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2,900 | a8a76e68-6ddd-11ea-a363-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-3-5-x-10-3-m-hno-3-solution | 2.46 | start physical_unit 10 11 ph none qc_end physical_unit 10 11 6 9 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] HNO3 solution"}] | [{"type":"physical unit","value":"2.46"}] | [{"type":"physical unit","value":"Molarity [OF] HNO3 solution [=] \\pu{3.5 × 10^(−3) M}"}] | <h1 class="questionTitle" itemprop="name">What is the #"pH"# of a #3.5 * 10^-3# #"M"# #"HNO"_3# solution?</h1> | null | 2.46 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to keep in mind here is that nitric acid is a <strong>strong acid</strong>, which means that it ionizes <em>completely</em> in aqueous solution to produce hydronium cations and nitrate anions.</p>
<blockquote>
<p><mathjax>#"HNO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "NO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>This tells you that the concentration of the hydronium cations in the solution will be <strong>equal</strong> to that of the acid. In other words, you can expect <strong>all the moles</strong> of nitric acid present in your sample to ionize. </p>
<p>You can thus say that you have</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 3.5 * 10^(-3)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p>Now, the <mathjax>#"pH"#</mathjax> of the solution is simply a measure of the concentration of hydronium cations. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>Plug in your value to find </p>
<blockquote>
<p><mathjax>#"pH" = - log(3.5 * 10^(-3))#</mathjax></p>
<p><mathjax>#"pH" = -log(3.5) - log(10^(-3))#</mathjax></p>
<p><mathjax>#"pH" = 3 log(10) - log(3.5)#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 2.46)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <em>decimal places</em> because you have two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> for the concentration of the acid. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 2.46#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to keep in mind here is that nitric acid is a <strong>strong acid</strong>, which means that it ionizes <em>completely</em> in aqueous solution to produce hydronium cations and nitrate anions.</p>
<blockquote>
<p><mathjax>#"HNO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "NO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>This tells you that the concentration of the hydronium cations in the solution will be <strong>equal</strong> to that of the acid. In other words, you can expect <strong>all the moles</strong> of nitric acid present in your sample to ionize. </p>
<p>You can thus say that you have</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 3.5 * 10^(-3)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p>Now, the <mathjax>#"pH"#</mathjax> of the solution is simply a measure of the concentration of hydronium cations. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>Plug in your value to find </p>
<blockquote>
<p><mathjax>#"pH" = - log(3.5 * 10^(-3))#</mathjax></p>
<p><mathjax>#"pH" = -log(3.5) - log(10^(-3))#</mathjax></p>
<p><mathjax>#"pH" = 3 log(10) - log(3.5)#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 2.46)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <em>decimal places</em> because you have two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> for the concentration of the acid. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the #"pH"# of a #3.5 * 10^-3# #"M"# #"HNO"_3# solution?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"pH" = 2.46#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The important thing to keep in mind here is that nitric acid is a <strong>strong acid</strong>, which means that it ionizes <em>completely</em> in aqueous solution to produce hydronium cations and nitrate anions.</p>
<blockquote>
<p><mathjax>#"HNO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "NO"_ (3(aq))^(-)#</mathjax></p>
</blockquote>
<p>This tells you that the concentration of the hydronium cations in the solution will be <strong>equal</strong> to that of the acid. In other words, you can expect <strong>all the moles</strong> of nitric acid present in your sample to ionize. </p>
<p>You can thus say that you have</p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = 3.5 * 10^(-3)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p>Now, the <mathjax>#"pH"#</mathjax> of the solution is simply a measure of the concentration of hydronium cations. </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>Plug in your value to find </p>
<blockquote>
<p><mathjax>#"pH" = - log(3.5 * 10^(-3))#</mathjax></p>
<p><mathjax>#"pH" = -log(3.5) - log(10^(-3))#</mathjax></p>
<p><mathjax>#"pH" = 3 log(10) - log(3.5)#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 2.46)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <em>decimal places</em> because you have two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> for the concentration of the acid. </p></div>
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</article> | What is the #"pH"# of a #3.5 * 10^-3# #"M"# #"HNO"_3# solution? | null |
2,901 | ac42e576-6ddd-11ea-a1db-ccda262736ce | https://socratic.org/questions/i-ve-got-a-car-with-an-internal-volume-of-12-000-l-if-i-drive-my-car-into-the-ri | 8571 L | start physical_unit 28 29 volume l qc_end physical_unit 28 29 9 10 volume qc_end physical_unit 28 29 35 36 pressure qc_end physical_unit 28 29 38 39 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"8571 L"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{12000 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.0 atm}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{1.4 atm}"}] | <h1 class="questionTitle" itemprop="name">I've got a car with an internal volume of #12000# #L#. If I drive my car into the river and it implodes, what will be the volume of the gas when the pressure goes from #1.0# #atm# to #1.4# #atm#?</h1> | null | 8571 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The answer is all above, but let me share a tip. It's possible to memorise Boyle's, Charles' and Gay-Lussac's laws and use them appropriately.</p>
<p>But I find it simpler to just memorise the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a>, and then leave out anything that does not change in a particular question:</p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<p>In this case, since temperature isn't mentioned, we simply leave it out and get Boyle's law.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>: <mathjax>#P_1V_1 = P_2V_2#</mathjax></p>
<p>Rearanging: <mathjax>#V_2 = (P_1V_1)/P_2 =(1.0xx12000)/1.4 = 8571#</mathjax> <mathjax>#L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The answer is all above, but let me share a tip. It's possible to memorise Boyle's, Charles' and Gay-Lussac's laws and use them appropriately.</p>
<p>But I find it simpler to just memorise the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a>, and then leave out anything that does not change in a particular question:</p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<p>In this case, since temperature isn't mentioned, we simply leave it out and get Boyle's law.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">I've got a car with an internal volume of #12000# #L#. If I drive my car into the river and it implodes, what will be the volume of the gas when the pressure goes from #1.0# #atm# to #1.4# #atm#?</h1>
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David G.
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May 21, 2017
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<div class="markdown"><p>We use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>: <mathjax>#P_1V_1 = P_2V_2#</mathjax></p>
<p>Rearanging: <mathjax>#V_2 = (P_1V_1)/P_2 =(1.0xx12000)/1.4 = 8571#</mathjax> <mathjax>#L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The answer is all above, but let me share a tip. It's possible to memorise Boyle's, Charles' and Gay-Lussac's laws and use them appropriately.</p>
<p>But I find it simpler to just memorise the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a>, and then leave out anything that does not change in a particular question:</p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<p>In this case, since temperature isn't mentioned, we simply leave it out and get Boyle's law.</p></div>
</div>
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</div>
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<a href="https://socratic.org/answers/427055" itemprop="url">Answer link</a>
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</article> | I've got a car with an internal volume of #12000# #L#. If I drive my car into the river and it implodes, what will be the volume of the gas when the pressure goes from #1.0# #atm# to #1.4# #atm#? | null |
2,902 | a973c46f-6ddd-11ea-8e76-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-of-all-ions-present-in-the-following-solutions-0-100-m | 3.00 M | start physical_unit 5 6 concentration mol/l qc_end physical_unit 15 15 12 13 mole qc_end physical_unit 19 19 17 18 volume qc_end end | [{"type":"physical unit","value":"Concentration [OF] all ions [IN] M"}] | [{"type":"physical unit","value":"3.00 M"}] | [{"type":"physical unit","value":"Mole [OF] Ca(NO3)2 [=] \\pu{0.100 mole}"},{"type":"physical unit","value":"Volume [OF] solution [=] \\pu{100.0 mL}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of all ions present in the following solutions: 0.100 mole of #Ca(NO_3)_2# in 100.0 mL solution?</h1> | null | 3.00 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here's our concentration for multiplying the solvated ions by.</p>
<p><mathjax>#(.100mol)/(.1000L) = 1.00M#</mathjax></p>
<p><mathjax>#Ca(NO_3)_2#</mathjax> is an ionic compound, thus it will completely dissolve in solution. Therefore,</p>
<p><mathjax>#Ca^(2+)#</mathjax> and <mathjax>#2NO_3^-#</mathjax> will form. </p>
<p><mathjax>#Ca^(2+)#</mathjax>: <mathjax>#1*1 = 1.00M#</mathjax> <br/>
<mathjax>#2NO_3^-#</mathjax>: <mathjax>#2*1 = 2.00M#</mathjax> </p>
<p>Summing all of these ions would give <mathjax>#3.00M#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#3.00M#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here's our concentration for multiplying the solvated ions by.</p>
<p><mathjax>#(.100mol)/(.1000L) = 1.00M#</mathjax></p>
<p><mathjax>#Ca(NO_3)_2#</mathjax> is an ionic compound, thus it will completely dissolve in solution. Therefore,</p>
<p><mathjax>#Ca^(2+)#</mathjax> and <mathjax>#2NO_3^-#</mathjax> will form. </p>
<p><mathjax>#Ca^(2+)#</mathjax>: <mathjax>#1*1 = 1.00M#</mathjax> <br/>
<mathjax>#2NO_3^-#</mathjax>: <mathjax>#2*1 = 2.00M#</mathjax> </p>
<p>Summing all of these ions would give <mathjax>#3.00M#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the concentration of all ions present in the following solutions: 0.100 mole of #Ca(NO_3)_2# in 100.0 mL solution?</h1>
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Al E.
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<div class="markdown"><p><mathjax>#3.00M#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here's our concentration for multiplying the solvated ions by.</p>
<p><mathjax>#(.100mol)/(.1000L) = 1.00M#</mathjax></p>
<p><mathjax>#Ca(NO_3)_2#</mathjax> is an ionic compound, thus it will completely dissolve in solution. Therefore,</p>
<p><mathjax>#Ca^(2+)#</mathjax> and <mathjax>#2NO_3^-#</mathjax> will form. </p>
<p><mathjax>#Ca^(2+)#</mathjax>: <mathjax>#1*1 = 1.00M#</mathjax> <br/>
<mathjax>#2NO_3^-#</mathjax>: <mathjax>#2*1 = 2.00M#</mathjax> </p>
<p>Summing all of these ions would give <mathjax>#3.00M#</mathjax>.</p></div>
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</article> | What is the concentration of all ions present in the following solutions: 0.100 mole of #Ca(NO_3)_2# in 100.0 mL solution? | null |
2,903 | a994ff22-6ddd-11ea-bc04-ccda262736ce | https://socratic.org/questions/whatare-the-molar-concentrations-of-all-ions-in-a-500-ml-saturated-solution-of-s | 1.34 × 10^(-5) M | start physical_unit 6 7 molarity mol/l qc_end physical_unit 13 13 10 11 volume qc_end end | [{"type":"physical unit","value":"Molar concentration [OF] all ions [IN] M"}] | [{"type":"physical unit","value":"1.34 × 10^(-5) M"}] | [{"type":"physical unit","value":"Volume [OF] silver chloride saturated solution [=] \\pu{500 mL}"},{"type":"physical unit","value":"Temperature [OF] silver chloride saturated solution [=] \\pu{25 degrees Celsius}"}] | <h1 class="questionTitle" itemprop="name">Whatare the molar concentrations of all ions in a 500 mL saturated solution of silver chloride at 25 degrees Celsius?</h1> | null | 1.34 × 10^(-5) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to use the <strong>solubility product constant</strong>, <mathjax>#K_(sp)#</mathjax>, of silver chloride to find the compound's <strong>molar solubility</strong> at <mathjax>#25^@"C"#</mathjax>. </p>
<p>You can find the value for the <mathjax>#K_(sp)#</mathjax> of silver chloride here</p>
<p><a href="http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm" rel="nofollow" target="_blank">http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm</a></p>
<p>So, you know that</p>
<blockquote>
<p><mathjax>#K_(sp) = 1.8 * 10^(-10)#</mathjax></p>
</blockquote>
<p>Silver chloride, <mathjax>#"AgCl"#</mathjax>, is considered <strong>insoluble</strong> in aqueous solution, which means that an equilibrium will be established between the undissolved solid and the dissolved ions.</p>
<p>Use an <strong>ICE table</strong> to find the <em>molar solubility</em> of silver chloride, <mathjax>#s#</mathjax>, at this temperature</p>
<blockquote>
<p><mathjax>#" ""AgCl"_text((s]) " "rightleftharpoons" " "Ag"_text((aq])^(+) " "+" " "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "color(white)(a)-" " " " " " " " " "0" " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "color(white)(a)-" " " " " " " "(+s)" " " " " "(+s)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "color(white)(a)-" " " " " " " " " "s" " " " " " " " " "s#</mathjax></p>
<p>By definition, the solubility product constant will be equal to </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Ag"^(+)] * ["Cl"^(-)]#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#1.8 * 10^(-10) = s * s = s^2#</mathjax></p>
</blockquote>
<p>This will give you</p>
<blockquote>
<p><mathjax>#s = sqrt(1.8 * 10^(-10)) = 1.34 * 10^(-5)#</mathjax></p>
</blockquote>
<p>This means that the concentrations of silver cations and of chloride anons in a <strong>saturated</strong> solution of silver chloride will be equal to </p>
<blockquote>
<p><mathjax>#["Ag"^(+)] = ["Cl"^(-)] = s = 1.34 * 10^(-5)"mol L"^(-1)#</mathjax></p>
</blockquote>
<p>Therefore, the concentration of both ions in your <mathjax>#"500-mL"#</mathjax> saturated silver chloride solution will be equal to </p>
<blockquote>
<p><mathjax>#["Ag"^(+)] = ["Cl"^(-)] = color(green)(|bar(ul(color(white)(a/a)1.34 * 10^(-5)"M"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p><strong>Do not</strong> get confused by the fact that you're dealing with a <mathjax>#"500-mL"#</mathjax> solution! </p>
<p>The molar solubility represents <strong>concentration</strong>, not <em>number of moles</em> of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. This means that the concentration of both ions in a saturated silver chloride solution will be equal to <mathjax>#1.34 * 10^(-5)"M"#</mathjax> <strong>regardless</strong> of the <em>volume</em> of said solution! </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1.34 * 10^(-5)"M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to use the <strong>solubility product constant</strong>, <mathjax>#K_(sp)#</mathjax>, of silver chloride to find the compound's <strong>molar solubility</strong> at <mathjax>#25^@"C"#</mathjax>. </p>
<p>You can find the value for the <mathjax>#K_(sp)#</mathjax> of silver chloride here</p>
<p><a href="http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm" rel="nofollow" target="_blank">http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm</a></p>
<p>So, you know that</p>
<blockquote>
<p><mathjax>#K_(sp) = 1.8 * 10^(-10)#</mathjax></p>
</blockquote>
<p>Silver chloride, <mathjax>#"AgCl"#</mathjax>, is considered <strong>insoluble</strong> in aqueous solution, which means that an equilibrium will be established between the undissolved solid and the dissolved ions.</p>
<p>Use an <strong>ICE table</strong> to find the <em>molar solubility</em> of silver chloride, <mathjax>#s#</mathjax>, at this temperature</p>
<blockquote>
<p><mathjax>#" ""AgCl"_text((s]) " "rightleftharpoons" " "Ag"_text((aq])^(+) " "+" " "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "color(white)(a)-" " " " " " " " " "0" " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "color(white)(a)-" " " " " " " "(+s)" " " " " "(+s)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "color(white)(a)-" " " " " " " " " "s" " " " " " " " " "s#</mathjax></p>
<p>By definition, the solubility product constant will be equal to </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Ag"^(+)] * ["Cl"^(-)]#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#1.8 * 10^(-10) = s * s = s^2#</mathjax></p>
</blockquote>
<p>This will give you</p>
<blockquote>
<p><mathjax>#s = sqrt(1.8 * 10^(-10)) = 1.34 * 10^(-5)#</mathjax></p>
</blockquote>
<p>This means that the concentrations of silver cations and of chloride anons in a <strong>saturated</strong> solution of silver chloride will be equal to </p>
<blockquote>
<p><mathjax>#["Ag"^(+)] = ["Cl"^(-)] = s = 1.34 * 10^(-5)"mol L"^(-1)#</mathjax></p>
</blockquote>
<p>Therefore, the concentration of both ions in your <mathjax>#"500-mL"#</mathjax> saturated silver chloride solution will be equal to </p>
<blockquote>
<p><mathjax>#["Ag"^(+)] = ["Cl"^(-)] = color(green)(|bar(ul(color(white)(a/a)1.34 * 10^(-5)"M"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p><strong>Do not</strong> get confused by the fact that you're dealing with a <mathjax>#"500-mL"#</mathjax> solution! </p>
<p>The molar solubility represents <strong>concentration</strong>, not <em>number of moles</em> of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. This means that the concentration of both ions in a saturated silver chloride solution will be equal to <mathjax>#1.34 * 10^(-5)"M"#</mathjax> <strong>regardless</strong> of the <em>volume</em> of said solution! </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Whatare the molar concentrations of all ions in a 500 mL saturated solution of silver chloride at 25 degrees Celsius?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#1.34 * 10^(-5)"M"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your strategy here will be to use the <strong>solubility product constant</strong>, <mathjax>#K_(sp)#</mathjax>, of silver chloride to find the compound's <strong>molar solubility</strong> at <mathjax>#25^@"C"#</mathjax>. </p>
<p>You can find the value for the <mathjax>#K_(sp)#</mathjax> of silver chloride here</p>
<p><a href="http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm" rel="nofollow" target="_blank">http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm</a></p>
<p>So, you know that</p>
<blockquote>
<p><mathjax>#K_(sp) = 1.8 * 10^(-10)#</mathjax></p>
</blockquote>
<p>Silver chloride, <mathjax>#"AgCl"#</mathjax>, is considered <strong>insoluble</strong> in aqueous solution, which means that an equilibrium will be established between the undissolved solid and the dissolved ions.</p>
<p>Use an <strong>ICE table</strong> to find the <em>molar solubility</em> of silver chloride, <mathjax>#s#</mathjax>, at this temperature</p>
<blockquote>
<p><mathjax>#" ""AgCl"_text((s]) " "rightleftharpoons" " "Ag"_text((aq])^(+) " "+" " "Cl"_text((aq])^(-)#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")" " " "color(white)(a)-" " " " " " " " " "0" " " " " " " " " "0#</mathjax><br/>
<mathjax>#color(purple)("C")" " " "color(white)(a)-" " " " " " " "(+s)" " " " " "(+s)#</mathjax><br/>
<mathjax>#color(purple)("E")" " " "color(white)(a)-" " " " " " " " " "s" " " " " " " " " "s#</mathjax></p>
<p>By definition, the solubility product constant will be equal to </p>
<blockquote>
<p><mathjax>#K_(sp) = ["Ag"^(+)] * ["Cl"^(-)]#</mathjax></p>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#1.8 * 10^(-10) = s * s = s^2#</mathjax></p>
</blockquote>
<p>This will give you</p>
<blockquote>
<p><mathjax>#s = sqrt(1.8 * 10^(-10)) = 1.34 * 10^(-5)#</mathjax></p>
</blockquote>
<p>This means that the concentrations of silver cations and of chloride anons in a <strong>saturated</strong> solution of silver chloride will be equal to </p>
<blockquote>
<p><mathjax>#["Ag"^(+)] = ["Cl"^(-)] = s = 1.34 * 10^(-5)"mol L"^(-1)#</mathjax></p>
</blockquote>
<p>Therefore, the concentration of both ions in your <mathjax>#"500-mL"#</mathjax> saturated silver chloride solution will be equal to </p>
<blockquote>
<p><mathjax>#["Ag"^(+)] = ["Cl"^(-)] = color(green)(|bar(ul(color(white)(a/a)1.34 * 10^(-5)"M"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p><strong>Do not</strong> get confused by the fact that you're dealing with a <mathjax>#"500-mL"#</mathjax> solution! </p>
<p>The molar solubility represents <strong>concentration</strong>, not <em>number of moles</em> of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. This means that the concentration of both ions in a saturated silver chloride solution will be equal to <mathjax>#1.34 * 10^(-5)"M"#</mathjax> <strong>regardless</strong> of the <em>volume</em> of said solution! </p></div>
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</article> | Whatare the molar concentrations of all ions in a 500 mL saturated solution of silver chloride at 25 degrees Celsius? | null |
2,904 | ab09a59a-6ddd-11ea-8268-ccda262736ce | https://socratic.org/questions/how-do-you-balance-srbr-2-nh-4-2co-3-srco-3-nh-4br | SrBr2 + (NH4)2CO3 -> SrCO3 + 2 NH4Br | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"SrBr2 + (NH4)2CO3 -> SrCO3 + 2 NH4Br"}] | [{"type":"chemical equation","value":"SrBr2 + (NH4)2CO3 -> SrCO3 + NH4Br"}] | <h1 class="questionTitle" itemprop="name">How do you balance #SrBr_2 + (NH_4)_2CO_3 -> SrCO_3 + NH_4Br#?</h1> | null | SrBr2 + (NH4)2CO3 -> SrCO3 + 2 NH4Br | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balance the chemical equation:<br/>
<mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + "NH"_4"Br"#</mathjax></p>
<p>The word equation is strontium bromide <mathjax>#("SrBr"_2")#</mathjax> plus ammonium carbonate <mathjax>#(("NH"_4)_2"CO"_3)"#</mathjax> produce strontium carbonate <mathjax>#("SrCO"_3)#</mathjax> plus ammonium bromide <mathjax>#("NH"_4"Br")#</mathjax>.</p>
<p>This is a double replacement reaction, in which the positive ions and negative ions switch partners. In this example, all of the reactants and products are <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>.</p>
<p>A balanced equation must have the same number of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> of each compound on both sides of the equation.</p>
<p>When balancing equations with ionic <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>, a good start is to look for and balance <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a>. They can be treated as a single quantity when balancing. This equation has two polyatomic ions; the ammonium ion <mathjax>#("NH"_4^(+)")#</mathjax>, and the carbonate ion <mathjax>#("CO"_3^(2-)")#</mathjax>.</p>
<p><mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + "NH"_4"Br"#</mathjax></p>
<p>There are two ammonium ions on the left side and one on the right side. We CANNOT change the formulas, but we can change the amount by adding coefficients. In the case of <mathjax>#"NH"_4"Br"#</mathjax>, we can add a coefficient of 2 in front of the compound.</p>
<p><mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + color(red)(2)"NH"_4"Br"#</mathjax></p>
<p>Now there is one carbonate ion on both sides of the equation, two bromide ions <mathjax>#("Br"^(-)")#</mathjax> on both sides, and one strontium ion <mathjax>#("Sr"^(2+)")#</mathjax> on both sides. So the equation is now balanced.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + color(red)(2)"NH"_4"Br"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balance the chemical equation:<br/>
<mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + "NH"_4"Br"#</mathjax></p>
<p>The word equation is strontium bromide <mathjax>#("SrBr"_2")#</mathjax> plus ammonium carbonate <mathjax>#(("NH"_4)_2"CO"_3)"#</mathjax> produce strontium carbonate <mathjax>#("SrCO"_3)#</mathjax> plus ammonium bromide <mathjax>#("NH"_4"Br")#</mathjax>.</p>
<p>This is a double replacement reaction, in which the positive ions and negative ions switch partners. In this example, all of the reactants and products are <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>.</p>
<p>A balanced equation must have the same number of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> of each compound on both sides of the equation.</p>
<p>When balancing equations with ionic <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>, a good start is to look for and balance <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a>. They can be treated as a single quantity when balancing. This equation has two polyatomic ions; the ammonium ion <mathjax>#("NH"_4^(+)")#</mathjax>, and the carbonate ion <mathjax>#("CO"_3^(2-)")#</mathjax>.</p>
<p><mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + "NH"_4"Br"#</mathjax></p>
<p>There are two ammonium ions on the left side and one on the right side. We CANNOT change the formulas, but we can change the amount by adding coefficients. In the case of <mathjax>#"NH"_4"Br"#</mathjax>, we can add a coefficient of 2 in front of the compound.</p>
<p><mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + color(red)(2)"NH"_4"Br"#</mathjax></p>
<p>Now there is one carbonate ion on both sides of the equation, two bromide ions <mathjax>#("Br"^(-)")#</mathjax> on both sides, and one strontium ion <mathjax>#("Sr"^(2+)")#</mathjax> on both sides. So the equation is now balanced.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance #SrBr_2 + (NH_4)_2CO_3 -> SrCO_3 + NH_4Br#?</h1>
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<div class="markdown"><p><mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + color(red)(2)"NH"_4"Br"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balance the chemical equation:<br/>
<mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + "NH"_4"Br"#</mathjax></p>
<p>The word equation is strontium bromide <mathjax>#("SrBr"_2")#</mathjax> plus ammonium carbonate <mathjax>#(("NH"_4)_2"CO"_3)"#</mathjax> produce strontium carbonate <mathjax>#("SrCO"_3)#</mathjax> plus ammonium bromide <mathjax>#("NH"_4"Br")#</mathjax>.</p>
<p>This is a double replacement reaction, in which the positive ions and negative ions switch partners. In this example, all of the reactants and products are <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>.</p>
<p>A balanced equation must have the same number of <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> of each compound on both sides of the equation.</p>
<p>When balancing equations with ionic <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>, a good start is to look for and balance <a href="https://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ions</a>. They can be treated as a single quantity when balancing. This equation has two polyatomic ions; the ammonium ion <mathjax>#("NH"_4^(+)")#</mathjax>, and the carbonate ion <mathjax>#("CO"_3^(2-)")#</mathjax>.</p>
<p><mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + "NH"_4"Br"#</mathjax></p>
<p>There are two ammonium ions on the left side and one on the right side. We CANNOT change the formulas, but we can change the amount by adding coefficients. In the case of <mathjax>#"NH"_4"Br"#</mathjax>, we can add a coefficient of 2 in front of the compound.</p>
<p><mathjax>#"SrBr"_2 + ("NH"_4)_2"CO"_3#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"SrCO"_3 + color(red)(2)"NH"_4"Br"#</mathjax></p>
<p>Now there is one carbonate ion on both sides of the equation, two bromide ions <mathjax>#("Br"^(-)")#</mathjax> on both sides, and one strontium ion <mathjax>#("Sr"^(2+)")#</mathjax> on both sides. So the equation is now balanced.</p></div>
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</article> | How do you balance #SrBr_2 + (NH_4)_2CO_3 -> SrCO_3 + NH_4Br#? | null |
2,905 | aaf3983e-6ddd-11ea-b0c3-ccda262736ce | https://socratic.org/questions/the-compound-adrenaline-contains-56-79-carbon-6-56-hydrogen-28-3-oxygen-and-8-28 | C8H11O3N | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"C8H11O3N"}] | [{"type":"physical unit","value":"Percent by mass [OF] carbon in the compound [=] \\pu{56.79%}"},{"type":"physical unit","value":"Percent by mass [OF] hydrogen in the compound [=] \\pu{6.56%}"},{"type":"physical unit","value":"Percent by mass [OF] oxygen in the compound [=] \\pu{28.3%}"},{"type":"physical unit","value":"Percent by mass [OF] nitrogen in the compound [=] \\pu{8.28%}"}] | <h1 class="questionTitle" itemprop="name">The compound adrenaline contains 56.79% carbon, 6.56% hydrogen, 28.3% oxygen, and 8.28% nitrogen by mass, what is its empirical formula?</h1> | null | C8H11O3N | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, as each percentage is over 100, multiply the values of carbon, hydrogen, oxygen and nitrogen by 100. You will obtain:</p>
<p>56.79 grams of carbon<br/>
6.56 grams of hydrogen<br/>
28.3 grams of oxygen<br/>
8.28 grams of nitrogen</p>
<p>Moles of carbon: 56.79/12.0107 = 4.728 moles <br/>
Moles of hydrogen: 6.56/1.00794 = 6.51 moles <br/>
Moles of oxygen: 28.3/15.9994 = 1.77 moles<br/>
Moles of nitrogen: 8.28/14.0067 = 0.59 moles </p>
<p>Divide all moles by the smallest value. In this case, it's 0.59 moles. </p>
<p>4.728/0.59 = 8 <br/>
6.51/0.59 = 11 <br/>
1.77/0.59 = 3 </p>
<p>Thus, the emperical formula is C8H11NO3</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Moles = mass/molar mass </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, as each percentage is over 100, multiply the values of carbon, hydrogen, oxygen and nitrogen by 100. You will obtain:</p>
<p>56.79 grams of carbon<br/>
6.56 grams of hydrogen<br/>
28.3 grams of oxygen<br/>
8.28 grams of nitrogen</p>
<p>Moles of carbon: 56.79/12.0107 = 4.728 moles <br/>
Moles of hydrogen: 6.56/1.00794 = 6.51 moles <br/>
Moles of oxygen: 28.3/15.9994 = 1.77 moles<br/>
Moles of nitrogen: 8.28/14.0067 = 0.59 moles </p>
<p>Divide all moles by the smallest value. In this case, it's 0.59 moles. </p>
<p>4.728/0.59 = 8 <br/>
6.51/0.59 = 11 <br/>
1.77/0.59 = 3 </p>
<p>Thus, the emperical formula is C8H11NO3</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The compound adrenaline contains 56.79% carbon, 6.56% hydrogen, 28.3% oxygen, and 8.28% nitrogen by mass, what is its empirical formula?</h1>
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<div class="markdown"><p>Moles = mass/molar mass </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, as each percentage is over 100, multiply the values of carbon, hydrogen, oxygen and nitrogen by 100. You will obtain:</p>
<p>56.79 grams of carbon<br/>
6.56 grams of hydrogen<br/>
28.3 grams of oxygen<br/>
8.28 grams of nitrogen</p>
<p>Moles of carbon: 56.79/12.0107 = 4.728 moles <br/>
Moles of hydrogen: 6.56/1.00794 = 6.51 moles <br/>
Moles of oxygen: 28.3/15.9994 = 1.77 moles<br/>
Moles of nitrogen: 8.28/14.0067 = 0.59 moles </p>
<p>Divide all moles by the smallest value. In this case, it's 0.59 moles. </p>
<p>4.728/0.59 = 8 <br/>
6.51/0.59 = 11 <br/>
1.77/0.59 = 3 </p>
<p>Thus, the emperical formula is C8H11NO3</p></div>
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Oct 24, 2015
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<div class="markdown"><p><mathjax>#C_8H_11O_3N#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>With this kind of problem, it is always safe to assume that you have 100g sample. It makes the mathematical equation easier.</p>
<p>Assume 100g sample. Therefore,<br/>
wt of C = 56.79 g<br/>
wt of H = 6.56 g<br/>
wt of O = 28.30 g<br/>
wt of N = 8.28 g</p>
<p>Since molecular and empirical formulas deal with number of moles rather than weight, you need to convert these number of grams into number of moles. How? By multiplying the weight with the corresponding atomic weight.</p>
<p>Hence,</p>
<p><mathjax>#56.79 cancel ("grams C")* "1 mole"/(12.0107 cancel ("grams C")) = "4.728 mol C"#</mathjax> </p>
<p><mathjax>#6.56 cancel ("grams H")* "1 mole"/(1.00794 cancel ("grams H")) = "6.508 mol H"#</mathjax> </p>
<p><mathjax>#28.30 cancel ("grams O")* "1 mole"/(15.9994 cancel ("grams O")) = "1.769 mol O"#</mathjax> </p>
<p><mathjax>#8.28 cancel ("grams N")* "1 mole"/(14.0067cancel ("grams N")) = "0.591 mol N"#</mathjax></p>
