<div class="markdown"><p>By the method of half-equations............</p>
<p><mathjax>#2Cr^(3+) + 2H_2O +3H_2O_2 rarr 2CrO_4^(2-) +10H^(+) #</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Chromic chloride"#</mathjax>, <mathjax>#CrCl_3#</mathjax>, is oxidized from <mathjax>#Cr(+III)#</mathjax> to <mathjax>#Cr(+VI)#</mathjax> in chromate:</p>
<p><mathjax>#Cr^(3+) + 4H_2O rarr stackrel(VI+)(Cr)O_4^(2-) +8H^(+)+ 3e^(-) #</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Is this balanced with respect to mass and charge? It is your problem not mine.</p>
<p>Hydrogen peroxide, <mathjax>#H_2O_2#</mathjax>, is reduced from <mathjax>#O^(-I)#</mathjax> to <mathjax>#O^(-II)#</mathjax>. </p>
<p><mathjax>#H_2O_2 +2H^(+) + 2e^(-) rarr 2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Again is the thing balanced?</p>
<p>To give the overall redox equation, we cross-multiply the individual redox equations: <mathjax>#2xx(i) + 3xx(ii)#</mathjax>:</p>
<p><mathjax>#2Cr^(3+) + cancel(8)2H_2O +3H_2O_2 +cancel(6H^(+)) + cancel(6e^(-)) rarr cancel(6H_2O)+2CrO_4^(2-) +cancel(16)10H^(+)+ cancel(6e^(-)) #</mathjax></p>
<p>And then we SUBTRACT the superfluous reagents from each side:</p>
<p><mathjax>#2Cr^(3+) + 2H_2O +3H_2O_2 rarr 2CrO_4^(2-) +10H^(+) #</mathjax></p>
<p>But you wanted the reaction done under basic conditions. And so we add <mathjax>#10xxHO^(-)#</mathjax> to BOTH SIDES OF THE REACTION:</p>
<p><mathjax>#2Cr^(3+) +3H_2O_2 +10HO^(-) rarr 2CrO_4^(2-) +8H_2O #</mathjax></p>
<p>Is this balanced with respect to mass and charge?</p></div>
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</article> | How do we represent the oxidation of #Cr^(3+)# ion to #CrO_4^(2-)# by hydrogen peroxide, using the method of half-equations? | null |
1,288 | aab42b7a-6ddd-11ea-9aef-ccda262736ce | https://socratic.org/questions/how-much-heat-is-needed-to-raise-the-temperature-of-50-g-of-a-substance-by-15-c- | 0.69 kJ | start physical_unit 14 14 heat_energy kj qc_end physical_unit 14 14 10 11 mass qc_end physical_unit 14 14 16 17 temperature qc_end physical_unit 14 14 23 26 specific_heat qc_end end | [{"type":"physical unit","value":"Needed heat [OF] the substance [IN] kJ"}] | [{"type":"physical unit","value":"0.69 kJ"}] | [{"type":"physical unit","value":"Mass [OF] the substance [=] \\pu{50 g}"},{"type":"physical unit","value":"Raised temperature [OF] the substance [=] \\pu{15 ℃ }"},{"type":"physical unit","value":"Specific heat [OF] the substance [=] \\pu{0.92 J/(g * ℃)}"}] | <h1 class="questionTitle" itemprop="name">How much heat is needed to raise the temperature of 50 g of a substance by 15°C when the specific heat is .92 J/g-°C?</h1> | null | 0.69 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the formula <mathjax>#q=mCΔT#</mathjax></p>
<p>q = heat absorbed or released, in joules (J)<br/>
m = mass<br/>
C = <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity <br/>
ΔT = change in temperature</p>
<p><strong>Plug known values into the formula.</strong></p>
<p><mathjax>#q = 50(.92)(15)#</mathjax><br/>
<mathjax>#q = 690#</mathjax></p>
<p><strong>690 joules of heat energy are needed.</strong> </p></div>
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<div class="markdown"><p>690 J</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the formula <mathjax>#q=mCΔT#</mathjax></p>
<p>q = heat absorbed or released, in joules (J)<br/>
m = mass<br/>
C = <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity <br/>
ΔT = change in temperature</p>
<p><strong>Plug known values into the formula.</strong></p>
<p><mathjax>#q = 50(.92)(15)#</mathjax><br/>
<mathjax>#q = 690#</mathjax></p>
<p><strong>690 joules of heat energy are needed.</strong> </p></div>
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<div class="markdown"><p>690 J</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Use the formula <mathjax>#q=mCΔT#</mathjax></p>
<p>q = heat absorbed or released, in joules (J)<br/>
m = mass<br/>
C = <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity <br/>
ΔT = change in temperature</p>
<p><strong>Plug known values into the formula.</strong></p>
<p><mathjax>#q = 50(.92)(15)#</mathjax><br/>
<mathjax>#q = 690#</mathjax></p>
<p><strong>690 joules of heat energy are needed.</strong> </p></div>
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</article> | How much heat is needed to raise the temperature of 50 g of a substance by 15°C when the specific heat is .92 J/g-°C? | null |
1,289 | a83e100d-6ddd-11ea-8cc4-ccda262736ce | https://socratic.org/questions/a-student-performs-an-experiment-that-produces-5-78-grams-of-silver-chloride-he- | 79.7% | start physical_unit 4 4 percent_yield none qc_end physical_unit 10 11 7 8 mass qc_end physical_unit 10 11 15 16 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"percent yield [OF] the experiment"}] | [{"type":"physical unit","value":"79.7%"}] | [{"type":"physical unit","value":"produced mass1 [OF] silver chloride [=] \\pu{5.78 grams}"},{"type":"physical unit","value":"Mass2 [OF] silver chloride [=] \\pu{7.25 grams}"},{"type":"other","value":"All of the limiting reactant reacted."}] | <h1 class="questionTitle" itemprop="name">A student performs an experiment that produces 5.78 grams of silver chloride. He calculated that 7.25 grams of silver chloride should have precipitated if all of the limiting reactant reacted. What is his percent yield?</h1> | null | 79.7% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/stoichiometry/percent-yield">Percent yield</a> is defined as (Actual Yield/Theoretical Yield) * <mathjax>#100#</mathjax></p>
<p>plug in your variables to get <mathjax>#(5.78g)/(7.25g)#</mathjax> * <mathjax>#100#</mathjax> = 79.7%</p></div>
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<div class="markdown"><p>79.7% yield</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/stoichiometry/percent-yield">Percent yield</a> is defined as (Actual Yield/Theoretical Yield) * <mathjax>#100#</mathjax></p>
<p>plug in your variables to get <mathjax>#(5.78g)/(7.25g)#</mathjax> * <mathjax>#100#</mathjax> = 79.7%</p></div>
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<h1 class="questionTitle" itemprop="name">A student performs an experiment that produces 5.78 grams of silver chloride. He calculated that 7.25 grams of silver chloride should have precipitated if all of the limiting reactant reacted. What is his percent yield?</h1>
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<div class="markdown"><p>79.7% yield</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="http://socratic.org/chemistry/stoichiometry/percent-yield">Percent yield</a> is defined as (Actual Yield/Theoretical Yield) * <mathjax>#100#</mathjax></p>
<p>plug in your variables to get <mathjax>#(5.78g)/(7.25g)#</mathjax> * <mathjax>#100#</mathjax> = 79.7%</p></div>
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</article> | A student performs an experiment that produces 5.78 grams of silver chloride. He calculated that 7.25 grams of silver chloride should have precipitated if all of the limiting reactant reacted. What is his percent yield? | null |
1,290 | a9a3d9d5-6ddd-11ea-9aab-ccda262736ce | https://socratic.org/questions/55862b80581e2a7bae15ca2e | 0.20 g | start physical_unit 2 2 mass g qc_end physical_unit 8 8 6 7 volume qc_end chemical_equation 15 21 qc_end end | [{"type":"physical unit","value":"Mass [OF] Ca [IN] g"}] | [{"type":"physical unit","value":"0.20 g"}] | [{"type":"physical unit","value":"Volume [OF] O2 [=] \\pu{56.8 mL}"},{"type":"chemical equation","value":"2 Ca + O2 -> 2 CaO"}] | <h1 class="questionTitle" itemprop="name">How much #"Ca"# is consumed when #"56.8 mL O"_2"# reacts with calcium to produce #"CaO"#?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"2Ca+O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CaO"#</mathjax></p></div>
</h2>
</div>