<p>Get the element with the smallest number of moles, (in this case, N) and use it as your denominator in determining the number of atoms for the other <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>.</p>
<p>Thus,</p>
<p><mathjax>#N = "0.591 mol" / "0.591 mol" = 1#</mathjax> </p>
<p><mathjax>#C = "4.728 mol" / "0.591 mol" = 8#</mathjax></p>
<p><mathjax>#H = "6.508 mol" / "0.591 mol" = 11.01 ~= 11#</mathjax></p>
<p><mathjax>#O = "1.769 mol" / "0.591 mol" = 2.99 ~= 3#</mathjax> </p>
<p>Therefore, the empirical formula for adrenaline is <mathjax>#C_8H_11O_3N#</mathjax>. Please take note, empirical formulas are estimations only and should not be confused with the molecular formula. </p></div>
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</article> | The compound adrenaline contains 56.79% carbon, 6.56% hydrogen, 28.3% oxygen, and 8.28% nitrogen by mass, what is its empirical formula? | null |
2,906 | a8ff1118-6ddd-11ea-b98c-ccda262736ce | https://socratic.org/questions/given-the-following-how-many-millimoles-mmol-of-the-non-limiting-reactant-will-r | 2.33 millimoles | start physical_unit 9 10 mole mmol qc_end physical_unit 21 21 32 33 mass qc_end physical_unit 28 29 40 41 molarity qc_end physical_unit 28 29 37 38 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] non-limiting reactant [IN] millimoles"}] | [{"type":"physical unit","value":"2.33 millimoles"}] | [{"type":"physical unit","value":"Mass [OF] iron [=] \\pu{0.485 grams}"},{"type":"physical unit","value":"Molarity [OF] copper(II) sulfate [=] \\pu{0.438 M}"},{"type":"physical unit","value":"Volume [OF] copper(II) sulfate [=] \\pu{14.5 mL}"},{"type":"other","value":"Iron is oxidized to Fe^2+ by a copper(II) sulfate solution."}] | <h1 class="questionTitle" itemprop="name">Given the following, how many millimoles (mmol) of the non-limiting reactant will remain unused at the end of the reaction?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>If iron is oxidized to <mathjax>#Fe^+2#</mathjax> by a copper(II) sulfate solution, and 0.485 grams of iron and 14.5 mL of 0.438M copper(II) sulfate react to form as much product as possible</p></div>
</h2>
</div>
</div> | 2.33 millimoles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this <strong><a href="https://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reaction</a></strong></p>
<blockquote>
<p><mathjax>#"Fe"_ ((s)) + "CuSO"_ (4(aq)) -> "FeSO"_ (4(aq)) + "Cu"_ ((s))#</mathjax></p>
</blockquote>
<p>Notice that iron and copper(II) sulfate react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, which means that your <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong> will be the reactant that has the <strong>smallest number of moles</strong> available. </p>
<p>Use the <strong>molar mass</strong> of iron </p>
<blockquote>
<p><mathjax>#M_("M Fe") = "55.845 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>to calculate how many moles of iron you have in your sample</p>
<blockquote>
<p><mathjax>#0.485 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "0.008685 moles Fe"#</mathjax></p>
</blockquote>
<p>Since the problem wants you to find the number of <em>millimoles</em> of the <strong>excess reactant</strong>, convert the number of moles of iron to millimoles</p>
<blockquote>
<p><mathjax>#0.008685 color(red)(cancel(color(black)("moles"))) * (10^3"mmol")/(1color(red)(cancel(color(black)("moles")))) = "8.685 mmoles Fe"#</mathjax></p>
</blockquote>
<p>Now, the problem provides you with the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>volume</em> of the copper(II) sulfate solution, so use this info to find the number of millimoles of copper(II) sulfate added to the reaction</p>
<blockquote>
<p><mathjax>#14.5 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * (0.438color(red)(cancel(color(black)("moles CuSO"_4))))/(1color(red)(cancel(color(black)("L")))) * (10^3"mmol")/(1color(red)(cancel(color(black)("mole")))) = "6.351 mmoles CuSO"_4#</mathjax></p>
</blockquote>
<p>As you can see, you have <strong>fewer</strong> millimoles of copper(II) sulfate than of iron metal, which means that copper(II) sulfate will be your <em><strong>limiting reagent</strong></em>. </p>
<p>On other words, iron metal will be <em>in excess</em>. </p>
<p>Since the reaction consumes the two reactants in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, you can say that after the reaction is complete, your solution will contain</p>
<blockquote>
<p><mathjax>#n_("CuSO"_4) = 6.351 - 6.351 = "0 mmoles CuSO"_4 ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_("Fe") = 8.685 - 6.351 = "2.334 mmoles Fe"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("no. of mmoles of Fe left" = 2.33)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2.33 mmoles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this <strong><a href="https://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reaction</a></strong></p>
<blockquote>
<p><mathjax>#"Fe"_ ((s)) + "CuSO"_ (4(aq)) -> "FeSO"_ (4(aq)) + "Cu"_ ((s))#</mathjax></p>
</blockquote>
<p>Notice that iron and copper(II) sulfate react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, which means that your <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong> will be the reactant that has the <strong>smallest number of moles</strong> available. </p>
<p>Use the <strong>molar mass</strong> of iron </p>
<blockquote>
<p><mathjax>#M_("M Fe") = "55.845 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>to calculate how many moles of iron you have in your sample</p>
<blockquote>
<p><mathjax>#0.485 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "0.008685 moles Fe"#</mathjax></p>
</blockquote>
<p>Since the problem wants you to find the number of <em>millimoles</em> of the <strong>excess reactant</strong>, convert the number of moles of iron to millimoles</p>
<blockquote>
<p><mathjax>#0.008685 color(red)(cancel(color(black)("moles"))) * (10^3"mmol")/(1color(red)(cancel(color(black)("moles")))) = "8.685 mmoles Fe"#</mathjax></p>
</blockquote>
<p>Now, the problem provides you with the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>volume</em> of the copper(II) sulfate solution, so use this info to find the number of millimoles of copper(II) sulfate added to the reaction</p>
<blockquote>
<p><mathjax>#14.5 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * (0.438color(red)(cancel(color(black)("moles CuSO"_4))))/(1color(red)(cancel(color(black)("L")))) * (10^3"mmol")/(1color(red)(cancel(color(black)("mole")))) = "6.351 mmoles CuSO"_4#</mathjax></p>
</blockquote>
<p>As you can see, you have <strong>fewer</strong> millimoles of copper(II) sulfate than of iron metal, which means that copper(II) sulfate will be your <em><strong>limiting reagent</strong></em>. </p>
<p>On other words, iron metal will be <em>in excess</em>. </p>
<p>Since the reaction consumes the two reactants in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, you can say that after the reaction is complete, your solution will contain</p>
<blockquote>
<p><mathjax>#n_("CuSO"_4) = 6.351 - 6.351 = "0 mmoles CuSO"_4 ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_("Fe") = 8.685 - 6.351 = "2.334 mmoles Fe"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("no. of mmoles of Fe left" = 2.33)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Given the following, how many millimoles (mmol) of the non-limiting reactant will remain unused at the end of the reaction?</h1>
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<div class="markdown"><p>If iron is oxidized to <mathjax>#Fe^+2#</mathjax> by a copper(II) sulfate solution, and 0.485 grams of iron and 14.5 mL of 0.438M copper(II) sulfate react to form as much product as possible</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2016-08-22T00:02:36" itemprop="dateCreated">
Aug 22, 2016
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<div class="markdown"><p><mathjax>#"2.33 mmoles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing the balanced chemical equation that describes this <strong><a href="https://socratic.org/chemistry/chemical-reactions/redox-reactions">redox reaction</a></strong></p>
<blockquote>
<p><mathjax>#"Fe"_ ((s)) + "CuSO"_ (4(aq)) -> "FeSO"_ (4(aq)) + "Cu"_ ((s))#</mathjax></p>
</blockquote>
<p>Notice that iron and copper(II) sulfate react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, which means that your <strong><a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a></strong> will be the reactant that has the <strong>smallest number of moles</strong> available. </p>
<p>Use the <strong>molar mass</strong> of iron </p>
<blockquote>
<p><mathjax>#M_("M Fe") = "55.845 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>to calculate how many moles of iron you have in your sample</p>
<blockquote>
<p><mathjax>#0.485 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = "0.008685 moles Fe"#</mathjax></p>
</blockquote>
<p>Since the problem wants you to find the number of <em>millimoles</em> of the <strong>excess reactant</strong>, convert the number of moles of iron to millimoles</p>
<blockquote>
<p><mathjax>#0.008685 color(red)(cancel(color(black)("moles"))) * (10^3"mmol")/(1color(red)(cancel(color(black)("moles")))) = "8.685 mmoles Fe"#</mathjax></p>
</blockquote>
<p>Now, the problem provides you with the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></em> and <em>volume</em> of the copper(II) sulfate solution, so use this info to find the number of millimoles of copper(II) sulfate added to the reaction</p>
<blockquote>
<p><mathjax>#14.5 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * (0.438color(red)(cancel(color(black)("moles CuSO"_4))))/(1color(red)(cancel(color(black)("L")))) * (10^3"mmol")/(1color(red)(cancel(color(black)("mole")))) = "6.351 mmoles CuSO"_4#</mathjax></p>
</blockquote>
<p>As you can see, you have <strong>fewer</strong> millimoles of copper(II) sulfate than of iron metal, which means that copper(II) sulfate will be your <em><strong>limiting reagent</strong></em>. </p>
<p>On other words, iron metal will be <em>in excess</em>. </p>
<p>Since the reaction consumes the two reactants in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, you can say that after the reaction is complete, your solution will contain</p>
<blockquote>
<p><mathjax>#n_("CuSO"_4) = 6.351 - 6.351 = "0 mmoles CuSO"_4 ->#</mathjax> <em><strong>completely consumed</strong></em></p>
<p><mathjax>#n_("Fe") = 8.685 - 6.351 = "2.334 mmoles Fe"#</mathjax></p>
</blockquote>
<p>Rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the answer will be </p>
<blockquote>
<p><mathjax>#color(green)(|bar(ul(color(white)(a/a)color(black)("no. of mmoles of Fe left" = 2.33)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | Given the following, how many millimoles (mmol) of the non-limiting reactant will remain unused at the end of the reaction? |
If iron is oxidized to #Fe^+2# by a copper(II) sulfate solution, and 0.485 grams of iron and 14.5 mL of 0.438M copper(II) sulfate react to form as much product as possible
|
2,907 | ac363370-6ddd-11ea-8cde-ccda262736ce | https://socratic.org/questions/5a272bbe11ef6b03e70d3550 | 2 Li + FeBr2 -> Fe + 2 LiBr | start chemical_equation qc_end substance 2 3 qc_end substance 5 6 qc_end substance 9 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 Li + FeBr2 -> Fe + 2 LiBr"}] | [{"type":"substance name","value":"Lithium metal"},{"type":"substance name","value":"Ferrous bromide"},{"type":"substance name","value":"Iron metal"}] | <h1 class="questionTitle" itemprop="name">How could #"lithium metal"# reduce #"ferrous bromide"# to give #"iron metal"#?</h1> | null | 2 Li + FeBr2 -> Fe + 2 LiBr | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Fe(II)#</mathjax> is reduced....</p>
<p><mathjax>#FeBr_2 +2e^(-) rarr Fe + 2Br^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Lithium metal"#</mathjax> is oxidized....</p>
<p><mathjax>#Li rarr Li^+ + e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>We add <mathjax>#(i)#</mathjax> and <mathjax>#(ii)#</mathjax> together in such a way that the electrons are retired from the equation....<mathjax>#(i)+2xx(ii)#</mathjax></p>
<p><mathjax>#2Li +FeBr_2 +2e^(-)rarr 2Li^+ +Fe + 2Br^(-)+ 2e^(-)#</mathjax></p>
<p>To give finally:</p>
<p><mathjax>#2Li +FeBr_2 rarr 2LiBr +Fe#</mathjax></p>
<p>Both charge and mass are balanced as is absolutely required (and this is why we employ the oxidation number method).</p>
<p>Do I win <mathjax>#£5-00?#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2Li + FeBr_2 rarr Fe + 2LiBr#</mathjax> is the stoichiometric equation....</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Fe(II)#</mathjax> is reduced....</p>
<p><mathjax>#FeBr_2 +2e^(-) rarr Fe + 2Br^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Lithium metal"#</mathjax> is oxidized....</p>
<p><mathjax>#Li rarr Li^+ + e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>We add <mathjax>#(i)#</mathjax> and <mathjax>#(ii)#</mathjax> together in such a way that the electrons are retired from the equation....<mathjax>#(i)+2xx(ii)#</mathjax></p>
<p><mathjax>#2Li +FeBr_2 +2e^(-)rarr 2Li^+ +Fe + 2Br^(-)+ 2e^(-)#</mathjax></p>
<p>To give finally:</p>
<p><mathjax>#2Li +FeBr_2 rarr 2LiBr +Fe#</mathjax></p>
<p>Both charge and mass are balanced as is absolutely required (and this is why we employ the oxidation number method).</p>
<p>Do I win <mathjax>#£5-00?#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How could #"lithium metal"# reduce #"ferrous bromide"# to give #"iron metal"#?</h1>
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<div class="markdown"><p><mathjax>#2Li + FeBr_2 rarr Fe + 2LiBr#</mathjax> is the stoichiometric equation....</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Fe(II)#</mathjax> is reduced....</p>
<p><mathjax>#FeBr_2 +2e^(-) rarr Fe + 2Br^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Lithium metal"#</mathjax> is oxidized....</p>
<p><mathjax>#Li rarr Li^+ + e^(-)#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>We add <mathjax>#(i)#</mathjax> and <mathjax>#(ii)#</mathjax> together in such a way that the electrons are retired from the equation....<mathjax>#(i)+2xx(ii)#</mathjax></p>
<p><mathjax>#2Li +FeBr_2 +2e^(-)rarr 2Li^+ +Fe + 2Br^(-)+ 2e^(-)#</mathjax></p>
<p>To give finally:</p>
<p><mathjax>#2Li +FeBr_2 rarr 2LiBr +Fe#</mathjax></p>
<p>Both charge and mass are balanced as is absolutely required (and this is why we employ the oxidation number method).</p>
<p>Do I win <mathjax>#£5-00?#</mathjax></p></div>
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</article> | How could #"lithium metal"# reduce #"ferrous bromide"# to give #"iron metal"#? | null |
2,908 | abdd9af1-6ddd-11ea-82c7-ccda262736ce | https://socratic.org/questions/how-do-you-find-the-value-of-oh-for-a-solution-with-a-ph-of-8-00 | 1.00 × 10^(-6) M | start physical_unit 10 10 [oh-] mol/l qc_end physical_unit 10 10 15 15 ph qc_end end | [{"type":"physical unit","value":"[OH-] [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"1.00 × 10^(-6) M"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{8.00}"}] | <h1 class="questionTitle" itemprop="name">How do you find the value of #[OH^-]# for a solution with a pH of 8.00?</h1> | null | 1.00 × 10^(-6) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Water undergoes so-called <mathjax>#"autoprotolysis"#</mathjax>......</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>At <mathjax>#298*K#</mathjax> we can measure the <mathjax>#"autoprotolysis"#</mathjax> very precisely....</p>
<p><mathjax>#[HO^-][H_3O^+]=10^-14=K_w#</mathjax></p>
<p>This is a mathematical equation which we can divide, multiply, or otherwise manipulate provided that we does it to both sides.....One think that we can do is take <mathjax>#log_10#</mathjax> of both sides....</p>
<p><mathjax>#log_10[HO^-]+log_10[H_3O^+]=log_10(10^-14)#</mathjax></p>
<p>And thus <mathjax>#log_10[HO^-]+log_10[H_3O^+]=-14#</mathjax></p>
<p>(Why? Because mathematically, <mathjax>#log_(a)a^y=y#</mathjax> by definition. It is the power to which we raise the <mathjax>#"base a"#</mathjax> to get <mathjax>#a^y#</mathjax>. With me?</p>
<p>And so <mathjax>#14=-log_10[HO^-]-log_10[H_3O^+]#</mathjax></p>
<p>BUT, <mathjax>#-log_10[H_3O^+]=pH#</mathjax>, and <mathjax>#-log_10[HO^-]=pOH#</mathjax> BY DEFINITION........</p>
<p>And so (at last!) <mathjax>#14=pH+pOH#</mathjax></p>
<p>And so (after all that!) if <mathjax>#pH=8#</mathjax>, then CLEARLY, <mathjax>#pOH=6#</mathjax>. And thus <mathjax>#[HO^-]=10^-6*mol*L^-1#</mathjax>.</p>
<p>Claro?</p>
<p>Note the logarithmic function is something that you learn about in mathematics; it is very powerful, and useful. Its application to chemistry is (I think) rather straightforward. If there is still an issue, voice it and someone will help you.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#[HO^-]=10^-6*mol*L^-1#</mathjax>. How does we know.........?</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Water undergoes so-called <mathjax>#"autoprotolysis"#</mathjax>......</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>At <mathjax>#298*K#</mathjax> we can measure the <mathjax>#"autoprotolysis"#</mathjax> very precisely....</p>
<p><mathjax>#[HO^-][H_3O^+]=10^-14=K_w#</mathjax></p>
<p>This is a mathematical equation which we can divide, multiply, or otherwise manipulate provided that we does it to both sides.....One think that we can do is take <mathjax>#log_10#</mathjax> of both sides....</p>
<p><mathjax>#log_10[HO^-]+log_10[H_3O^+]=log_10(10^-14)#</mathjax></p>
<p>And thus <mathjax>#log_10[HO^-]+log_10[H_3O^+]=-14#</mathjax></p>
<p>(Why? Because mathematically, <mathjax>#log_(a)a^y=y#</mathjax> by definition. It is the power to which we raise the <mathjax>#"base a"#</mathjax> to get <mathjax>#a^y#</mathjax>. With me?</p>
<p>And so <mathjax>#14=-log_10[HO^-]-log_10[H_3O^+]#</mathjax></p>
<p>BUT, <mathjax>#-log_10[H_3O^+]=pH#</mathjax>, and <mathjax>#-log_10[HO^-]=pOH#</mathjax> BY DEFINITION........</p>
<p>And so (at last!) <mathjax>#14=pH+pOH#</mathjax></p>
<p>And so (after all that!) if <mathjax>#pH=8#</mathjax>, then CLEARLY, <mathjax>#pOH=6#</mathjax>. And thus <mathjax>#[HO^-]=10^-6*mol*L^-1#</mathjax>.</p>
<p>Claro?</p>
<p>Note the logarithmic function is something that you learn about in mathematics; it is very powerful, and useful. Its application to chemistry is (I think) rather straightforward. If there is still an issue, voice it and someone will help you.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you find the value of #[OH^-]# for a solution with a pH of 8.00?</h1>
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<div class="markdown"><p><mathjax>#[HO^-]=10^-6*mol*L^-1#</mathjax>. How does we know.........?</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Water undergoes so-called <mathjax>#"autoprotolysis"#</mathjax>......</p>
<p><mathjax>#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#</mathjax></p>
<p>At <mathjax>#298*K#</mathjax> we can measure the <mathjax>#"autoprotolysis"#</mathjax> very precisely....</p>
<p><mathjax>#[HO^-][H_3O^+]=10^-14=K_w#</mathjax></p>
<p>This is a mathematical equation which we can divide, multiply, or otherwise manipulate provided that we does it to both sides.....One think that we can do is take <mathjax>#log_10#</mathjax> of both sides....</p>
<p><mathjax>#log_10[HO^-]+log_10[H_3O^+]=log_10(10^-14)#</mathjax></p>
<p>And thus <mathjax>#log_10[HO^-]+log_10[H_3O^+]=-14#</mathjax></p>
<p>(Why? Because mathematically, <mathjax>#log_(a)a^y=y#</mathjax> by definition. It is the power to which we raise the <mathjax>#"base a"#</mathjax> to get <mathjax>#a^y#</mathjax>. With me?</p>
<p>And so <mathjax>#14=-log_10[HO^-]-log_10[H_3O^+]#</mathjax></p>
<p>BUT, <mathjax>#-log_10[H_3O^+]=pH#</mathjax>, and <mathjax>#-log_10[HO^-]=pOH#</mathjax> BY DEFINITION........</p>
<p>And so (at last!) <mathjax>#14=pH+pOH#</mathjax></p>
<p>And so (after all that!) if <mathjax>#pH=8#</mathjax>, then CLEARLY, <mathjax>#pOH=6#</mathjax>. And thus <mathjax>#[HO^-]=10^-6*mol*L^-1#</mathjax>.</p>
<p>Claro?</p>
<p>Note the logarithmic function is something that you learn about in mathematics; it is very powerful, and useful. Its application to chemistry is (I think) rather straightforward. If there is still an issue, voice it and someone will help you.</p></div>
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</article> | How do you find the value of #[OH^-]# for a solution with a pH of 8.00? | null |
2,909 | aa0cd6f4-6ddd-11ea-9866-ccda262736ce | https://socratic.org/questions/5971ac97b72cff04d50e0ffe | 6.02 × 10^25 | start physical_unit 5 6 number none qc_end physical_unit 5 5 9 10 mole qc_end end | [{"type":"physical unit","value":"Number [OF] sulfur atoms"}] | [{"type":"physical unit","value":"6.02 × 10^25"}] | [{"type":"physical unit","value":"Mole [OF] sulfur [=] \\pu{100 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the number of sulfur atoms in a #100*mol# of sulfur?</h1> | null | 6.02 × 10^25 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Number of sulfur atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100xxN_A#</mathjax>...where <mathjax>#N_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Avogadro's number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And thus <mathjax>#"Number of sulfur atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100*molxxN_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^2*molxx6.022xx10^23*mol^-1=6.022xx10^25#</mathjax> <mathjax>#"sulfur atoms"#</mathjax>.</p>
<p>How many sulfur atoms in <mathjax>#100*mol#</mathjax> of <mathjax>#Na_2^(+)S_2O_3^(2-)#</mathjax>, <mathjax>#"sodium thiosulfate"#</mathjax>?</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Number of sulfur atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100xxN_A#</mathjax>...where...</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Number of sulfur atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100xxN_A#</mathjax>...where <mathjax>#N_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Avogadro's number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And thus <mathjax>#"Number of sulfur atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100*molxxN_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^2*molxx6.022xx10^23*mol^-1=6.022xx10^25#</mathjax> <mathjax>#"sulfur atoms"#</mathjax>.</p>
<p>How many sulfur atoms in <mathjax>#100*mol#</mathjax> of <mathjax>#Na_2^(+)S_2O_3^(2-)#</mathjax>, <mathjax>#"sodium thiosulfate"#</mathjax>?</p></div>
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<div class="markdown"><p><mathjax>#"Number of sulfur atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100xxN_A#</mathjax>...where...</p></div>
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<div class="markdown"><p><mathjax>#"Number of sulfur atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100xxN_A#</mathjax>...where <mathjax>#N_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Avogadro's number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p>
<p>And thus <mathjax>#"Number of sulfur atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#100*molxxN_A#</mathjax> <mathjax>#=#</mathjax> <mathjax>#10^2*molxx6.022xx10^23*mol^-1=6.022xx10^25#</mathjax> <mathjax>#"sulfur atoms"#</mathjax>.</p>
<p>How many sulfur atoms in <mathjax>#100*mol#</mathjax> of <mathjax>#Na_2^(+)S_2O_3^(2-)#</mathjax>, <mathjax>#"sodium thiosulfate"#</mathjax>?</p></div>
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</article> | What is the number of sulfur atoms in a #100*mol# of sulfur? | null |
2,910 | aab4fa2c-6ddd-11ea-af5f-ccda262736ce | https://socratic.org/questions/a-140-5g-sample-of-niso4-xh2o-was-heated-until-no-further-decrease-in-mass-was-o | 7 | start physical_unit 30 30 value none qc_end physical_unit 3 5 1 2 mass qc_end physical_unit 19 22 24 25 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Value [OF] x"}] | [{"type":"physical unit","value":"7"}] | [{"type":"physical unit","value":"Mass [OF] NiSO4.xH2O sample [=] \\pu{140.5 g}"},{"type":"physical unit","value":"Mass [OF] the anhydrous (dry) salt [=] \\pu{77.5 g}"},{"type":"other","value":"A sample of NiSO4.xH2O was heated until no further decrease in mass was observed."}] | <h1 class="questionTitle" itemprop="name">A #"140.5-g"# sample of #"NiSO"_4 * x "H"_2"O"# was heated until no further decrease in mass was observed. The mass of the anhydrous (dry) salt is #"77.5 g"#. Determine the value of #x# in the formula?</h1> | null | 7 | <div class="answerDescription">
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<h1 class="questionTitle" itemprop="name">A #"140.5-g"# sample of #"NiSO"_4 * x "H"_2"O"# was heated until no further decrease in mass was observed. The mass of the anhydrous (dry) salt is #"77.5 g"#. Determine the value of #x# in the formula?</h1>
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<div class="markdown"><p><mathjax>#x=7#</mathjax>....</p></div>
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<div class="markdown"><p>We start with the HYDRATE of a nickel salt...</p>
<p><mathjax>#NiSO_4*n(H_2O)(s)+Delta rarr NiSO_4(s) + nH_2O(g)uarr#</mathjax></p>
<p>We heat the nickel hydrate such that we are left with nickel sulfate, and the mass loss represents the moles of water...</p>
<p><mathjax>#"Moles of nickel sulfate (anhydrous)"=(77.5*g)/(154.75*g*mol^-1)=0.501*mol#</mathjax>...</p>
<p>But the mass LOST represented the water present....</p>
<p><mathjax>#"Moles of water lost"=(140.5*g-77.5*g)/(18.01*g*mol^-1)=3.50*mol#</mathjax>...and so we divide thru by the LOWEST molar quantity to get...</p>
<p><mathjax>#(NiSO_4)_((0.501*mol)/(0.501*mol))*(OH_2)_((3.50*mol)/(0.501*mol))-=(NiSO_4)*(OH_2)_7#</mathjax>..</p>
<p>And so we got...<mathjax>#NiSO_4*(OH_2)_7#</mathjax>...a salt of nickel sulfate with SEVEN waters of crystallization....</p></div>
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</article> | A #"140.5-g"# sample of #"NiSO"_4 * x "H"_2"O"# was heated until no further decrease in mass was observed. The mass of the anhydrous (dry) salt is #"77.5 g"#. Determine the value of #x# in the formula? | null |
2,911 | a88c97c0-6ddd-11ea-9180-ccda262736ce | https://socratic.org/questions/56ef416b11ef6b0cdb5bdc8b | 1008 L | start physical_unit 25 25 volume l qc_end physical_unit 6 6 0 1 mass qc_end chemical_equation 15 20 qc_end physical_unit 25 25 27 28 temperature qc_end end | [{"type":"physical unit","value":"Total volume [OF] gases [IN] L"}] | [{"type":"physical unit","value":"1008 L"}] | [{"type":"physical unit","value":"Mass [OF] NH4NO2 [=] \\pu{320 g}"},{"type":"chemical equation","value":"NH4NO2 -> N2 + 2 H2O"},{"type":"physical unit","value":"Temperature [OF] gases [=] \\pu{819 K}"}] | <h1 class="questionTitle" itemprop="name">#320# #g# of solid ammonium nitrite, #NH_4NO_2#, decomposes when heated according to the balanced equation: #NH_4NO_2 -> N_2+2H_2O#. What total volume of gases at #819# #K# is emitted by this reaction?</h1> | null | 1008 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The molar mass of <mathjax>#NH_4NO_2#</mathjax> is <mathjax>#64#</mathjax> <mathjax>#gmol^-1#</mathjax> (two <mathjax>#N#</mathjax> at <mathjax>#14#</mathjax>, four <mathjax>#H#</mathjax> at <mathjax>#1#</mathjax>, two <mathjax>#O#</mathjax> at <mathjax>#16#</mathjax>).</p>
<p>Find the number of moles of <mathjax>#NH_4NO_2#</mathjax>:</p>
<p><mathjax>#n=m/M=320/64=5#</mathjax> <mathjax>#mol#</mathjax></p>
<p>From the balanced equation, each mole of <mathjax>#NH_4NO_2#</mathjax> yields 1 mole of <mathjax>#N_2#</mathjax> and 2 moles of <mathjax>#H_2O#</mathjax> (which is definitely a gas at <mathjax>#819#</mathjax> <mathjax>#K#</mathjax>), for a total of 3 moles of gas. </p>
<p>For ideal gases (which we can treat these real gases as for our purposes), 1 mole of any gas at STP (<mathjax>#273#</mathjax> <mathjax>#K#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>) occupies <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax>.</p>
<p>That means we have <mathjax>#3xx5 = 15#</mathjax> <mathjax>#mol#</mathjax> of product gases, and they occupy a volume of <mathjax>#336#</mathjax> <mathjax>#L#</mathjax> at STP.</p>
<p>We need to change the temperature from <mathjax>#273#</mathjax> <mathjax>#K#</mathjax> to <mathjax>#819#</mathjax> <mathjax>#K#</mathjax>, and the pressure stays the same at <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>, so:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p>Rearranging:</p>
<p><mathjax>#V_2=V_1T_2/T_1 = 336 xx 819/273 = 1008#</mathjax> <mathjax>#L#</mathjax></p></div>
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<div class="markdown"><p>Each mole of <mathjax>#NH_4NO_2#</mathjax> produces <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> of gases total (<mathjax>#N_2#</mathjax> and <mathjax>#H_2O#</mathjax>), and we have <mathjax>#5#</mathjax> <mathjax>#mol#</mathjax>. So <mathjax>#3xx5 = 15#</mathjax> <mathjax>#mol#</mathjax> of gas at STP is <mathjax>#336#</mathjax> <mathjax>#L#</mathjax>. Converting to <mathjax>#819#</mathjax> <mathjax>#K#</mathjax> gives a volume of <mathjax>#1008#</mathjax> <mathjax>#L#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of <mathjax>#NH_4NO_2#</mathjax> is <mathjax>#64#</mathjax> <mathjax>#gmol^-1#</mathjax> (two <mathjax>#N#</mathjax> at <mathjax>#14#</mathjax>, four <mathjax>#H#</mathjax> at <mathjax>#1#</mathjax>, two <mathjax>#O#</mathjax> at <mathjax>#16#</mathjax>).</p>
<p>Find the number of moles of <mathjax>#NH_4NO_2#</mathjax>:</p>
<p><mathjax>#n=m/M=320/64=5#</mathjax> <mathjax>#mol#</mathjax></p>
<p>From the balanced equation, each mole of <mathjax>#NH_4NO_2#</mathjax> yields 1 mole of <mathjax>#N_2#</mathjax> and 2 moles of <mathjax>#H_2O#</mathjax> (which is definitely a gas at <mathjax>#819#</mathjax> <mathjax>#K#</mathjax>), for a total of 3 moles of gas. </p>
<p>For ideal gases (which we can treat these real gases as for our purposes), 1 mole of any gas at STP (<mathjax>#273#</mathjax> <mathjax>#K#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>) occupies <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax>.</p>
<p>That means we have <mathjax>#3xx5 = 15#</mathjax> <mathjax>#mol#</mathjax> of product gases, and they occupy a volume of <mathjax>#336#</mathjax> <mathjax>#L#</mathjax> at STP.</p>
<p>We need to change the temperature from <mathjax>#273#</mathjax> <mathjax>#K#</mathjax> to <mathjax>#819#</mathjax> <mathjax>#K#</mathjax>, and the pressure stays the same at <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>, so:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p>Rearranging:</p>
<p><mathjax>#V_2=V_1T_2/T_1 = 336 xx 819/273 = 1008#</mathjax> <mathjax>#L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">#320# #g# of solid ammonium nitrite, #NH_4NO_2#, decomposes when heated according to the balanced equation: #NH_4NO_2 -> N_2+2H_2O#. What total volume of gases at #819# #K# is emitted by this reaction?</h1>
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<div class="markdown"><p>Each mole of <mathjax>#NH_4NO_2#</mathjax> produces <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> of gases total (<mathjax>#N_2#</mathjax> and <mathjax>#H_2O#</mathjax>), and we have <mathjax>#5#</mathjax> <mathjax>#mol#</mathjax>. So <mathjax>#3xx5 = 15#</mathjax> <mathjax>#mol#</mathjax> of gas at STP is <mathjax>#336#</mathjax> <mathjax>#L#</mathjax>. Converting to <mathjax>#819#</mathjax> <mathjax>#K#</mathjax> gives a volume of <mathjax>#1008#</mathjax> <mathjax>#L#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The molar mass of <mathjax>#NH_4NO_2#</mathjax> is <mathjax>#64#</mathjax> <mathjax>#gmol^-1#</mathjax> (two <mathjax>#N#</mathjax> at <mathjax>#14#</mathjax>, four <mathjax>#H#</mathjax> at <mathjax>#1#</mathjax>, two <mathjax>#O#</mathjax> at <mathjax>#16#</mathjax>).</p>
<p>Find the number of moles of <mathjax>#NH_4NO_2#</mathjax>:</p>
<p><mathjax>#n=m/M=320/64=5#</mathjax> <mathjax>#mol#</mathjax></p>
<p>From the balanced equation, each mole of <mathjax>#NH_4NO_2#</mathjax> yields 1 mole of <mathjax>#N_2#</mathjax> and 2 moles of <mathjax>#H_2O#</mathjax> (which is definitely a gas at <mathjax>#819#</mathjax> <mathjax>#K#</mathjax>), for a total of 3 moles of gas. </p>
<p>For ideal gases (which we can treat these real gases as for our purposes), 1 mole of any gas at STP (<mathjax>#273#</mathjax> <mathjax>#K#</mathjax> and <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>) occupies <mathjax>#22.4#</mathjax> <mathjax>#L#</mathjax>.</p>
<p>That means we have <mathjax>#3xx5 = 15#</mathjax> <mathjax>#mol#</mathjax> of product gases, and they occupy a volume of <mathjax>#336#</mathjax> <mathjax>#L#</mathjax> at STP.</p>
<p>We need to change the temperature from <mathjax>#273#</mathjax> <mathjax>#K#</mathjax> to <mathjax>#819#</mathjax> <mathjax>#K#</mathjax>, and the pressure stays the same at <mathjax>#1#</mathjax> <mathjax>#atm#</mathjax>, so:</p>
<p><mathjax>#V_1/T_1=V_2/T_2#</mathjax></p>
<p>Rearranging:</p>
<p><mathjax>#V_2=V_1T_2/T_1 = 336 xx 819/273 = 1008#</mathjax> <mathjax>#L#</mathjax></p></div>
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</article> | #320# #g# of solid ammonium nitrite, #NH_4NO_2#, decomposes when heated according to the balanced equation: #NH_4NO_2 -> N_2+2H_2O#. What total volume of gases at #819# #K# is emitted by this reaction? | null |
2,912 | ab695c6d-6ddd-11ea-9046-ccda262736ce | https://socratic.org/questions/5a29ddf6b72cff3911cf992e | 4.34 × 10^25 | start physical_unit 2 5 number none qc_end physical_unit 14 14 10 11 mole qc_end end | [{"type":"physical unit","value":"Number [OF] iron and chlorine atoms"}] | [{"type":"physical unit","value":"4.34 × 10^25"}] | [{"type":"physical unit","value":"Mole [OF] FeCl3 [=] \\pu{18 mol}"}] | <h1 class="questionTitle" itemprop="name">How many iron and chlorine atoms are there in an #18*mol# quantity of #FeCl_3#?</h1> | null | 4.34 × 10^25 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is simply another collective number like a dozen, or a gross, or a score; just as there are <mathjax>#"12 items per dozen"#</mathjax>, there are <mathjax>#6.022xx10^23#</mathjax> <mathjax>#"items per mole"#</mathjax>. Chemists often use the abbreviation <mathjax>#N_A#</mathjax> to represent this number.</p>
<p>And so we gots an <mathjax>#"18 mole quantity"#</mathjax> of <mathjax>#FeCl_3#</mathjax>. Do you agree that CLEARLY there are <mathjax>#54*mol#</mathjax> of chloride ions, and <mathjax>#"18 mole"#</mathjax> of <mathjax>#Fe^(3+)#</mathjax> ions?</p>
<p>What is the mass of this molar quantity?</p>
<p>And why should we use such an absurdly large number like a mole? If the answer is not immediately apparent, then what is the MASS of <mathjax>#6.022xx10^23#</mathjax> <mathjax>#"individual hydrogen atoms?"#</mathjax> The mole is thus the link between the sub-micro world or atoms and molecules, to the macro world of grams and litres.....</p></div>
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<div class="markdown"><p>How many eggses in 18 dozen eggs? Are there not 216 individual eggs?</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/the-mole-concept/the-mole">The mole</a> is simply another collective number like a dozen, or a gross, or a score; just as there are <mathjax>#"12 items per dozen"#</mathjax>, there are <mathjax>#6.022xx10^23#</mathjax> <mathjax>#"items per mole"#</mathjax>. Chemists often use the abbreviation <mathjax>#N_A#</mathjax> to represent this number.</p>
<p>And so we gots an <mathjax>#"18 mole quantity"#</mathjax> of <mathjax>#FeCl_3#</mathjax>. Do you agree that CLEARLY there are <mathjax>#54*mol#</mathjax> of chloride ions, and <mathjax>#"18 mole"#</mathjax> of <mathjax>#Fe^(3+)#</mathjax> ions?</p>
<p>What is the mass of this molar quantity?</p>
<p>And why should we use such an absurdly large number like a mole? If the answer is not immediately apparent, then what is the MASS of <mathjax>#6.022xx10^23#</mathjax> <mathjax>#"individual hydrogen atoms?"#</mathjax> The mole is thus the link between the sub-micro world or atoms and molecules, to the macro world of grams and litres.....</p></div>
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<h1 class="questionTitle" itemprop="name">How many iron and chlorine atoms are there in an #18*mol# quantity of #FeCl_3#?</h1>
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<div class="markdown"><p>How many eggses in 18 dozen eggs? Are there not 216 individual eggs?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> is simply another collective number like a dozen, or a gross, or a score; just as there are <mathjax>#"12 items per dozen"#</mathjax>, there are <mathjax>#6.022xx10^23#</mathjax> <mathjax>#"items per mole"#</mathjax>. Chemists often use the abbreviation <mathjax>#N_A#</mathjax> to represent this number.</p>
<p>And so we gots an <mathjax>#"18 mole quantity"#</mathjax> of <mathjax>#FeCl_3#</mathjax>. Do you agree that CLEARLY there are <mathjax>#54*mol#</mathjax> of chloride ions, and <mathjax>#"18 mole"#</mathjax> of <mathjax>#Fe^(3+)#</mathjax> ions?</p>
<p>What is the mass of this molar quantity?</p>
<p>And why should we use such an absurdly large number like a mole? If the answer is not immediately apparent, then what is the MASS of <mathjax>#6.022xx10^23#</mathjax> <mathjax>#"individual hydrogen atoms?"#</mathjax> The mole is thus the link between the sub-micro world or atoms and molecules, to the macro world of grams and litres.....</p></div>
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</article> | How many iron and chlorine atoms are there in an #18*mol# quantity of #FeCl_3#? | null |
2,913 | ac3d14ba-6ddd-11ea-acc3-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-solution-that-has-an-oh-concentration-of-1-10-9-m | 5.00 | start physical_unit 6 6 ph none qc_end physical_unit 10 10 13 16 concentration qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"5.00"}] | [{"type":"physical unit","value":"Concentration [OF] OH- [=] \\pu{1 × 10^(-9) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a solution that has an #OH^-# concentration of #1*10^-9# #M#?</h1> | null | 5.00 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH+pOH=14#</mathjax></p>
<p><mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1xx10^-9#</mathjax>, and <mathjax>#pOH=-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-9)=9#</mathjax></p>
<p>And thus <mathjax>#pH=5#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH=5#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#pH+pOH=14#</mathjax></p>
<p><mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1xx10^-9#</mathjax>, and <mathjax>#pOH=-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-9)=9#</mathjax></p>
<p>And thus <mathjax>#pH=5#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH=5#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#pH+pOH=14#</mathjax></p>
<p><mathjax>#[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1xx10^-9#</mathjax>, and <mathjax>#pOH=-log_10[HO^-]#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-(-9)=9#</mathjax></p>
<p>And thus <mathjax>#pH=5#</mathjax></p></div>
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</article> | What is the pH of a solution that has an #OH^-# concentration of #1*10^-9# #M#? | null |
2,914 | a85cd7de-6ddd-11ea-8005-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-phosphorous-trichloride | PCl3 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] phosphorous trichloride [IN] default"}] | [{"type":"chemical equation","value":"PCl3"}] | [{"type":"substance name","value":"Phosphorous trichloride"}] | <h1 class="questionTitle" itemprop="name">What is the formula for phosphorous trichloride?</h1> | null | PCl3 | <div class="answerDescription">
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<div class="markdown"><p><mathjax>#"Phosphorus trichloride"#</mathjax> is a colourless, volatile liquid, with a normal boiling point of <mathjax>#76#</mathjax> <mathjax>#"^@C#</mathjax>. It react with violently with water, and is a staple precursor in organophosphorus chemistry.</p></div>
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<div class="markdown"><p><mathjax>#PCl_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Phosphorus trichloride"#</mathjax> is a colourless, volatile liquid, with a normal boiling point of <mathjax>#76#</mathjax> <mathjax>#"^@C#</mathjax>. It react with violently with water, and is a staple precursor in organophosphorus chemistry.</p></div>
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<div class="markdown"><p><mathjax>#PCl_3#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Phosphorus trichloride"#</mathjax> is a colourless, volatile liquid, with a normal boiling point of <mathjax>#76#</mathjax> <mathjax>#"^@C#</mathjax>. It react with violently with water, and is a staple precursor in organophosphorus chemistry.</p></div>
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</article> | What is the formula for phosphorous trichloride? | null |
2,915 | ac95c68c-6ddd-11ea-ab3e-ccda262736ce | https://socratic.org/questions/how-many-grams-would-13-2-moles-of-lead-ii-nitrate-weigh | 4372.10 grams | start physical_unit 7 9 mass g qc_end physical_unit 7 9 4 5 mole qc_end end | [{"type":"physical unit","value":"Weight [OF] lead (II) nitrate [IN] grams"}] | [{"type":"physical unit","value":"4372.10 grams"}] | [{"type":"physical unit","value":"Mole [OF] lead (II) nitrate [=] \\pu{13.2 moles}"}] | <h1 class="questionTitle" itemprop="name">How many grams would 13.2 moles of lead (II) nitrate weigh? </h1> | null | 4372.10 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And how did we get this.....we summed up the individual atomic masses in <mathjax>#Pb(NO_3)_2#</mathjax>:</p>
<p><mathjax>#{1xx207.2+2xx14.01+6xx15.999}*g*mol^-1=331.22*g*mol^-1#</mathjax> </p>
<p>.................as required...........</p>
<p>From where did I get the individual atomic masses? Did I just happen to know them?</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well <mathjax>#"lead nitrate"#</mathjax> has an equivalent mass of <mathjax>#331.2*g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And how did we get this.....we summed up the individual atomic masses in <mathjax>#Pb(NO_3)_2#</mathjax>:</p>
<p><mathjax>#{1xx207.2+2xx14.01+6xx15.999}*g*mol^-1=331.22*g*mol^-1#</mathjax> </p>
<p>.................as required...........</p>
<p>From where did I get the individual atomic masses? Did I just happen to know them?</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams would 13.2 moles of lead (II) nitrate weigh? </h1>
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<div class="markdown"><p>Well <mathjax>#"lead nitrate"#</mathjax> has an equivalent mass of <mathjax>#331.2*g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And how did we get this.....we summed up the individual atomic masses in <mathjax>#Pb(NO_3)_2#</mathjax>:</p>
<p><mathjax>#{1xx207.2+2xx14.01+6xx15.999}*g*mol^-1=331.22*g*mol^-1#</mathjax> </p>
<p>.................as required...........</p>
<p>From where did I get the individual atomic masses? Did I just happen to know them?</p></div>
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</article> | How many grams would 13.2 moles of lead (II) nitrate weigh? | null |
2,916 | a84bf7e8-6ddd-11ea-8293-ccda262736ce | https://socratic.org/questions/1-50-ml-of-a-0-0500-m-solution-of-calcium-chloride-cacl2-was-diluted-with-water- | 0.53 ppm | start physical_unit 26 27 concentration ppm qc_end physical_unit 6 6 0 1 volume qc_end physical_unit 10 10 4 5 concentration qc_end physical_unit 6 6 19 20 volume qc_end substance 14 14 qc_end end | [{"type":"physical unit","value":"Concentration [OF] Cl- ions [IN] ppm"}] | [{"type":"physical unit","value":"0.53 ppm"}] | [{"type":"physical unit","value":"Volume1 [OF] CaCl2 solution [=] \\pu{1.50 mL}"},{"type":"physical unit","value":"Concentration [OF] CaCl2 solution [=] \\pu{0.0500 M}"},{"type":"physical unit","value":"Volume2 [OF] CaCl2 solution [=] \\pu{10.0 L}"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">#"1.50 mL"# of a #"0.0500-M"# solution of calcium chloride, #"CaCl"_2#, was diluted with water to a volume of #"10.0 L"#. What is the concentration of #"Cl"^(–)# ions in the diluted solution in ppm?</h1> | null | 0.53 ppm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that a solution's concentration in <strong>parts per million</strong>, or <strong>ppm</strong>, is supposed to tell you the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present for every</p>
<blockquote>
<p><mathjax>#1,000,000 = 10^6#</mathjax></p>
</blockquote>
<p>grams of solution. Because you'll have a very, very small amount of calcium chloride in the diluted solution, you can assume that the total mass of the solution will be equal to the mass of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, i.e. the mass of water.</p>
<p>This will allow you to use the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of water to find the total mass of the solution, since</p>
<blockquote>
<p><mathjax>#"mass of solution " ~~ " mass of water"#</mathjax></p>
</blockquote>
<p>Since the problem doesn't provide a value for the density of water, you can <em>approximate</em> it to be equal to <mathjax>#"1 g mL"^(-1)#</mathjax>. </p>
<p>Now, the first thing to do here is to figure out exactly how many <em>grams</em> of chloride anions are present in the diluted solution. </p>
<p>Calcium chloride is <strong>soluble</strong> in water, which means that you have</p>
<blockquote>
<p><mathjax>#"CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This tells you that an aqueous solution of calcium chloride has</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = color(red)(2) * ["CaCl"_2]#</mathjax></p>
<blockquote>
<p><em>This is the case because <strong>every mole</strong> of calcium chloride that dissociates produces</em> <mathjax>#color(red)(2)#</mathjax> <em><strong>moles</strong> of chloride anions</em>.</p>
</blockquote>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = color(red)(2) * "0.0500 M" = "0.100 M"#</mathjax></p>
</blockquote>
<p>Use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the <strong>initial solution</strong> to determine how many <em>moles</em> of chloride anions it contains</p>
<blockquote>
<p><mathjax>#1.50 color(red)(cancel(color(black)("mL solution"))) * "0.100 moles Cl"^(-)/(10^3color(red)(cancel(color(black)("mL solution")))) = 1.5 * 10^(-4)#</mathjax> <mathjax>#"moles Cl"^(-)#</mathjax></p>
</blockquote>
<p>As you know, the <em>mass</em> of an ion is approximately <strong>equal</strong> to the mass of the neutral atom, so use the <strong>molar mass</strong> of elemental chlorine to determine how many <em>grams</em> of chloride anions are present in the initial solution</p>
<blockquote>
<p><mathjax>#1.5 * 10^(-4) color(red)(cancel(color(black)("moles Cl"^(-)))) * "35.453 g"/(1color(red)(cancel(color(black)("mole Cl"^(-))))) = 5.318 * 10^(-3)#</mathjax> <mathjax>#"g"#</mathjax></p>
</blockquote>
<p>Now, after you <strong>dilute</strong> this initial solution, the mass of chloride anions <strong>remains unchanged</strong>--this is the case because you're diluting the solution by adding <em>water</em>. </p>
<p>Use the density of water to find the total mass of the diluted solution</p>
<blockquote>
<p><mathjax>#10.0 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = 10^4#</mathjax> <mathjax>#"g"#</mathjax></p>
</blockquote>
<p>So, you know that <mathjax>#10^4#</mathjax> <mathjax>#"g"#</mathjax> of solution contain <mathjax>#5.318 * 10^(-3)#</mathjax> <mathjax>#"g"#</mathjax> of chloride anions, so all you have to do now is figure out how many grams of chloride anions are present in <mathjax>#10^6#</mathjax> <mathjax>#"g"#</mathjax> of this solution</p>
<blockquote>
<p><mathjax>#10^6 color(red)(cancel(color(black)("g solution"))) * (5.318 * 10^(-3)color(white)(.)"g Cl"^(-))/(10^4color(red)(cancel(color(black)("g solution")))) = "0.5318 g Cl"^(-)#</mathjax></p>
</blockquote>
<p>You can thus say that the concentration of the chloride anions in <strong>ppm</strong> will be equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("ppm concnetration = 0.532 ppm Cl"^(-))))#</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 your values. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.532 ppm Cl"^(-)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that a solution's concentration in <strong>parts per million</strong>, or <strong>ppm</strong>, is supposed to tell you the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present for every</p>
<blockquote>
<p><mathjax>#1,000,000 = 10^6#</mathjax></p>
</blockquote>
<p>grams of solution. Because you'll have a very, very small amount of calcium chloride in the diluted solution, you can assume that the total mass of the solution will be equal to the mass of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, i.e. the mass of water.</p>
<p>This will allow you to use the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of water to find the total mass of the solution, since</p>
<blockquote>
<p><mathjax>#"mass of solution " ~~ " mass of water"#</mathjax></p>
</blockquote>
<p>Since the problem doesn't provide a value for the density of water, you can <em>approximate</em> it to be equal to <mathjax>#"1 g mL"^(-1)#</mathjax>. </p>
<p>Now, the first thing to do here is to figure out exactly how many <em>grams</em> of chloride anions are present in the diluted solution. </p>
<p>Calcium chloride is <strong>soluble</strong> in water, which means that you have</p>
<blockquote>
<p><mathjax>#"CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This tells you that an aqueous solution of calcium chloride has</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = color(red)(2) * ["CaCl"_2]#</mathjax></p>
<blockquote>
<p><em>This is the case because <strong>every mole</strong> of calcium chloride that dissociates produces</em> <mathjax>#color(red)(2)#</mathjax> <em><strong>moles</strong> of chloride anions</em>.</p>
</blockquote>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = color(red)(2) * "0.0500 M" = "0.100 M"#</mathjax></p>
</blockquote>
<p>Use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the <strong>initial solution</strong> to determine how many <em>moles</em> of chloride anions it contains</p>
<blockquote>
<p><mathjax>#1.50 color(red)(cancel(color(black)("mL solution"))) * "0.100 moles Cl"^(-)/(10^3color(red)(cancel(color(black)("mL solution")))) = 1.5 * 10^(-4)#</mathjax> <mathjax>#"moles Cl"^(-)#</mathjax></p>
</blockquote>
<p>As you know, the <em>mass</em> of an ion is approximately <strong>equal</strong> to the mass of the neutral atom, so use the <strong>molar mass</strong> of elemental chlorine to determine how many <em>grams</em> of chloride anions are present in the initial solution</p>
<blockquote>
<p><mathjax>#1.5 * 10^(-4) color(red)(cancel(color(black)("moles Cl"^(-)))) * "35.453 g"/(1color(red)(cancel(color(black)("mole Cl"^(-))))) = 5.318 * 10^(-3)#</mathjax> <mathjax>#"g"#</mathjax></p>
</blockquote>
<p>Now, after you <strong>dilute</strong> this initial solution, the mass of chloride anions <strong>remains unchanged</strong>--this is the case because you're diluting the solution by adding <em>water</em>. </p>
<p>Use the density of water to find the total mass of the diluted solution</p>
<blockquote>
<p><mathjax>#10.0 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = 10^4#</mathjax> <mathjax>#"g"#</mathjax></p>
</blockquote>
<p>So, you know that <mathjax>#10^4#</mathjax> <mathjax>#"g"#</mathjax> of solution contain <mathjax>#5.318 * 10^(-3)#</mathjax> <mathjax>#"g"#</mathjax> of chloride anions, so all you have to do now is figure out how many grams of chloride anions are present in <mathjax>#10^6#</mathjax> <mathjax>#"g"#</mathjax> of this solution</p>
<blockquote>
<p><mathjax>#10^6 color(red)(cancel(color(black)("g solution"))) * (5.318 * 10^(-3)color(white)(.)"g Cl"^(-))/(10^4color(red)(cancel(color(black)("g solution")))) = "0.5318 g Cl"^(-)#</mathjax></p>
</blockquote>
<p>You can thus say that the concentration of the chloride anions in <strong>ppm</strong> will be equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("ppm concnetration = 0.532 ppm Cl"^(-))))#</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 your values. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">#"1.50 mL"# of a #"0.0500-M"# solution of calcium chloride, #"CaCl"_2#, was diluted with water to a volume of #"10.0 L"#. What is the concentration of #"Cl"^(–)# ions in the diluted solution in ppm?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2017-07-27T13:53:00" itemprop="dateCreated">
Jul 27, 2017
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<div class="markdown"><p><mathjax>#"0.532 ppm Cl"^(-)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that a solution's concentration in <strong>parts per million</strong>, or <strong>ppm</strong>, is supposed to tell you the number of grams of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present for every</p>
<blockquote>
<p><mathjax>#1,000,000 = 10^6#</mathjax></p>
</blockquote>
<p>grams of solution. Because you'll have a very, very small amount of calcium chloride in the diluted solution, you can assume that the total mass of the solution will be equal to the mass of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, i.e. the mass of water.</p>
<p>This will allow you to use the <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a></strong> of water to find the total mass of the solution, since</p>
<blockquote>
<p><mathjax>#"mass of solution " ~~ " mass of water"#</mathjax></p>
</blockquote>
<p>Since the problem doesn't provide a value for the density of water, you can <em>approximate</em> it to be equal to <mathjax>#"1 g mL"^(-1)#</mathjax>. </p>
<p>Now, the first thing to do here is to figure out exactly how many <em>grams</em> of chloride anions are present in the diluted solution. </p>
<p>Calcium chloride is <strong>soluble</strong> in water, which means that you have</p>
<blockquote>
<p><mathjax>#"CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>This tells you that an aqueous solution of calcium chloride has</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = color(red)(2) * ["CaCl"_2]#</mathjax></p>
<blockquote>
<p><em>This is the case because <strong>every mole</strong> of calcium chloride that dissociates produces</em> <mathjax>#color(red)(2)#</mathjax> <em><strong>moles</strong> of chloride anions</em>.</p>
</blockquote>
</blockquote>
<p>In your case, you will have</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = color(red)(2) * "0.0500 M" = "0.100 M"#</mathjax></p>
</blockquote>
<p>Use the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> and volume of the <strong>initial solution</strong> to determine how many <em>moles</em> of chloride anions it contains</p>
<blockquote>
<p><mathjax>#1.50 color(red)(cancel(color(black)("mL solution"))) * "0.100 moles Cl"^(-)/(10^3color(red)(cancel(color(black)("mL solution")))) = 1.5 * 10^(-4)#</mathjax> <mathjax>#"moles Cl"^(-)#</mathjax></p>
</blockquote>
<p>As you know, the <em>mass</em> of an ion is approximately <strong>equal</strong> to the mass of the neutral atom, so use the <strong>molar mass</strong> of elemental chlorine to determine how many <em>grams</em> of chloride anions are present in the initial solution</p>
<blockquote>
<p><mathjax>#1.5 * 10^(-4) color(red)(cancel(color(black)("moles Cl"^(-)))) * "35.453 g"/(1color(red)(cancel(color(black)("mole Cl"^(-))))) = 5.318 * 10^(-3)#</mathjax> <mathjax>#"g"#</mathjax></p>
</blockquote>
<p>Now, after you <strong>dilute</strong> this initial solution, the mass of chloride anions <strong>remains unchanged</strong>--this is the case because you're diluting the solution by adding <em>water</em>. </p>
<p>Use the density of water to find the total mass of the diluted solution</p>
<blockquote>
<p><mathjax>#10.0 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = 10^4#</mathjax> <mathjax>#"g"#</mathjax></p>
</blockquote>
<p>So, you know that <mathjax>#10^4#</mathjax> <mathjax>#"g"#</mathjax> of solution contain <mathjax>#5.318 * 10^(-3)#</mathjax> <mathjax>#"g"#</mathjax> of chloride anions, so all you have to do now is figure out how many grams of chloride anions are present in <mathjax>#10^6#</mathjax> <mathjax>#"g"#</mathjax> of this solution</p>
<blockquote>
<p><mathjax>#10^6 color(red)(cancel(color(black)("g solution"))) * (5.318 * 10^(-3)color(white)(.)"g Cl"^(-))/(10^4color(red)(cancel(color(black)("g solution")))) = "0.5318 g Cl"^(-)#</mathjax></p>
</blockquote>
<p>You can thus say that the concentration of the chloride anions in <strong>ppm</strong> will be equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("ppm concnetration = 0.532 ppm Cl"^(-))))#</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 your values. </p></div>
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</article> | #"1.50 mL"# of a #"0.0500-M"# solution of calcium chloride, #"CaCl"_2#, was diluted with water to a volume of #"10.0 L"#. What is the concentration of #"Cl"^(–)# ions in the diluted solution in ppm? | null |
2,917 | a8867be8-6ddd-11ea-b9bb-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-the-potassium-permanganate-solution | 0.02 mol/L | start physical_unit 6 8 molarity mol/l qc_end physical_unit 62 63 57 58 volume qc_end physical_unit 62 63 60 61 molarity qc_end physical_unit 6 8 65 66 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"molarity [OF] potassium permanganate solution [IN] mol/L"}] | [{"type":"physical unit","value":"0.02 mol/L"}] | [{"type":"physical unit","value":"Volume [OF] oxalic acid solution samples [=] \\pu{25.00 ml}"},{"type":"physical unit","value":"Volume [OF] sodium hydroxide [=] \\pu{32.15 ml}"},{"type":"physical unit","value":"molarity [OF] sodium hydroxide [=] \\pu{0.1050 M}"},{"type":"physical unit","value":"Volume [OF] potassium permanganate solution [=] \\pu{28.12 ml}"},{"type":"other","value":"An oxalic acid solution exact concentration is determined by an acid-base titration. Then, the oxalic acid solution is used to determine the concentration of a potassium permanganate solutionby a redox titration."}] | <h1 class="questionTitle" itemprop="name">What is the molarity of the potassium permanganate solution?
</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>In order to standardize an oxalic acid solution, its exact concentration is determined by an acid-base titration. Then, the oxalic acid solution is used to determine the concentration of a potassium permanganate solutionby a redox titration. The titration of 25.00 ml samples of the oxalic acid solution requires 32.15 ml of 0.1050M sodium hydroxide and 28.12 ml of the potassium permanganate solution. </p></div>
</h2>
</div>
</div> | 0.02 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxalic (ethandioic) acid is diprotic and can be standardised by titrating against sodium hydroxide solution of known concentration:</p>
<p><mathjax>#sf((COOH)_2+2NaOHrarr(COONa)_2+2H_2O)#</mathjax></p>
<p>Concentration = moles 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>So we can get the no. of moles of sodium hydroxide:</p>
<p><mathjax>#sf(nNaOH=0.1050xx32.15/1000=3.3757xx10^(-3))#</mathjax></p>
<p>From the equation we can see from <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that the number of moles of oxalic acid is given by:</p>
<p><mathjax>#sf(n(COOH)_2=(3.3757xx10^(-3))/2=1.688xx10^(-3))#</mathjax></p>
<p>Since <mathjax>#sf(c=n/v)#</mathjax> we get:</p>
<p><mathjax>#sf([(COOH)_2]=(1.688xx10^(-3))/(25.00/1000)=0.0675color(white)(x)"mol/l")#</mathjax></p>
<p>Note I have converted ml into litre by dividing by 1000.</p>
<p>Now a redox titration is carried out to determine the concentration of the manganate(VII) solution.</p>
<p>The 1/2 equations are:</p>
<p><mathjax>#sf(C_2O_4^(2-)rarr2CO_2+2e" "color(red)((1))#</mathjax> </p>
<p><mathjax>#sf(MnO_4^(-)+8H^(+)+5erarrMn^(2+)+4H_2O" "color(red)((2))#</mathjax></p>
<p>To get the electrons to balance we need to x <mathjax>#sf(color(red)((1)))#</mathjax> by 5 and x <mathjax>#sf(color(red)((2)))#</mathjax> by 2 then add both sides of each 1/2 equation together <mathjax>#sf(rArr)#</mathjax></p>
<p><mathjax>#sf(5C_2O_4^(2-)+2MnO_4^(-)+16H^(+)+cancel(10e)rarr2Mn^(2+)+8H_2O+10CO_2+cancel(10e))#</mathjax></p>
<p>This tells us that 2 moles of Mn(VII) react with 5 moles of oxalic acid.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(n(COOH)_2=cxxv=0.06752xx25.00/1000=1.688xx10^(-3)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(nMnO_4^(-)=1.688xx10^(-3)xx2/5=0.6752xx10^(-3))#</mathjax></p>
<p><mathjax>#sf(c=n/v)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf([MnO_4^(-)]=(0.6752xx10^(-3))/(28.12/1000)=0.0240color(white)(x)"mol/l")#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#sf(0.0240color(white)(x)"mol/l")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxalic (ethandioic) acid is diprotic and can be standardised by titrating against sodium hydroxide solution of known concentration:</p>
<p><mathjax>#sf((COOH)_2+2NaOHrarr(COONa)_2+2H_2O)#</mathjax></p>
<p>Concentration = moles 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>So we can get the no. of moles of sodium hydroxide:</p>
<p><mathjax>#sf(nNaOH=0.1050xx32.15/1000=3.3757xx10^(-3))#</mathjax></p>
<p>From the equation we can see from <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that the number of moles of oxalic acid is given by:</p>
<p><mathjax>#sf(n(COOH)_2=(3.3757xx10^(-3))/2=1.688xx10^(-3))#</mathjax></p>
<p>Since <mathjax>#sf(c=n/v)#</mathjax> we get:</p>
<p><mathjax>#sf([(COOH)_2]=(1.688xx10^(-3))/(25.00/1000)=0.0675color(white)(x)"mol/l")#</mathjax></p>
<p>Note I have converted ml into litre by dividing by 1000.</p>
<p>Now a redox titration is carried out to determine the concentration of the manganate(VII) solution.</p>
<p>The 1/2 equations are:</p>
<p><mathjax>#sf(C_2O_4^(2-)rarr2CO_2+2e" "color(red)((1))#</mathjax> </p>
<p><mathjax>#sf(MnO_4^(-)+8H^(+)+5erarrMn^(2+)+4H_2O" "color(red)((2))#</mathjax></p>
<p>To get the electrons to balance we need to x <mathjax>#sf(color(red)((1)))#</mathjax> by 5 and x <mathjax>#sf(color(red)((2)))#</mathjax> by 2 then add both sides of each 1/2 equation together <mathjax>#sf(rArr)#</mathjax></p>
<p><mathjax>#sf(5C_2O_4^(2-)+2MnO_4^(-)+16H^(+)+cancel(10e)rarr2Mn^(2+)+8H_2O+10CO_2+cancel(10e))#</mathjax></p>
<p>This tells us that 2 moles of Mn(VII) react with 5 moles of oxalic acid.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(n(COOH)_2=cxxv=0.06752xx25.00/1000=1.688xx10^(-3)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(nMnO_4^(-)=1.688xx10^(-3)xx2/5=0.6752xx10^(-3))#</mathjax></p>
<p><mathjax>#sf(c=n/v)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf([MnO_4^(-)]=(0.6752xx10^(-3))/(28.12/1000)=0.0240color(white)(x)"mol/l")#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of the potassium permanganate solution?
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<div class="markdown"><p>In order to standardize an oxalic acid solution, its exact concentration is determined by an acid-base titration. Then, the oxalic acid solution is used to determine the concentration of a potassium permanganate solutionby a redox titration. The titration of 25.00 ml samples of the oxalic acid solution requires 32.15 ml of 0.1050M sodium hydroxide and 28.12 ml of the potassium permanganate solution. </p></div>
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Michael
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<div class="markdown"><p><mathjax>#sf(0.0240color(white)(x)"mol/l")#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxalic (ethandioic) acid is diprotic and can be standardised by titrating against sodium hydroxide solution of known concentration:</p>
<p><mathjax>#sf((COOH)_2+2NaOHrarr(COONa)_2+2H_2O)#</mathjax></p>
<p>Concentration = moles 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>So we can get the no. of moles of sodium hydroxide:</p>
<p><mathjax>#sf(nNaOH=0.1050xx32.15/1000=3.3757xx10^(-3))#</mathjax></p>
<p>From the equation we can see from <a href="https://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio that the number of moles of oxalic acid is given by:</p>
<p><mathjax>#sf(n(COOH)_2=(3.3757xx10^(-3))/2=1.688xx10^(-3))#</mathjax></p>
<p>Since <mathjax>#sf(c=n/v)#</mathjax> we get:</p>
<p><mathjax>#sf([(COOH)_2]=(1.688xx10^(-3))/(25.00/1000)=0.0675color(white)(x)"mol/l")#</mathjax></p>
<p>Note I have converted ml into litre by dividing by 1000.</p>
<p>Now a redox titration is carried out to determine the concentration of the manganate(VII) solution.</p>
<p>The 1/2 equations are:</p>
<p><mathjax>#sf(C_2O_4^(2-)rarr2CO_2+2e" "color(red)((1))#</mathjax> </p>
<p><mathjax>#sf(MnO_4^(-)+8H^(+)+5erarrMn^(2+)+4H_2O" "color(red)((2))#</mathjax></p>
<p>To get the electrons to balance we need to x <mathjax>#sf(color(red)((1)))#</mathjax> by 5 and x <mathjax>#sf(color(red)((2)))#</mathjax> by 2 then add both sides of each 1/2 equation together <mathjax>#sf(rArr)#</mathjax></p>
<p><mathjax>#sf(5C_2O_4^(2-)+2MnO_4^(-)+16H^(+)+cancel(10e)rarr2Mn^(2+)+8H_2O+10CO_2+cancel(10e))#</mathjax></p>
<p>This tells us that 2 moles of Mn(VII) react with 5 moles of oxalic acid.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(n(COOH)_2=cxxv=0.06752xx25.00/1000=1.688xx10^(-3)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(nMnO_4^(-)=1.688xx10^(-3)xx2/5=0.6752xx10^(-3))#</mathjax></p>
<p><mathjax>#sf(c=n/v)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf([MnO_4^(-)]=(0.6752xx10^(-3))/(28.12/1000)=0.0240color(white)(x)"mol/l")#</mathjax></p></div>
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</article> | What is the molarity of the potassium permanganate solution?
|
In order to standardize an oxalic acid solution, its exact concentration is determined by an acid-base titration. Then, the oxalic acid solution is used to determine the concentration of a potassium permanganate solutionby a redox titration. The titration of 25.00 ml samples of the oxalic acid solution requires 32.15 ml of 0.1050M sodium hydroxide and 28.12 ml of the potassium permanganate solution.