</div> | 0.20 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the balanced chemical equation for this reaction</p>
<p><mathjax>#color(red)(2)Ca_((s)) + O_(2(g)) -> 2CaO_((s))#</mathjax></p>
<p>Notice that you have a <mathjax>#color(red)(2):1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between calcium and oxygen. This means that, regardless of how many moles of oxygen reacted, the reaction needed <strong>twice as many moles</strong> of calcium. </p>
<p>Since you're at <strong>STP</strong> conditions, you can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a>. More specifically, you know that at STP conditions, which imply a pressure of <strong>100 kPa</strong> and a temperature of <strong>273.15 K</strong>, <em>1 mole</em> of any ideal gas occupies exactly <strong>22.7 L</strong>.</p>
<p>This means that you can determine how many moles of oxygen reacted by using its volume</p>
<p><mathjax>#56.8cancel("mL") * (1cancel("L"))/(1000cancel("mL")) * "1 mole"/(22.7cancel("L")) = "0.002502 moles"#</mathjax> <mathjax>#O_2#</mathjax></p>
<p>This means that the reaction also consumed </p>
<p><mathjax>#0.002502cancel("moles"O_2) * (color(red)(2)"moles Ca")/(1cancel("mole"O_2)) = "0.005004 moles"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>To get the mass of calcium that contained this many moles, use calcium's molar mass</p>
<p><mathjax>#0.005004cancel("moles") * "40.078 g"/(1cancel("mole")) = "0.2006 g"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the volume of oxygen, the answer will be </p>
<p><mathjax>#m_(Ca) = color(green)("0.201 g")#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>Many textbooks and online sources still list STP conditions as a pressure of 1 atm and temperature of 273.15, which would make the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> equal to 22.4 L, not 22.7 L.</em></p>
<p><em>If this is the value you're supposed to use, simply redo all the calculations using 22.4 L instead of 22.7 L. The answer should come out to be 0.203 g of calcium</em>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>This reaction will consume <strong>0.201 g</strong> of calcium.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the balanced chemical equation for this reaction</p>
<p><mathjax>#color(red)(2)Ca_((s)) + O_(2(g)) -> 2CaO_((s))#</mathjax></p>
<p>Notice that you have a <mathjax>#color(red)(2):1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between calcium and oxygen. This means that, regardless of how many moles of oxygen reacted, the reaction needed <strong>twice as many moles</strong> of calcium. </p>
<p>Since you're at <strong>STP</strong> conditions, you can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a>. More specifically, you know that at STP conditions, which imply a pressure of <strong>100 kPa</strong> and a temperature of <strong>273.15 K</strong>, <em>1 mole</em> of any ideal gas occupies exactly <strong>22.7 L</strong>.</p>
<p>This means that you can determine how many moles of oxygen reacted by using its volume</p>
<p><mathjax>#56.8cancel("mL") * (1cancel("L"))/(1000cancel("mL")) * "1 mole"/(22.7cancel("L")) = "0.002502 moles"#</mathjax> <mathjax>#O_2#</mathjax></p>
<p>This means that the reaction also consumed </p>
<p><mathjax>#0.002502cancel("moles"O_2) * (color(red)(2)"moles Ca")/(1cancel("mole"O_2)) = "0.005004 moles"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>To get the mass of calcium that contained this many moles, use calcium's molar mass</p>
<p><mathjax>#0.005004cancel("moles") * "40.078 g"/(1cancel("mole")) = "0.2006 g"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the volume of oxygen, the answer will be </p>
<p><mathjax>#m_(Ca) = color(green)("0.201 g")#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>Many textbooks and online sources still list STP conditions as a pressure of 1 atm and temperature of 273.15, which would make the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> equal to 22.4 L, not 22.7 L.</em></p>
<p><em>If this is the value you're supposed to use, simply redo all the calculations using 22.4 L instead of 22.7 L. The answer should come out to be 0.203 g of calcium</em>.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How much #"Ca"# is consumed when #"56.8 mL O"_2"# reacts with calcium to produce #"CaO"#?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#"2Ca+O"_2"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CaO"#</mathjax></p></div>
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<div class="markdown"><p>This reaction will consume <strong>0.201 g</strong> of calcium.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by taking a look at the balanced chemical equation for this reaction</p>
<p><mathjax>#color(red)(2)Ca_((s)) + O_(2(g)) -> 2CaO_((s))#</mathjax></p>
<p>Notice that you have a <mathjax>#color(red)(2):1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between calcium and oxygen. This means that, regardless of how many moles of oxygen reacted, the reaction needed <strong>twice as many moles</strong> of calcium. </p>
<p>Since you're at <strong>STP</strong> conditions, you can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas at STP</a>. More specifically, you know that at STP conditions, which imply a pressure of <strong>100 kPa</strong> and a temperature of <strong>273.15 K</strong>, <em>1 mole</em> of any ideal gas occupies exactly <strong>22.7 L</strong>.</p>
<p>This means that you can determine how many moles of oxygen reacted by using its volume</p>
<p><mathjax>#56.8cancel("mL") * (1cancel("L"))/(1000cancel("mL")) * "1 mole"/(22.7cancel("L")) = "0.002502 moles"#</mathjax> <mathjax>#O_2#</mathjax></p>
<p>This means that the reaction also consumed </p>
<p><mathjax>#0.002502cancel("moles"O_2) * (color(red)(2)"moles Ca")/(1cancel("mole"O_2)) = "0.005004 moles"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>To get the mass of calcium that contained this many moles, use calcium's molar mass</p>
<p><mathjax>#0.005004cancel("moles") * "40.078 g"/(1cancel("mole")) = "0.2006 g"#</mathjax> <mathjax>#Ca#</mathjax></p>
<p>Rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you gave for the volume of oxygen, the answer will be </p>
<p><mathjax>#m_(Ca) = color(green)("0.201 g")#</mathjax></p>
<p><strong>SIDE NOTE</strong> <em>Many textbooks and online sources still list STP conditions as a pressure of 1 atm and temperature of 273.15, which would make the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a> equal to 22.4 L, not 22.7 L.</em></p>
<p><em>If this is the value you're supposed to use, simply redo all the calculations using 22.4 L instead of 22.7 L. The answer should come out to be 0.203 g of calcium</em>.</p></div>
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Meave60
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<div class="markdown"><p><mathjax>#0.203"g Ca"#</mathjax> are consumed when <mathjax>#56.8"mL O"_2#</mathjax> reacts with calcium to produce calcium oxide, <mathjax>#"CaO"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"2Ca+O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CaO"#</mathjax></p>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio of <mathjax>#"Ca"#</mathjax> to <mathjax>#"O"_2#</mathjax> is <mathjax>#(2 "mol Ca")/(1 "mol O"_2")#</mathjax>.</p>
<p>In order to solve this problem, we need to convert the volume of oxygen to liters. Then convert the volume to moles using the <a href="http://socratic.org/chemistry/the-behavior-of-gases/molar-volume-of-a-gas-224-l-at-stp">molar volume of a gas</a>. Next we will multiply the moles of oxygen times the mole ratio from the equation to get moles of calcium. Finally, we will multiply the moles of calcium times its molar mass to get the mass of calcium consumed.</p>
<p><strong>Convert volume to liters.</strong></p>
<p><mathjax>#56.8color(red)cancel(color(black)("mL O"_2))xx("1L O"_2)/(1000color(red)cancel(color(black)("mL O"_2")))#</mathjax><mathjax>#=0.0568"mol O"_2"#</mathjax></p>
<p><strong>Convert volume in liters to moles.</strong></p>
<p>The volume of one mole of a gas is its molar volume. At STP the molar volume of an ideal gas is <mathjax>#22.414"L/mol"#</mathjax>.</p>
<p><mathjax>#0.0568color(red)cancel(color(black)("L O"_2))xx(1"mol O"_2)/(22.414color(red)cancel(color(black)("L O"_2)))=0.0025341"mol O"_2#</mathjax></p>
<p><strong>Determine the moles of Ca consumed.</strong></p>
<p>Multiply the moles of <mathjax>#"O"_2#</mathjax> times the previously determined mole ratio between calcium and oxygen.</p>
<p><mathjax>#0.0025341color(red)cancel(color(black)("mol O"_2))xx("2mol Ca")/(1color(red)cancel(color(black)("mol O"_2)))=0.0050682"mol Ca"#</mathjax></p>
<p><strong>Convert moles of Ca consumed to mass of Ca consumed.</strong></p>
<p>We need the molar mass of <mathjax>#"Ca"#</mathjax>, which is <mathjax>#40.078"g/mol"#</mathjax>. (This is its atomic weight on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in grams/mole.)</p>
<p><mathjax>#0.0050682color(red)cancel(color(black)("mol Ca"))xx(40.078"g Ca")/(1color(red)cancel(color(black)("mol Ca")))=0.203"g Ca"#</mathjax> (rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a>)</p></div>
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</article> | How much #"Ca"# is consumed when #"56.8 mL O"_2"# reacts with calcium to produce #"CaO"#? |
#"2Ca+O"_2"##rarr##"2CaO"#
|
1,291 | ac90bff5-6ddd-11ea-aa18-ccda262736ce | https://socratic.org/questions/how-do-you-balance-lial-oh-4-h-2o-lioh-al-oh-3-h-2o | LiAl(OH)4 + H2O -> LiOH + Al(OH)3 + H2O | start chemical_equation qc_end chemical_equation 4 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"LiAl(OH)4 + H2O -> LiOH + Al(OH)3 + H2O"}] | [{"type":"chemical equation","value":"LiAl(OH)4 + H2O -> LiOH + Al(OH)3 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance #LiAl(OH)_4 + H_2O -> LiOH + Al(OH)_3 + H_2O#?</h1> | null | LiAl(OH)4 + H2O -> LiOH + Al(OH)3 + H2O | <div class="answerDescription">
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<div class="markdown"><p>It's already balanced. Here, let's see what happens when we check what's on each side.</p>
<blockquote>
<p><mathjax>#"LiAl"("OH")_4 + cancel("H"_2"O") -> "LiOH" + "Al"("OH")_3 + cancel("H"_2"O")#</mathjax></p>
<p><mathjax>#color(green)("Li")color(highlight)("Al")color(blue)(("OH")_4) -> color(green)("Li")color(blue)("OH") + color(highlight)("Al")color(blue)(("OH")_3)#</mathjax></p>
</blockquote>
<p>Yep, it's fine the way it is. One lithium on each side, one aluminum on each side, and four <mathjax>#"OH"#</mathjax> groups on each side. </p>
<p>As for the charges, <mathjax>#"Al"("OH")_4^(-)#</mathjax> is balanced by <mathjax>#"Li"^(+)#</mathjax>, <mathjax>#"OH"^(-)#</mathjax> is balanced by <mathjax>#"Li"^(+)#</mathjax>, and <mathjax>#3xx"OH"^(-)#</mathjax> is balanced by <mathjax>#"Al"^(3+)#</mathjax>.</p></div>
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<div class="markdown"><p>It's already balanced. Here, let's see what happens when we check what's on each side.</p>
<blockquote>
<p><mathjax>#"LiAl"("OH")_4 + cancel("H"_2"O") -> "LiOH" + "Al"("OH")_3 + cancel("H"_2"O")#</mathjax></p>
<p><mathjax>#color(green)("Li")color(highlight)("Al")color(blue)(("OH")_4) -> color(green)("Li")color(blue)("OH") + color(highlight)("Al")color(blue)(("OH")_3)#</mathjax></p>
</blockquote>
<p>Yep, it's fine the way it is. One lithium on each side, one aluminum on each side, and four <mathjax>#"OH"#</mathjax> groups on each side. </p>
<p>As for the charges, <mathjax>#"Al"("OH")_4^(-)#</mathjax> is balanced by <mathjax>#"Li"^(+)#</mathjax>, <mathjax>#"OH"^(-)#</mathjax> is balanced by <mathjax>#"Li"^(+)#</mathjax>, and <mathjax>#3xx"OH"^(-)#</mathjax> is balanced by <mathjax>#"Al"^(3+)#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #LiAl(OH)_4 + H_2O -> LiOH + Al(OH)_3 + H_2O#?</h1>
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Truong-Son N.
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<div class="markdown"><p>It's already balanced. Here, let's see what happens when we check what's on each side.</p>
<blockquote>
<p><mathjax>#"LiAl"("OH")_4 + cancel("H"_2"O") -> "LiOH" + "Al"("OH")_3 + cancel("H"_2"O")#</mathjax></p>
<p><mathjax>#color(green)("Li")color(highlight)("Al")color(blue)(("OH")_4) -> color(green)("Li")color(blue)("OH") + color(highlight)("Al")color(blue)(("OH")_3)#</mathjax></p>
</blockquote>
<p>Yep, it's fine the way it is. One lithium on each side, one aluminum on each side, and four <mathjax>#"OH"#</mathjax> groups on each side. </p>
<p>As for the charges, <mathjax>#"Al"("OH")_4^(-)#</mathjax> is balanced by <mathjax>#"Li"^(+)#</mathjax>, <mathjax>#"OH"^(-)#</mathjax> is balanced by <mathjax>#"Li"^(+)#</mathjax>, and <mathjax>#3xx"OH"^(-)#</mathjax> is balanced by <mathjax>#"Al"^(3+)#</mathjax>.</p></div>
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anor277
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<div class="markdown"><p>Do you mean the reaction of lithium tetrahydroaluminate (<mathjax>#LiAlH_4#</mathjax>) with water?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#LiAlH_4(s) + 4H_2O(l) rarr LiAl(OH)_4(aq) + 4H_2(g)uarr#</mathjax></p>
<p>Lithium tetrahydroaluminate is an important hydride transfer reagent, and a very common reductant in organic chemistry. The reduction reaction is performed in THF (<mathjax>#C_4H_8O#</mathjax>) or ether (both <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> must be dry). </p>
<p><mathjax>#LiAlH_4#</mathjax> can transfer up to 4 hydrides (with commercial grades usually it transfers about 3). When your organic reduction is finished, the mixture is worked up with water. This can get pretty violent (especially if the lithal has been added 1:1, as is quite commonly done).</p></div>
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</article> | How do you balance #LiAl(OH)_4 + H_2O -> LiOH + Al(OH)_3 + H_2O#? | null |
1,292 | acbede4c-6ddd-11ea-bb42-ccda262736ce | https://socratic.org/questions/a-solution-contains-0-274-m-sodium-hypochlorite-and-0-146-m-hypochlorous-acid-wh | 7.73 | start physical_unit 17 18 ph none qc_end physical_unit 5 6 3 4 molarity qc_end physical_unit 10 11 8 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"7.73"}] | [{"type":"physical unit","value":"Molarity [OF] sodium hypochlorite solution [=] \\pu{0.274 M}"},{"type":"physical unit","value":"Molarity [OF] hypochlorous acid solution [=] \\pu{0.146 M}"}] | <h1 class="questionTitle" itemprop="name">A solution contains 0.274 M sodium hypochlorite and 0.146 M hypochlorous acid. What is the pH of the solution?</h1> | null | 7.73 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice for this problem will be the <strong>Henderson - Hasselbalch equation</strong>, which, for a <a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a> that consists of a <em>weak acid</em> and its <em>conjugate base</em>, allows you to calculate buffer <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> by </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))#</mathjax></p>
</blockquote>
<p>In your case, the buffer consists of <em>hypochlorous acid</em>, <mathjax>#"HClO"#</mathjax>, a weak acid, and its conjugate base, the <em>hypochlorite anion</em>, <mathjax>#"ClO"^(-)#</mathjax>, its conjugate base. </p>
<p>The hypochlorite anions are deliver to the solution via sodium hypochlorite, <mathjax>#"NaClO"#</mathjax>, a salt of the hypochlorite ion.</p>
<blockquote>
<p><mathjax>#"NaClO"_text((aq]) -> "Na"_text((aq])^(+) + "ClO"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>The salt dissociates in a <mathjax>#1:1#</mathjax> ratio with the hypochlorite ion, which means that you have</p>
<blockquote>
<p><mathjax>#["NaClO"] = ["ClO"^(-)]#</mathjax></p>
</blockquote>