|
2,918 | abf33a49-6ddd-11ea-aefe-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-8-moles-of-sodium-chloride | 467.52 g | start physical_unit 8 9 mass g qc_end physical_unit 8 9 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] sodium chloride [IN] g"}] | [{"type":"physical unit","value":"467.52 g"}] | [{"type":"physical unit","value":"Mole [OF] sodium chloride [=] \\pu{8 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 8 moles of sodium chloride?</h1> | null | 467.52 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Multiply the given moles of sodium chloride times the molar mass of sodium chloride.</p>
<p>Molar mass of NaCl is <mathjax>#"58.44g/mol"#</mathjax>. <a href="https://www.ncbi.nlm.nih.gov/pccompound?term=NaCl" rel="nofollow" target="_blank">https://www.ncbi.nlm.nih.gov/pccompound?term=NaCl</a></p>
<p><mathjax>#8cancel"mol NaCl"xx(58.44"g NaCl")/(1cancel"mol NaCl")="467.52 g NaCl"="500 g NaCl"#</mathjax> to one significant figure. </p></div>
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<div class="markdown"><p>The mass of eight moles of sodium chloride is <mathjax>#~~500"g"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Multiply the given moles of sodium chloride times the molar mass of sodium chloride.</p>
<p>Molar mass of NaCl is <mathjax>#"58.44g/mol"#</mathjax>. <a href="https://www.ncbi.nlm.nih.gov/pccompound?term=NaCl" rel="nofollow" target="_blank">https://www.ncbi.nlm.nih.gov/pccompound?term=NaCl</a></p>
<p><mathjax>#8cancel"mol NaCl"xx(58.44"g NaCl")/(1cancel"mol NaCl")="467.52 g NaCl"="500 g NaCl"#</mathjax> to one significant figure. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of 8 moles of sodium chloride?</h1>
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<div class="markdown"><p>The mass of eight moles of sodium chloride is <mathjax>#~~500"g"#</mathjax>.</p></div>
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<div class="markdown"><p>Multiply the given moles of sodium chloride times the molar mass of sodium chloride.</p>
<p>Molar mass of NaCl is <mathjax>#"58.44g/mol"#</mathjax>. <a href="https://www.ncbi.nlm.nih.gov/pccompound?term=NaCl" rel="nofollow" target="_blank">https://www.ncbi.nlm.nih.gov/pccompound?term=NaCl</a></p>
<p><mathjax>#8cancel"mol NaCl"xx(58.44"g NaCl")/(1cancel"mol NaCl")="467.52 g NaCl"="500 g NaCl"#</mathjax> to one significant figure. </p></div>
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</article> | What is the mass of 8 moles of sodium chloride? | null |
2,919 | a99c821f-6ddd-11ea-afd0-ccda262736ce | https://socratic.org/questions/how-do-you-balance-c-6h-6-h-2-c-6h-12 | C6H6 + 3 H2 -> C6H12 | start chemical_equation qc_end chemical_equation 4 8 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"C6H6 + 3 H2 -> C6H12"}] | [{"type":"chemical equation","value":"C6H6 + H2 -> C6H12"}] | <h1 class="questionTitle" itemprop="name">How do you balance #C_6H_6 + H_2 -> C_6H_12#?</h1> | null | C6H6 + 3 H2 -> C6H12 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balancing equations is more of an algorithm than a precise process most of the time. You try one thing, it doesn't work, you try another. It's the way some things go. </p>
<p>To begin with, its helpful to know the amounts of things on each side, how much carbon and hydrogen is reacted and formed afterwards.</p>
<p>On the left we have <mathjax>#6#</mathjax> carbon and <mathjax>#8#</mathjax> hydrogen, while on the right we have <mathjax>#6#</mathjax> carbon and <mathjax>#12#</mathjax> hydrogen. You want all the numbers for each element to be the same. Carbon is already there, six on both sides, it's just hydrogen that you have to sort out. </p>
<p>You need four more hydrogen to make <mathjax>#8#</mathjax> into <mathjax>#12#</mathjax>, preferably without changing the number of carbons, though sometimes this is unavoidable and you have to do a few extra steps to sort this new problem, or try something else. </p>
<p>Luckily you have some elemental hydrogen, <mathjax>#H_2#</mathjax>, which means you can change the number of hydrogen without affecting anything else. When balancing equations, elemental molecules are your friend. </p>
<p>To get <mathjax>#4#</mathjax> more hydrogen atoms, you need <mathjax>#2#</mathjax> more <mathjax>#H_2#</mathjax> molecules, which makes <mathjax>#3#</mathjax> in total, rendering the whole equation</p>
<p><mathjax>#C_6H_6 + 3H_2 -> C_6H_12#</mathjax></p>
<p>There is <mathjax>#6#</mathjax> carbon and <mathjax>#12#</mathjax> hydrogen on the left, and <mathjax>#6#</mathjax> carbon and <mathjax>#12#</mathjax> hydrogen on the right. The equation is balanced. </p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#C_6H_6 + 3H_2 -> C_6H_12#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balancing equations is more of an algorithm than a precise process most of the time. You try one thing, it doesn't work, you try another. It's the way some things go. </p>
<p>To begin with, its helpful to know the amounts of things on each side, how much carbon and hydrogen is reacted and formed afterwards.</p>
<p>On the left we have <mathjax>#6#</mathjax> carbon and <mathjax>#8#</mathjax> hydrogen, while on the right we have <mathjax>#6#</mathjax> carbon and <mathjax>#12#</mathjax> hydrogen. You want all the numbers for each element to be the same. Carbon is already there, six on both sides, it's just hydrogen that you have to sort out. </p>
<p>You need four more hydrogen to make <mathjax>#8#</mathjax> into <mathjax>#12#</mathjax>, preferably without changing the number of carbons, though sometimes this is unavoidable and you have to do a few extra steps to sort this new problem, or try something else. </p>
<p>Luckily you have some elemental hydrogen, <mathjax>#H_2#</mathjax>, which means you can change the number of hydrogen without affecting anything else. When balancing equations, elemental molecules are your friend. </p>
<p>To get <mathjax>#4#</mathjax> more hydrogen atoms, you need <mathjax>#2#</mathjax> more <mathjax>#H_2#</mathjax> molecules, which makes <mathjax>#3#</mathjax> in total, rendering the whole equation</p>
<p><mathjax>#C_6H_6 + 3H_2 -> C_6H_12#</mathjax></p>
<p>There is <mathjax>#6#</mathjax> carbon and <mathjax>#12#</mathjax> hydrogen on the left, and <mathjax>#6#</mathjax> carbon and <mathjax>#12#</mathjax> hydrogen on the right. The equation is balanced. </p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #C_6H_6 + H_2 -> C_6H_12#?</h1>
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<div class="markdown"><p><mathjax>#C_6H_6 + 3H_2 -> C_6H_12#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Balancing equations is more of an algorithm than a precise process most of the time. You try one thing, it doesn't work, you try another. It's the way some things go. </p>
<p>To begin with, its helpful to know the amounts of things on each side, how much carbon and hydrogen is reacted and formed afterwards.</p>
<p>On the left we have <mathjax>#6#</mathjax> carbon and <mathjax>#8#</mathjax> hydrogen, while on the right we have <mathjax>#6#</mathjax> carbon and <mathjax>#12#</mathjax> hydrogen. You want all the numbers for each element to be the same. Carbon is already there, six on both sides, it's just hydrogen that you have to sort out. </p>
<p>You need four more hydrogen to make <mathjax>#8#</mathjax> into <mathjax>#12#</mathjax>, preferably without changing the number of carbons, though sometimes this is unavoidable and you have to do a few extra steps to sort this new problem, or try something else. </p>
<p>Luckily you have some elemental hydrogen, <mathjax>#H_2#</mathjax>, which means you can change the number of hydrogen without affecting anything else. When balancing equations, elemental molecules are your friend. </p>
<p>To get <mathjax>#4#</mathjax> more hydrogen atoms, you need <mathjax>#2#</mathjax> more <mathjax>#H_2#</mathjax> molecules, which makes <mathjax>#3#</mathjax> in total, rendering the whole equation</p>
<p><mathjax>#C_6H_6 + 3H_2 -> C_6H_12#</mathjax></p>
<p>There is <mathjax>#6#</mathjax> carbon and <mathjax>#12#</mathjax> hydrogen on the left, and <mathjax>#6#</mathjax> carbon and <mathjax>#12#</mathjax> hydrogen on the right. The equation is balanced. </p></div>
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</article> | How do you balance #C_6H_6 + H_2 -> C_6H_12#? | null |
2,920 | a9b4ffee-6ddd-11ea-b31f-ccda262736ce | https://socratic.org/questions/25-5-g-of-liquid-benzene-c-6h-6-loses-200-j-of-heat-as-it-cools-what-is-the-temp | 5 ℃ | start physical_unit 5 5 temperature °c qc_end physical_unit 5 5 0 1 mass qc_end physical_unit 5 5 7 8 heat_energy qc_end physical_unit 5 5 21 24 cp qc_end end | [{"type":"physical unit","value":"Decreased temperature [OF] C6H6 [IN] ℃"}] | [{"type":"physical unit","value":"5 ℃"}] | [{"type":"physical unit","value":"Mass [OF] C6H6 [=] \\pu{25.5 g}"},{"type":"physical unit","value":"Lost heat [OF] C6H6 [=] \\pu{200 J}"},{"type":"physical unit","value":"Cp [OF] C6H6 [=] \\pu{1740 J/(kg * ℃)}"}] | <h1 class="questionTitle" itemprop="name">25.5 g of liquid Benzene (#C_6H_6#) loses 200 J of heat as it cools. What is the temperature decrease? </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Cp(<mathjax>#C_6H_6#</mathjax>) = 1740 J/<mathjax>#kg*C#</mathjax></p></div>
</h2>
</div>
</div> | 5 ℃ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the equation:</p>
<p><mathjax>#q=mcDeltaT#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#q#</mathjax> is energy, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#DeltaT#</mathjax> is the change in temperature.</p>
<p><strong>Known</strong></p>
<p><mathjax>#q="200 J"#</mathjax></p>
<p><mathjax>#m="25.5 g"="0.0255 kg"#</mathjax> <br/>
(Mass needs to be in kg due to the units used for the specific heat capacity.)</p>
<p><mathjax>#c_"C6H6"=(1740"J")/("kg"*""^@"C")#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#DeltaT#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate <mathjax>#DeltaT#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#DeltaT=q/(m*c)#</mathjax></p>
<p><mathjax>#DeltaT=(200color(red)cancel(color(black)("J")))/((0.0255color(red)cancel(color(black)("kg")))xx(1740color(red)cancel(color(black)("J")))/(color(red)cancel(color(black)("kg"))*""^@"C"))="5"^@"C"#</mathjax> (rounded to one significant figure)</p>
<p>The temperature decrease is <mathjax>#~~"5"^@"C"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The temperature decrease is <mathjax>#~~"5"^@"C"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the equation:</p>
<p><mathjax>#q=mcDeltaT#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#q#</mathjax> is energy, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#DeltaT#</mathjax> is the change in temperature.</p>
<p><strong>Known</strong></p>
<p><mathjax>#q="200 J"#</mathjax></p>
<p><mathjax>#m="25.5 g"="0.0255 kg"#</mathjax> <br/>
(Mass needs to be in kg due to the units used for the specific heat capacity.)</p>
<p><mathjax>#c_"C6H6"=(1740"J")/("kg"*""^@"C")#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#DeltaT#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate <mathjax>#DeltaT#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#DeltaT=q/(m*c)#</mathjax></p>
<p><mathjax>#DeltaT=(200color(red)cancel(color(black)("J")))/((0.0255color(red)cancel(color(black)("kg")))xx(1740color(red)cancel(color(black)("J")))/(color(red)cancel(color(black)("kg"))*""^@"C"))="5"^@"C"#</mathjax> (rounded to one significant figure)</p>
<p>The temperature decrease is <mathjax>#~~"5"^@"C"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">25.5 g of liquid Benzene (#C_6H_6#) loses 200 J of heat as it cools. What is the temperature decrease? </h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Cp(<mathjax>#C_6H_6#</mathjax>) = 1740 J/<mathjax>#kg*C#</mathjax></p></div>
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<div class="markdown"><p>The temperature decrease is <mathjax>#~~"5"^@"C"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the equation:</p>
<p><mathjax>#q=mcDeltaT#</mathjax>,</p>
<p>where:</p>
<p><mathjax>#q#</mathjax> is energy, <mathjax>#m#</mathjax> is mass, <mathjax>#c#</mathjax> is <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity, and <mathjax>#DeltaT#</mathjax> is the change in temperature.</p>
<p><strong>Known</strong></p>
<p><mathjax>#q="200 J"#</mathjax></p>
<p><mathjax>#m="25.5 g"="0.0255 kg"#</mathjax> <br/>
(Mass needs to be in kg due to the units used for the specific heat capacity.)</p>
<p><mathjax>#c_"C6H6"=(1740"J")/("kg"*""^@"C")#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#DeltaT#</mathjax></p>
<p><strong>Solution</strong></p>
<p>Rearrange the equation to isolate <mathjax>#DeltaT#</mathjax>. Plug in the known values and solve.</p>
<p><mathjax>#DeltaT=q/(m*c)#</mathjax></p>
<p><mathjax>#DeltaT=(200color(red)cancel(color(black)("J")))/((0.0255color(red)cancel(color(black)("kg")))xx(1740color(red)cancel(color(black)("J")))/(color(red)cancel(color(black)("kg"))*""^@"C"))="5"^@"C"#</mathjax> (rounded to one significant figure)</p>
<p>The temperature decrease is <mathjax>#~~"5"^@"C"#</mathjax>.</p></div>
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</article> | 25.5 g of liquid Benzene (#C_6H_6#) loses 200 J of heat as it cools. What is the temperature decrease? |
Cp(#C_6H_6#) = 1740 J/#kg*C#
|
2,921 | a8b863d8-6ddd-11ea-82f1-ccda262736ce | https://socratic.org/questions/58fc88ca7c0149323f27a294 | 8.5 | start physical_unit 14 14 ph none qc_end physical_unit 7 8 3 5 kb qc_end physical_unit 16 16 19 22 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] NaNO2(aq) solution"}] | [{"type":"physical unit","value":"8.5"}] | [{"type":"physical unit","value":"Kb [OF] nitrite ion [=] \\pu{2.22 × 10^(-11)}"},{"type":"physical unit","value":"Molarity [OF] NaNO2(aq) solution [=] \\pu{4.5 × 10^(-4) mol/L}"}] | <h1 class="questionTitle" itemprop="name">Given #K_b=2.22xx10^-11# for nitrite ion, what is #pH# for a solution of #NaNO_2(aq)# that is #4.5xx10^-4*mol*L^-1# with respect to the salt?</h1> | null | 8.5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that sodium nitrite will dissociate completely in aqueous solution to produce sodium cations and nitrite anions, <mathjax>#"NO"_2^(-)#</mathjax>.</p>
<p>The nitrite anions will react with water to form nitrous acid, <mathjax>#"HNO"_2#</mathjax>, and produce hydroxide anions, so right from the start, you should expect the <mathjax>#"pH"#</mathjax> of the solution to be <mathjax>#>7#</mathjax>. </p>
<blockquote>
<p><mathjax>#"NO"_ (2(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HNO"_ (2(aq)) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>For an aqueous solution at <mathjax>#25^@"C"#</mathjax>, the <strong>base dissociation constant</strong>, <mathjax>#K_b#</mathjax>, is equal to </p>
<blockquote>
<p><mathjax>#K_b = (1 * 10^(-14))/K_a#</mathjax></p>
</blockquote>
<p>with <mathjax>#K_a#</mathjax> being the <strong>acid dissociation constant</strong> of the base's conjugate acid. </p>
<p>The problem provides you with the acid dissociation constant of nitrous acid, so you can say that the nitrite anion will have</p>
<blockquote>
<p><mathjax>#K_b = (1 * 10^(-14))/(4.5 * 10^(-4)) = 2.22 * 10^(-11)#</mathjax></p>
</blockquote>
<p>By definition, the base dissociation constant is equal to</p>
<blockquote>
<p><mathjax>#K_b = (["HNO"_2] * ["OH"^(-)])/(["NO"_2^(-)])#</mathjax></p>
</blockquote>
<p>Now, notice that every more of nitrite anions <strong>that react</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of nitrous acid and <mathjax>#1#</mathjax> <strong>mole</strong> of hydroxide anions.</p>
<p>This means that if you take <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> to be the concentration of nitrite anions <strong>that react</strong>, you can say that, at equilibrium, the solution will contain</p>
<blockquote>
<p><mathjax>#["HNO"_2] = xcolor(white)(.)"M"#</mathjax></p>
<p><mathjax>#["OH"^(-)] = xcolor(white)(.)"M"#</mathjax></p>
</blockquote>
<p>The solution will also contain </p>
<blockquote>
<p><mathjax>#["NO"_2^(-)] = "0.5 M" - xcolor(white)(.)"M"#</mathjax></p>
<blockquote>
<p>This basically means that in order for the reaction to <strong>produce</strong> <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of nitrous acid and of hydroxide anions, it must <strong>consume</strong> <mathjax>#x#</mathjax> #"M"3 of nitrite anions. </p>
</blockquote>
</blockquote>
<p>Plug this into the expression of the base dissociation constant to get</p>
<blockquote>
<p><mathjax>#K_b = (x * x)/(0.5 - x)#</mathjax></p>
<p><mathjax>#2.22 * 10^(-11) = x^2/(0.5 - x)#</mathjax></p>
</blockquote>
<p>Now, because the value of the base dissociation constant is so small compared to the initial concentration of the nitrite anions, you can say that</p>
<blockquote>
<p><mathjax>#0.5 - x ~~ 0.5#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#2.22 * 10^(-11) = x^2/0.5#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#x = sqrt(0.5 * 2.22 * 10^(-11)) = 3.33 * 10^(-6)#</mathjax></p>
</blockquote>
<p>The resulting solution will have</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 3.33 * 10^(-6)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p>which means that its <mathjax>#"[pH](https://socratic.org/chemistry/acids-and-bases/the-ph-concept)"#</mathjax> will be equal to</p>
<blockquote>
<p><mathjax>#"pH" = 14 - [-log(["OH"^(-)])]#</mathjax></p>
<p><mathjax>#"pH" = 14 - [-log(3.33 * 10^(-6))]#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 8.5)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong>decimal place</strong>, the number of s<a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">ig figs</a>** you have for the concentration of the nitrite anions. </p>
<p>As predicted, the <mathjax>#"pH"#</mathjax> of the solution is indeed <mathjax>#>7#</mathjax>.</p>
<p><strong>SIDE NOTE</strong> <em>If you want, you can redo the calculations by taking into account the fact that the solution already contains a small concentration of hydroxide anions produced by the auto-ionization of water</em>.</p>
<p><em>In that case, you would have</em></p>
<blockquote>
<p><mathjax>#["OH"^(-)] = (1 * 10^(-7) + x)color(white)(.)"M"#</mathjax></p>
<p><mathjax>#["HNO"_2] = xcolor(white)(.)"M"#</mathjax></p>
<p><mathjax>#["NO"_2^(-)] = (0.5 - x)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p><em>with</em></p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 3.28 * 10^(-6)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p><em>Once again, you will have</em></p>
<blockquote>
<p><mathjax>#"pH" = 8.516 ~~ 8.5 ->#</mathjax> <em>rounded to one decimal place</em></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 8.5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that sodium nitrite will dissociate completely in aqueous solution to produce sodium cations and nitrite anions, <mathjax>#"NO"_2^(-)#</mathjax>.</p>
<p>The nitrite anions will react with water to form nitrous acid, <mathjax>#"HNO"_2#</mathjax>, and produce hydroxide anions, so right from the start, you should expect the <mathjax>#"pH"#</mathjax> of the solution to be <mathjax>#>7#</mathjax>. </p>
<blockquote>
<p><mathjax>#"NO"_ (2(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HNO"_ (2(aq)) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>For an aqueous solution at <mathjax>#25^@"C"#</mathjax>, the <strong>base dissociation constant</strong>, <mathjax>#K_b#</mathjax>, is equal to </p>
<blockquote>
<p><mathjax>#K_b = (1 * 10^(-14))/K_a#</mathjax></p>
</blockquote>
<p>with <mathjax>#K_a#</mathjax> being the <strong>acid dissociation constant</strong> of the base's conjugate acid. </p>
<p>The problem provides you with the acid dissociation constant of nitrous acid, so you can say that the nitrite anion will have</p>
<blockquote>
<p><mathjax>#K_b = (1 * 10^(-14))/(4.5 * 10^(-4)) = 2.22 * 10^(-11)#</mathjax></p>
</blockquote>
<p>By definition, the base dissociation constant is equal to</p>
<blockquote>
<p><mathjax>#K_b = (["HNO"_2] * ["OH"^(-)])/(["NO"_2^(-)])#</mathjax></p>
</blockquote>
<p>Now, notice that every more of nitrite anions <strong>that react</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of nitrous acid and <mathjax>#1#</mathjax> <strong>mole</strong> of hydroxide anions.</p>
<p>This means that if you take <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> to be the concentration of nitrite anions <strong>that react</strong>, you can say that, at equilibrium, the solution will contain</p>
<blockquote>
<p><mathjax>#["HNO"_2] = xcolor(white)(.)"M"#</mathjax></p>
<p><mathjax>#["OH"^(-)] = xcolor(white)(.)"M"#</mathjax></p>
</blockquote>
<p>The solution will also contain </p>
<blockquote>
<p><mathjax>#["NO"_2^(-)] = "0.5 M" - xcolor(white)(.)"M"#</mathjax></p>
<blockquote>
<p>This basically means that in order for the reaction to <strong>produce</strong> <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of nitrous acid and of hydroxide anions, it must <strong>consume</strong> <mathjax>#x#</mathjax> #"M"3 of nitrite anions. </p>
</blockquote>
</blockquote>
<p>Plug this into the expression of the base dissociation constant to get</p>
<blockquote>
<p><mathjax>#K_b = (x * x)/(0.5 - x)#</mathjax></p>
<p><mathjax>#2.22 * 10^(-11) = x^2/(0.5 - x)#</mathjax></p>
</blockquote>
<p>Now, because the value of the base dissociation constant is so small compared to the initial concentration of the nitrite anions, you can say that</p>
<blockquote>
<p><mathjax>#0.5 - x ~~ 0.5#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#2.22 * 10^(-11) = x^2/0.5#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#x = sqrt(0.5 * 2.22 * 10^(-11)) = 3.33 * 10^(-6)#</mathjax></p>
</blockquote>
<p>The resulting solution will have</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 3.33 * 10^(-6)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p>which means that its <mathjax>#"[pH](https://socratic.org/chemistry/acids-and-bases/the-ph-concept)"#</mathjax> will be equal to</p>
<blockquote>
<p><mathjax>#"pH" = 14 - [-log(["OH"^(-)])]#</mathjax></p>
<p><mathjax>#"pH" = 14 - [-log(3.33 * 10^(-6))]#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 8.5)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong>decimal place</strong>, the number of s<a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">ig figs</a>** you have for the concentration of the nitrite anions. </p>
<p>As predicted, the <mathjax>#"pH"#</mathjax> of the solution is indeed <mathjax>#>7#</mathjax>.</p>
<p><strong>SIDE NOTE</strong> <em>If you want, you can redo the calculations by taking into account the fact that the solution already contains a small concentration of hydroxide anions produced by the auto-ionization of water</em>.</p>
<p><em>In that case, you would have</em></p>
<blockquote>
<p><mathjax>#["OH"^(-)] = (1 * 10^(-7) + x)color(white)(.)"M"#</mathjax></p>
<p><mathjax>#["HNO"_2] = xcolor(white)(.)"M"#</mathjax></p>
<p><mathjax>#["NO"_2^(-)] = (0.5 - x)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p><em>with</em></p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 3.28 * 10^(-6)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p><em>Once again, you will have</em></p>
<blockquote>
<p><mathjax>#"pH" = 8.516 ~~ 8.5 ->#</mathjax> <em>rounded to one decimal place</em></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Given #K_b=2.22xx10^-11# for nitrite ion, what is #pH# for a solution of #NaNO_2(aq)# that is #4.5xx10^-4*mol*L^-1# with respect to the salt?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2018-01-03T14:26:58" itemprop="dateCreated">
Jan 3, 2018
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<div class="markdown"><p><mathjax>#"pH" = 8.5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that sodium nitrite will dissociate completely in aqueous solution to produce sodium cations and nitrite anions, <mathjax>#"NO"_2^(-)#</mathjax>.</p>
<p>The nitrite anions will react with water to form nitrous acid, <mathjax>#"HNO"_2#</mathjax>, and produce hydroxide anions, so right from the start, you should expect the <mathjax>#"pH"#</mathjax> of the solution to be <mathjax>#>7#</mathjax>. </p>
<blockquote>
<p><mathjax>#"NO"_ (2(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HNO"_ (2(aq)) + "OH"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>For an aqueous solution at <mathjax>#25^@"C"#</mathjax>, the <strong>base dissociation constant</strong>, <mathjax>#K_b#</mathjax>, is equal to </p>
<blockquote>
<p><mathjax>#K_b = (1 * 10^(-14))/K_a#</mathjax></p>
</blockquote>
<p>with <mathjax>#K_a#</mathjax> being the <strong>acid dissociation constant</strong> of the base's conjugate acid. </p>
<p>The problem provides you with the acid dissociation constant of nitrous acid, so you can say that the nitrite anion will have</p>
<blockquote>
<p><mathjax>#K_b = (1 * 10^(-14))/(4.5 * 10^(-4)) = 2.22 * 10^(-11)#</mathjax></p>
</blockquote>
<p>By definition, the base dissociation constant is equal to</p>
<blockquote>
<p><mathjax>#K_b = (["HNO"_2] * ["OH"^(-)])/(["NO"_2^(-)])#</mathjax></p>
</blockquote>
<p>Now, notice that every more of nitrite anions <strong>that react</strong> produces <mathjax>#1#</mathjax> <strong>mole</strong> of nitrous acid and <mathjax>#1#</mathjax> <strong>mole</strong> of hydroxide anions.</p>
<p>This means that if you take <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> to be the concentration of nitrite anions <strong>that react</strong>, you can say that, at equilibrium, the solution will contain</p>
<blockquote>
<p><mathjax>#["HNO"_2] = xcolor(white)(.)"M"#</mathjax></p>
<p><mathjax>#["OH"^(-)] = xcolor(white)(.)"M"#</mathjax></p>
</blockquote>
<p>The solution will also contain </p>
<blockquote>
<p><mathjax>#["NO"_2^(-)] = "0.5 M" - xcolor(white)(.)"M"#</mathjax></p>
<blockquote>
<p>This basically means that in order for the reaction to <strong>produce</strong> <mathjax>#x#</mathjax> <mathjax>#"M"#</mathjax> of nitrous acid and of hydroxide anions, it must <strong>consume</strong> <mathjax>#x#</mathjax> #"M"3 of nitrite anions. </p>
</blockquote>
</blockquote>
<p>Plug this into the expression of the base dissociation constant to get</p>
<blockquote>
<p><mathjax>#K_b = (x * x)/(0.5 - x)#</mathjax></p>
<p><mathjax>#2.22 * 10^(-11) = x^2/(0.5 - x)#</mathjax></p>
</blockquote>
<p>Now, because the value of the base dissociation constant is so small compared to the initial concentration of the nitrite anions, you can say that</p>
<blockquote>
<p><mathjax>#0.5 - x ~~ 0.5#</mathjax></p>
</blockquote>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#2.22 * 10^(-11) = x^2/0.5#</mathjax></p>
</blockquote>
<p>which gets you </p>
<blockquote>
<p><mathjax>#x = sqrt(0.5 * 2.22 * 10^(-11)) = 3.33 * 10^(-6)#</mathjax></p>
</blockquote>
<p>The resulting solution will have</p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 3.33 * 10^(-6)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p>which means that its <mathjax>#"[pH](https://socratic.org/chemistry/acids-and-bases/the-ph-concept)"#</mathjax> will be equal to</p>
<blockquote>
<p><mathjax>#"pH" = 14 - [-log(["OH"^(-)])]#</mathjax></p>
<p><mathjax>#"pH" = 14 - [-log(3.33 * 10^(-6))]#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = 8.5)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to one <strong>decimal place</strong>, the number of s<a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">ig figs</a>** you have for the concentration of the nitrite anions. </p>
<p>As predicted, the <mathjax>#"pH"#</mathjax> of the solution is indeed <mathjax>#>7#</mathjax>.</p>
<p><strong>SIDE NOTE</strong> <em>If you want, you can redo the calculations by taking into account the fact that the solution already contains a small concentration of hydroxide anions produced by the auto-ionization of water</em>.</p>
<p><em>In that case, you would have</em></p>
<blockquote>
<p><mathjax>#["OH"^(-)] = (1 * 10^(-7) + x)color(white)(.)"M"#</mathjax></p>
<p><mathjax>#["HNO"_2] = xcolor(white)(.)"M"#</mathjax></p>
<p><mathjax>#["NO"_2^(-)] = (0.5 - x)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p><em>with</em></p>
<blockquote>
<p><mathjax>#["OH"^(-)] = 3.28 * 10^(-6)color(white)(.)"M"#</mathjax></p>
</blockquote>
<p><em>Once again, you will have</em></p>
<blockquote>
<p><mathjax>#"pH" = 8.516 ~~ 8.5 ->#</mathjax> <em>rounded to one decimal place</em></p>
</blockquote></div>
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</article> | Given #K_b=2.22xx10^-11# for nitrite ion, what is #pH# for a solution of #NaNO_2(aq)# that is #4.5xx10^-4*mol*L^-1# with respect to the salt? | null |
2,922 | a92df1b8-6ddd-11ea-b9e4-ccda262736ce | https://socratic.org/questions/the-pressure-on-a-sample-of-an-ideal-gas-is-increased-from-715-mmhg-to-3-55-atm- | 129 mL | start physical_unit 25 26 volume ml qc_end physical_unit 4 8 12 13 pressure qc_end physical_unit 4 8 15 16 pressure qc_end c_other constant_temperature qc_end physical_unit 25 26 28 29 volume qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"129 mL"}] | [{"type":"physical unit","value":"Pressure1 [OF] an ideal gas sample [=] \\pu{715 mmHg}"},{"type":"physical unit","value":"Pressure2 [OF] an ideal gas sample [=] \\pu{3.55 atm}"},{"type":"other","value":"ConstantTemperature"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{485 mL}"}] | <h1 class="questionTitle" itemprop="name">The pressure on a sample of an ideal gas is increased from 715 mmHg to 3.55 atm at constant temperature. If the initial volume of the gas is 485 mL, what is the final volume of the gas? </h1> | null | 129 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant <mathjax>#T#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax>; this is Boyle's law.</p>
<p>Now <mathjax>#P_1=(715*mm*Hg)/(760*mm*Hg*atm^-1)=0.941*atm#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(0.941*cancel(atm)xx485*mL)/(3.55*cancel(atm))=129*mL#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The gas is compressed to a volume of <mathjax>#129*mL#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant <mathjax>#T#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax>; this is Boyle's law.</p>
<p>Now <mathjax>#P_1=(715*mm*Hg)/(760*mm*Hg*atm^-1)=0.941*atm#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(0.941*cancel(atm)xx485*mL)/(3.55*cancel(atm))=129*mL#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The pressure on a sample of an ideal gas is increased from 715 mmHg to 3.55 atm at constant temperature. If the initial volume of the gas is 485 mL, what is the final volume of the gas? </h1>
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<div class="markdown"><p>The gas is compressed to a volume of <mathjax>#129*mL#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant <mathjax>#T#</mathjax>, <mathjax>#P_1V_1=P_2V_2#</mathjax>; this is Boyle's law.</p>
<p>Now <mathjax>#P_1=(715*mm*Hg)/(760*mm*Hg*atm^-1)=0.941*atm#</mathjax></p>
<p><mathjax>#V_2=(P_1V_1)/P_2=(0.941*cancel(atm)xx485*mL)/(3.55*cancel(atm))=129*mL#</mathjax></p></div>
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</article> | The pressure on a sample of an ideal gas is increased from 715 mmHg to 3.55 atm at constant temperature. If the initial volume of the gas is 485 mL, what is the final volume of the gas? | null |
2,923 | ad210335-6ddd-11ea-aa4f-ccda262736ce | https://socratic.org/questions/what-coefficient-would-the-o-2-have-after-balancing-c-4h-8-o-2-co-2-h-2o | 6 | start physical_unit 4 4 coefficient none qc_end chemical_equation 8 14 qc_end end | [{"type":"physical unit","value":"Coefficient [OF] O2"}] | [{"type":"physical unit","value":"6"}] | [{"type":"chemical equation","value":"C4H8 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">What coefficient would the #O_2# have after balancing #C_4H_8 + O_2 -> CO_2 + H_2O#?</h1> | null | 6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balancement of the reaction requires the conservation of mass, each atom of carbon in <mathjax>#C_4H_8#</mathjax> consumes one <mathjax>#O_2#</mathjax> molecule, and every two atoms of hydrogen one oxygen atom is consumed.<br/>
So 4 molecules of <mathjax>#O_2#</mathjax> used by carbon add to 2 molecules of <mathjax>#O_2#</mathjax> used by hydrogen, in total 6 molecules of <mathjax>#O_2#</mathjax> <br/>
The balanced equation is:<br/>
<mathjax>#C_4H_8 + 6 O_2 rarr 4CO_2 + 4H_2O#</mathjax></p></div>
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<div class="markdown"><p>the coefficient is 6</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balancement of the reaction requires the conservation of mass, each atom of carbon in <mathjax>#C_4H_8#</mathjax> consumes one <mathjax>#O_2#</mathjax> molecule, and every two atoms of hydrogen one oxygen atom is consumed.<br/>
So 4 molecules of <mathjax>#O_2#</mathjax> used by carbon add to 2 molecules of <mathjax>#O_2#</mathjax> used by hydrogen, in total 6 molecules of <mathjax>#O_2#</mathjax> <br/>
The balanced equation is:<br/>
<mathjax>#C_4H_8 + 6 O_2 rarr 4CO_2 + 4H_2O#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What coefficient would the #O_2# have after balancing #C_4H_8 + O_2 -> CO_2 + H_2O#?</h1>
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<div class="markdown"><p>the coefficient is 6</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The balancement of the reaction requires the conservation of mass, each atom of carbon in <mathjax>#C_4H_8#</mathjax> consumes one <mathjax>#O_2#</mathjax> molecule, and every two atoms of hydrogen one oxygen atom is consumed.<br/>
So 4 molecules of <mathjax>#O_2#</mathjax> used by carbon add to 2 molecules of <mathjax>#O_2#</mathjax> used by hydrogen, in total 6 molecules of <mathjax>#O_2#</mathjax> <br/>
The balanced equation is:<br/>
<mathjax>#C_4H_8 + 6 O_2 rarr 4CO_2 + 4H_2O#</mathjax></p></div>
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</article> | What coefficient would the #O_2# have after balancing #C_4H_8 + O_2 -> CO_2 + H_2O#? | null |
2,924 | a866b1f6-6ddd-11ea-941c-ccda262736ce | https://socratic.org/questions/a-50-ml-sample-of-0-200-m-sodium-hydroxide-is-titrated-with-0-200-m-hcl-what-is- | 13.3 | start physical_unit 7 8 ph none qc_end physical_unit 7 8 1 2 volume qc_end physical_unit 7 8 5 6 molarity qc_end physical_unit 14 14 5 6 molarity qc_end end | [{"type":"physical unit","value":"Initial pH [OF] sodium hydroxide sample"}] | [{"type":"physical unit","value":"13.3"}] | [{"type":"physical unit","value":"Volume [OF] sodium hydroxide sample [=] \\pu{50 mL}"},{"type":"physical unit","value":"Molarity [OF] sodium hydroxide sample [=] \\pu{0.200 M}"},{"type":"physical unit","value":"Molarity [OF] HCl [=] \\pu{0.200 M}"}] | <h1 class="questionTitle" itemprop="name">A 50 mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M HCl. What is the initial pH?</h1> | null | 13.3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is well known that in aqueous solution, that</p>
<p><mathjax>#pH + pOH=14#</mathjax></p>
<p><mathjax>#pH=14-pOH#</mathjax></p>
<p><mathjax>#pOH=-log_(10)[HO^-]=-log_10(0.200)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.70#</mathjax></p>
<p><mathjax>#14-0.70=13.3#</mathjax></p>
<p><mathjax>#pH#</mathjax> should reasonably be high inasmuch as we measured it BEFORE the start of the titration. Given this, the information regarding the acid was a distractor. What will the <mathjax>#pH#</mathjax> be when the <mathjax>#50.0*mL#</mathjax> volume of <mathjax>#HCl#</mathjax> is added?</p></div>
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<div class="markdown"><p><mathjax>#pH=13.3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is well known that in aqueous solution, that</p>
<p><mathjax>#pH + pOH=14#</mathjax></p>
<p><mathjax>#pH=14-pOH#</mathjax></p>
<p><mathjax>#pOH=-log_(10)[HO^-]=-log_10(0.200)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.70#</mathjax></p>
<p><mathjax>#14-0.70=13.3#</mathjax></p>
<p><mathjax>#pH#</mathjax> should reasonably be high inasmuch as we measured it BEFORE the start of the titration. Given this, the information regarding the acid was a distractor. What will the <mathjax>#pH#</mathjax> be when the <mathjax>#50.0*mL#</mathjax> volume of <mathjax>#HCl#</mathjax> is added?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A 50 mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M HCl. What is the initial pH?</h1>
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Aug 22, 2016
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<div class="markdown"><p><mathjax>#pH=13.3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>It is well known that in aqueous solution, that</p>
<p><mathjax>#pH + pOH=14#</mathjax></p>
<p><mathjax>#pH=14-pOH#</mathjax></p>
<p><mathjax>#pOH=-log_(10)[HO^-]=-log_10(0.200)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.70#</mathjax></p>
<p><mathjax>#14-0.70=13.3#</mathjax></p>
<p><mathjax>#pH#</mathjax> should reasonably be high inasmuch as we measured it BEFORE the start of the titration. Given this, the information regarding the acid was a distractor. What will the <mathjax>#pH#</mathjax> be when the <mathjax>#50.0*mL#</mathjax> volume of <mathjax>#HCl#</mathjax> is added?</p></div>
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</article> | A 50 mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M HCl. What is the initial pH? | null |
2,925 | ab2605b6-6ddd-11ea-bce1-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-for-a-compound-that-is-33-36-calcium-26-69-sulfur- | CaSO3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"CaSO3"}] | [{"type":"physical unit","value":"Percent [OF] calcium in the compound [=] \\pu{33.36%}"},{"type":"physical unit","value":"Percent [OF] sulfur in the compound [=] \\pu{26.69%}"},{"type":"physical unit","value":"Percent [OF] oxygen in the compound [=] \\pu{40.00%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound that is 33.36% calcium, 26.69% sulfur, and 40.00% oxygen? </h1> | null | CaSO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>1:</strong> We have to do is <strong>assume 100g</strong>.</p>
<p>Therefore: <br/>
- Calcium: <mathjax>#33.36% -> 33.36g#</mathjax> <br/>
- Sulfur: <mathjax>#26.69% -> 26.69g#</mathjax> <br/>
- Oxygen: <mathjax>#40.00% -> 40.00g#</mathjax> </p>
<p>We do this because the percent is in relation to 100%. By assuming 100%, all the values can be viewed as values in grams. </p>
<p><strong>2:</strong> We find the mols of each element. </p>
<p>Now that we have the mass, the molar mass is the individual molar mass of each element. </p>
<blockquote>
<p><mathjax>#n_"C" = m/M#</mathjax> </p>
<p><mathjax>#=33.36/40.08#</mathjax></p>
<p><mathjax>#=0.832335329#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#n_"S" = m/M#</mathjax> </p>
<p><mathjax>#=26.69/32.07#</mathjax></p>
<p><mathjax>#=0.83224197#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#n_"O" = m/M#</mathjax></p>
<p><mathjax>#=40.00/16.00#</mathjax></p>
<p><mathjax>#=2.5#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p><strong>3:</strong> Now we take the lowest value of mols and divide every other mol by that. </p>
<p>Calcium has the smallest mol: <mathjax>#0.832335329#</mathjax></p>
<p>Calcium: </p>
<blockquote>
<p><mathjax>#=0.832335329/0.832335329#</mathjax></p>
<p><mathjax>#=1#</mathjax></p>
</blockquote>
<p>Sulfur: </p>
<blockquote>
<p><mathjax>#=0.83224197/0.832335329#</mathjax></p>
<p><mathjax>#=1#</mathjax></p>
</blockquote>
<p>Oxygen: </p>
<blockquote>
<p><mathjax>#=2.5/0.832335329#</mathjax></p>
<p><mathjax>#=3#</mathjax></p>
</blockquote>
<p>Therefore, the empirical formula is <mathjax>#CaSO_"3"#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The empirical formula is <mathjax>#CaSO_"3"#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>1:</strong> We have to do is <strong>assume 100g</strong>.</p>
<p>Therefore: <br/>
- Calcium: <mathjax>#33.36% -> 33.36g#</mathjax> <br/>
- Sulfur: <mathjax>#26.69% -> 26.69g#</mathjax> <br/>
- Oxygen: <mathjax>#40.00% -> 40.00g#</mathjax> </p>
<p>We do this because the percent is in relation to 100%. By assuming 100%, all the values can be viewed as values in grams. </p>
<p><strong>2:</strong> We find the mols of each element. </p>
<p>Now that we have the mass, the molar mass is the individual molar mass of each element. </p>
<blockquote>
<p><mathjax>#n_"C" = m/M#</mathjax> </p>
<p><mathjax>#=33.36/40.08#</mathjax></p>
<p><mathjax>#=0.832335329#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#n_"S" = m/M#</mathjax> </p>
<p><mathjax>#=26.69/32.07#</mathjax></p>
<p><mathjax>#=0.83224197#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#n_"O" = m/M#</mathjax></p>
<p><mathjax>#=40.00/16.00#</mathjax></p>
<p><mathjax>#=2.5#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p><strong>3:</strong> Now we take the lowest value of mols and divide every other mol by that. </p>
<p>Calcium has the smallest mol: <mathjax>#0.832335329#</mathjax></p>
<p>Calcium: </p>
<blockquote>
<p><mathjax>#=0.832335329/0.832335329#</mathjax></p>
<p><mathjax>#=1#</mathjax></p>
</blockquote>
<p>Sulfur: </p>
<blockquote>
<p><mathjax>#=0.83224197/0.832335329#</mathjax></p>
<p><mathjax>#=1#</mathjax></p>
</blockquote>
<p>Oxygen: </p>
<blockquote>
<p><mathjax>#=2.5/0.832335329#</mathjax></p>
<p><mathjax>#=3#</mathjax></p>
</blockquote>
<p>Therefore, the empirical formula is <mathjax>#CaSO_"3"#</mathjax>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the empirical formula for a compound that is 33.36% calcium, 26.69% sulfur, and 40.00% oxygen? </h1>
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<div class="markdown"><p>The empirical formula is <mathjax>#CaSO_"3"#</mathjax>. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>1:</strong> We have to do is <strong>assume 100g</strong>.</p>
<p>Therefore: <br/>
- Calcium: <mathjax>#33.36% -> 33.36g#</mathjax> <br/>
- Sulfur: <mathjax>#26.69% -> 26.69g#</mathjax> <br/>
- Oxygen: <mathjax>#40.00% -> 40.00g#</mathjax> </p>
<p>We do this because the percent is in relation to 100%. By assuming 100%, all the values can be viewed as values in grams. </p>
<p><strong>2:</strong> We find the mols of each element. </p>
<p>Now that we have the mass, the molar mass is the individual molar mass of each element. </p>
<blockquote>
<p><mathjax>#n_"C" = m/M#</mathjax> </p>
<p><mathjax>#=33.36/40.08#</mathjax></p>
<p><mathjax>#=0.832335329#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#n_"S" = m/M#</mathjax> </p>
<p><mathjax>#=26.69/32.07#</mathjax></p>
<p><mathjax>#=0.83224197#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax>#n_"O" = m/M#</mathjax></p>
<p><mathjax>#=40.00/16.00#</mathjax></p>
<p><mathjax>#=2.5#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p><strong>3:</strong> Now we take the lowest value of mols and divide every other mol by that. </p>
<p>Calcium has the smallest mol: <mathjax>#0.832335329#</mathjax></p>
<p>Calcium: </p>
<blockquote>
<p><mathjax>#=0.832335329/0.832335329#</mathjax></p>
<p><mathjax>#=1#</mathjax></p>
</blockquote>
<p>Sulfur: </p>
<blockquote>
<p><mathjax>#=0.83224197/0.832335329#</mathjax></p>
<p><mathjax>#=1#</mathjax></p>
</blockquote>
<p>Oxygen: </p>
<blockquote>
<p><mathjax>#=2.5/0.832335329#</mathjax></p>
<p><mathjax>#=3#</mathjax></p>
</blockquote>
<p>Therefore, the empirical formula is <mathjax>#CaSO_"3"#</mathjax>. </p></div>
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</article> | What is the empirical formula for a compound that is 33.36% calcium, 26.69% sulfur, and 40.00% oxygen? | null |
2,926 | ace66850-6ddd-11ea-b4e6-ccda262736ce | https://socratic.org/questions/how-many-milli-grams-of-fe0-9o-reacts-completely-with-10-ml-0-1-m-kmno4-solution | 400 milli grams | start physical_unit 5 5 mass mg qc_end physical_unit 18 18 20 21 molar_mass qc_end c_other OTHER qc_end c_other OTHER qc_end physical_unit 13 14 9 10 volume qc_end physical_unit 13 14 11 12 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] Fe0.9O [IN] milli grams"}] | [{"type":"physical unit","value":"400 milli grams"}] | [{"type":"physical unit","value":"Molar mass [OF] Fe [=] \\pu{56 g/mol}"},{"type":"other","value":"React completely."},{"type":"other","value":"In acidic conditions.."},{"type":"physical unit","value":"Volume [OF] KMnO4 solution [=] \\pu{10 mL}"},{"type":"physical unit","value":"Molarity [OF] KMnO4 solution [=] \\pu{0.1 M}"}] | <h1 class="questionTitle" itemprop="name">How many milli grams of #"Fe"_0.9"O"# reacts completely with 10 mL 0.1 M KMnO4 solution in acidic conditions. (Fe = 56)?