<p>You can find hypochlorous acid's <mathjax>#pK_a#</mathjax> here</p>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>So, before plugging in your values into the H - H equation, try to predict what you expect the result to be. </p>
<p>Notice that at <strong>equal concentrations</strong> of weak acid and conjugate base, the pH of the buffer is equal to the acid's <mathjax>#pK_a#</mathjax>. </p>
<p>In your case, the concentration of conjugate base is <strong>higher</strong> than the concentration of the weak acid, which means that the pH of the solution will be a little <strong>higher</strong> than the acid's <mathjax>#pK_a#</mathjax>. </p>
<p>So, the pH of the buffer will be equal to </p>
<blockquote>
<p><mathjax>#"pH" = 7.46 + log( (0.274color(red)(cancel(color(black)("M"))))/(0.146color(red)(cancel(color(black)("M"))))) = color(green)(7.73)#</mathjax></p>
</blockquote>
<p>Indeed, at these concentrations of weak acid and conjugate base, the buffer's pH is higher than the acid's <mathjax>#pK_a#</mathjax>.</p>
<p>
<iframe src="https://www.youtube.com/embed/Ub7eLn6ddvg?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div class="markdown"><p><mathjax>#7.73#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice for this problem will be the <strong>Henderson - Hasselbalch equation</strong>, which, for a <a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a> that consists of a <em>weak acid</em> and its <em>conjugate base</em>, allows you to calculate buffer <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> by </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))#</mathjax></p>
</blockquote>
<p>In your case, the buffer consists of <em>hypochlorous acid</em>, <mathjax>#"HClO"#</mathjax>, a weak acid, and its conjugate base, the <em>hypochlorite anion</em>, <mathjax>#"ClO"^(-)#</mathjax>, its conjugate base. </p>
<p>The hypochlorite anions are deliver to the solution via sodium hypochlorite, <mathjax>#"NaClO"#</mathjax>, a salt of the hypochlorite ion.</p>
<blockquote>
<p><mathjax>#"NaClO"_text((aq]) -> "Na"_text((aq])^(+) + "ClO"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>The salt dissociates in a <mathjax>#1:1#</mathjax> ratio with the hypochlorite ion, which means that you have</p>
<blockquote>
<p><mathjax>#["NaClO"] = ["ClO"^(-)]#</mathjax></p>
</blockquote>
<p>You can find hypochlorous acid's <mathjax>#pK_a#</mathjax> here</p>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>So, before plugging in your values into the H - H equation, try to predict what you expect the result to be. </p>
<p>Notice that at <strong>equal concentrations</strong> of weak acid and conjugate base, the pH of the buffer is equal to the acid's <mathjax>#pK_a#</mathjax>. </p>
<p>In your case, the concentration of conjugate base is <strong>higher</strong> than the concentration of the weak acid, which means that the pH of the solution will be a little <strong>higher</strong> than the acid's <mathjax>#pK_a#</mathjax>. </p>
<p>So, the pH of the buffer will be equal to </p>
<blockquote>
<p><mathjax>#"pH" = 7.46 + log( (0.274color(red)(cancel(color(black)("M"))))/(0.146color(red)(cancel(color(black)("M"))))) = color(green)(7.73)#</mathjax></p>
</blockquote>
<p>Indeed, at these concentrations of weak acid and conjugate base, the buffer's pH is higher than the acid's <mathjax>#pK_a#</mathjax>.</p>
<p>
<iframe src="https://www.youtube.com/embed/Ub7eLn6ddvg?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<h1 class="questionTitle" itemprop="name">A solution contains 0.274 M sodium hypochlorite and 0.146 M hypochlorous acid. What is the pH of the solution?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#7.73#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice for this problem will be the <strong>Henderson - Hasselbalch equation</strong>, which, for a <a href="http://socratic.org/chemistry/reactions-in-solution/buffer-theory">buffer solution</a> that consists of a <em>weak acid</em> and its <em>conjugate base</em>, allows you to calculate buffer <a href="http://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> by </p>
<blockquote>
<p><mathjax>#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))#</mathjax></p>
</blockquote>
<p>In your case, the buffer consists of <em>hypochlorous acid</em>, <mathjax>#"HClO"#</mathjax>, a weak acid, and its conjugate base, the <em>hypochlorite anion</em>, <mathjax>#"ClO"^(-)#</mathjax>, its conjugate base. </p>
<p>The hypochlorite anions are deliver to the solution via sodium hypochlorite, <mathjax>#"NaClO"#</mathjax>, a salt of the hypochlorite ion.</p>
<blockquote>
<p><mathjax>#"NaClO"_text((aq]) -> "Na"_text((aq])^(+) + "ClO"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p>The salt dissociates in a <mathjax>#1:1#</mathjax> ratio with the hypochlorite ion, which means that you have</p>
<blockquote>
<p><mathjax>#["NaClO"] = ["ClO"^(-)]#</mathjax></p>
</blockquote>
<p>You can find hypochlorous acid's <mathjax>#pK_a#</mathjax> here</p>
<p><a href="http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf" rel="nofollow" target="_blank">http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf</a></p>
<p>So, before plugging in your values into the H - H equation, try to predict what you expect the result to be. </p>
<p>Notice that at <strong>equal concentrations</strong> of weak acid and conjugate base, the pH of the buffer is equal to the acid's <mathjax>#pK_a#</mathjax>. </p>
<p>In your case, the concentration of conjugate base is <strong>higher</strong> than the concentration of the weak acid, which means that the pH of the solution will be a little <strong>higher</strong> than the acid's <mathjax>#pK_a#</mathjax>. </p>
<p>So, the pH of the buffer will be equal to </p>
<blockquote>
<p><mathjax>#"pH" = 7.46 + log( (0.274color(red)(cancel(color(black)("M"))))/(0.146color(red)(cancel(color(black)("M"))))) = color(green)(7.73)#</mathjax></p>
</blockquote>
<p>Indeed, at these concentrations of weak acid and conjugate base, the buffer's pH is higher than the acid's <mathjax>#pK_a#</mathjax>.</p>
<p>
<iframe src="https://www.youtube.com/embed/Ub7eLn6ddvg?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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</article> | A solution contains 0.274 M sodium hypochlorite and 0.146 M hypochlorous acid. What is the pH of the solution? | null |
1,293 | acbee752-6ddd-11ea-893e-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-10-2m-hcl-solution | 2.00 | start physical_unit 7 8 ph none qc_end physical_unit 7 8 5 6 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] HCl solution"}] | [{"type":"physical unit","value":"2.00"}] | [{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{10^(-2) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of #10^(-2)# #"M HCl"# solution?</h1> | null | 2.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Because</p>
<p><mathjax>#["H"^(+)]= 10^-"pH"#</mathjax></p>
<p>Therefore, the exponent of <mathjax>#10#</mathjax> is the pH and this makes sense because the <mathjax>#"pH"#</mathjax> <mathjax>#(2)#</mathjax> is less than <mathjax>#7#</mathjax>, which indicates that it should be an acid. This is true because <mathjax>#"HCl"#</mathjax> is a strong acid, i.e. </p>
<p><mathjax>#["H"^(+)] = ["HCl"]#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Because</p>
<p><mathjax>#["H"^(+)]= 10^-"pH"#</mathjax></p>
<p>Therefore, the exponent of <mathjax>#10#</mathjax> is the pH and this makes sense because the <mathjax>#"pH"#</mathjax> <mathjax>#(2)#</mathjax> is less than <mathjax>#7#</mathjax>, which indicates that it should be an acid. This is true because <mathjax>#"HCl"#</mathjax> is a strong acid, i.e. </p>
<p><mathjax>#["H"^(+)] = ["HCl"]#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the pH of #10^(-2)# #"M HCl"# solution?</h1>
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Queen
</a>·<a class="topContributorLink" href="/users/stefan-zdre">
Stefan V.
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<span class="dateCreated" datetime="2018-04-06T00:54:50" itemprop="dateCreated">
Apr 6, 2018
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<div class="markdown"><p><mathjax>#2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Because</p>
<p><mathjax>#["H"^(+)]= 10^-"pH"#</mathjax></p>
<p>Therefore, the exponent of <mathjax>#10#</mathjax> is the pH and this makes sense because the <mathjax>#"pH"#</mathjax> <mathjax>#(2)#</mathjax> is less than <mathjax>#7#</mathjax>, which indicates that it should be an acid. This is true because <mathjax>#"HCl"#</mathjax> is a strong acid, i.e. </p>
<p><mathjax>#["H"^(+)] = ["HCl"]#</mathjax></p></div>
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Nam D.
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<span class="dateCreated" datetime="2018-04-06T11:33:37" itemprop="dateCreated">
Apr 6, 2018
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<div class="markdown"><p><mathjax>#2#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hydrochloric acid is a strong acid which dissociates into single hydrogen <mathjax>#(H^+)#</mathjax> ions in an aqueous solution. </p>
<p>The <mathjax>#"pH"#</mathjax> of a solution is given by the equation,</p>
<p><mathjax>#"pH"=-log[H^+]#</mathjax></p>
<blockquote>
<ul>
<li><mathjax>#[H^+]#</mathjax> is the hydrogen ion concentration in terms of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></li>
</ul>
</blockquote>
<p>And so, we got:</p>
<p><mathjax>#"pH"=-log[10^-2]#</mathjax></p>
<p><mathjax>#=2#</mathjax></p></div>
</div>
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</article> | What is the pH of #10^(-2)# #"M HCl"# solution? | null |
1,294 | abcc2c4e-6ddd-11ea-afdb-ccda262736ce | https://socratic.org/questions/how-do-you-balance-h2-o2-h2o | 2 H2 + O2 -> 2 H2O | start chemical_equation qc_end chemical_equation 4 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 H2 + O2 -> 2 H2O"}] | [{"type":"chemical equation","value":"H2 + O2 -> H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance H2+O2-->H2O?
</h1> | null | 2 H2 + O2 -> 2 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With <mathjax>#O_2#</mathjax> on the left side the right side will need to have an even multiple of <mathjax>#H_2O#</mathjax> to balance the oxygen.</p>
<p>This means that there will be 4 times some multiple of hydrogen on the right side.</p>
<p>The simplest way to to have 4 times a multiple of hydrogen from the let side using <mathjax>#H_2#</mathjax> is with <mathjax>#2H_2#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2H_2 + O_2 rarr 2H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With <mathjax>#O_2#</mathjax> on the left side the right side will need to have an even multiple of <mathjax>#H_2O#</mathjax> to balance the oxygen.</p>
<p>This means that there will be 4 times some multiple of hydrogen on the right side.</p>
<p>The simplest way to to have 4 times a multiple of hydrogen from the let side using <mathjax>#H_2#</mathjax> is with <mathjax>#2H_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance H2+O2-->H2O?
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Alan P.
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<div class="markdown"><p><mathjax>#2H_2 + O_2 rarr 2H_2O#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With <mathjax>#O_2#</mathjax> on the left side the right side will need to have an even multiple of <mathjax>#H_2O#</mathjax> to balance the oxygen.</p>
<p>This means that there will be 4 times some multiple of hydrogen on the right side.</p>
<p>The simplest way to to have 4 times a multiple of hydrogen from the let side using <mathjax>#H_2#</mathjax> is with <mathjax>#2H_2#</mathjax></p></div>
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</article> | How do you balance H2+O2-->H2O?
| null |
1,295 | ab18fccc-6ddd-11ea-8893-ccda262736ce | https://socratic.org/questions/an-ideal-gas-has-a-volume-of-2-28-l-at-279-k-and-1-07-atm-what-is-the-pressure-w | 2.61 atm | start physical_unit 1 2 pressure atm qc_end physical_unit 1 2 7 8 volume qc_end physical_unit 1 2 10 11 temperature qc_end physical_unit 1 2 13 14 pressure qc_end physical_unit 1 2 23 24 volume qc_end physical_unit 1 2 29 30 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the ideal gas [IN] atm"}] | [{"type":"physical unit","value":"2.61 atm"}] | [{"type":"physical unit","value":"Volume1 [OF] the ideal gas [=] \\pu{2.28 L}"},{"type":"physical unit","value":"Temperature1 [OF] the ideal gas [=] \\pu{279 K}"},{"type":"physical unit","value":"Pressure1 [OF] the ideal gas [=] \\pu{1.07 atm}"},{"type":"physical unit","value":"Volume2 [OF] the ideal gas [=] \\pu{1.03 L}"},{"type":"physical unit","value":"Temperature2 [OF] the ideal gas [=] \\pu{307 K}"}] | <h1 class="questionTitle" itemprop="name">An ideal gas has a volume of 2.28 L at 279 K and 1.07 atm. What is the pressure when the volume is 1.03 L and the temperature is 307 K?</h1> | null | 2.61 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we use the first set of data to calculate the <strong>number of moles</strong> using the <strong>ideal gas equation</strong>:</p>
<p><mathjax>#"pV " = " nRT"#</mathjax></p>
<p>Where:</p>
<ul>
<li><mathjax>#"p"#</mathjax> is <strong>pressure</strong> in <strong>pascals</strong> (<mathjax>#"Pa"#</mathjax>)</li>
<li><mathjax>#"V"#</mathjax> is <strong>volume</strong> in <strong>cubic metres</strong> ( <mathjax>#"m"^3#</mathjax>)</li>
<li><mathjax>#"n"#</mathjax> is the <strong>number of moles</strong></li>
<li><mathjax>#"R"#</mathjax> is the <strong>gas constant</strong> = <mathjax>#8.314#</mathjax></li>
<li><mathjax>#"T"#</mathjax> is the <strong>temperature</strong> in <strong>Kelvin</strong> (<mathjax>#"K"#</mathjax>)</li>
</ul>
<p>First, convert your given values into workable units:</p>
<ul>
<li><mathjax>#1"L" = 0.001"m"^3, :. 2.28"L" = 0.00228"m"^3#</mathjax></li>
<li><mathjax>#1"atm" = 101325"Pa", :. 1.07"atm" = 108417.8"Pa"#</mathjax></li>
</ul>
<p>Second, rearrange the equation to solve for <strong>moles</strong>:</p>
<p><mathjax>#"n "=" ""pV"/"RT"#</mathjax></p>
<p>Next, substitute in your given values and calculate the number of moles:</p>
<p><mathjax>#"n "=" "(108417.8"Pa" * 0.00228"m"^3)/(8.314 * 279"K")#</mathjax></p>
<p><mathjax>#"n "=" "0.1065color(red)(666)255" moles"#</mathjax></p>
<p>We can then move onto calculating the new pressure value. The first thing to do here is to, again, convert non-compliant units into ones that are accepted by the equation:</p>
<ul>
<li><mathjax>#1"L" = 0.001"m"^3, :. 1.03"L" = 0.00103"m"^3#</mathjax></li>
</ul>
<p>Then we rearrange the equation to solve for <strong>pressure</strong>:</p>
<p><mathjax>#"p "=" ""nRT"/"V"#</mathjax></p>
<p>And substituting in our values, we get:</p>
<p><mathjax>#"p "=(0.1065666255*8.314 * 307"K")/(0.00103"m"^3)#</mathjax></p>
<p><mathjax>#"p "=264078.0989" Pa" = 2.61" atm to 3 significant figures"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"p"= 2.61" atm to 3 significant figures"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we use the first set of data to calculate the <strong>number of moles</strong> using the <strong>ideal gas equation</strong>:</p>
<p><mathjax>#"pV " = " nRT"#</mathjax></p>
<p>Where:</p>
<ul>
<li><mathjax>#"p"#</mathjax> is <strong>pressure</strong> in <strong>pascals</strong> (<mathjax>#"Pa"#</mathjax>)</li>
<li><mathjax>#"V"#</mathjax> is <strong>volume</strong> in <strong>cubic metres</strong> ( <mathjax>#"m"^3#</mathjax>)</li>
<li><mathjax>#"n"#</mathjax> is the <strong>number of moles</strong></li>
<li><mathjax>#"R"#</mathjax> is the <strong>gas constant</strong> = <mathjax>#8.314#</mathjax></li>
<li><mathjax>#"T"#</mathjax> is the <strong>temperature</strong> in <strong>Kelvin</strong> (<mathjax>#"K"#</mathjax>)</li>
</ul>