thanking you in anticipation.</h1> | null | 400 milli grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Nonstoichiometric <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a></strong></p>
<p><mathjax>#"Fe"_0.9"O"#</mathjax> is a <strong>nonstoichiometric compound</strong>.</p>
<p>Some of the <mathjax>#"Fe"^"2+"#</mathjax> ions have been oxidized to <mathjax>#"Fe"^"3+"#</mathjax>.</p>
<p>Thus, to balance the charge, the compound contains two <mathjax>#"Fe"^"3+"#</mathjax> ions for every three "missing" <mathjax>#"Fe"^"2+"#</mathjax> ions.</p>
<p>There is still the same mass of <mathjax>#"Fe"#</mathjax>, but only 90 % of it is in the +2 oxidation state.</p>
<blockquote></blockquote>
<p><strong><a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">Stoichiometry</a></strong></p>
<p>The system will behave as if the reaction goes only to 90 % completion.</p>
<p>The equation for the reaction is</p>
<p><mathjax>#"5FeO"color(white)(l) + "KMnO"_4 + "18H"^"+" → "5Fe"^"3+" + "Mn"^"2+" + "K"^"+" + "9H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Calculations</strong></p>
<p><mathjax>#"Moles of KMnO"_4 color(white)(l)"used" = 0.010 color(red)(cancel(color(black)("L KMnO"_4))) × "0.1 mol MnO"_4/(1 color(red)(cancel(color(black)("L KMnO"_4)))) = "0.001 mol KMnO"_4#</mathjax></p>
<p><mathjax>#"Moles of FeO reacted" = 0.001 color(red)(cancel(color(black)("mol KMnO"_4))) × "5 mol FeO"/(1 color(red)(cancel(color(black)("mol KMnO"_4)))) = "0.005 mol FeO"#</mathjax></p>
<p><mathjax>#"Moles of Fe"_0.9"O" = 0.005 color(red)(cancel(color(black)("mol FeO"))) × ("100 mol Fe"_0.9"O")/(90 color(red)(cancel(color(black)("mol FeO")))) = "0.0056 mol Fe"_0.9"O"#</mathjax></p>
<p><mathjax>#"Mass of Fe"_0.9"O" = 0.0056 color(red)(cancel(color(black)("mol Fe"_0.9"O"))) × ("71.84 g Fe"_0.9"O")/(1 color(red)(cancel(color(black)("mol Fe"_0.9"O")))) = "0.4 g Fe"_0.9"O" = "400 mg Fe"_0.9"O"#</mathjax></p>
<p><strong>Note</strong>: The answer can have only 1 significant figure, because that is all you gave for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the <mathjax>#"KMnO"_4#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The <mathjax>#"KMnO"_4#</mathjax> will react with 400 mg of <mathjax>#"Fe"_0.9"O"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Nonstoichiometric <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a></strong></p>
<p><mathjax>#"Fe"_0.9"O"#</mathjax> is a <strong>nonstoichiometric compound</strong>.</p>
<p>Some of the <mathjax>#"Fe"^"2+"#</mathjax> ions have been oxidized to <mathjax>#"Fe"^"3+"#</mathjax>.</p>
<p>Thus, to balance the charge, the compound contains two <mathjax>#"Fe"^"3+"#</mathjax> ions for every three "missing" <mathjax>#"Fe"^"2+"#</mathjax> ions.</p>
<p>There is still the same mass of <mathjax>#"Fe"#</mathjax>, but only 90 % of it is in the +2 oxidation state.</p>
<blockquote></blockquote>
<p><strong><a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">Stoichiometry</a></strong></p>
<p>The system will behave as if the reaction goes only to 90 % completion.</p>
<p>The equation for the reaction is</p>
<p><mathjax>#"5FeO"color(white)(l) + "KMnO"_4 + "18H"^"+" → "5Fe"^"3+" + "Mn"^"2+" + "K"^"+" + "9H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Calculations</strong></p>
<p><mathjax>#"Moles of KMnO"_4 color(white)(l)"used" = 0.010 color(red)(cancel(color(black)("L KMnO"_4))) × "0.1 mol MnO"_4/(1 color(red)(cancel(color(black)("L KMnO"_4)))) = "0.001 mol KMnO"_4#</mathjax></p>
<p><mathjax>#"Moles of FeO reacted" = 0.001 color(red)(cancel(color(black)("mol KMnO"_4))) × "5 mol FeO"/(1 color(red)(cancel(color(black)("mol KMnO"_4)))) = "0.005 mol FeO"#</mathjax></p>
<p><mathjax>#"Moles of Fe"_0.9"O" = 0.005 color(red)(cancel(color(black)("mol FeO"))) × ("100 mol Fe"_0.9"O")/(90 color(red)(cancel(color(black)("mol FeO")))) = "0.0056 mol Fe"_0.9"O"#</mathjax></p>
<p><mathjax>#"Mass of Fe"_0.9"O" = 0.0056 color(red)(cancel(color(black)("mol Fe"_0.9"O"))) × ("71.84 g Fe"_0.9"O")/(1 color(red)(cancel(color(black)("mol Fe"_0.9"O")))) = "0.4 g Fe"_0.9"O" = "400 mg Fe"_0.9"O"#</mathjax></p>
<p><strong>Note</strong>: The answer can have only 1 significant figure, because that is all you gave for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the <mathjax>#"KMnO"_4#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many milli grams of #"Fe"_0.9"O"# reacts completely with 10 mL 0.1 M KMnO4 solution in acidic conditions. (Fe = 56)?
thanking you in anticipation.</h1>
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Ernest Z.
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<div class="markdown"><p>The <mathjax>#"KMnO"_4#</mathjax> will react with 400 mg of <mathjax>#"Fe"_0.9"O"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Nonstoichiometric <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a></strong></p>
<p><mathjax>#"Fe"_0.9"O"#</mathjax> is a <strong>nonstoichiometric compound</strong>.</p>
<p>Some of the <mathjax>#"Fe"^"2+"#</mathjax> ions have been oxidized to <mathjax>#"Fe"^"3+"#</mathjax>.</p>
<p>Thus, to balance the charge, the compound contains two <mathjax>#"Fe"^"3+"#</mathjax> ions for every three "missing" <mathjax>#"Fe"^"2+"#</mathjax> ions.</p>
<p>There is still the same mass of <mathjax>#"Fe"#</mathjax>, but only 90 % of it is in the +2 oxidation state.</p>
<blockquote></blockquote>
<p><strong><a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">Stoichiometry</a></strong></p>
<p>The system will behave as if the reaction goes only to 90 % completion.</p>
<p>The equation for the reaction is</p>
<p><mathjax>#"5FeO"color(white)(l) + "KMnO"_4 + "18H"^"+" → "5Fe"^"3+" + "Mn"^"2+" + "K"^"+" + "9H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Calculations</strong></p>
<p><mathjax>#"Moles of KMnO"_4 color(white)(l)"used" = 0.010 color(red)(cancel(color(black)("L KMnO"_4))) × "0.1 mol MnO"_4/(1 color(red)(cancel(color(black)("L KMnO"_4)))) = "0.001 mol KMnO"_4#</mathjax></p>
<p><mathjax>#"Moles of FeO reacted" = 0.001 color(red)(cancel(color(black)("mol KMnO"_4))) × "5 mol FeO"/(1 color(red)(cancel(color(black)("mol KMnO"_4)))) = "0.005 mol FeO"#</mathjax></p>
<p><mathjax>#"Moles of Fe"_0.9"O" = 0.005 color(red)(cancel(color(black)("mol FeO"))) × ("100 mol Fe"_0.9"O")/(90 color(red)(cancel(color(black)("mol FeO")))) = "0.0056 mol Fe"_0.9"O"#</mathjax></p>
<p><mathjax>#"Mass of Fe"_0.9"O" = 0.0056 color(red)(cancel(color(black)("mol Fe"_0.9"O"))) × ("71.84 g Fe"_0.9"O")/(1 color(red)(cancel(color(black)("mol Fe"_0.9"O")))) = "0.4 g Fe"_0.9"O" = "400 mg Fe"_0.9"O"#</mathjax></p>
<p><strong>Note</strong>: The answer can have only 1 significant figure, because that is all you gave for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the <mathjax>#"KMnO"_4#</mathjax>.</p></div>
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</article> | How many milli grams of #"Fe"_0.9"O"# reacts completely with 10 mL 0.1 M KMnO4 solution in acidic conditions. (Fe = 56)?
thanking you in anticipation. | null |
2,927 | ad094e82-6ddd-11ea-af4e-ccda262736ce | https://socratic.org/questions/the-elemental-mass-percent-composition-of-succinic-acid-is-c-40-68-h-5-12-and-o- | C2H3O2 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] succinic acid [IN] empirical"}] | [{"type":"chemical equation","value":"C2H3O2"}] | [{"type":"physical unit","value":"Mass percent composition [OF] C in succinic acid [=] \\pu{40.68%}"},{"type":"physical unit","value":"Mass percent composition [OF] H in succinic acid [=] \\pu{5.12%}"},{"type":"physical unit","value":"Mass percent composition [OF] O in succinic acid [=] \\pu{54.19%}"}] | <h1 class="questionTitle" itemprop="name">The elemental mass percent composition of succinic acid is C 40.68% , H 5.12% , and O 54.19%. What is the empirical formula of succinic acid? </h1> | null | C2H3O2 | <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/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of succinic acid, you must first determine what the <em>mole ratio</em> between the atoms that make up the molecule is. </p>
<p>To do that, you'll use the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">mass percent composition</a> of carbon, hydrogen, and oxygen. </p>
<p>A mass <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of <strong>40.68%</strong> for carbon means that, for <em>every</em> 100 g of succinic acid, you'll get <strong>40.68 g</strong> of carbon. </p>
<p>The same is true for the other two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> - every 100 g of succinic acid will contain <strong>5.12 g</strong> of hydrogen and <strong>54.19 g</strong> of oxygen.</p>
<p>To make the calculations easier, assume that you're dealing with a <strong>100-g</strong> sample of succinic acid. </p>
<p>To determine how many moles of each element are present in the sample, use carbon, hydrogen, and oxygen's molar masses.</p>
<p><mathjax>#"For C": (40.68cancel("g"))/(12.011cancel("g")/"mol") = "3.387 moles"#</mathjax></p>
<p><mathjax>#"For H": (5.12 cancel("g"))/(1.001cancel("g")/"mol") = "5.115 moles"#</mathjax></p>
<p><mathjax>#"For O": (54.19cancel("g"))/(15.9994cancel("g")/"mol") = "3.387 moles"#</mathjax></p>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio these elements have in the compound can be determined by dividing these values by the smallest of the group.</p>
<p><mathjax>#"For C": (3.387cancel("moles"))/(3.387cancel("moles")) = 1#</mathjax></p>
<p><mathjax>#"For H": (5.115cancel("moles"))/(3.387cancel("moles")) = 1.51 ~= 1.5#</mathjax></p>
<p><mathjax>#"For O": (3.387cancel("moles"))/(3.387cancel("moles")) = 1#</mathjax></p>
<p>The first draft of the empirical formula will be</p>
<p><mathjax>#C_1H_1.5O_1#</mathjax></p>
<p>SInce you can't have fractional subscripts, multiply every subscript by <mathjax>#color(blue)(2)#</mathjax> to get the <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of succinic acid</p>
<p><mathjax>#C_(color(blue)(2) * 1)H_(color(blue)(2) * 1.5)O_(color(blue)(2) * 1) => color(green)(C_2H_3O_2)#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The empirical formula of succinic acid is <mathjax>#C_2H_3O_2#</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/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of succinic acid, you must first determine what the <em>mole ratio</em> between the atoms that make up the molecule is. </p>
<p>To do that, you'll use the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">mass percent composition</a> of carbon, hydrogen, and oxygen. </p>
<p>A mass <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of <strong>40.68%</strong> for carbon means that, for <em>every</em> 100 g of succinic acid, you'll get <strong>40.68 g</strong> of carbon. </p>
<p>The same is true for the other two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> - every 100 g of succinic acid will contain <strong>5.12 g</strong> of hydrogen and <strong>54.19 g</strong> of oxygen.</p>
<p>To make the calculations easier, assume that you're dealing with a <strong>100-g</strong> sample of succinic acid. </p>
<p>To determine how many moles of each element are present in the sample, use carbon, hydrogen, and oxygen's molar masses.</p>
<p><mathjax>#"For C": (40.68cancel("g"))/(12.011cancel("g")/"mol") = "3.387 moles"#</mathjax></p>
<p><mathjax>#"For H": (5.12 cancel("g"))/(1.001cancel("g")/"mol") = "5.115 moles"#</mathjax></p>
<p><mathjax>#"For O": (54.19cancel("g"))/(15.9994cancel("g")/"mol") = "3.387 moles"#</mathjax></p>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio these elements have in the compound can be determined by dividing these values by the smallest of the group.</p>
<p><mathjax>#"For C": (3.387cancel("moles"))/(3.387cancel("moles")) = 1#</mathjax></p>
<p><mathjax>#"For H": (5.115cancel("moles"))/(3.387cancel("moles")) = 1.51 ~= 1.5#</mathjax></p>
<p><mathjax>#"For O": (3.387cancel("moles"))/(3.387cancel("moles")) = 1#</mathjax></p>
<p>The first draft of the empirical formula will be</p>
<p><mathjax>#C_1H_1.5O_1#</mathjax></p>
<p>SInce you can't have fractional subscripts, multiply every subscript by <mathjax>#color(blue)(2)#</mathjax> to get the <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of succinic acid</p>
<p><mathjax>#C_(color(blue)(2) * 1)H_(color(blue)(2) * 1.5)O_(color(blue)(2) * 1) => color(green)(C_2H_3O_2)#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The elemental mass percent composition of succinic acid is C 40.68% , H 5.12% , and O 54.19%. What is the empirical formula of succinic acid? </h1>
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Stefan V.
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<div class="markdown"><p>The empirical formula of succinic acid is <mathjax>#C_2H_3O_2#</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/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of succinic acid, you must first determine what the <em>mole ratio</em> between the atoms that make up the molecule is. </p>
<p>To do that, you'll use the <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">mass percent composition</a> of carbon, hydrogen, and oxygen. </p>
<p>A mass <a href="http://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a> of <strong>40.68%</strong> for carbon means that, for <em>every</em> 100 g of succinic acid, you'll get <strong>40.68 g</strong> of carbon. </p>
<p>The same is true for the other two <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> - every 100 g of succinic acid will contain <strong>5.12 g</strong> of hydrogen and <strong>54.19 g</strong> of oxygen.</p>
<p>To make the calculations easier, assume that you're dealing with a <strong>100-g</strong> sample of succinic acid. </p>
<p>To determine how many moles of each element are present in the sample, use carbon, hydrogen, and oxygen's molar masses.</p>
<p><mathjax>#"For C": (40.68cancel("g"))/(12.011cancel("g")/"mol") = "3.387 moles"#</mathjax></p>
<p><mathjax>#"For H": (5.12 cancel("g"))/(1.001cancel("g")/"mol") = "5.115 moles"#</mathjax></p>
<p><mathjax>#"For O": (54.19cancel("g"))/(15.9994cancel("g")/"mol") = "3.387 moles"#</mathjax></p>
<p><a href="http://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio these elements have in the compound can be determined by dividing these values by the smallest of the group.</p>
<p><mathjax>#"For C": (3.387cancel("moles"))/(3.387cancel("moles")) = 1#</mathjax></p>
<p><mathjax>#"For H": (5.115cancel("moles"))/(3.387cancel("moles")) = 1.51 ~= 1.5#</mathjax></p>
<p><mathjax>#"For O": (3.387cancel("moles"))/(3.387cancel("moles")) = 1#</mathjax></p>
<p>The first draft of the empirical formula will be</p>
<p><mathjax>#C_1H_1.5O_1#</mathjax></p>
<p>SInce you can't have fractional subscripts, multiply every subscript by <mathjax>#color(blue)(2)#</mathjax> to get the <a href="http://socratic.org/chemistry/the-mole-concept/empirical-and-molecular-formulas">empirical formula</a> of succinic acid</p>
<p><mathjax>#C_(color(blue)(2) * 1)H_(color(blue)(2) * 1.5)O_(color(blue)(2) * 1) => color(green)(C_2H_3O_2)#</mathjax></p></div>
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</article> | The elemental mass percent composition of succinic acid is C 40.68% , H 5.12% , and O 54.19%. What is the empirical formula of succinic acid? | null |
2,928 | ab44b70a-6ddd-11ea-a528-ccda262736ce | https://socratic.org/questions/the-velocity-of-nitrogen-is-55-6-cm-s-determine-the-the-rate-at-which-hydrogen-s | 50.76 cm/s | start physical_unit 13 14 percent cm/s qc_end physical_unit 3 3 5 6 velocity qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Rate [OF] hydrogen sulfide [IN] cm/s"}] | [{"type":"physical unit","value":"50.76 cm/s"}] | [{"type":"physical unit","value":"Velocity [OF] nitrogen [=] \\pu{55.6 cm/s}"},{"type":"other","value":"Under these same experimental conditions."}] | <h1 class="questionTitle" itemprop="name">The velocity of nitrogen is 55.6 cm/s. Determine the the rate at which hydrogen sulfide would travel under these same experimental conditions?
</h1> | null | 50.76 cm/s | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I assume that by <em>velocity</em> you mean <strong>root-mean-square speed</strong>, <mathjax>#v_"rms"#</mathjax>, which for an ideal gas kept under a temperature <mathjax>#T#</mathjax> is equal to </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)v_"rms" = sqrt((3RT)/M_M)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<p><mathjax>#R#</mathjax> - the universal gas constant<br/>
<mathjax>#M_M#</mathjax> - the <strong>molar mass</strong> of the gas</p>
</blockquote>
<p>Now, the root-mean-square speed is usually expressed is <em>meters per second</em>, <mathjax>#"m s"^(-1)#</mathjax>, but you don't have to convert it because you're going to express the root-mean-square speed of hydrogen sulfide, <mathjax>#"H"_2"S"#</mathjax>, <strong>relative</strong> to that of nitrogen gas, <mathjax>#"N"_2#</mathjax>.</p>
<p>So, you know that both gases are kept under the same conditions, which can only mean that they have the same absolute temperature <mathjax>#T#</mathjax>.</p>
<p>You can thus say that</p>
<blockquote>
<p><mathjax>#v_("rms N"_ 2) = sqrt( (3RT)/M_("M N"_2)) ->#</mathjax> <em>the root-mean-squares speed for <strong>nitrogen gas</strong></em></p>
<p><mathjax>#v_("rms H"_ 2"S") = sqrt((3RT)/(M_("M H"_2"S"))) ->#</mathjax> the <em>root-mean-square speed for <strong>hydrogen sulfide</strong></em></p>
</blockquote>
<p>Divide these two equations to get </p>
<blockquote>
<p><mathjax>#v_("rms H"_ 2"S")/v_("rms N"_ 2) = sqrt( color(red)(cancel(color(black)(3RT)))/M_("M H"_2"S") * M_("M N"_2)/color(red)(cancel(color(black)(3RT))))#</mathjax></p>
<p><mathjax>#v_("rms H"_ 2"S")/v_("rms N"_ 2) = sqrt(M_("M N"_ 2)/M_("M H"_2"S"))#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#v_("rms H"_ 2"S") = v_("rms N"_2) * sqrt(M_("M N"_ 2)/M_("M H"_2"S"))#</mathjax></p>
</blockquote>
<p>The <strong>molar masses</strong> of the two gases are</p>
<blockquote>
<p><mathjax>#M_("M N"_2) = "28.0134 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M H"_2"S") = "34.089 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#v_("rms H"_2"S") = "56 cm s"^(-1) * sqrt((28.0134 color(red)(cancel(color(black)("g mol"^(-1)))))/(34.089color(red)(cancel(color(black)("g mol"^(-1))))))#</mathjax></p>
<p><mathjax>#v_("rms H"_2"S") = color(green)(|bar(ul(color(white)(a/a)color(black)("51 cm s"^(-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>
<p><em>Now, does the result make sense?</em></p>
<p>The root-mean-square speed is <strong>inversely proportional</strong> to the square root of the molar mass of the gas, which means that the <strong>heavier</strong> the molar mass, the <strong>lower</strong> the root-mean-square speed for gases kept <em>under the same</em> absolute temperature <mathjax>#T#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"51 cm s"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I assume that by <em>velocity</em> you mean <strong>root-mean-square speed</strong>, <mathjax>#v_"rms"#</mathjax>, which for an ideal gas kept under a temperature <mathjax>#T#</mathjax> is equal to </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)v_"rms" = sqrt((3RT)/M_M)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<p><mathjax>#R#</mathjax> - the universal gas constant<br/>
<mathjax>#M_M#</mathjax> - the <strong>molar mass</strong> of the gas</p>
</blockquote>
<p>Now, the root-mean-square speed is usually expressed is <em>meters per second</em>, <mathjax>#"m s"^(-1)#</mathjax>, but you don't have to convert it because you're going to express the root-mean-square speed of hydrogen sulfide, <mathjax>#"H"_2"S"#</mathjax>, <strong>relative</strong> to that of nitrogen gas, <mathjax>#"N"_2#</mathjax>.</p>
<p>So, you know that both gases are kept under the same conditions, which can only mean that they have the same absolute temperature <mathjax>#T#</mathjax>.</p>
<p>You can thus say that</p>
<blockquote>
<p><mathjax>#v_("rms N"_ 2) = sqrt( (3RT)/M_("M N"_2)) ->#</mathjax> <em>the root-mean-squares speed for <strong>nitrogen gas</strong></em></p>
<p><mathjax>#v_("rms H"_ 2"S") = sqrt((3RT)/(M_("M H"_2"S"))) ->#</mathjax> the <em>root-mean-square speed for <strong>hydrogen sulfide</strong></em></p>
</blockquote>
<p>Divide these two equations to get </p>
<blockquote>
<p><mathjax>#v_("rms H"_ 2"S")/v_("rms N"_ 2) = sqrt( color(red)(cancel(color(black)(3RT)))/M_("M H"_2"S") * M_("M N"_2)/color(red)(cancel(color(black)(3RT))))#</mathjax></p>
<p><mathjax>#v_("rms H"_ 2"S")/v_("rms N"_ 2) = sqrt(M_("M N"_ 2)/M_("M H"_2"S"))#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#v_("rms H"_ 2"S") = v_("rms N"_2) * sqrt(M_("M N"_ 2)/M_("M H"_2"S"))#</mathjax></p>
</blockquote>
<p>The <strong>molar masses</strong> of the two gases are</p>
<blockquote>
<p><mathjax>#M_("M N"_2) = "28.0134 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M H"_2"S") = "34.089 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#v_("rms H"_2"S") = "56 cm s"^(-1) * sqrt((28.0134 color(red)(cancel(color(black)("g mol"^(-1)))))/(34.089color(red)(cancel(color(black)("g mol"^(-1))))))#</mathjax></p>
<p><mathjax>#v_("rms H"_2"S") = color(green)(|bar(ul(color(white)(a/a)color(black)("51 cm s"^(-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>
<p><em>Now, does the result make sense?</em></p>
<p>The root-mean-square speed is <strong>inversely proportional</strong> to the square root of the molar mass of the gas, which means that the <strong>heavier</strong> the molar mass, the <strong>lower</strong> the root-mean-square speed for gases kept <em>under the same</em> absolute temperature <mathjax>#T#</mathjax>. </p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">The velocity of nitrogen is 55.6 cm/s. Determine the the rate at which hydrogen sulfide would travel under these same experimental conditions?
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Stefan V.
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Jun 17, 2016
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<div class="markdown"><p><mathjax>#"51 cm s"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I assume that by <em>velocity</em> you mean <strong>root-mean-square speed</strong>, <mathjax>#v_"rms"#</mathjax>, which for an ideal gas kept under a temperature <mathjax>#T#</mathjax> is equal to </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)v_"rms" = sqrt((3RT)/M_M)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<p><mathjax>#R#</mathjax> - the universal gas constant<br/>
<mathjax>#M_M#</mathjax> - the <strong>molar mass</strong> of the gas</p>
</blockquote>
<p>Now, the root-mean-square speed is usually expressed is <em>meters per second</em>, <mathjax>#"m s"^(-1)#</mathjax>, but you don't have to convert it because you're going to express the root-mean-square speed of hydrogen sulfide, <mathjax>#"H"_2"S"#</mathjax>, <strong>relative</strong> to that of nitrogen gas, <mathjax>#"N"_2#</mathjax>.</p>
<p>So, you know that both gases are kept under the same conditions, which can only mean that they have the same absolute temperature <mathjax>#T#</mathjax>.</p>
<p>You can thus say that</p>
<blockquote>
<p><mathjax>#v_("rms N"_ 2) = sqrt( (3RT)/M_("M N"_2)) ->#</mathjax> <em>the root-mean-squares speed for <strong>nitrogen gas</strong></em></p>
<p><mathjax>#v_("rms H"_ 2"S") = sqrt((3RT)/(M_("M H"_2"S"))) ->#</mathjax> the <em>root-mean-square speed for <strong>hydrogen sulfide</strong></em></p>
</blockquote>
<p>Divide these two equations to get </p>
<blockquote>
<p><mathjax>#v_("rms H"_ 2"S")/v_("rms N"_ 2) = sqrt( color(red)(cancel(color(black)(3RT)))/M_("M H"_2"S") * M_("M N"_2)/color(red)(cancel(color(black)(3RT))))#</mathjax></p>
<p><mathjax>#v_("rms H"_ 2"S")/v_("rms N"_ 2) = sqrt(M_("M N"_ 2)/M_("M H"_2"S"))#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#v_("rms H"_ 2"S") = v_("rms N"_2) * sqrt(M_("M N"_ 2)/M_("M H"_2"S"))#</mathjax></p>
</blockquote>
<p>The <strong>molar masses</strong> of the two gases are</p>
<blockquote>
<p><mathjax>#M_("M N"_2) = "28.0134 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M H"_2"S") = "34.089 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>You will thus have</p>
<blockquote>
<p><mathjax>#v_("rms H"_2"S") = "56 cm s"^(-1) * sqrt((28.0134 color(red)(cancel(color(black)("g mol"^(-1)))))/(34.089color(red)(cancel(color(black)("g mol"^(-1))))))#</mathjax></p>
<p><mathjax>#v_("rms H"_2"S") = color(green)(|bar(ul(color(white)(a/a)color(black)("51 cm s"^(-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>
<p><em>Now, does the result make sense?</em></p>
<p>The root-mean-square speed is <strong>inversely proportional</strong> to the square root of the molar mass of the gas, which means that the <strong>heavier</strong> the molar mass, the <strong>lower</strong> the root-mean-square speed for gases kept <em>under the same</em> absolute temperature <mathjax>#T#</mathjax>. </p></div>
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</article> | The velocity of nitrogen is 55.6 cm/s. Determine the the rate at which hydrogen sulfide would travel under these same experimental conditions?
| null |
2,929 | a9b7e368-6ddd-11ea-a556-ccda262736ce | https://socratic.org/questions/boyle-s-law-7 | 0.32 L | start physical_unit 5 5 volume l qc_end physical_unit 5 5 11 12 pressure qc_end physical_unit 5 5 16 17 volume qc_end physical_unit 5 5 24 25 pressure qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"0.32 L"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.47 atm}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{1.62 L}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{7.42 am}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">At a constant temperature, a gas that exerted a pressure of 1.47 atm and that occupied 1.62 L is compressed until its pressure is 7.42 am. What is the final volume?</h1> | null | 0.32 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From Boyle’s Law</p>
<blockquote>
<p><mathjax>#"P"_1"V"_1 = "P"_2"V"_2#</mathjax></p>
</blockquote>
<p><mathjax>#"V"_2 = ("P"_1"V"_1)/("P"_2) = (1.47 cancel"atm" × "1.62 L")/(7.42 cancel"atm") = "0.32 L"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.32 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From Boyle’s Law</p>
<blockquote>
<p><mathjax>#"P"_1"V"_1 = "P"_2"V"_2#</mathjax></p>
</blockquote>
<p><mathjax>#"V"_2 = ("P"_1"V"_1)/("P"_2) = (1.47 cancel"atm" × "1.62 L")/(7.42 cancel"atm") = "0.32 L"#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">At a constant temperature, a gas that exerted a pressure of 1.47 atm and that occupied 1.62 L is compressed until its pressure is 7.42 am. What is the final volume?</h1>
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Junaid Mirza
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<div class="markdown"><p><mathjax>#"0.32 L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From Boyle’s Law</p>
<blockquote>
<p><mathjax>#"P"_1"V"_1 = "P"_2"V"_2#</mathjax></p>
</blockquote>
<p><mathjax>#"V"_2 = ("P"_1"V"_1)/("P"_2) = (1.47 cancel"atm" × "1.62 L")/(7.42 cancel"atm") = "0.32 L"#</mathjax></p></div>
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</article> | At a constant temperature, a gas that exerted a pressure of 1.47 atm and that occupied 1.62 L is compressed until its pressure is 7.42 am. What is the final volume? | null |
2,930 | abe62bc4-6ddd-11ea-8360-ccda262736ce | https://socratic.org/questions/how-do-you-balance-agno3-nh4-2cro4-ag2cro4-nh4no3 | 2 AgNO3 + (NH4)2CrO4 -> Ag2CrO4 + 2 NH4NO3 | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 AgNO3 + (NH4)2CrO4 -> Ag2CrO4 + 2 NH4NO3"}] | [{"type":"chemical equation","value":"AgNO3 + (NH4)2CrO4 -> Ag2CrO4 + NH4NO3"}] | <h1 class="questionTitle" itemprop="name">How do you balance ___AgNO3+___(NH4)2CrO4>___Ag2CrO4+___NH4NO3?