<p>First, convert your given values into workable units:</p>
<ul>
<li><mathjax>#1"L" = 0.001"m"^3, :. 2.28"L" = 0.00228"m"^3#</mathjax></li>
<li><mathjax>#1"atm" = 101325"Pa", :. 1.07"atm" = 108417.8"Pa"#</mathjax></li>
</ul>
<p>Second, rearrange the equation to solve for <strong>moles</strong>:</p>
<p><mathjax>#"n "=" ""pV"/"RT"#</mathjax></p>
<p>Next, substitute in your given values and calculate the number of moles:</p>
<p><mathjax>#"n "=" "(108417.8"Pa" * 0.00228"m"^3)/(8.314 * 279"K")#</mathjax></p>
<p><mathjax>#"n "=" "0.1065color(red)(666)255" moles"#</mathjax></p>
<p>We can then move onto calculating the new pressure value. The first thing to do here is to, again, convert non-compliant units into ones that are accepted by the equation:</p>
<ul>
<li><mathjax>#1"L" = 0.001"m"^3, :. 1.03"L" = 0.00103"m"^3#</mathjax></li>
</ul>
<p>Then we rearrange the equation to solve for <strong>pressure</strong>:</p>
<p><mathjax>#"p "=" ""nRT"/"V"#</mathjax></p>
<p>And substituting in our values, we get:</p>
<p><mathjax>#"p "=(0.1065666255*8.314 * 307"K")/(0.00103"m"^3)#</mathjax></p>
<p><mathjax>#"p "=264078.0989" Pa" = 2.61" atm to 3 significant figures"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">An ideal gas has a volume of 2.28 L at 279 K and 1.07 atm. What is the pressure when the volume is 1.03 L and the temperature is 307 K?</h1>
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Owen Bell
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<span class="dateCreated" datetime="2015-12-29T22:19:58" itemprop="dateCreated">
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<div class="markdown"><p><mathjax>#"p"= 2.61" atm to 3 significant figures"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, we use the first set of data to calculate the <strong>number of moles</strong> using the <strong>ideal gas equation</strong>:</p>
<p><mathjax>#"pV " = " nRT"#</mathjax></p>
<p>Where:</p>
<ul>
<li><mathjax>#"p"#</mathjax> is <strong>pressure</strong> in <strong>pascals</strong> (<mathjax>#"Pa"#</mathjax>)</li>
<li><mathjax>#"V"#</mathjax> is <strong>volume</strong> in <strong>cubic metres</strong> ( <mathjax>#"m"^3#</mathjax>)</li>
<li><mathjax>#"n"#</mathjax> is the <strong>number of moles</strong></li>
<li><mathjax>#"R"#</mathjax> is the <strong>gas constant</strong> = <mathjax>#8.314#</mathjax></li>
<li><mathjax>#"T"#</mathjax> is the <strong>temperature</strong> in <strong>Kelvin</strong> (<mathjax>#"K"#</mathjax>)</li>
</ul>
<p>First, convert your given values into workable units:</p>
<ul>
<li><mathjax>#1"L" = 0.001"m"^3, :. 2.28"L" = 0.00228"m"^3#</mathjax></li>
<li><mathjax>#1"atm" = 101325"Pa", :. 1.07"atm" = 108417.8"Pa"#</mathjax></li>
</ul>
<p>Second, rearrange the equation to solve for <strong>moles</strong>:</p>
<p><mathjax>#"n "=" ""pV"/"RT"#</mathjax></p>
<p>Next, substitute in your given values and calculate the number of moles:</p>
<p><mathjax>#"n "=" "(108417.8"Pa" * 0.00228"m"^3)/(8.314 * 279"K")#</mathjax></p>
<p><mathjax>#"n "=" "0.1065color(red)(666)255" moles"#</mathjax></p>
<p>We can then move onto calculating the new pressure value. The first thing to do here is to, again, convert non-compliant units into ones that are accepted by the equation:</p>
<ul>
<li><mathjax>#1"L" = 0.001"m"^3, :. 1.03"L" = 0.00103"m"^3#</mathjax></li>
</ul>
<p>Then we rearrange the equation to solve for <strong>pressure</strong>:</p>
<p><mathjax>#"p "=" ""nRT"/"V"#</mathjax></p>
<p>And substituting in our values, we get:</p>
<p><mathjax>#"p "=(0.1065666255*8.314 * 307"K")/(0.00103"m"^3)#</mathjax></p>
<p><mathjax>#"p "=264078.0989" Pa" = 2.61" atm to 3 significant figures"#</mathjax></p></div>
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</article> | An ideal gas has a volume of 2.28 L at 279 K and 1.07 atm. What is the pressure when the volume is 1.03 L and the temperature is 307 K? | null |
1,296 | abe47e16-6ddd-11ea-b249-ccda262736ce | https://socratic.org/questions/57fc365711ef6b3d72383a0c | 228.00 mL | start physical_unit 3 3 volume ml qc_end physical_unit 13 13 8 9 volume qc_end physical_unit 13 13 11 12 molarity qc_end physical_unit 13 13 16 17 volume qc_end physical_unit 13 13 19 20 molarity qc_end end | [{"type":"physical unit","value":"Volume [OF] water [IN] mL"}] | [{"type":"physical unit","value":"228.00 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] solution [=] \\pu{57 mL}"},{"type":"physical unit","value":"Molarity1 [OF] solution [=] \\pu{0.500 mol/L}"},{"type":"physical unit","value":"Volume2 [OF] solution [=] \\pu{285 mL}"},{"type":"physical unit","value":"Molarity2 [OF] solution [=] \\pu{0.100 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What volume of water must be added to #57*mL# of #0.500*mol*L^-1# solution to make #285*mL# of #0.100*mol*L^-1# solution?</h1> | null | 228.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The product <mathjax>#C_1V_1#</mathjax> gives the starting molar quantity; this is clear from the dimensional product, i.e. <mathjax>#mol*L^-1xxL=mol#</mathjax> as req'd.</p>
<p>From the given relationship, </p>
<p><mathjax>#V_2=(C_1V_1)/C_2=(57.0xx10^-3Lxx0.500*cancel(mol*L^-1))/(0.100*cancel(mol*L^-1))=285*mL#</mathjax></p>
<p>But we already have a <mathjax>#57*mL#</mathjax> volume, so we add appropriately. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#C_1V_1=C_2V_2#</mathjax>, and thus we must add <mathjax>#228*mL#</mathjax> to the mother solution. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The product <mathjax>#C_1V_1#</mathjax> gives the starting molar quantity; this is clear from the dimensional product, i.e. <mathjax>#mol*L^-1xxL=mol#</mathjax> as req'd.</p>
<p>From the given relationship, </p>
<p><mathjax>#V_2=(C_1V_1)/C_2=(57.0xx10^-3Lxx0.500*cancel(mol*L^-1))/(0.100*cancel(mol*L^-1))=285*mL#</mathjax></p>
<p>But we already have a <mathjax>#57*mL#</mathjax> volume, so we add appropriately. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume of water must be added to #57*mL# of #0.500*mol*L^-1# solution to make #285*mL# of #0.100*mol*L^-1# solution?</h1>
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<div class="markdown"><p><mathjax>#C_1V_1=C_2V_2#</mathjax>, and thus we must add <mathjax>#228*mL#</mathjax> to the mother solution. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The product <mathjax>#C_1V_1#</mathjax> gives the starting molar quantity; this is clear from the dimensional product, i.e. <mathjax>#mol*L^-1xxL=mol#</mathjax> as req'd.</p>
<p>From the given relationship, </p>
<p><mathjax>#V_2=(C_1V_1)/C_2=(57.0xx10^-3Lxx0.500*cancel(mol*L^-1))/(0.100*cancel(mol*L^-1))=285*mL#</mathjax></p>
<p>But we already have a <mathjax>#57*mL#</mathjax> volume, so we add appropriately. </p></div>
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</article> | What volume of water must be added to #57*mL# of #0.500*mol*L^-1# solution to make #285*mL# of #0.100*mol*L^-1# solution? | null |
1,297 | a8db04da-6ddd-11ea-bf28-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-9-04-10-26-atoms-of-zinc | 1.50 × 10^3 moles | start physical_unit 10 10 mole mol qc_end end | [{"type":"physical unit","value":"Mole [OF] zinc [IN] moles"}] | [{"type":"physical unit","value":"1.50 × 10^3 moles"}] | [{"type":"physical unit","value":"Number [OF] zinc atoms [=] \\pu{9.04 × 10^26}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in #9.04*10^26# atoms of zinc?</h1> | null | 1.50 × 10^3 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem basically wants you to find a way to <em>convert</em> between the number of atoms present in the sample and the number of <strong>moles</strong> they are equivalent to. </p>
<p>To convert between <em>atoms</em> and <em>moles</em> we use something called <strong>Avogadro's constant</strong>, which basically acts as the definition of a mole.</p>