</h1> | null | 2 AgNO3 + (NH4)2CrO4 -> Ag2CrO4 + 2 NH4NO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing down the <em>unbalanced</em> chemical equation for this <a href="http://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a></p>
<blockquote>
<p><mathjax>#"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + "NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>In this reaction, <em>silver nitrate</em>, <mathjax>#"AgNO"_3#</mathjax>, will react with <em>ammonium chromate</em>, <mathjax>#("NH"_4)_2"CrO"_4#</mathjax>, both <strong>soluble</strong> <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>, to form <em>Silver chromate</em>, <mathjax>#"Ag"_2"CrO"_4#</mathjax>, an <strong>insoluble</strong> solid, and <em>ammonium nitrate</em>, <mathjax>#"NH"_4"CO"_3#</mathjax>, also a <strong>soluble</strong> compound. </p>
<p>When you're dealing with <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, you can treat <em>cations</em> and <em>anions</em> <strong>as units</strong>. This will allow you to balance the chemical equation faster than if you went by atoms alone. </p>
<p>So, the <em>cations</em> are</p>
<blockquote>
<ul>
<li><mathjax>#"Ag"^(+)#</mathjax></li>
<li><mathjax>#"NH"_4^(+)#</mathjax></li>
</ul>
</blockquote>
<p>and the <em>anions</em> are</p>
<blockquote>
<ul>
<li><mathjax>#"NO"_3^(-)#</mathjax></li>
<li><mathjax>#"CrO"_4^(2-)#</mathjax></li>
</ul>
</blockquote>
<p>So, start with the silver cations. Notice that you have <strong>one</strong> silver cation on the reactants' side and <strong>two</strong> on the products' side. Multiply the <em>silver nitrate</em> by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#color(red)(2)"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_color(red)(2)"CrO"_text(4(s]) darr + "NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>Now look at the ammonium cations. You have <strong>two</strong> ammonium cations on the reactants' side, and only <strong>one</strong> on the products' side. </p>
<p>Multiply the ammonium nitrate by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#color(red)(2)"AgNO"_text(3(aq]) + ("NH"_4)_color(green)(2)"CrO"_text(4(aq]) -> "Ag"_color(red)(2)"CrO"_text(4(s]) darr + color(green)(2)"NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>The nitrate aniuons are balanced, since you have <strong>two</strong> on the products' side and <strong>two</strong> on the reactants' side. </p>
<p>The same can be said for the chromate anions, since you have <strong>one</strong> on the reactants's side and <strong>one</strong> on the products' side. </p>
<p>Therefore, the balanced chemical equation for this reaction is </p>
<blockquote>
<p><mathjax>#2"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + 2"NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + 2"NH"_4"NO"_text(3(aq])#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing down the <em>unbalanced</em> chemical equation for this <a href="http://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a></p>
<blockquote>
<p><mathjax>#"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + "NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>In this reaction, <em>silver nitrate</em>, <mathjax>#"AgNO"_3#</mathjax>, will react with <em>ammonium chromate</em>, <mathjax>#("NH"_4)_2"CrO"_4#</mathjax>, both <strong>soluble</strong> <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>, to form <em>Silver chromate</em>, <mathjax>#"Ag"_2"CrO"_4#</mathjax>, an <strong>insoluble</strong> solid, and <em>ammonium nitrate</em>, <mathjax>#"NH"_4"CO"_3#</mathjax>, also a <strong>soluble</strong> compound. </p>
<p>When you're dealing with <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, you can treat <em>cations</em> and <em>anions</em> <strong>as units</strong>. This will allow you to balance the chemical equation faster than if you went by atoms alone. </p>
<p>So, the <em>cations</em> are</p>
<blockquote>
<ul>
<li><mathjax>#"Ag"^(+)#</mathjax></li>
<li><mathjax>#"NH"_4^(+)#</mathjax></li>
</ul>
</blockquote>
<p>and the <em>anions</em> are</p>
<blockquote>
<ul>
<li><mathjax>#"NO"_3^(-)#</mathjax></li>
<li><mathjax>#"CrO"_4^(2-)#</mathjax></li>
</ul>
</blockquote>
<p>So, start with the silver cations. Notice that you have <strong>one</strong> silver cation on the reactants' side and <strong>two</strong> on the products' side. Multiply the <em>silver nitrate</em> by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#color(red)(2)"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_color(red)(2)"CrO"_text(4(s]) darr + "NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>Now look at the ammonium cations. You have <strong>two</strong> ammonium cations on the reactants' side, and only <strong>one</strong> on the products' side. </p>
<p>Multiply the ammonium nitrate by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#color(red)(2)"AgNO"_text(3(aq]) + ("NH"_4)_color(green)(2)"CrO"_text(4(aq]) -> "Ag"_color(red)(2)"CrO"_text(4(s]) darr + color(green)(2)"NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>The nitrate aniuons are balanced, since you have <strong>two</strong> on the products' side and <strong>two</strong> on the reactants' side. </p>
<p>The same can be said for the chromate anions, since you have <strong>one</strong> on the reactants's side and <strong>one</strong> on the products' side. </p>
<p>Therefore, the balanced chemical equation for this reaction is </p>
<blockquote>
<p><mathjax>#2"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + 2"NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you balance ___AgNO3+___(NH4)2CrO4>___Ag2CrO4+___NH4NO3?
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Stefan V.
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<div class="markdown"><p><mathjax>#2"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + 2"NH"_4"NO"_text(3(aq])#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Start by writing down the <em>unbalanced</em> chemical equation for this <a href="http://socratic.org/chemistry/chemical-reactions/double-replacement-reactions">double replacement reaction</a></p>
<blockquote>
<p><mathjax>#"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + "NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>In this reaction, <em>silver nitrate</em>, <mathjax>#"AgNO"_3#</mathjax>, will react with <em>ammonium chromate</em>, <mathjax>#("NH"_4)_2"CrO"_4#</mathjax>, both <strong>soluble</strong> <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a>, to form <em>Silver chromate</em>, <mathjax>#"Ag"_2"CrO"_4#</mathjax>, an <strong>insoluble</strong> solid, and <em>ammonium nitrate</em>, <mathjax>#"NH"_4"CO"_3#</mathjax>, also a <strong>soluble</strong> compound. </p>
<p>When you're dealing with <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compounds</a>, you can treat <em>cations</em> and <em>anions</em> <strong>as units</strong>. This will allow you to balance the chemical equation faster than if you went by atoms alone. </p>
<p>So, the <em>cations</em> are</p>
<blockquote>
<ul>
<li><mathjax>#"Ag"^(+)#</mathjax></li>
<li><mathjax>#"NH"_4^(+)#</mathjax></li>
</ul>
</blockquote>
<p>and the <em>anions</em> are</p>
<blockquote>
<ul>
<li><mathjax>#"NO"_3^(-)#</mathjax></li>
<li><mathjax>#"CrO"_4^(2-)#</mathjax></li>
</ul>
</blockquote>
<p>So, start with the silver cations. Notice that you have <strong>one</strong> silver cation on the reactants' side and <strong>two</strong> on the products' side. Multiply the <em>silver nitrate</em> by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#color(red)(2)"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_color(red)(2)"CrO"_text(4(s]) darr + "NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>Now look at the ammonium cations. You have <strong>two</strong> ammonium cations on the reactants' side, and only <strong>one</strong> on the products' side. </p>
<p>Multiply the ammonium nitrate by <mathjax>#2#</mathjax> to get</p>
<blockquote>
<p><mathjax>#color(red)(2)"AgNO"_text(3(aq]) + ("NH"_4)_color(green)(2)"CrO"_text(4(aq]) -> "Ag"_color(red)(2)"CrO"_text(4(s]) darr + color(green)(2)"NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote>
<p>The nitrate aniuons are balanced, since you have <strong>two</strong> on the products' side and <strong>two</strong> on the reactants' side. </p>
<p>The same can be said for the chromate anions, since you have <strong>one</strong> on the reactants's side and <strong>one</strong> on the products' side. </p>
<p>Therefore, the balanced chemical equation for this reaction is </p>
<blockquote>
<p><mathjax>#2"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + 2"NH"_4"NO"_text(3(aq])#</mathjax></p>
</blockquote></div>
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</article> | How do you balance ___AgNO3+___(NH4)2CrO4>___Ag2CrO4+___NH4NO3?
| null |
2,931 | aa692912-6ddd-11ea-8b76-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-0-500-m-solution-of-acetylsalicylic-acid-pk-3-52 | 1.91 | start physical_unit 8 8 ph none qc_end physical_unit 10 11 6 7 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] acetylsalicylic acid solution"}] | [{"type":"physical unit","value":"1.91"}] | [{"type":"physical unit","value":"Molarity [OF] acetylsalicylic acid solution [=] \\pu{0.500 M}"},{"type":"physical unit","value":"pKa [OF] acetylsalicylic acid solution [=] \\pu{3.52}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a 0.500 M solution of acetylsalicylic acid, pK 3.52?</h1> | null | 1.91 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a weak acid:</p>
<p><mathjax>#sf(pH=1/2(pK_a-loga))#</mathjax></p>
<p>Where <mathjax>#sf(a)#</mathjax> is the concentration of the acid.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pH=1/2[3.52-log(0.5)]#</mathjax></p>
<p><mathjax>#sf(pH=1/2[3.52-(-0.301)]=1.91)#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#sf(1.91)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a weak acid:</p>
<p><mathjax>#sf(pH=1/2(pK_a-loga))#</mathjax></p>
<p>Where <mathjax>#sf(a)#</mathjax> is the concentration of the acid.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pH=1/2[3.52-log(0.5)]#</mathjax></p>
<p><mathjax>#sf(pH=1/2[3.52-(-0.301)]=1.91)#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a 0.500 M solution of acetylsalicylic acid, pK 3.52?</h1>
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<a class="topContributorPic" href="/users/michael-2"><img alt="" class="" src="https://lh3.googleusercontent.com/-gCv6FQRhls0/AAAAAAAAAAI/AAAAAAAAAAA/AAnnY7oOJS05Ylqn3KuDSW0LfnbOk7FezQ/mo/photo.jpg?sz=50" title=""/></a>
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Michael
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Sep 19, 2016
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<div class="markdown"><p><mathjax>#sf(1.91)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For a weak acid:</p>
<p><mathjax>#sf(pH=1/2(pK_a-loga))#</mathjax></p>
<p>Where <mathjax>#sf(a)#</mathjax> is the concentration of the acid.</p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(pH=1/2[3.52-log(0.5)]#</mathjax></p>
<p><mathjax>#sf(pH=1/2[3.52-(-0.301)]=1.91)#</mathjax></p></div>
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</article> | What is the pH of a 0.500 M solution of acetylsalicylic acid, pK 3.52? | null |
2,932 | a9fd8e7e-6ddd-11ea-bb78-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-that-contains-75-7-arsenic-and-24-3- | As2O3 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"As2O3"}] | [{"type":"physical unit","value":"Percentage [OF] arsenic in the compound [=] \\pu{75.7%}"},{"type":"physical unit","value":"Percentage [OF] oxygen in the compound [=] \\pu{24.3%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound that contains 75.7% arsenic and 24.3% oxygen?</h1> | null | As2O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Empirical formula is the <strong>simplest formula</strong> for a compound.</p>
<p>Arsenic's symbol is <mathjax>#As#</mathjax> while oxygen's symbol is <mathjax>#O#</mathjax></p>
<p>Firstly, we need to <strong>make an assumption</strong> that the <strong>mass of the compound is 100g</strong> so that it is easier for us to <strong>convert the percentage of each of the element into its respective mass</strong> .</p>
<p>Based on the assumption;</p>
<p>Compound contain <mathjax>#75.7%#</mathjax> arsenic;</p>
<p><mathjax>#75.7/100xx100g=75.7g#</mathjax></p>
<p>Compound contain <mathjax>#24.3%#</mathjax> oxygen;</p>
<p><mathjax>#24.3/100xx100g=24.3g#</mathjax></p>
<p>Then, we <strong>convert each of the element mass into the number of moles</strong> by dividing it with its respective molar mass;</p>
<p>To calculate the number of moles of arsenic;</p>
<p><mathjax>#(75.7g)/(74.9 gquadmol^-1)=1.01mol#</mathjax></p>
<p>To calculate the number of moles of oxygen;</p>
<p><mathjax>#(24.3g)/(16.0 gquadmol^-1)=1.52mol#</mathjax></p>
<p>To find the relation between arsenic and oxygen, we <strong>divide each <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>' number of moles with the smallest number of moles available</strong> .</p>
<p>For arsenic; <mathjax>#(1.01mol)/(1.01mol)=1.00#</mathjax></p>
<p>For oxygen; <mathjax>#(1.52mol)/(1.01mol)=1.50#</mathjax></p>
<p><strong>Convert both values into whole number</strong>, which is by multiplying with <mathjax>#2#</mathjax>;</p>
<p>For arsenic; <mathjax>#1.00xx2=2#</mathjax></p>
<p>For oxygen; <mathjax>#1.50xx2=3#</mathjax></p>
<p><strong>Put the values all together with respective element's symbol</strong> and we get the empirical formula of <mathjax>#As_2O_3#</mathjax>, namely arsenic trioxide.</p></div>
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<div>
<div class="markdown"><p><mathjax>#As_2O_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Empirical formula is the <strong>simplest formula</strong> for a compound.</p>
<p>Arsenic's symbol is <mathjax>#As#</mathjax> while oxygen's symbol is <mathjax>#O#</mathjax></p>
<p>Firstly, we need to <strong>make an assumption</strong> that the <strong>mass of the compound is 100g</strong> so that it is easier for us to <strong>convert the percentage of each of the element into its respective mass</strong> .</p>
<p>Based on the assumption;</p>
<p>Compound contain <mathjax>#75.7%#</mathjax> arsenic;</p>
<p><mathjax>#75.7/100xx100g=75.7g#</mathjax></p>
<p>Compound contain <mathjax>#24.3%#</mathjax> oxygen;</p>
<p><mathjax>#24.3/100xx100g=24.3g#</mathjax></p>
<p>Then, we <strong>convert each of the element mass into the number of moles</strong> by dividing it with its respective molar mass;</p>
<p>To calculate the number of moles of arsenic;</p>
<p><mathjax>#(75.7g)/(74.9 gquadmol^-1)=1.01mol#</mathjax></p>
<p>To calculate the number of moles of oxygen;</p>
<p><mathjax>#(24.3g)/(16.0 gquadmol^-1)=1.52mol#</mathjax></p>
<p>To find the relation between arsenic and oxygen, we <strong>divide each <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>' number of moles with the smallest number of moles available</strong> .</p>
<p>For arsenic; <mathjax>#(1.01mol)/(1.01mol)=1.00#</mathjax></p>
<p>For oxygen; <mathjax>#(1.52mol)/(1.01mol)=1.50#</mathjax></p>
<p><strong>Convert both values into whole number</strong>, which is by multiplying with <mathjax>#2#</mathjax>;</p>
<p>For arsenic; <mathjax>#1.00xx2=2#</mathjax></p>
<p>For oxygen; <mathjax>#1.50xx2=3#</mathjax></p>
<p><strong>Put the values all together with respective element's symbol</strong> and we get the empirical formula of <mathjax>#As_2O_3#</mathjax>, namely arsenic trioxide.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound that contains 75.7% arsenic and 24.3% oxygen?</h1>
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<span class="dateCreated" datetime="2016-04-14T03:13:03" itemprop="dateCreated">
Apr 14, 2016
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<div class="markdown"><p><mathjax>#As_2O_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Empirical formula is the <strong>simplest formula</strong> for a compound.</p>
<p>Arsenic's symbol is <mathjax>#As#</mathjax> while oxygen's symbol is <mathjax>#O#</mathjax></p>
<p>Firstly, we need to <strong>make an assumption</strong> that the <strong>mass of the compound is 100g</strong> so that it is easier for us to <strong>convert the percentage of each of the element into its respective mass</strong> .</p>
<p>Based on the assumption;</p>
<p>Compound contain <mathjax>#75.7%#</mathjax> arsenic;</p>
<p><mathjax>#75.7/100xx100g=75.7g#</mathjax></p>
<p>Compound contain <mathjax>#24.3%#</mathjax> oxygen;</p>
<p><mathjax>#24.3/100xx100g=24.3g#</mathjax></p>
<p>Then, we <strong>convert each of the element mass into the number of moles</strong> by dividing it with its respective molar mass;</p>
<p>To calculate the number of moles of arsenic;</p>
<p><mathjax>#(75.7g)/(74.9 gquadmol^-1)=1.01mol#</mathjax></p>
<p>To calculate the number of moles of oxygen;</p>
<p><mathjax>#(24.3g)/(16.0 gquadmol^-1)=1.52mol#</mathjax></p>
<p>To find the relation between arsenic and oxygen, we <strong>divide each <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>' number of moles with the smallest number of moles available</strong> .</p>
<p>For arsenic; <mathjax>#(1.01mol)/(1.01mol)=1.00#</mathjax></p>
<p>For oxygen; <mathjax>#(1.52mol)/(1.01mol)=1.50#</mathjax></p>
<p><strong>Convert both values into whole number</strong>, which is by multiplying with <mathjax>#2#</mathjax>;</p>
<p>For arsenic; <mathjax>#1.00xx2=2#</mathjax></p>
<p>For oxygen; <mathjax>#1.50xx2=3#</mathjax></p>
<p><strong>Put the values all together with respective element's symbol</strong> and we get the empirical formula of <mathjax>#As_2O_3#</mathjax>, namely arsenic trioxide.</p></div>
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</article> | What is the empirical formula of a compound that contains 75.7% arsenic and 24.3% oxygen? | null |
2,933 | a8a63ec2-6ddd-11ea-957b-ccda262736ce | https://socratic.org/questions/a-4-44-l-container-holds-15-4-g-of-oxygen-at-22-55-c-what-is-the-pressure | 2.63 atm | start physical_unit 8 8 pressure atm qc_end physical_unit 3 3 1 2 volume qc_end physical_unit 8 8 5 6 mass qc_end physical_unit 8 8 10 11 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] oxygen [IN] atm"}] | [{"type":"physical unit","value":"2.63 atm"}] | [{"type":"physical unit","value":"Volume [OF] container [=] \\pu{4.44 L}"},{"type":"physical unit","value":"Mass [OF] oxygen [=] \\pu{15.4 g}"},{"type":"physical unit","value":"Temperature [OF] oxygen [=] \\pu{22.55 ℃}"}] | <h1 class="questionTitle" itemprop="name">A 4.44 L container holds 15.4 g of oxygen at 22.55°C. What is the pressure?</h1> | null | 2.63 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P=(nRT)/V=((15.4*g)/(32.00*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx295.7*K)/(4.44*L)=#</mathjax></p>
<p><mathjax>#2.63*atm#</mathjax>. </p>
<p>Do the units cancel out to give an answer in <mathjax>#atm#</mathjax>? Do they? Don't assume that I haven't made a mistake!</p>
<p>How did I do this? Well for (i) I know that oxygen is a binuclear gas, i.e. <mathjax>#O_2#</mathjax>; in fact all the elemental gases (save for the Noble Gases) are BINUCLEAR.....and thus I used a molecular mass of <mathjax>#32.00*g*mol^-1#</mathjax> for dioxygen gas.......</p>
<p>And (ii) I knew the gas constant was............... <mathjax>#0.0821*L*atm*K^-1*mol^-1#</mathjax>. </p>
<p>And (iii) I knew the relationship between <mathjax>#""^@C#</mathjax> and <mathjax>#"degrees Kelvin"#</mathjax>.</p>
<p>These data (well (ii) and (iii)) need not be remembered, as they should be provided (with various units) in a chemistry exam. You still need to know how to use the appropriate gas constant, <mathjax>#R#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use the Ideal Gas Equation to get...........<mathjax>#P=2.63*atm#</mathjax></p></div>
</div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P=(nRT)/V=((15.4*g)/(32.00*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx295.7*K)/(4.44*L)=#</mathjax></p>
<p><mathjax>#2.63*atm#</mathjax>. </p>
<p>Do the units cancel out to give an answer in <mathjax>#atm#</mathjax>? Do they? Don't assume that I haven't made a mistake!</p>
<p>How did I do this? Well for (i) I know that oxygen is a binuclear gas, i.e. <mathjax>#O_2#</mathjax>; in fact all the elemental gases (save for the Noble Gases) are BINUCLEAR.....and thus I used a molecular mass of <mathjax>#32.00*g*mol^-1#</mathjax> for dioxygen gas.......</p>
<p>And (ii) I knew the gas constant was............... <mathjax>#0.0821*L*atm*K^-1*mol^-1#</mathjax>. </p>
<p>And (iii) I knew the relationship between <mathjax>#""^@C#</mathjax> and <mathjax>#"degrees Kelvin"#</mathjax>.</p>
<p>These data (well (ii) and (iii)) need not be remembered, as they should be provided (with various units) in a chemistry exam. You still need to know how to use the appropriate gas constant, <mathjax>#R#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">A 4.44 L container holds 15.4 g of oxygen at 22.55°C. What is the pressure?</h1>
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<div class="markdown"><p>We use the Ideal Gas Equation to get...........<mathjax>#P=2.63*atm#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P=(nRT)/V=((15.4*g)/(32.00*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx295.7*K)/(4.44*L)=#</mathjax></p>
<p><mathjax>#2.63*atm#</mathjax>. </p>
<p>Do the units cancel out to give an answer in <mathjax>#atm#</mathjax>? Do they? Don't assume that I haven't made a mistake!</p>
<p>How did I do this? Well for (i) I know that oxygen is a binuclear gas, i.e. <mathjax>#O_2#</mathjax>; in fact all the elemental gases (save for the Noble Gases) are BINUCLEAR.....and thus I used a molecular mass of <mathjax>#32.00*g*mol^-1#</mathjax> for dioxygen gas.......</p>
<p>And (ii) I knew the gas constant was............... <mathjax>#0.0821*L*atm*K^-1*mol^-1#</mathjax>. </p>
<p>And (iii) I knew the relationship between <mathjax>#""^@C#</mathjax> and <mathjax>#"degrees Kelvin"#</mathjax>.</p>
<p>These data (well (ii) and (iii)) need not be remembered, as they should be provided (with various units) in a chemistry exam. You still need to know how to use the appropriate gas constant, <mathjax>#R#</mathjax>.</p></div>
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</article> | A 4.44 L container holds 15.4 g of oxygen at 22.55°C. What is the pressure? | null |
2,934 | ac30dc5a-6ddd-11ea-9f5b-ccda262736ce | https://socratic.org/questions/5802c242b72cff0ea417726b | 2.51 × 10^21 | start physical_unit 2 3 number none qc_end physical_unit 8 8 6 7 mass qc_end end | [{"type":"physical unit","value":"Number [OF] carbon atoms"}] | [{"type":"physical unit","value":"2.51 × 10^21"}] | [{"type":"physical unit","value":"Mass [OF] diamond [=] \\pu{3.25 carat}"}] | <h1 class="questionTitle" itemprop="name">How many carbon atoms in a #3.25*"carat"# diamond?</h1> | null | 2.51 × 10^21 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"1 carat"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"200 mg"#</mathjax></p>
<p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax></p>
<p><mathjax>#"Number of moles of carbon"#</mathjax> </p>
<p><mathjax>#=#</mathjax> <mathjax>#(3.25 cancel" carat"xx200xx10^-3*cancelg*cancel"carat"^-1)/(12.011*cancelg*mol^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.054*mol#</mathjax>, and we mulitply this by <mathjax>#"Avogadro's number"#</mathjax> to get the number of carbon atoms. </p>
<p><mathjax>#"Number of carbon atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.054*molxx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>And we go thru the same process for the aspirin sample:</p>
<p><mathjax>#(0.75*g)/(180.16*g*mol^-1)xx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#"How many aspirin molecules?"#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Number of carbon atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.054*molxx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"1 carat"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"200 mg"#</mathjax></p>
<p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax></p>
<p><mathjax>#"Number of moles of carbon"#</mathjax> </p>
<p><mathjax>#=#</mathjax> <mathjax>#(3.25 cancel" carat"xx200xx10^-3*cancelg*cancel"carat"^-1)/(12.011*cancelg*mol^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.054*mol#</mathjax>, and we mulitply this by <mathjax>#"Avogadro's number"#</mathjax> to get the number of carbon atoms. </p>
<p><mathjax>#"Number of carbon atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.054*molxx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>And we go thru the same process for the aspirin sample:</p>
<p><mathjax>#(0.75*g)/(180.16*g*mol^-1)xx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#"How many aspirin molecules?"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many carbon atoms in a #3.25*"carat"# diamond?</h1>
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Oct 16, 2016
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<div class="markdown"><p><mathjax>#"Number of carbon atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.054*molxx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"1 carat"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"200 mg"#</mathjax></p>
<p><mathjax>#"Number of moles"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Mass"/"Molar mass"#</mathjax></p>
<p><mathjax>#"Number of moles of carbon"#</mathjax> </p>
<p><mathjax>#=#</mathjax> <mathjax>#(3.25 cancel" carat"xx200xx10^-3*cancelg*cancel"carat"^-1)/(12.011*cancelg*mol^-1)#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#0.054*mol#</mathjax>, and we mulitply this by <mathjax>#"Avogadro's number"#</mathjax> to get the number of carbon atoms. </p>
<p><mathjax>#"Number of carbon atoms"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.054*molxx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax></p>
<p>And we go thru the same process for the aspirin sample:</p>
<p><mathjax>#(0.75*g)/(180.16*g*mol^-1)xx6.022xx10^23*mol^-1#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#"How many aspirin molecules?"#</mathjax></p></div>
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</article> | How many carbon atoms in a #3.25*"carat"# diamond? | null |
2,935 | ac492774-6ddd-11ea-b4da-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-titanium-iv-chloride | TiCl4 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] titanium(IV) chloride [IN] default"}] | [{"type":"chemical equation","value":"TiCl4"}] | [{"type":"substance name","value":"Titanium(IV) chloride"}] | <h1 class="questionTitle" itemprop="name">What is the formula for titanium(IV) chloride?</h1> | null | TiCl4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Titanic chloride"#</mathjax> is a volatile liquid, <mathjax>#"boiling point, 136.4 "^@C#</mathjax>.</p>
<p>The <mathjax>#(IV)#</mathjax> refers to the oxidation state of the metal, i.e. we have formally a <mathjax>#Ti^(4+)#</mathjax> centre bound to 4 chloride ligands, and hence a <mathjax>#+IV#</mathjax> oxidation state.</p>
<p>Titanic chloride reacts with moist air according to the following rxn:</p>
<p><mathjax>#TiCl_4(l) + 2H_2O(l) rarr TiO_2(s) + 4HCl(aq)#</mathjax></p>
<p>You don't tend to see it as much these days, but have you ever seen a light aircraft writing messages in the sky, i.e. skywriting? The airplane would carry a canister of titanic chloride, which could be opened and shut during the plane's maneouvers, and write messages in the sky by means of the solid <mathjax>#TiO_2#</mathjax> formed in the moist air. </p>
<p>Note that <mathjax>#TiO_2#</mathjax>, <mathjax>#"titanic oxide"#</mathjax>, is used in vast quantities by the paint industry to produce white paint, that is <mathjax>#"whiter than white"#</mathjax> due to the <mathjax>#TiO_2#</mathjax> content. </p></div>
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<div>
<div class="markdown"><p>For <mathjax>#"titanic chloride"#</mathjax>? <mathjax>#TiCl_4#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Titanic chloride"#</mathjax> is a volatile liquid, <mathjax>#"boiling point, 136.4 "^@C#</mathjax>.</p>
<p>The <mathjax>#(IV)#</mathjax> refers to the oxidation state of the metal, i.e. we have formally a <mathjax>#Ti^(4+)#</mathjax> centre bound to 4 chloride ligands, and hence a <mathjax>#+IV#</mathjax> oxidation state.</p>
<p>Titanic chloride reacts with moist air according to the following rxn:</p>
<p><mathjax>#TiCl_4(l) + 2H_2O(l) rarr TiO_2(s) + 4HCl(aq)#</mathjax></p>
<p>You don't tend to see it as much these days, but have you ever seen a light aircraft writing messages in the sky, i.e. skywriting? The airplane would carry a canister of titanic chloride, which could be opened and shut during the plane's maneouvers, and write messages in the sky by means of the solid <mathjax>#TiO_2#</mathjax> formed in the moist air. </p>
<p>Note that <mathjax>#TiO_2#</mathjax>, <mathjax>#"titanic oxide"#</mathjax>, is used in vast quantities by the paint industry to produce white paint, that is <mathjax>#"whiter than white"#</mathjax> due to the <mathjax>#TiO_2#</mathjax> content. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for titanium(IV) chloride?</h1>
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anor277
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Nov 3, 2016
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<div class="markdown"><p>For <mathjax>#"titanic chloride"#</mathjax>? <mathjax>#TiCl_4#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Titanic chloride"#</mathjax> is a volatile liquid, <mathjax>#"boiling point, 136.4 "^@C#</mathjax>.</p>
<p>The <mathjax>#(IV)#</mathjax> refers to the oxidation state of the metal, i.e. we have formally a <mathjax>#Ti^(4+)#</mathjax> centre bound to 4 chloride ligands, and hence a <mathjax>#+IV#</mathjax> oxidation state.</p>
<p>Titanic chloride reacts with moist air according to the following rxn:</p>
<p><mathjax>#TiCl_4(l) + 2H_2O(l) rarr TiO_2(s) + 4HCl(aq)#</mathjax></p>
<p>You don't tend to see it as much these days, but have you ever seen a light aircraft writing messages in the sky, i.e. skywriting? The airplane would carry a canister of titanic chloride, which could be opened and shut during the plane's maneouvers, and write messages in the sky by means of the solid <mathjax>#TiO_2#</mathjax> formed in the moist air. </p>
<p>Note that <mathjax>#TiO_2#</mathjax>, <mathjax>#"titanic oxide"#</mathjax>, is used in vast quantities by the paint industry to produce white paint, that is <mathjax>#"whiter than white"#</mathjax> due to the <mathjax>#TiO_2#</mathjax> content. </p></div>
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</article> | What is the formula for titanium(IV) chloride? | null |
2,936 | acb227e6-6ddd-11ea-8ac9-ccda262736ce | https://socratic.org/questions/the-boiling-point-of-a-solution-containing-10-44g-og-an-unknown-nonelectrolyte-i | 15.5 g/mol | start physical_unit 29 30 molar_mass g/mol qc_end physical_unit 11 12 7 8 mass qc_end physical_unit 17 18 14 15 mass qc_end physical_unit 5 5 20 22 boiling_point_temperature qc_end end | [{"type":"physical unit","value":"molar mass [OF] the solute [IN] g/mol"}] | [{"type":"physical unit","value":"15.5 g/mol"}] | [{"type":"physical unit","value":"Mass [OF] the unknown nonelectrolyte [=] \\pu{10.44 g}"},{"type":"physical unit","value":"Mass [OF] acetic acid [=] \\pu{50 g}"},{"type":"physical unit","value":"Boiling point [OF] the solution [=] \\pu{159.2 degree celsius}"}] | <h1 class="questionTitle" itemprop="name">The boiling point of a solution containing 10.44g og an unknown nonelectrolyte in 50g of acetic acid is 159.2 degree celsius. What is the molar mass of the solute?</h1> | null | 15.5 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>molar mass</strong> of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, given some known information about boiling-point elevation.</p>
<blockquote></blockquote>
<p>To do this, we'll be using the equation</p>
<blockquote>
<p><mathjax>#ul(DeltaT_b = imK_b#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li>
<p><mathjax>#DeltaT_b#</mathjax> is the <strong>change in boiling point</strong> of the solution</p>
</li>
<li>
<p><mathjax>#i#</mathjax> is the <strong>van't Hoff factor</strong> of the solute, which is <mathjax>#1#</mathjax> since it is a <strong><em>nonelectrolyte</em></strong></p>
</li>
<li>
<p><mathjax>#m#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution</p>
</li>
<li>
<p><mathjax>#K_b#</mathjax> is the <strong>molal boiling-point elevation constant for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong> (acetic acid), which is <mathjax>#3.07#</mathjax> <mathjax>#""^"o""C/"m#</mathjax> (from Wikipedia).</p>
<blockquote></blockquote>
</li>
</ul>
<p>We're given that the <strong>final boiling point</strong> is <mathjax>#159.2#</mathjax> <mathjax>#""^"o""C"#</mathjax>, and according to Wikipedia, the <strong>normal boiling point</strong> of acetic acid is <mathjax>#117.9#</mathjax> <mathjax>#""^"o""C"#</mathjax>, so the <strong>change in boiling point</strong> <mathjax>#DeltaT_b#</mathjax> is</p>
<blockquote>
<p><mathjax>#DeltaT_b = 159.2#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#- 117.9#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#= ul(41.3color(white)(l)""^"o""C"#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Plugging in known values:</p>
<blockquote>
<p><mathjax>#41.3color(white)(l)""^"o""C" = (1)m(3.07color(white)(l)""^"o""C/"m)#</mathjax></p>
</blockquote>
<p>The <strong>molality</strong> (moles of solute per kilogram of solvent) is</p>
<blockquote>
<p><mathjax>#color(red)(ul(m = 13.5color(white)(l)"mol/kg"#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Now that we know the <strong>molality</strong> of the solution, we can use the given value of <mathjax>#50#</mathjax> <mathjax>#"g acetic acid"#</mathjax> and the molality equation to find the number of <strong>moles</strong> of the solute present in solution:</p>
<blockquote>
<p><mathjax>#"mol solute" = ("molality")("kg solvent")#</mathjax></p>
</blockquote>
<p>So</p>
<blockquote>
<p><mathjax>#"mol solute" = (color(red)(13.5color(white)(l)"mol/")cancel(color(red)("kg")))(0.050cancel("kg")) = color(green)(ul(0.673color(white)(l)"mol solute"#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Finally, we can use the given number of <strong>grams</strong> of solute and this to find its <strong>molar mass</strong>:</p>
<blockquote>
<p><mathjax>#color(blue)("molar mass") = (10.44color(white)(l)"g")/(color(green)(0.673color(white)(l)"mol")) = color(blue)(ulbar(|stackrel(" ")(" "15.5color(white)(l)"g/mol"" ")|)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#15.5#</mathjax> <mathjax>#"g/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>molar mass</strong> of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, given some known information about boiling-point elevation.</p>
<blockquote></blockquote>
<p>To do this, we'll be using the equation</p>
<blockquote>
<p><mathjax>#ul(DeltaT_b = imK_b#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li>
<p><mathjax>#DeltaT_b#</mathjax> is the <strong>change in boiling point</strong> of the solution</p>
</li>
<li>
<p><mathjax>#i#</mathjax> is the <strong>van't Hoff factor</strong> of the solute, which is <mathjax>#1#</mathjax> since it is a <strong><em>nonelectrolyte</em></strong></p>
</li>
<li>
<p><mathjax>#m#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution</p>
</li>
<li>
<p><mathjax>#K_b#</mathjax> is the <strong>molal boiling-point elevation constant for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong> (acetic acid), which is <mathjax>#3.07#</mathjax> <mathjax>#""^"o""C/"m#</mathjax> (from Wikipedia).</p>
<blockquote></blockquote>
</li>
</ul>
<p>We're given that the <strong>final boiling point</strong> is <mathjax>#159.2#</mathjax> <mathjax>#""^"o""C"#</mathjax>, and according to Wikipedia, the <strong>normal boiling point</strong> of acetic acid is <mathjax>#117.9#</mathjax> <mathjax>#""^"o""C"#</mathjax>, so the <strong>change in boiling point</strong> <mathjax>#DeltaT_b#</mathjax> is</p>
<blockquote>
<p><mathjax>#DeltaT_b = 159.2#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#- 117.9#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#= ul(41.3color(white)(l)""^"o""C"#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Plugging in known values:</p>
<blockquote>
<p><mathjax>#41.3color(white)(l)""^"o""C" = (1)m(3.07color(white)(l)""^"o""C/"m)#</mathjax></p>
</blockquote>
<p>The <strong>molality</strong> (moles of solute per kilogram of solvent) is</p>
<blockquote>
<p><mathjax>#color(red)(ul(m = 13.5color(white)(l)"mol/kg"#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Now that we know the <strong>molality</strong> of the solution, we can use the given value of <mathjax>#50#</mathjax> <mathjax>#"g acetic acid"#</mathjax> and the molality equation to find the number of <strong>moles</strong> of the solute present in solution:</p>
<blockquote>
<p><mathjax>#"mol solute" = ("molality")("kg solvent")#</mathjax></p>
</blockquote>
<p>So</p>
<blockquote>
<p><mathjax>#"mol solute" = (color(red)(13.5color(white)(l)"mol/")cancel(color(red)("kg")))(0.050cancel("kg")) = color(green)(ul(0.673color(white)(l)"mol solute"#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Finally, we can use the given number of <strong>grams</strong> of solute and this to find its <strong>molar mass</strong>:</p>
<blockquote>
<p><mathjax>#color(blue)("molar mass") = (10.44color(white)(l)"g")/(color(green)(0.673color(white)(l)"mol")) = color(blue)(ulbar(|stackrel(" ")(" "15.5color(white)(l)"g/mol"" ")|)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The boiling point of a solution containing 10.44g og an unknown nonelectrolyte in 50g of acetic acid is 159.2 degree celsius. What is the molar mass of the solute?</h1>
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Nathan L.
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Aug 17, 2017
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<div class="markdown"><p><mathjax>#15.5#</mathjax> <mathjax>#"g/mol"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We're asked to find the <strong>molar mass</strong> of the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, given some known information about boiling-point elevation.</p>
<blockquote></blockquote>
<p>To do this, we'll be using the equation</p>
<blockquote>
<p><mathjax>#ul(DeltaT_b = imK_b#</mathjax></p>
</blockquote>
<p>where</p>
<ul>
<li>
<p><mathjax>#DeltaT_b#</mathjax> is the <strong>change in boiling point</strong> of the solution</p>
</li>
<li>
<p><mathjax>#i#</mathjax> is the <strong>van't Hoff factor</strong> of the solute, which is <mathjax>#1#</mathjax> since it is a <strong><em>nonelectrolyte</em></strong></p>
</li>
<li>
<p><mathjax>#m#</mathjax> is the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molality">molality</a></strong> of the solution</p>
</li>
<li>
<p><mathjax>#K_b#</mathjax> is the <strong>molal boiling-point elevation constant for the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong> (acetic acid), which is <mathjax>#3.07#</mathjax> <mathjax>#""^"o""C/"m#</mathjax> (from Wikipedia).</p>
<blockquote></blockquote>
</li>
</ul>
<p>We're given that the <strong>final boiling point</strong> is <mathjax>#159.2#</mathjax> <mathjax>#""^"o""C"#</mathjax>, and according to Wikipedia, the <strong>normal boiling point</strong> of acetic acid is <mathjax>#117.9#</mathjax> <mathjax>#""^"o""C"#</mathjax>, so the <strong>change in boiling point</strong> <mathjax>#DeltaT_b#</mathjax> is</p>
<blockquote>
<p><mathjax>#DeltaT_b = 159.2#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#- 117.9#</mathjax> <mathjax>#""^"o""C"#</mathjax> <mathjax>#= ul(41.3color(white)(l)""^"o""C"#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Plugging in known values:</p>
<blockquote>
<p><mathjax>#41.3color(white)(l)""^"o""C" = (1)m(3.07color(white)(l)""^"o""C/"m)#</mathjax></p>
</blockquote>
<p>The <strong>molality</strong> (moles of solute per kilogram of solvent) is</p>
<blockquote>
<p><mathjax>#color(red)(ul(m = 13.5color(white)(l)"mol/kg"#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Now that we know the <strong>molality</strong> of the solution, we can use the given value of <mathjax>#50#</mathjax> <mathjax>#"g acetic acid"#</mathjax> and the molality equation to find the number of <strong>moles</strong> of the solute present in solution:</p>
<blockquote>
<p><mathjax>#"mol solute" = ("molality")("kg solvent")#</mathjax></p>
</blockquote>
<p>So</p>
<blockquote>
<p><mathjax>#"mol solute" = (color(red)(13.5color(white)(l)"mol/")cancel(color(red)("kg")))(0.050cancel("kg")) = color(green)(ul(0.673color(white)(l)"mol solute"#</mathjax></p>
<blockquote></blockquote>
</blockquote>
<p>Finally, we can use the given number of <strong>grams</strong> of solute and this to find its <strong>molar mass</strong>:</p>
<blockquote>
<p><mathjax>#color(blue)("molar mass") = (10.44color(white)(l)"g")/(color(green)(0.673color(white)(l)"mol")) = color(blue)(ulbar(|stackrel(" ")(" "15.5color(white)(l)"g/mol"" ")|)#</mathjax></p>
</blockquote></div>
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</article> | The boiling point of a solution containing 10.44g og an unknown nonelectrolyte in 50g of acetic acid is 159.2 degree celsius. What is the molar mass of the solute? | null |
2,937 | aae62f37-6ddd-11ea-87fa-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-1-5-mol-c-5h-12 | 108.2 g | start physical_unit 7 7 mass g qc_end physical_unit 7 7 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] C5H12 [IN] g"}] | [{"type":"physical unit","value":"108.2 g"}] | [{"type":"physical unit","value":"Mole [OF] C5H12 [=] \\pu{1.5 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 1.5 mol #C_5H_12?#</h1> | null | 108.2 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Pentane has a molar mass of <mathjax>#72.15*g*mol^-1#</mathjax>. A <mathjax>#1.5#</mathjax> <mathjax>#mol#</mathjax> quantity has a mass of <mathjax>#1.5*cancel(mol)xx72.15*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p></div>
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<div class="markdown"><p>Fill er up with <mathjax>#108.2#</mathjax> <mathjax>#g#</mathjax> petrol.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Pentane has a molar mass of <mathjax>#72.15*g*mol^-1#</mathjax>. A <mathjax>#1.5#</mathjax> <mathjax>#mol#</mathjax> quantity has a mass of <mathjax>#1.5*cancel(mol)xx72.15*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p></div>
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<div class="markdown"><p>Fill er up with <mathjax>#108.2#</mathjax> <mathjax>#g#</mathjax> petrol.</p></div>
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<div class="markdown"><p>Pentane has a molar mass of <mathjax>#72.15*g*mol^-1#</mathjax>. A <mathjax>#1.5#</mathjax> <mathjax>#mol#</mathjax> quantity has a mass of <mathjax>#1.5*cancel(mol)xx72.15*g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p></div>
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</article> | What is the mass of 1.5 mol #C_5H_12?# | null |
2,938 | a917521c-6ddd-11ea-9c8b-ccda262736ce | https://socratic.org/questions/how-many-oxygen-atoms-are-in-al-2-so-4 | 12 | start physical_unit 2 3 number none qc_end chemical_equation 6 6 qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"12"}] | [{"type":"chemical equation","value":"Al2(SO4)3"}] | <h1 class="questionTitle" itemprop="name">How many oxygen atoms are in #Al_2(SO_4)_3#?</h1> | null | 12 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In <mathjax>#Al_2(SO_4)_3#</mathjax>, the subscript of three next to the <mathjax>#SO_4#</mathjax> polyatomic ion refers to the number of that same ion is in this molecule. Since there are four atoms of oxygen in each <mathjax>#SO_4#</mathjax>, indicated by the subscript of four next to the oxygen, and there are three <mathjax>#SO_4#</mathjax> ions, <mathjax>#4*3= 12#</mathjax> oxygen atoms present in <mathjax>#Al_2(SO_4)_3#</mathjax>.</p></div>
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<div class="markdown"><p>There are 12 oxygen atoms in <mathjax>#Al_2(SO_4)_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In <mathjax>#Al_2(SO_4)_3#</mathjax>, the subscript of three next to the <mathjax>#SO_4#</mathjax> polyatomic ion refers to the number of that same ion is in this molecule. Since there are four atoms of oxygen in each <mathjax>#SO_4#</mathjax>, indicated by the subscript of four next to the oxygen, and there are three <mathjax>#SO_4#</mathjax> ions, <mathjax>#4*3= 12#</mathjax> oxygen atoms present in <mathjax>#Al_2(SO_4)_3#</mathjax>.</p></div>
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<div class="markdown"><p>There are 12 oxygen atoms in <mathjax>#Al_2(SO_4)_3#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In <mathjax>#Al_2(SO_4)_3#</mathjax>, the subscript of three next to the <mathjax>#SO_4#</mathjax> polyatomic ion refers to the number of that same ion is in this molecule. Since there are four atoms of oxygen in each <mathjax>#SO_4#</mathjax>, indicated by the subscript of four next to the oxygen, and there are three <mathjax>#SO_4#</mathjax> ions, <mathjax>#4*3= 12#</mathjax> oxygen atoms present in <mathjax>#Al_2(SO_4)_3#</mathjax>.</p></div>
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</article> | How many oxygen atoms are in #Al_2(SO_4)_3#? | null |
2,939 | a8d9cc5c-6ddd-11ea-80c9-ccda262736ce | https://socratic.org/questions/if-you-have-7-molecules-of-so-2-gas-in-2-617-10-5-molecules-of-air-what-is-the-c | 27 ppm | start physical_unit 6 6 concentration ppm qc_end end | [{"type":"physical unit","value":"Concentration [OF] SO2 [IN] ppm"}] | [{"type":"physical unit","value":"27 ppm"}] | [{"type":"physical unit","value":"Number [OF] SO2 molecules [=] \\pu{7}"},{"type":"physical unit","value":"Number [OF] air molecules [=] \\pu{2.617 × 10^5}"}] | <h1 class="questionTitle" itemprop="name">If you have 7 molecules of #SO_2# gas in #2.617 * 10^5# molecules of air, what is the concentration of #SO_2# in ppm?</h1> | null | 27 ppm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is use the <em>definition</em> of <strong>parts per million</strong>, or <strong>ppm</strong>. </p>
<p>You can use <em>parts per million</em> to express the concentration of a solution that contains very, very small amounts of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. As its name suggests, this way of expressing concentration uses <em>parts of solute</em> per <strong>one million parts of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong>. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"ppm" = "parts of solute"/"parts of solvent" xx 10^6color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In order to have a <mathjax>#"1 ppm"#</mathjax> solution, you need to have a solution that contains <strong>one part</strong> of solute <strong>for every</strong> <mathjax>#10^6#</mathjax> <strong>parts</strong> of solvent. </p>
<p>Now, you can consider air to be your <em>solution</em> and sulfur dioxide, <mathjax>#"SO"_2#</mathjax>, to be your solute. Since you have <em>significantly fewer</em> molecules of sulfur dioxide than the total number of molecules of air, you can say that you'll have <em>approximately</em> <mathjax>#2.617 * 10^5#</mathjax> <strong>parts</strong> of solvent. </p>
<p>This means that the concentration of sulfur dioxide in <strong>ppm</strong> will be </p>
<blockquote>
<p><mathjax>#"ppm SO"_2 = (7 color(red)(cancel(color(black)("molecules"))))/(2.617 * 10^5color(red)(cancel(color(black)("molecules")))) xx 10^6 = color(green)(|bar(ul(color(white)(a/a)"27 ppm"color(white)(a/a)|)))#</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>. </p>
<p>So, this tells you that out of <mathjax>#10^6#</mathjax> <strong>molecules of air</strong>, <mathjax>#27#</mathjax> <strong>molecules</strong> will be molecules of sulfur dioxide. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"27 ppm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is use the <em>definition</em> of <strong>parts per million</strong>, or <strong>ppm</strong>. </p>
<p>You can use <em>parts per million</em> to express the concentration of a solution that contains very, very small amounts of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. As its name suggests, this way of expressing concentration uses <em>parts of solute</em> per <strong>one million parts of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong>. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"ppm" = "parts of solute"/"parts of solvent" xx 10^6color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In order to have a <mathjax>#"1 ppm"#</mathjax> solution, you need to have a solution that contains <strong>one part</strong> of solute <strong>for every</strong> <mathjax>#10^6#</mathjax> <strong>parts</strong> of solvent. </p>
<p>Now, you can consider air to be your <em>solution</em> and sulfur dioxide, <mathjax>#"SO"_2#</mathjax>, to be your solute. Since you have <em>significantly fewer</em> molecules of sulfur dioxide than the total number of molecules of air, you can say that you'll have <em>approximately</em> <mathjax>#2.617 * 10^5#</mathjax> <strong>parts</strong> of solvent. </p>
<p>This means that the concentration of sulfur dioxide in <strong>ppm</strong> will be </p>
<blockquote>
<p><mathjax>#"ppm SO"_2 = (7 color(red)(cancel(color(black)("molecules"))))/(2.617 * 10^5color(red)(cancel(color(black)("molecules")))) xx 10^6 = color(green)(|bar(ul(color(white)(a/a)"27 ppm"color(white)(a/a)|)))#</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>. </p>
<p>So, this tells you that out of <mathjax>#10^6#</mathjax> <strong>molecules of air</strong>, <mathjax>#27#</mathjax> <strong>molecules</strong> will be molecules of sulfur dioxide. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If you have 7 molecules of #SO_2# gas in #2.617 * 10^5# molecules of air, what is the concentration of #SO_2# in ppm?</h1>
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<div class="markdown"><p><mathjax>#"27 ppm"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here is use the <em>definition</em> of <strong>parts per million</strong>, or <strong>ppm</strong>. </p>
<p>You can use <em>parts per million</em> to express the concentration of a solution that contains very, very small amounts of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>. As its name suggests, this way of expressing concentration uses <em>parts of solute</em> per <strong>one million parts of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></strong>. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)"ppm" = "parts of solute"/"parts of solvent" xx 10^6color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>In order to have a <mathjax>#"1 ppm"#</mathjax> solution, you need to have a solution that contains <strong>one part</strong> of solute <strong>for every</strong> <mathjax>#10^6#</mathjax> <strong>parts</strong> of solvent. </p>
<p>Now, you can consider air to be your <em>solution</em> and sulfur dioxide, <mathjax>#"SO"_2#</mathjax>, to be your solute. Since you have <em>significantly fewer</em> molecules of sulfur dioxide than the total number of molecules of air, you can say that you'll have <em>approximately</em> <mathjax>#2.617 * 10^5#</mathjax> <strong>parts</strong> of solvent. </p>
<p>This means that the concentration of sulfur dioxide in <strong>ppm</strong> will be </p>
<blockquote>
<p><mathjax>#"ppm SO"_2 = (7 color(red)(cancel(color(black)("molecules"))))/(2.617 * 10^5color(red)(cancel(color(black)("molecules")))) xx 10^6 = color(green)(|bar(ul(color(white)(a/a)"27 ppm"color(white)(a/a)|)))#</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>. </p>
<p>So, this tells you that out of <mathjax>#10^6#</mathjax> <strong>molecules of air</strong>, <mathjax>#27#</mathjax> <strong>molecules</strong> will be molecules of sulfur dioxide. </p></div>
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</article> | If you have 7 molecules of #SO_2# gas in #2.617 * 10^5# molecules of air, what is the concentration of #SO_2# in ppm? | null |
2,940 | a9133402-6ddd-11ea-a80c-ccda262736ce | https://socratic.org/questions/599ee920b72cff6f3d25d545 | 250 | start physical_unit 9 10 equilibrium_constant_k none qc_end chemical_equation 11 16 qc_end physical_unit 16 16 21 22 equilibrium_concentration qc_end physical_unit 11 11 25 26 equilibrium_concentration qc_end physical_unit 14 14 30 31 equilibrium_concentration qc_end end | [{"type":"physical unit","value":"Equilibrium constant Kc [OF] the reaction"}] | [{"type":"physical unit","value":"250"}] | [{"type":"chemical equation","value":"X + 2 Y <=> Z"},{"type":"physical unit","value":"Equilibrium concentration [OF] Z [=] \\pu{0.216 M}"},{"type":"physical unit","value":"Equilibrium concentration [OF] X [=] \\pu{0.06 M}"},{"type":"physical unit","value":"Equilibrium concentration [OF] Y [=] \\pu{0.12 M}"}] | <h1 class="questionTitle" itemprop="name">How do you determine the equilibrium constant #K_c# for the reaction #X + 2Y rightleftharpoons Z#? The equilibrium concentrations are #"0.216 M"# for #Z#, #"0.06 M"# for #X#, and #"0.12 M"# for #Y#.</h1> | null | 250 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equilibrium constant is simply a ratio between two <em>multiplications</em>. </p>
<p>In the <strong>numerator</strong>, you have the multiplication of the <strong>equilibrium product concentrations</strong> raised to the power of the products' respective stoichiometric coefficients. </p>
<p>In the <strong>denominator</strong>, you have the multiplication of the <strong>equilibrium reactant concentrations</strong> raised to the power of the reactants' respective stoichiometric coefficients. </p>
<p>In your case, the equilibrium reaction looks like this</p>
<blockquote>
<p><mathjax>#"X" + 2"Y" rightleftharpoons "Z"#</mathjax></p>
</blockquote>
<p>which means that you have</p>
<blockquote>
<ul>
<li><mathjax>#"Products:" #</mathjax></li>
</ul>
<blockquote>
<blockquote>
<p><mathjax>#"Z " -> " no coefficient = coefficient of 1"#</mathjax></p>
</blockquote>
</blockquote>
<ul>
<li><mathjax>#"Reactants:"#</mathjax> </li>
</ul>
<blockquote>
<blockquote>
<p><mathjax>#"X " -> " no coefficient = coefficient of 1"#</mathjax> </p>
<p><mathjax>#"Y " -> " coefficient of 2"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>By definition, the equilibrium constant for this reaction will be</p>
<blockquote>
<p><mathjax>#K_c = (["Z"]^1)/(["X"]^1 * ["Y"]^2)#</mathjax></p>
</blockquote>
<p>which is equivalent to</p>
<blockquote>
<p><mathjax>#K_c = (["Z"])/(["X"] * ["Y"]^2)#</mathjax></p>
</blockquote>
<p>Plug in the values you have for the equilibrium concentrations of the three chemical species to find the value of the equilibrium constant--I'll leave the answer <em>without added units</em></p>
<blockquote>
<p><mathjax>#K_c = (0.216)/(0.06 * 0.12^2) = color(darkgreen)(ul(color(black)(250)))#</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 equilibrium concentration of <mathjax>#"X"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#K_c = 250#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equilibrium constant is simply a ratio between two <em>multiplications</em>. </p>
<p>In the <strong>numerator</strong>, you have the multiplication of the <strong>equilibrium product concentrations</strong> raised to the power of the products' respective stoichiometric coefficients. </p>
<p>In the <strong>denominator</strong>, you have the multiplication of the <strong>equilibrium reactant concentrations</strong> raised to the power of the reactants' respective stoichiometric coefficients. </p>
<p>In your case, the equilibrium reaction looks like this</p>
<blockquote>
<p><mathjax>#"X" + 2"Y" rightleftharpoons "Z"#</mathjax></p>
</blockquote>
<p>which means that you have</p>
<blockquote>
<ul>
<li><mathjax>#"Products:" #</mathjax></li>
</ul>
<blockquote>
<blockquote>
<p><mathjax>#"Z " -> " no coefficient = coefficient of 1"#</mathjax></p>
</blockquote>
</blockquote>
<ul>
<li><mathjax>#"Reactants:"#</mathjax> </li>
</ul>
<blockquote>
<blockquote>
<p><mathjax>#"X " -> " no coefficient = coefficient of 1"#</mathjax> </p>
<p><mathjax>#"Y " -> " coefficient of 2"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>By definition, the equilibrium constant for this reaction will be</p>
<blockquote>
<p><mathjax>#K_c = (["Z"]^1)/(["X"]^1 * ["Y"]^2)#</mathjax></p>
</blockquote>
<p>which is equivalent to</p>
<blockquote>
<p><mathjax>#K_c = (["Z"])/(["X"] * ["Y"]^2)#</mathjax></p>
</blockquote>
<p>Plug in the values you have for the equilibrium concentrations of the three chemical species to find the value of the equilibrium constant--I'll leave the answer <em>without added units</em></p>
<blockquote>
<p><mathjax>#K_c = (0.216)/(0.06 * 0.12^2) = color(darkgreen)(ul(color(black)(250)))#</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 equilibrium concentration of <mathjax>#"X"#</mathjax>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you determine the equilibrium constant #K_c# for the reaction #X + 2Y rightleftharpoons Z#? The equilibrium concentrations are #"0.216 M"# for #Z#, #"0.06 M"# for #X#, and #"0.12 M"# for #Y#.</h1>
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Truong-Son N.
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<div class="markdown"><p><mathjax>#K_c = 250#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equilibrium constant is simply a ratio between two <em>multiplications</em>. </p>
<p>In the <strong>numerator</strong>, you have the multiplication of the <strong>equilibrium product concentrations</strong> raised to the power of the products' respective stoichiometric coefficients. </p>
<p>In the <strong>denominator</strong>, you have the multiplication of the <strong>equilibrium reactant concentrations</strong> raised to the power of the reactants' respective stoichiometric coefficients. </p>
<p>In your case, the equilibrium reaction looks like this</p>
<blockquote>
<p><mathjax>#"X" + 2"Y" rightleftharpoons "Z"#</mathjax></p>
</blockquote>
<p>which means that you have</p>
<blockquote>
<ul>
<li><mathjax>#"Products:" #</mathjax></li>
</ul>
<blockquote>
<blockquote>
<p><mathjax>#"Z " -> " no coefficient = coefficient of 1"#</mathjax></p>
</blockquote>
</blockquote>
<ul>
<li><mathjax>#"Reactants:"#</mathjax> </li>
</ul>
<blockquote>
<blockquote>
<p><mathjax>#"X " -> " no coefficient = coefficient of 1"#</mathjax> </p>
<p><mathjax>#"Y " -> " coefficient of 2"#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>By definition, the equilibrium constant for this reaction will be</p>
<blockquote>
<p><mathjax>#K_c = (["Z"]^1)/(["X"]^1 * ["Y"]^2)#</mathjax></p>
</blockquote>
<p>which is equivalent to</p>
<blockquote>
<p><mathjax>#K_c = (["Z"])/(["X"] * ["Y"]^2)#</mathjax></p>
</blockquote>
<p>Plug in the values you have for the equilibrium concentrations of the three chemical species to find the value of the equilibrium constant--I'll leave the answer <em>without added units</em></p>
<blockquote>
<p><mathjax>#K_c = (0.216)/(0.06 * 0.12^2) = color(darkgreen)(ul(color(black)(250)))#</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 equilibrium concentration of <mathjax>#"X"#</mathjax>.</p></div>
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</article> | How do you determine the equilibrium constant #K_c# for the reaction #X + 2Y rightleftharpoons Z#? The equilibrium concentrations are #"0.216 M"# for #Z#, #"0.06 M"# for #X#, and #"0.12 M"# for #Y#. | null |
2,941 | a8c35e8a-6ddd-11ea-aa27-ccda262736ce | https://socratic.org/questions/what-is-the-molecular-mass-in-amu-of-a-gaseous-element-if-8-79-g-occupies-4-54-l | 83.8 amu | start physical_unit 9 10 molecular_weight amu qc_end physical_unit 9 10 12 13 mass qc_end physical_unit 9 10 21 22 temperature qc_end physical_unit 9 10 15 16 volume qc_end physical_unit 9 10 18 19 pressure qc_end end | [{"type":"physical unit","value":"Molecular mass [OF] the gaseous element [IN] amu"}] | [{"type":"physical unit","value":"83.8 amu"}] | [{"type":"physical unit","value":"Mass [OF] the gaseous element [=] \\pu{8.79 g}"},{"type":"physical unit","value":"Temperature [OF] the gaseous element [=] \\pu{37.6 ℃}"},{"type":"physical unit","value":"Volume [OF] the gaseous element [=] \\pu{4.54 L}"},{"type":"physical unit","value":"Pressure [OF] the gaseous element [=] \\pu{447.77 torr}"}] | <h1 class="questionTitle" itemprop="name">What is the molecular mass (in amu) of a gaseous element if 8.79 g occupies 4.54 L at 447.77 torr and 37.6 °C?</h1> | null | 83.8 amu | <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 = nRT color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>, we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#M = (mRT)/(PV)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#m = "8.79 g"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(37.6 + 273.15) K" = "310.75 K"#</mathjax><br/>
<mathjax>#P = 447.77 color(red)(cancel(color(black)("torr"))) × "1 atm"/(760 color(red)(cancel(color(black)("torr")))) = "0.589 17 atm"#</mathjax><br/>
<mathjax>#V = "4.54 L"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = ("8.79 g" × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 310.75 color(red)(cancel(color(black)("K"))))/("0.589 17" color(red)(cancel(color(black)("atm"))) × 4.54 color(red)(cancel(color(black)("L")))) = "83.8 g/mol"#</mathjax></p>
<p>∴ The molecular mass is 83.8 u.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The molecular mass is 83.8 u.</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 = nRT color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>, we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#M = (mRT)/(PV)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#m = "8.79 g"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(37.6 + 273.15) K" = "310.75 K"#</mathjax><br/>
<mathjax>#P = 447.77 color(red)(cancel(color(black)("torr"))) × "1 atm"/(760 color(red)(cancel(color(black)("torr")))) = "0.589 17 atm"#</mathjax><br/>
<mathjax>#V = "4.54 L"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = ("8.79 g" × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 310.75 color(red)(cancel(color(black)("K"))))/("0.589 17" color(red)(cancel(color(black)("atm"))) × 4.54 color(red)(cancel(color(black)("L")))) = "83.8 g/mol"#</mathjax></p>
<p>∴ The molecular mass is 83.8 u.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molecular mass (in amu) of a gaseous element if 8.79 g occupies 4.54 L at 447.77 torr and 37.6 °C?</h1>
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<div class="markdown"><p>The molecular mass is 83.8 u.</p></div>
</div>
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<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 = nRT color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>Since <mathjax>#n = "mass"/"molar mass" = m/M#</mathjax>, we can write the Ideal Gas Law as</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a)PV = m/MRTcolor(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this to get</p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax>#M = (mRT)/(PV)#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p><mathjax>#m = "8.79 g"#</mathjax><br/>
<mathjax>#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#</mathjax><br/>
<mathjax>#T = "(37.6 + 273.15) K" = "310.75 K"#</mathjax><br/>
<mathjax>#P = 447.77 color(red)(cancel(color(black)("torr"))) × "1 atm"/(760 color(red)(cancel(color(black)("torr")))) = "0.589 17 atm"#</mathjax><br/>
<mathjax>#V = "4.54 L"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#M = ("8.79 g" × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 310.75 color(red)(cancel(color(black)("K"))))/("0.589 17" color(red)(cancel(color(black)("atm"))) × 4.54 color(red)(cancel(color(black)("L")))) = "83.8 g/mol"#</mathjax></p>
<p>∴ The molecular mass is 83.8 u.</p></div>
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</article> | What is the molecular mass (in amu) of a gaseous element if 8.79 g occupies 4.54 L at 447.77 torr and 37.6 °C? | null |
2,942 | ab1f8cf0-6ddd-11ea-b166-ccda262736ce | https://socratic.org/questions/what-is-the-h-in-a-solution-that-has-a-ph-of-3-35 | 4.47 × 10^(-4) M | start physical_unit 6 6 [h+] mol/l qc_end physical_unit 6 6 12 12 ph qc_end end | [{"type":"physical unit","value":"[H+] [OF] the solution [IN] M"}] | [{"type":"physical unit","value":"4.47 × 10^(-4) M"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{3.35}"}] | <h1 class="questionTitle" itemprop="name">What is the #[H^+]# in a solution that has a pH of 3.35?</h1> | null | 4.47 × 10^(-4) M | <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 is usually found by the following expression:</p>
<p><mathjax>#pH=-log[H^+]#</mathjax></p>
<p>Therefore, to find the concentration of <mathjax>#H^+#</mathjax> we can rearrange this expression and thus, </p>
<p><mathjax>#[H^+]=10^(-pH)#</mathjax></p>
<p><mathjax>#=>[H^+]=10^(-3.35)=4.47xx10^(-4)M#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#[H^+]=4.47xx10^(-4)M#</mathjax></p></div>
</div>
</div>
<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 is usually found by the following expression:</p>
<p><mathjax>#pH=-log[H^+]#</mathjax></p>
<p>Therefore, to find the concentration of <mathjax>#H^+#</mathjax> we can rearrange this expression and thus, </p>
<p><mathjax>#[H^+]=10^(-pH)#</mathjax></p>
<p><mathjax>#=>[H^+]=10^(-3.35)=4.47xx10^(-4)M#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the #[H^+]# in a solution that has a pH of 3.35?</h1>
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<div class="markdown"><p><mathjax>#[H^+]=4.47xx10^(-4)M#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of a solution is usually found by the following expression:</p>
<p><mathjax>#pH=-log[H^+]#</mathjax></p>
<p>Therefore, to find the concentration of <mathjax>#H^+#</mathjax> we can rearrange this expression and thus, </p>
<p><mathjax>#[H^+]=10^(-pH)#</mathjax></p>
<p><mathjax>#=>[H^+]=10^(-3.35)=4.47xx10^(-4)M#</mathjax></p></div>
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</article> | What is the #[H^+]# in a solution that has a pH of 3.35? | null |
2,943 | aca85d98-6ddd-11ea-a85d-ccda262736ce | https://socratic.org/questions/lithium-and-nitrogen-react-to-produce-lithium-nitride-6li-s-n-2-g-2li-3n-s-how-m | 0.24 moles | start physical_unit 6 7 mole mol qc_end physical_unit 0 0 24 25 mole qc_end chemical_equation 8 14 qc_end end | [{"type":"physical unit","value":"Mole [OF] lithium nitride [IN] moles"}] | [{"type":"physical unit","value":"0.24 moles"}] | [{"type":"physical unit","value":"Mole [OF] lithium [=] \\pu{0.710 mol}"},{"type":"chemical equation","value":"6 Li(s) + N2(g) -> 2 Li3N(s)"}] | <h1 class="questionTitle" itemprop="name">Lithium and nitrogen react to produce lithium nitride: #6Li(s) + N_2(g) -> 2Li_3N(s)#. How many moles of lithium nitride are produced when 0.710 mol of lithium react in this fashion?</h1> | null | 0.24 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since the given information is in moles, we need only look at the ratio between the chemicals:</p>
<p>Six moles of Li will produce 2 moles of lithium nitride. So, the ratio is 6:2 or 3:1.</p>
<p>Therefore, we can set up a proportion:</p>
<p><mathjax>#(Li)/(Li_3N) = 3/1 = 0.710/x#</mathjax></p>
<p>This tells us that in the general case, we need three times as much Li as the amount of <mathjax>#Li_3N#</mathjax> to be produced, and that the amount of <mathjax>#Li_3N#</mathjax> produced in this case must be in the same ratio. so, we can solve </p>
<p><mathjax>#3/1 = 0.710/x#</mathjax></p>
<p>Cross-multiply</p>
<p><mathjax>#3x = 0.710#</mathjax></p>
<p><mathjax>#x= 0.710 -: 3 = 0.237#</mathjax> mol of <mathjax>#Li_3N#</mathjax> produced</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We will obtain 0.237 mol of <mathjax>#Li_3N#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since the given information is in moles, we need only look at the ratio between the chemicals:</p>
<p>Six moles of Li will produce 2 moles of lithium nitride. So, the ratio is 6:2 or 3:1.</p>
<p>Therefore, we can set up a proportion:</p>
<p><mathjax>#(Li)/(Li_3N) = 3/1 = 0.710/x#</mathjax></p>
<p>This tells us that in the general case, we need three times as much Li as the amount of <mathjax>#Li_3N#</mathjax> to be produced, and that the amount of <mathjax>#Li_3N#</mathjax> produced in this case must be in the same ratio. so, we can solve </p>
<p><mathjax>#3/1 = 0.710/x#</mathjax></p>
<p>Cross-multiply</p>
<p><mathjax>#3x = 0.710#</mathjax></p>
<p><mathjax>#x= 0.710 -: 3 = 0.237#</mathjax> mol of <mathjax>#Li_3N#</mathjax> produced</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Lithium and nitrogen react to produce lithium nitride: #6Li(s) + N_2(g) -> 2Li_3N(s)#. How many moles of lithium nitride are produced when 0.710 mol of lithium react in this fashion?</h1>
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<div class="markdown"><p>We will obtain 0.237 mol of <mathjax>#Li_3N#</mathjax></p></div>
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<div class="markdown"><p>Since the given information is in moles, we need only look at the ratio between the chemicals:</p>
<p>Six moles of Li will produce 2 moles of lithium nitride. So, the ratio is 6:2 or 3:1.</p>
<p>Therefore, we can set up a proportion:</p>
<p><mathjax>#(Li)/(Li_3N) = 3/1 = 0.710/x#</mathjax></p>
<p>This tells us that in the general case, we need three times as much Li as the amount of <mathjax>#Li_3N#</mathjax> to be produced, and that the amount of <mathjax>#Li_3N#</mathjax> produced in this case must be in the same ratio. so, we can solve </p>
<p><mathjax>#3/1 = 0.710/x#</mathjax></p>
<p>Cross-multiply</p>
<p><mathjax>#3x = 0.710#</mathjax></p>
<p><mathjax>#x= 0.710 -: 3 = 0.237#</mathjax> mol of <mathjax>#Li_3N#</mathjax> produced</p></div>
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</article> | Lithium and nitrogen react to produce lithium nitride: #6Li(s) + N_2(g) -> 2Li_3N(s)#. How many moles of lithium nitride are produced when 0.710 mol of lithium react in this fashion? | null |
2,944 | a9428322-6ddd-11ea-b267-ccda262736ce | https://socratic.org/questions/consider-the-reaction-when-aqueous-solutions-of-iron-iii-chloride-and-silver-i-n | Ag+(aq) + Cl-(aq) -> AgCl(s) | start chemical_equation qc_end substance 10 11 qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] the net ionic equation"}] | [{"type":"chemical equation","value":"Ag+(aq) + Cl-(aq) -> AgCl(s)"}] | [{"type":"substance name","value":"Iron(III) chloride aqueous solutions"},{"type":"substance name","value":"Silver(I) nitrate"},{"type":"other","value":"Label aq, l,g,s."}] | <h1 class="questionTitle" itemprop="name">Consider the reaction when aqueous solutions of iron(III) chloride and silver(I) nitrate are combined. The net ionic equation for this reaction is:? label aq, l,g,s</h1> | null | Ag+(aq) + Cl-(aq) -> AgCl(s) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The evidence that a double replacement reaction has happened is the formation of a solid, liquid or gas from aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>. In this case, silver chloride is a precipitate.</p></div>
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<div class="markdown"><p>Reaction:<br/>
<mathjax>#FeCl_3(aq)+3AgNO_3(aq)->Fe(NO_3)_3(aq)+3AgCl(s)#</mathjax></p>
<p>Net ionic:</p>
<p><mathjax>#Ag^(+)(aq)+Cl^(-)(aq)-> AgCl(s)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The evidence that a double replacement reaction has happened is the formation of a solid, liquid or gas from aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>. In this case, silver chloride is a precipitate.</p></div>
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<h1 class="questionTitle" itemprop="name">Consider the reaction when aqueous solutions of iron(III) chloride and silver(I) nitrate are combined. The net ionic equation for this reaction is:? label aq, l,g,s</h1>
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<div class="markdown"><p>Reaction:<br/>
<mathjax>#FeCl_3(aq)+3AgNO_3(aq)->Fe(NO_3)_3(aq)+3AgCl(s)#</mathjax></p>
<p>Net ionic:</p>
<p><mathjax>#Ag^(+)(aq)+Cl^(-)(aq)-> AgCl(s)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The evidence that a double replacement reaction has happened is the formation of a solid, liquid or gas from aqueous <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a>. In this case, silver chloride is a precipitate.</p></div>
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</article> | Consider the reaction when aqueous solutions of iron(III) chloride and silver(I) nitrate are combined. The net ionic equation for this reaction is:? label aq, l,g,s | null |
2,945 | aca6d79f-6ddd-11ea-babc-ccda262736ce | https://socratic.org/questions/using-the-periodic-table-how-would-you-determine-the-number-of-neutrons-in-16-o | 8 | start physical_unit 11 11 number none qc_end chemical_equation 13 13 qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Number [OF] neutrons"}] | [{"type":"physical unit","value":"8"}] | [{"type":"chemical equation","value":"^16O"},{"type":"other","value":"Using the periodic table."}] | <h1 class="questionTitle" itemprop="name">Using the periodic table, how would you determine the number of neutrons in 16 O?</h1> | null | 8 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How does this help us?</p>
<p>Well, we have the <mathjax>#""^16O#</mathjax> isotope. The nucleus of the isotope contains 8 protons (i.e. <mathjax>#Z=8#</mathjax>) unequivocally: this number of protons is what defines the element as an oxygen atom. But along with the 8 protons, the nucleus contains another 8 MASSIVE particles, i.e. 8 neutrons.......fundamental, massive <mathjax>#"nucular particles"#</mathjax> of NEUTRAL charge; the sum of the massive particles, neutrons, and protons, <mathjax>#8+8=16#</mathjax> gives the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mass-number">mass number</a> of the isotope, which (as here) is commonly superscripted on the left of the element symbol. The <mathjax>#""^16O#</mathjax> isotope is the common one, and is approx. <mathjax>#99.8%#</mathjax> abundant (check this figure!). </p>
<p><mathjax>#""^17O#</mathjax>, and <mathjax>#""^18O#</mathjax> <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/isotopes">isotopes</a> are the next most common isotopes. Can you tell me here (i) how many protons in each isotope; and (ii) how many neutrons in each isotope?</p></div>
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<div class="markdown"><p>Well <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a> tells me UNEQUIVOCALLY, that the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a>, <mathjax>#Z#</mathjax>, for oxygen is <mathjax>#8#</mathjax>, and thus the nucleus contains 8 neutrons.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How does this help us?</p>
<p>Well, we have the <mathjax>#""^16O#</mathjax> isotope. The nucleus of the isotope contains 8 protons (i.e. <mathjax>#Z=8#</mathjax>) unequivocally: this number of protons is what defines the element as an oxygen atom. But along with the 8 protons, the nucleus contains another 8 MASSIVE particles, i.e. 8 neutrons.......fundamental, massive <mathjax>#"nucular particles"#</mathjax> of NEUTRAL charge; the sum of the massive particles, neutrons, and protons, <mathjax>#8+8=16#</mathjax> gives the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mass-number">mass number</a> of the isotope, which (as here) is commonly superscripted on the left of the element symbol. The <mathjax>#""^16O#</mathjax> isotope is the common one, and is approx. <mathjax>#99.8%#</mathjax> abundant (check this figure!). </p>
<p><mathjax>#""^17O#</mathjax>, and <mathjax>#""^18O#</mathjax> <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/isotopes">isotopes</a> are the next most common isotopes. Can you tell me here (i) how many protons in each isotope; and (ii) how many neutrons in each isotope?</p></div>
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<h1 class="questionTitle" itemprop="name">Using the periodic table, how would you determine the number of neutrons in 16 O?</h1>
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<div class="markdown"><p>Well <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the Periodic Table</a> tells me UNEQUIVOCALLY, that the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a>, <mathjax>#Z#</mathjax>, for oxygen is <mathjax>#8#</mathjax>, and thus the nucleus contains 8 neutrons.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>How does this help us?</p>
<p>Well, we have the <mathjax>#""^16O#</mathjax> isotope. The nucleus of the isotope contains 8 protons (i.e. <mathjax>#Z=8#</mathjax>) unequivocally: this number of protons is what defines the element as an oxygen atom. But along with the 8 protons, the nucleus contains another 8 MASSIVE particles, i.e. 8 neutrons.......fundamental, massive <mathjax>#"nucular particles"#</mathjax> of NEUTRAL charge; the sum of the massive particles, neutrons, and protons, <mathjax>#8+8=16#</mathjax> gives the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mass-number">mass number</a> of the isotope, which (as here) is commonly superscripted on the left of the element symbol. The <mathjax>#""^16O#</mathjax> isotope is the common one, and is approx. <mathjax>#99.8%#</mathjax> abundant (check this figure!). </p>
<p><mathjax>#""^17O#</mathjax>, and <mathjax>#""^18O#</mathjax> <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/isotopes">isotopes</a> are the next most common isotopes. Can you tell me here (i) how many protons in each isotope; and (ii) how many neutrons in each isotope?</p></div>
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Apr 4, 2017
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<div class="markdown"><p><mathjax>#8#</mathjax> neutrons</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mass-number">mass number</a> is the number on the top-left of each element symbol that represents the sum of the number of protons and neutrons.<br/>
as stated in the question, this is <mathjax>#16#</mathjax> for <mathjax>#16O#</mathjax>.<br/>
the proton number is the number of protons and the number of electrons in the atom (not added together). this is on the bottom-left of each element symbol. <br/>
this is <mathjax>#8#</mathjax> for oxygen, no matter the number of neutrons.<br/>
number of neutrons = mass number - proton number<br/>
<mathjax>#= 16 - 8#</mathjax><br/>
<mathjax>#=8#</mathjax></p>
<p>therefore there are 8 neutrons.</p></div>
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</article> | Using the periodic table, how would you determine the number of neutrons in 16 O? | null |
2,946 | aa368602-6ddd-11ea-8f21-ccda262736ce | https://socratic.org/questions/what-is-the-correct-formula-of-a-compound-that-has-ten-oxygen-atoms-and-four-pho | P4O10 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] default"}] | [{"type":"chemical equation","value":"P4O10"}] | [{"type":"physical unit","value":"Number [OF] oxygen atoms [=] \\pu{10}"},{"type":"physical unit","value":"Number [OF] phosphorus atoms [=] \\pu{4}"}] | <h1 class="questionTitle" itemprop="name">What is the correct formula of a compound that has ten oxygen atoms and four phosphorus atoms? </h1> | null | P4O10 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The number of atoms of each element in the compound is represented by <strong>subscripts</strong> following the symbol of each element.</p>
<p>Therefore, we can write a compound with <mathjax>#color(red)(4#</mathjax> atoms of <mathjax>#sfcolor(red)("phosphorus"#</mathjax> and <mathjax>#color(blue)(10#</mathjax> atoms of <mathjax>#sfcolor(blue)("oxygen"#</mathjax> as</p>
<p><mathjax>#color(red)("P"_4)color(blue)("O"_10)#</mathjax></p>
<p>(<mathjax>#"P"#</mathjax> is the symbol for phosphorus, and <mathjax>#"O"#</mathjax> is the symbol for oxygen.)</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"P"_4"O"_10#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The number of atoms of each element in the compound is represented by <strong>subscripts</strong> following the symbol of each element.</p>
<p>Therefore, we can write a compound with <mathjax>#color(red)(4#</mathjax> atoms of <mathjax>#sfcolor(red)("phosphorus"#</mathjax> and <mathjax>#color(blue)(10#</mathjax> atoms of <mathjax>#sfcolor(blue)("oxygen"#</mathjax> as</p>
<p><mathjax>#color(red)("P"_4)color(blue)("O"_10)#</mathjax></p>
<p>(<mathjax>#"P"#</mathjax> is the symbol for phosphorus, and <mathjax>#"O"#</mathjax> is the symbol for oxygen.)</p></div>
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<h1 class="questionTitle" itemprop="name">What is the correct formula of a compound that has ten oxygen atoms and four phosphorus atoms? </h1>
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Nathan L.