<p>More specifically, in order to have <strong>one mole</strong> of an element you need <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of that element. You can thus use this number as a conversion factor to take you from atoms to moles or vice versa.</p>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#9.04 * 10^(26)color(red)(cancel(color(black)("atoms Zn"))) * overbrace("1 mole Zn"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Zn")))))^(color(blue)("Avogadro's constant"))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= color(green)(bar(ul(|color(white)(a/a)color(black)(1.50 * 10^3"moles Zn")color(white)(a/a)|)))#</mathjax> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</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>#1.50 * 10^3"moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem basically wants you to find a way to <em>convert</em> between the number of atoms present in the sample and the number of <strong>moles</strong> they are equivalent to. </p>
<p>To convert between <em>atoms</em> and <em>moles</em> we use something called <strong>Avogadro's constant</strong>, which basically acts as the definition of a mole.</p>
<p>More specifically, in order to have <strong>one mole</strong> of an element you need <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of that element. You can thus use this number as a conversion factor to take you from atoms to moles or vice versa.</p>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#9.04 * 10^(26)color(red)(cancel(color(black)("atoms Zn"))) * overbrace("1 mole Zn"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Zn")))))^(color(blue)("Avogadro's constant"))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= color(green)(bar(ul(|color(white)(a/a)color(black)(1.50 * 10^3"moles Zn")color(white)(a/a)|)))#</mathjax> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are in #9.04*10^26# atoms of zinc?</h1>
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Stefan V.
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Oct 19, 2016
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<div class="markdown"><p><mathjax>#1.50 * 10^3"moles"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The problem basically wants you to find a way to <em>convert</em> between the number of atoms present in the sample and the number of <strong>moles</strong> they are equivalent to. </p>
<p>To convert between <em>atoms</em> and <em>moles</em> we use something called <strong>Avogadro's constant</strong>, which basically acts as the definition of a mole.</p>
<p>More specifically, in order to have <strong>one mole</strong> of an element you need <mathjax>#6.022 * 10^(23)#</mathjax> <strong>atoms</strong> of that element. You can thus use this number as a conversion factor to take you from atoms to moles or vice versa.</p>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#9.04 * 10^(26)color(red)(cancel(color(black)("atoms Zn"))) * overbrace("1 mole Zn"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Zn")))))^(color(blue)("Avogadro's constant"))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#= color(green)(bar(ul(|color(white)(a/a)color(black)(1.50 * 10^3"moles Zn")color(white)(a/a)|)))#</mathjax> </p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | How many moles are in #9.04*10^26# atoms of zinc? | null |
1,298 | a9243c68-6ddd-11ea-b211-ccda262736ce | https://socratic.org/questions/100-0-ml-of-a-ph-2-600-solution-of-hydrochloric-acid-is-diluted-with-pure-water--1 | 2.90 | start physical_unit 27 29 ph none qc_end physical_unit 7 7 0 1 volume qc_end substance 14 15 qc_end physical_unit 7 7 21 22 volume qc_end end | [{"type":"physical unit","value":"pH2 [OF] the new solution"}] | [{"type":"physical unit","value":"2.90"}] | [{"type":"physical unit","value":"Volume1 [OF] hydrochloric acid solution [=] \\pu{100.0 mL}"},{"type":"physical unit","value":"pH1 [OF] hydrochloric acid solution [=] \\pu{2.600}"},{"type":"substance name","value":"Pure water"},{"type":"physical unit","value":"Volume2 [OF] hydrochloric acid solution [=] \\pu{200.0 mL}"}] | <h1 class="questionTitle" itemprop="name">100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with
pure water to a new volume of 200.0 mL. Calculate the pH of the new solution?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with<br/>
pure water to a new volume of 200.0 mL. Calculate the pH of the new solution?</p></div>
</h2>
</div>
</div> | 2.90 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>pH is defined as 'minus the logarithm (in base 10) of the concentration of <mathjax>#H^+#</mathjax>'. To find the original concentration of <mathjax>#H^+#</mathjax>, we take the pH, change its sign from positive to negative and raise 10 to that power. </p>
<p>That means the initial concentration of <mathjax>#H^+#</mathjax> (sometimes written as [<mathjax>#H^+#</mathjax>]) is <mathjax>#10^(-2.6)#</mathjax> = <mathjax>#2.51 xx 10^-3#</mathjax> <mathjax>#molL^-1#</mathjax>.</p>
<p>If we double the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> (water) and leave the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> (HCl) constant, the concentration of the solution will halve, because <mathjax>#C=n/V#</mathjax>, concentration equals number of moles of solute divided by volume of solution.</p>
<p>The new concentration of <mathjax>#H^+#</mathjax> will be half of <mathjax>#2.51 xx 10^-3#</mathjax> = <mathjax>#1.26 xx 10^-3#</mathjax> <mathjax>#molL^-1#</mathjax>.</p>
<p>To find the pH of this final solution, we reverse the process. We take the logarithm (to base 10):</p>
<p><mathjax>#log_10(1.26 xx 10^-3)#</mathjax> = -2.90 </p>
<p>Then we change the sign back to positive, and the new pH is 2.90.</p>
<p>(Note: because HCl is a 'strong acid', I have assumed throughout that it is completely dissociated in solution)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use logarithms and powers of 10 to convert between <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> numbers and concentrations of <mathjax>#H^+#</mathjax> in acids.</p>
<p>The pH changes from 2.6 to 2.9. It moves closer to a neutral pH of 7.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>pH is defined as 'minus the logarithm (in base 10) of the concentration of <mathjax>#H^+#</mathjax>'. To find the original concentration of <mathjax>#H^+#</mathjax>, we take the pH, change its sign from positive to negative and raise 10 to that power. </p>
<p>That means the initial concentration of <mathjax>#H^+#</mathjax> (sometimes written as [<mathjax>#H^+#</mathjax>]) is <mathjax>#10^(-2.6)#</mathjax> = <mathjax>#2.51 xx 10^-3#</mathjax> <mathjax>#molL^-1#</mathjax>.</p>
<p>If we double the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> (water) and leave the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> (HCl) constant, the concentration of the solution will halve, because <mathjax>#C=n/V#</mathjax>, concentration equals number of moles of solute divided by volume of solution.</p>
<p>The new concentration of <mathjax>#H^+#</mathjax> will be half of <mathjax>#2.51 xx 10^-3#</mathjax> = <mathjax>#1.26 xx 10^-3#</mathjax> <mathjax>#molL^-1#</mathjax>.</p>
<p>To find the pH of this final solution, we reverse the process. We take the logarithm (to base 10):</p>
<p><mathjax>#log_10(1.26 xx 10^-3)#</mathjax> = -2.90 </p>
<p>Then we change the sign back to positive, and the new pH is 2.90.</p>
<p>(Note: because HCl is a 'strong acid', I have assumed throughout that it is completely dissociated in solution)</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with
pure water to a new volume of 200.0 mL. Calculate the pH of the new solution?</h1>
<div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with<br/>
pure water to a new volume of 200.0 mL. Calculate the pH of the new solution?</p></div>
</h2>
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David G.
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<span class="dateCreated" datetime="2016-09-18T00:00:40" itemprop="dateCreated">
Sep 18, 2016
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<div class="markdown"><p>We use logarithms and powers of 10 to convert between <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> numbers and concentrations of <mathjax>#H^+#</mathjax> in acids.</p>
<p>The pH changes from 2.6 to 2.9. It moves closer to a neutral pH of 7.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>pH is defined as 'minus the logarithm (in base 10) of the concentration of <mathjax>#H^+#</mathjax>'. To find the original concentration of <mathjax>#H^+#</mathjax>, we take the pH, change its sign from positive to negative and raise 10 to that power. </p>
<p>That means the initial concentration of <mathjax>#H^+#</mathjax> (sometimes written as [<mathjax>#H^+#</mathjax>]) is <mathjax>#10^(-2.6)#</mathjax> = <mathjax>#2.51 xx 10^-3#</mathjax> <mathjax>#molL^-1#</mathjax>.</p>
<p>If we double the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a> (water) and leave the amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> (HCl) constant, the concentration of the solution will halve, because <mathjax>#C=n/V#</mathjax>, concentration equals number of moles of solute divided by volume of solution.</p>
<p>The new concentration of <mathjax>#H^+#</mathjax> will be half of <mathjax>#2.51 xx 10^-3#</mathjax> = <mathjax>#1.26 xx 10^-3#</mathjax> <mathjax>#molL^-1#</mathjax>.</p>
<p>To find the pH of this final solution, we reverse the process. We take the logarithm (to base 10):</p>
<p><mathjax>#log_10(1.26 xx 10^-3)#</mathjax> = -2.90 </p>
<p>Then we change the sign back to positive, and the new pH is 2.90.</p>
<p>(Note: because HCl is a 'strong acid', I have assumed throughout that it is completely dissociated in solution)</p></div>
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Stefan V.
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Sep 18, 2016
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<div class="markdown"><p><mathjax>#2.901#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here's a quick way of solving this problem. You know that the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution is given by the <em>concentration</em> of hydronium cations, <mathjax>#"H"_3"O"^(+)#</mathjax>.</p>
<p>When you <strong>double</strong> the volume of the solution, you essentially <strong>halve</strong> the concentration of hydronium cations. This means that if you take <mathjax>#["H"_3"O"^(+)]_0#</mathjax> to be initial concentration of hydronium cations, you will have</p>
<blockquote>
<p><mathjax>#["H"_ 3"O"^(+)]_ "dil" = (["H"_ 3"O"^(+)]_0)/2 ->#</mathjax> <em>the concentration <strong>after the dilution</strong></em></p>
</blockquote>
<p>You know that</p>
<blockquote>
<p><mathjax>#color(purple)(bar(ul(|color(white)(a/a)color(black)("pH" = - log(["H"_3"O"^(+)]))color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You can thus say that <strong>after</strong> the solution is diluted, its pH will be</p>
<blockquote>
<p><mathjax>#"pH"_ "dil" = - log(["H"_ 3"O"^(+)]_"dil")#</mathjax></p>
</blockquote>
<p>This, in turn, is equivalent to </p>
<blockquote>
<p><mathjax>#"pH"_ "dil" = - log((["H"_ 3"O"^(+)]_0)/2)#</mathjax></p>
<p><mathjax>#"pH"_ "dil" = -[log(["H"_ 3"O"^(+)]_ 0) - log2]#</mathjax></p>
<p><mathjax>#"pH"_ "dil" = log2 - log(["H"_3"O"^(+)]_0)#</mathjax></p>
</blockquote>
<p>But since</p>
<blockquote>
<p><mathjax>#"pH"_ 0 = - log(["H"_ 3"O"^(+)]_0) = 2.600#</mathjax></p>
</blockquote>
<p>you will have</p>
<blockquote>
<p><mathjax>#"pH"_ "dil" = log2 + "pH"_ 0#</mathjax></p>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("pH" _ "dil" = 2.600 + log 2 = 2.901)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <em>decimal places</em>.</p>
<p>Finally, does this result make sense?</p>
<p>Diluting the solution means <strong>decreasing</strong> its concentration of hydronium cations, which in turn implies <strong>increasing</strong> the pH of the solution, i.e. making it <em>less acidic</em>. </p></div>
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</article> | 100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with
pure water to a new volume of 200.0 mL. Calculate the pH of the new solution? |
100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with
pure water to a new volume of 200.0 mL. Calculate the pH of the new solution?
|
1,299 | aaec1e26-6ddd-11ea-bc49-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-4-50-10-22-atoms-of-gold-au | 14.72 g | start physical_unit 11 11 mass g qc_end end | [{"type":"physical unit","value":"Mass [OF] Au [IN] g"}] | [{"type":"physical unit","value":"14.72 g"}] | [{"type":"physical unit","value":"Number [OF] Au atoms [=] \\pu{4.50 × 10^22}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of #4.50 * 10^22# atoms of gold, #Au#?</h1> | null | 14.72 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of gold has a mass of <mathjax>#196.97*g*mol^-1#</mathjax>. By definition there are <mathjax>#"Avogadro's number"#</mathjax> of gold atoms that constitute this mass, i.e. <mathjax>#N_A=6.022xx10^23#</mathjax>.</p>
<p>So we take the quotient to get an answer in moles, and mulitply this by the molar mass to get the mass:</p>
<p><mathjax>#(4.50xx10^22" gold atoms")/(6.022xx10^23" gold atoms per mole")xx196.97*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax></p></div>
</div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#(4.50xx10^22" gold atoms")/(6.022xx10^23" gold atoms per mole")xx196.97*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>One mole of gold has a mass of <mathjax>#196.97*g*mol^-1#</mathjax>. By definition there are <mathjax>#"Avogadro's number"#</mathjax> of gold atoms that constitute this mass, i.e. <mathjax>#N_A=6.022xx10^23#</mathjax>.</p>
<p>So we take the quotient to get an answer in moles, and mulitply this by the molar mass to get the mass:</p>
<p><mathjax>#(4.50xx10^22" gold atoms")/(6.022xx10^23" gold atoms per mole")xx196.97*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#(4.50xx10^22" gold atoms")/(6.022xx10^23" gold atoms per mole")xx196.97*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>One mole of gold has a mass of <mathjax>#196.97*g*mol^-1#</mathjax>. By definition there are <mathjax>#"Avogadro's number"#</mathjax> of gold atoms that constitute this mass, i.e. <mathjax>#N_A=6.022xx10^23#</mathjax>.</p>
<p>So we take the quotient to get an answer in moles, and mulitply this by the molar mass to get the mass:</p>
<p><mathjax>#(4.50xx10^22" gold atoms")/(6.022xx10^23" gold atoms per mole")xx196.97*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax></p></div>
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</article> | What is the mass of #4.50 * 10^22# atoms of gold, #Au#? | null |
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