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Jun 24, 2017
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<div class="markdown"><p><mathjax>#"P"_4"O"_10#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The number of atoms of each element in the compound is represented by <strong>subscripts</strong> following the symbol of each element.</p>
<p>Therefore, we can write a compound with <mathjax>#color(red)(4#</mathjax> atoms of <mathjax>#sfcolor(red)("phosphorus"#</mathjax> and <mathjax>#color(blue)(10#</mathjax> atoms of <mathjax>#sfcolor(blue)("oxygen"#</mathjax> as</p>
<p><mathjax>#color(red)("P"_4)color(blue)("O"_10)#</mathjax></p>
<p>(<mathjax>#"P"#</mathjax> is the symbol for phosphorus, and <mathjax>#"O"#</mathjax> is the symbol for oxygen.)</p></div>
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</article> | What is the correct formula of a compound that has ten oxygen atoms and four phosphorus atoms? | null |
2,947 | a95c89c0-6ddd-11ea-8a10-ccda262736ce | https://socratic.org/questions/68-ml-of-a-0-28-m-cacl2-solution-is-added-to-92-ml-of-a-0-46-m-cacl2-solution-de | 0.38 M | start physical_unit 23 25 concentration mol/l qc_end physical_unit 6 7 0 1 volume qc_end physical_unit 6 7 4 5 molarity qc_end physical_unit 6 7 11 12 volume qc_end physical_unit 6 7 15 16 molarity qc_end end | [{"type":"physical unit","value":"Concentration [OF] the combined solution [IN] M"}] | [{"type":"physical unit","value":"0.38 M"}] | [{"type":"physical unit","value":"Volume1 [OF] CaCl2 solution [=] \\pu{68 mL}"},{"type":"physical unit","value":"Molarity1 [OF] CaCl2 solution [=] \\pu{0.28 M}"},{"type":"physical unit","value":"Volume2 [OF] CaCl2 solution [=] \\pu{92 mL}"},{"type":"physical unit","value":"Molarity2 [OF] CaCl2 solution [=] \\pu{0.46 M}"}] | <h1 class="questionTitle" itemprop="name">68 mL of a 0.28 M CaCl2 solution is added to 92 mL of a 0.46 M CaCl2 solution. Determine the concentration of the combined solution. Show all work with units.
How would you solve this?</h1> | null | 0.38 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The basic definition of concentration is as <em>amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> per unit volume.</em> That is..........</p>
<p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>Most of the time we want to assess <mathjax>#"moles of solute"#</mathjax>, and this is simply the product:</p>
<p><mathjax>#"Moles of solute"="Volume"xx"concentration"#</mathjax>. I would get familiar with this expression, because you will use it a lot.</p>
<p>.........And in problems like these we must assume (REASONABLY!) that the volumes are additive. And so to address your problem (finally!), we solve the quotient:</p>
<p><mathjax>#(0.092*Lxx0.46*mol*L^-1+0.068*Lxx0.28*mol*L^-1)/((92+68)xx10^-3*L)#</mathjax></p>
<p><mathjax>#~=0.4*mol*L^-1#</mathjax></p>
<p>Do the units in the quotient cancel to give an answer in <mathjax>#mol*L^-1#</mathjax>? It is your problem not mine. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>I hope I would solve it correctly.</p>
<p>......I finally get a concentration with respect to <mathjax>#CaCl_2#</mathjax> of <mathjax>#~=0.4*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The basic definition of concentration is as <em>amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> per unit volume.</em> That is..........</p>
<p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>Most of the time we want to assess <mathjax>#"moles of solute"#</mathjax>, and this is simply the product:</p>
<p><mathjax>#"Moles of solute"="Volume"xx"concentration"#</mathjax>. I would get familiar with this expression, because you will use it a lot.</p>
<p>.........And in problems like these we must assume (REASONABLY!) that the volumes are additive. And so to address your problem (finally!), we solve the quotient:</p>
<p><mathjax>#(0.092*Lxx0.46*mol*L^-1+0.068*Lxx0.28*mol*L^-1)/((92+68)xx10^-3*L)#</mathjax></p>
<p><mathjax>#~=0.4*mol*L^-1#</mathjax></p>
<p>Do the units in the quotient cancel to give an answer in <mathjax>#mol*L^-1#</mathjax>? It is your problem not mine. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">68 mL of a 0.28 M CaCl2 solution is added to 92 mL of a 0.46 M CaCl2 solution. Determine the concentration of the combined solution. Show all work with units.
How would you solve this?</h1>
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anor277
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May 8, 2017
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<div class="markdown"><p>I hope I would solve it correctly.</p>
<p>......I finally get a concentration with respect to <mathjax>#CaCl_2#</mathjax> of <mathjax>#~=0.4*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The basic definition of concentration is as <em>amount of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> per unit volume.</em> That is..........</p>
<p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax>.</p>
<p>Most of the time we want to assess <mathjax>#"moles of solute"#</mathjax>, and this is simply the product:</p>
<p><mathjax>#"Moles of solute"="Volume"xx"concentration"#</mathjax>. I would get familiar with this expression, because you will use it a lot.</p>
<p>.........And in problems like these we must assume (REASONABLY!) that the volumes are additive. And so to address your problem (finally!), we solve the quotient:</p>
<p><mathjax>#(0.092*Lxx0.46*mol*L^-1+0.068*Lxx0.28*mol*L^-1)/((92+68)xx10^-3*L)#</mathjax></p>
<p><mathjax>#~=0.4*mol*L^-1#</mathjax></p>
<p>Do the units in the quotient cancel to give an answer in <mathjax>#mol*L^-1#</mathjax>? It is your problem not mine. </p></div>
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Ernest Z.
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<div class="markdown"><p>The concentration of the combined solution is 0.38 mol/L.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The steps to follow are:</p>
<ol>
<li>Calculate the number of moles in each solution.</li>
<li>Use the total moles and the total volume to calculate the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the combined solution.</li>
</ol>
<blockquote></blockquote>
<p><strong>1. Number of moles in each solution</strong></p>
<p><strong>(a)</strong> Solution 1</p>
<p><mathjax>#"Moles of CaCl"_2 = 0.068 color(red)(cancel(color(black)("L solution"))) × ("0.28 mol CaCl"_2)/(1 color(red)(cancel(color(black)("L solution")))) = "0.0190 mol CaCl"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>(b)</strong> Solution2</p>
<p><mathjax>#"Moles of CaCl"_2 = 0.092 color(red)(cancel(color(black)("L solution"))) × ("0.46 mol CaCl"_2)/(1 color(red)(cancel(color(black)("L solution")))) = "0.0423 mol CaCl"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>Molarity of combined <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a></strong></p>
<p><mathjax>#"Total moles" = "(0.0190 + 0.0423) mol" = "0.0613 mol"#</mathjax></p>
<p><mathjax>#"Total volume" = "(68 + 92) mL" = "160 mL" = "0.160 L"#</mathjax></p>
<p><mathjax>#"Molarity" = "moles"/"litres" = "0.0613 mol"/"0.160 L" = "0.38 mol/L"#</mathjax></p></div>
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</article> | 68 mL of a 0.28 M CaCl2 solution is added to 92 mL of a 0.46 M CaCl2 solution. Determine the concentration of the combined solution. Show all work with units.
How would you solve this? | null |
2,948 | a9a2ef5e-6ddd-11ea-9055-ccda262736ce | https://socratic.org/questions/in-the-formula-h-2o-2-h-2o-o-2-how-many-grams-of-o-2-are-produced-from-the-decom | 32.00 grams | start physical_unit 7 7 mass g qc_end chemical_equation 3 7 qc_end physical_unit 3 3 19 20 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] O2 [IN] grams"}] | [{"type":"physical unit","value":"32.00 grams"}] | [{"type":"chemical equation","value":"H2O2 -> H2O + O2"},{"type":"physical unit","value":"Mass [OF] H2O2 [=] \\pu{68 g}"}] | <h1 class="questionTitle" itemprop="name">In the formula #H_2O_2 -> H_2O + O_2#, how many grams of #O_2# are produced from the decomposition of 68 g of #H_2O_2#? </h1> | null | 32.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Your equation is not correctly balanced</strong>. The correct equation is as follows:</p>
<p><mathjax>#2"H"_2"O"_2 -> 2"H"_2"O" + "O"_2#</mathjax></p>
<p>First, calculate the <strong>moles of</strong> <mathjax>#"H"_2"O"_2#</mathjax> <strong>reacting</strong>. In order to do this, we must evaluate the <strong>relative molecular mass</strong> (<mathjax>#"M"_r#</mathjax>) of hydrogen peroxide:</p>
<p><mathjax>#"M"_r#</mathjax> (<mathjax>#"H"_2"O"_2#</mathjax>) <mathjax>#= 2xx1 + 2xx16 = 34#</mathjax></p>
<p><mathjax>#"mol" = "m"/"M"_r = 68/34 = 2#</mathjax> <mathjax>#"moles"#</mathjax></p>
<p>Next, we must compare the moles of <strong>hydrogen peroxide</strong> and <strong>oxygen gas</strong>. According to the corrected equation, this <strong>mole ratio</strong> is as shown below:</p>
<p><mathjax>#"mol"#</mathjax> (<mathjax>#"H"_2"O"_2#</mathjax>) : <mathjax>#"mol"#</mathjax> (<mathjax>#"O"_2#</mathjax>)<br/>
<mathjax>#" "2" ":" "1#</mathjax></p>
<p>Thus, the moles of oxygen produced in this reaction will be exactly <em>half</em> the moles of hydrogen peroxide reacting.</p>
<p><mathjax>#:.#</mathjax> <mathjax>#"mol"#</mathjax> (<mathjax>#"O"_2#</mathjax>)<mathjax># = ("mol"("H"_2"O"_2))/2 = 2/2 = 1#</mathjax> <mathjax>#"mole"#</mathjax></p>
<p>Finally, convert from moles into mass, given that <mathjax>#"M"_r#</mathjax> (<mathjax>#"O"_2#</mathjax>)<mathjax># = 32#</mathjax> :</p>
<p><mathjax>#"mol" = "m"/"M"_r => "m" = "mol"xx"M"_r = 1 xx 32 = 32#</mathjax> <mathjax>#"g"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"mass"#</mathjax> (<mathjax>#"O"_2#</mathjax>)<mathjax># = 32#</mathjax> <mathjax>#"g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Your equation is not correctly balanced</strong>. The correct equation is as follows:</p>
<p><mathjax>#2"H"_2"O"_2 -> 2"H"_2"O" + "O"_2#</mathjax></p>
<p>First, calculate the <strong>moles of</strong> <mathjax>#"H"_2"O"_2#</mathjax> <strong>reacting</strong>. In order to do this, we must evaluate the <strong>relative molecular mass</strong> (<mathjax>#"M"_r#</mathjax>) of hydrogen peroxide:</p>
<p><mathjax>#"M"_r#</mathjax> (<mathjax>#"H"_2"O"_2#</mathjax>) <mathjax>#= 2xx1 + 2xx16 = 34#</mathjax></p>
<p><mathjax>#"mol" = "m"/"M"_r = 68/34 = 2#</mathjax> <mathjax>#"moles"#</mathjax></p>
<p>Next, we must compare the moles of <strong>hydrogen peroxide</strong> and <strong>oxygen gas</strong>. According to the corrected equation, this <strong>mole ratio</strong> is as shown below:</p>
<p><mathjax>#"mol"#</mathjax> (<mathjax>#"H"_2"O"_2#</mathjax>) : <mathjax>#"mol"#</mathjax> (<mathjax>#"O"_2#</mathjax>)<br/>
<mathjax>#" "2" ":" "1#</mathjax></p>
<p>Thus, the moles of oxygen produced in this reaction will be exactly <em>half</em> the moles of hydrogen peroxide reacting.</p>
<p><mathjax>#:.#</mathjax> <mathjax>#"mol"#</mathjax> (<mathjax>#"O"_2#</mathjax>)<mathjax># = ("mol"("H"_2"O"_2))/2 = 2/2 = 1#</mathjax> <mathjax>#"mole"#</mathjax></p>
<p>Finally, convert from moles into mass, given that <mathjax>#"M"_r#</mathjax> (<mathjax>#"O"_2#</mathjax>)<mathjax># = 32#</mathjax> :</p>
<p><mathjax>#"mol" = "m"/"M"_r => "m" = "mol"xx"M"_r = 1 xx 32 = 32#</mathjax> <mathjax>#"g"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">In the formula #H_2O_2 -> H_2O + O_2#, how many grams of #O_2# are produced from the decomposition of 68 g of #H_2O_2#? </h1>
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Owen Bell
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Jan 4, 2016
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<div class="markdown"><p><mathjax>#"mass"#</mathjax> (<mathjax>#"O"_2#</mathjax>)<mathjax># = 32#</mathjax> <mathjax>#"g"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Your equation is not correctly balanced</strong>. The correct equation is as follows:</p>
<p><mathjax>#2"H"_2"O"_2 -> 2"H"_2"O" + "O"_2#</mathjax></p>
<p>First, calculate the <strong>moles of</strong> <mathjax>#"H"_2"O"_2#</mathjax> <strong>reacting</strong>. In order to do this, we must evaluate the <strong>relative molecular mass</strong> (<mathjax>#"M"_r#</mathjax>) of hydrogen peroxide:</p>
<p><mathjax>#"M"_r#</mathjax> (<mathjax>#"H"_2"O"_2#</mathjax>) <mathjax>#= 2xx1 + 2xx16 = 34#</mathjax></p>
<p><mathjax>#"mol" = "m"/"M"_r = 68/34 = 2#</mathjax> <mathjax>#"moles"#</mathjax></p>
<p>Next, we must compare the moles of <strong>hydrogen peroxide</strong> and <strong>oxygen gas</strong>. According to the corrected equation, this <strong>mole ratio</strong> is as shown below:</p>
<p><mathjax>#"mol"#</mathjax> (<mathjax>#"H"_2"O"_2#</mathjax>) : <mathjax>#"mol"#</mathjax> (<mathjax>#"O"_2#</mathjax>)<br/>
<mathjax>#" "2" ":" "1#</mathjax></p>
<p>Thus, the moles of oxygen produced in this reaction will be exactly <em>half</em> the moles of hydrogen peroxide reacting.</p>
<p><mathjax>#:.#</mathjax> <mathjax>#"mol"#</mathjax> (<mathjax>#"O"_2#</mathjax>)<mathjax># = ("mol"("H"_2"O"_2))/2 = 2/2 = 1#</mathjax> <mathjax>#"mole"#</mathjax></p>
<p>Finally, convert from moles into mass, given that <mathjax>#"M"_r#</mathjax> (<mathjax>#"O"_2#</mathjax>)<mathjax># = 32#</mathjax> :</p>
<p><mathjax>#"mol" = "m"/"M"_r => "m" = "mol"xx"M"_r = 1 xx 32 = 32#</mathjax> <mathjax>#"g"#</mathjax></p></div>
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</article> | In the formula #H_2O_2 -> H_2O + O_2#, how many grams of #O_2# are produced from the decomposition of 68 g of #H_2O_2#? | null |
2,949 | a9874dfb-6ddd-11ea-a187-ccda262736ce | https://socratic.org/questions/how-do-you-balance-h3po4-mg-oh-mg3-po4-2-h20 | 2 H3PO4 + 3 Mg(OH) -> Mg3(PO4)2 + 6 H2O | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 H3PO4 + 3 Mg(OH) -> Mg3(PO4)2 + 6 H2O"}] | [{"type":"chemical equation","value":"H3PO4 + Mg(OH) -> Mg3(PO4)2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance H3PO4 +Mg(OH)-->Mg3(PO4)2 +H20?</h1> | null | 2 H3PO4 + 3 Mg(OH) -> Mg3(PO4)2 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Please note that since magnesium is in group 2 it has a valency of 2+ and so the formula for magnesium hydroxide is <mathjax>#Mg(OH)_2#</mathjax> [Mistake in question]</p>
<p>This is a typical <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction of an acid with a base to form a salt and water. The reaction is exothermic, gives off heat, <mathjax>#Delta H<0#</mathjax>, and may be balanced by adding balancing numbers in front, ie adding molecules, in order to ensure that the total number of atoms of each element is the same on the left and right hand sides of the equation.</p>
<p>Doing so we obtain :</p>
<p><mathjax>#2H_3PO_4+3Mg(OH)_2->Mg_3(PO4)_2+6H_2O#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#2H_3PO_4+3Mg(OH)_2->Mg_3(PO4)_2+6H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Please note that since magnesium is in group 2 it has a valency of 2+ and so the formula for magnesium hydroxide is <mathjax>#Mg(OH)_2#</mathjax> [Mistake in question]</p>
<p>This is a typical <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction of an acid with a base to form a salt and water. The reaction is exothermic, gives off heat, <mathjax>#Delta H<0#</mathjax>, and may be balanced by adding balancing numbers in front, ie adding molecules, in order to ensure that the total number of atoms of each element is the same on the left and right hand sides of the equation.</p>
<p>Doing so we obtain :</p>
<p><mathjax>#2H_3PO_4+3Mg(OH)_2->Mg_3(PO4)_2+6H_2O#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you balance H3PO4 +Mg(OH)-->Mg3(PO4)2 +H20?</h1>
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<div class="markdown"><p><mathjax>#2H_3PO_4+3Mg(OH)_2->Mg_3(PO4)_2+6H_2O#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Please note that since magnesium is in group 2 it has a valency of 2+ and so the formula for magnesium hydroxide is <mathjax>#Mg(OH)_2#</mathjax> [Mistake in question]</p>
<p>This is a typical <a href="http://socratic.org/chemistry/reactions-in-solution/neutralization">neutralization</a> reaction of an acid with a base to form a salt and water. The reaction is exothermic, gives off heat, <mathjax>#Delta H<0#</mathjax>, and may be balanced by adding balancing numbers in front, ie adding molecules, in order to ensure that the total number of atoms of each element is the same on the left and right hand sides of the equation.</p>
<p>Doing so we obtain :</p>
<p><mathjax>#2H_3PO_4+3Mg(OH)_2->Mg_3(PO4)_2+6H_2O#</mathjax></p></div>
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</article> | How do you balance H3PO4 +Mg(OH)-->Mg3(PO4)2 +H20? | null |
2,950 | ab70a106-6ddd-11ea-b5f9-ccda262736ce | https://socratic.org/questions/what-is-the-volume-occupied-by-20-7-g-of-argon-gas-at-a-pressure-of-1-04-atm-and | 16.75 L | start physical_unit 9 10 volume l qc_end physical_unit 9 10 6 7 mass qc_end physical_unit 9 10 15 16 pressure qc_end physical_unit 9 10 21 22 temperature qc_end end | [{"type":"physical unit","value":"Volume [OF] argon gas [IN] L"}] | [{"type":"physical unit","value":"16.75 L"}] | [{"type":"physical unit","value":"Mass [OF] argon gas [=] \\pu{20.7 g}"},{"type":"physical unit","value":"Pressure [OF] argon gas [=] \\pu{1.04 atm}"},{"type":"physical unit","value":"Temperature [OF] argon gas [=] \\pu{409 K}"}] | <h1 class="questionTitle" itemprop="name">What is the volume occupied by 20.7 g of argon gas at a pressure of 1.04 atm and a temperature of 409 K? </h1> | null | 16.75 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Pressure is moderate, temperature is elevated, and the gas is inert, so we would guess the ideal gas equation will give a good approximation:</p>
<p><mathjax>#V#</mathjax> <mathjax>#=#</mathjax> <mathjax>#([(20.7*cancelg)/(39.9*cancelg*cancel(mol^-1))]*0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx409cancelK)/(1.04*cancel(atm))#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#L#</mathjax></p>
<p>The difficult part of this question is choosing the appropriate gas constant <mathjax>#R#</mathjax>. Here, the use of <br/>
<mathjax>#R#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0821*L*atm*K^-1*mol^-1#</mathjax> gives appropriate units. </p>
<p>You should (I think) always be quoted these constants in an examination, but you have to be able to use them dimensionally. If you cancel out the units in tne above exprression you get an answer in <mathjax>#"litres"#</mathjax> AS REQUIRED for volume. Of course, in an examination you will not be required to reproduce this dimensional treatment. I include it as a method of self-checking my answer. I wanted an answer in <mathjax>#"litres"#</mathjax>, I got an answer in <mathjax>#"litres"#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#V#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(nRT)/P#</mathjax> <mathjax>#=#</mathjax> <mathjax>#???#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Pressure is moderate, temperature is elevated, and the gas is inert, so we would guess the ideal gas equation will give a good approximation:</p>
<p><mathjax>#V#</mathjax> <mathjax>#=#</mathjax> <mathjax>#([(20.7*cancelg)/(39.9*cancelg*cancel(mol^-1))]*0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx409cancelK)/(1.04*cancel(atm))#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#L#</mathjax></p>
<p>The difficult part of this question is choosing the appropriate gas constant <mathjax>#R#</mathjax>. Here, the use of <br/>
<mathjax>#R#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0821*L*atm*K^-1*mol^-1#</mathjax> gives appropriate units. </p>
<p>You should (I think) always be quoted these constants in an examination, but you have to be able to use them dimensionally. If you cancel out the units in tne above exprression you get an answer in <mathjax>#"litres"#</mathjax> AS REQUIRED for volume. Of course, in an examination you will not be required to reproduce this dimensional treatment. I include it as a method of self-checking my answer. I wanted an answer in <mathjax>#"litres"#</mathjax>, I got an answer in <mathjax>#"litres"#</mathjax>.</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the volume occupied by 20.7 g of argon gas at a pressure of 1.04 atm and a temperature of 409 K? </h1>
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anor277
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Feb 28, 2016
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<div class="markdown"><p><mathjax>#V#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(nRT)/P#</mathjax> <mathjax>#=#</mathjax> <mathjax>#???#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Pressure is moderate, temperature is elevated, and the gas is inert, so we would guess the ideal gas equation will give a good approximation:</p>
<p><mathjax>#V#</mathjax> <mathjax>#=#</mathjax> <mathjax>#([(20.7*cancelg)/(39.9*cancelg*cancel(mol^-1))]*0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx409cancelK)/(1.04*cancel(atm))#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#L#</mathjax></p>
<p>The difficult part of this question is choosing the appropriate gas constant <mathjax>#R#</mathjax>. Here, the use of <br/>
<mathjax>#R#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.0821*L*atm*K^-1*mol^-1#</mathjax> gives appropriate units. </p>
<p>You should (I think) always be quoted these constants in an examination, but you have to be able to use them dimensionally. If you cancel out the units in tne above exprression you get an answer in <mathjax>#"litres"#</mathjax> AS REQUIRED for volume. Of course, in an examination you will not be required to reproduce this dimensional treatment. I include it as a method of self-checking my answer. I wanted an answer in <mathjax>#"litres"#</mathjax>, I got an answer in <mathjax>#"litres"#</mathjax>.</p></div>
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</article> | What is the volume occupied by 20.7 g of argon gas at a pressure of 1.04 atm and a temperature of 409 K? | null |
2,951 | ab71b22d-6ddd-11ea-9e59-ccda262736ce | https://socratic.org/questions/how-many-moles-are-in-2-9-g-of-magnesium-hydroxide | 0.05 moles | start physical_unit 8 9 mole mol qc_end physical_unit 8 9 5 6 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] magnesium hydroxide [IN] moles"}] | [{"type":"physical unit","value":"0.05 moles"}] | [{"type":"physical unit","value":"Mass [OF] magnesium hydroxide [=] \\pu{2.9 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles are in 2.9 g of magnesium hydroxide?</h1> | null | 0.05 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus........</p>
<p><mathjax>#"Moles of"#</mathjax> <mathjax>#"Mg(OH)"_2=("2.9"*"g")/("58.32"*g*mol^-1)=0.0497*mol#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"Moles"="Mass"/"Molar mass"~=0.05*mol*L^-1#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus........</p>
<p><mathjax>#"Moles of"#</mathjax> <mathjax>#"Mg(OH)"_2=("2.9"*"g")/("58.32"*g*mol^-1)=0.0497*mol#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How many moles are in 2.9 g of magnesium hydroxide?</h1>
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<div class="markdown"><p><mathjax>#"Moles"="Mass"/"Molar mass"~=0.05*mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And thus........</p>
<p><mathjax>#"Moles of"#</mathjax> <mathjax>#"Mg(OH)"_2=("2.9"*"g")/("58.32"*g*mol^-1)=0.0497*mol#</mathjax>.</p></div>
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</article> | How many moles are in 2.9 g of magnesium hydroxide? | null |
2,952 | abe679ae-6ddd-11ea-b2e3-ccda262736ce | https://socratic.org/questions/balance-the-following-redox-reaction-in-acidic-solution | 3 P4(s) + 20 NO3- + 8 H+ + 8 H2O -> 12 H2PO4-(aq) + 20 NO(g) | start chemical_equation qc_end chemical_equation 8 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the redox reaction"}] | [{"type":"chemical equation","value":"3 P4(s) + 20 NO3- + 8 H+ + 8 H2O -> 12 H2PO4-(aq) + 20 NO(g) "}] | [{"type":"chemical equation","value":"P4(s) + NO3- -> H2PO4-(aq) + NO(g) "}] | <h1 class="questionTitle" itemprop="name">Balance the following redox reaction in acidic solution?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p><mathjax>#\sf{P_4(s)+NO_3^(-)\toH_2PO_4^(-)(aq)+NO(g)}#</mathjax></p>
<p>I have the oxidation, but not the reduction.<br/>
Answer key tells me there is <mathjax>#\tt{2H_2O}#</mathjax> but that doesn't match with the <mathjax>#\tt{NO_3^(-)}#</mathjax> on the other end of the arror.</p>
<p>Plus, why add <mathjax>#\tt{3e^-}#</mathjax>? Bounded oxygen typically has a charge of -2, right? So three oxygens equals six electrons...?</p></div>
</h2>
</div>
</div> | 3 P4(s) + 20 NO3- + 8 H+ + 8 H2O -> 12 H2PO4-(aq) + 20 NO(g) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1/4P_4(s)+ 4H_2O rarr HPO_4^(2-)+7H^+ +5e^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Charge and mass are balanced, so this is kosher. </p>
<p><strong>Nitrate <mathjax>#N(+V)#</mathjax> is REDUCED to <mathjax>#NO#</mathjax> <mathjax>#N(+II)#</mathjax>..</strong> . (Oxygen is slightly more electronegative than nitrogen, so it FORMALLY gets the electrons of the bonds. What is the oxidation number of oxygen in the REAL molecule, <mathjax>#OF_2#</mathjax>?)</p>
<p><mathjax>#NO_3^(-) + +4H^+ +3e^(-) rarr NO + 2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>We add <mathjax>#3xx(i)+5xx(ii)#</mathjax> to get..</p>
<p><mathjax>#3/4P_4(s)+ 5NO_3^(-) +20H^+ +15e^(-)+12H_2O rarr 3HPO_4^(2-)+21H^+ +15e^(-)+5NO + 10H_2O#</mathjax></p>
<p>...and cancel away...</p>
<p><mathjax>#3/4P_4(s)+ 5NO_3^(-) +2H_2O rarr 3HPO_4^(2-) +5NO+1H^+#</mathjax></p>
<p>Is this balanced? Should it be? I ask because anybody can butcher the arifmetik. We know that if it is NOT balanced with respect to mass and charge, it cannot be accepted as representation of reality. </p>
<p>Oxygen is not involved in redox here...phosphorus is, and nitrogen ONLY.... And all I have really done is to assign oxidation states. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><strong>And so elemental phosphorus <mathjax>#P(0)#</mathjax> is oxidized to phosphoric acid <mathjax>#P(V)#</mathjax>, a five electron oxidation....</strong></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1/4P_4(s)+ 4H_2O rarr HPO_4^(2-)+7H^+ +5e^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Charge and mass are balanced, so this is kosher. </p>
<p><strong>Nitrate <mathjax>#N(+V)#</mathjax> is REDUCED to <mathjax>#NO#</mathjax> <mathjax>#N(+II)#</mathjax>..</strong> . (Oxygen is slightly more electronegative than nitrogen, so it FORMALLY gets the electrons of the bonds. What is the oxidation number of oxygen in the REAL molecule, <mathjax>#OF_2#</mathjax>?)</p>
<p><mathjax>#NO_3^(-) + +4H^+ +3e^(-) rarr NO + 2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>We add <mathjax>#3xx(i)+5xx(ii)#</mathjax> to get..</p>
<p><mathjax>#3/4P_4(s)+ 5NO_3^(-) +20H^+ +15e^(-)+12H_2O rarr 3HPO_4^(2-)+21H^+ +15e^(-)+5NO + 10H_2O#</mathjax></p>
<p>...and cancel away...</p>
<p><mathjax>#3/4P_4(s)+ 5NO_3^(-) +2H_2O rarr 3HPO_4^(2-) +5NO+1H^+#</mathjax></p>
<p>Is this balanced? Should it be? I ask because anybody can butcher the arifmetik. We know that if it is NOT balanced with respect to mass and charge, it cannot be accepted as representation of reality. </p>
<p>Oxygen is not involved in redox here...phosphorus is, and nitrogen ONLY.... And all I have really done is to assign oxidation states. </p></div>
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<h1 class="questionTitle" itemprop="name">Balance the following redox reaction in acidic solution?</h1>
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<div class="markdown"><p><mathjax>#\sf{P_4(s)+NO_3^(-)\toH_2PO_4^(-)(aq)+NO(g)}#</mathjax></p>
<p>I have the oxidation, but not the reduction.<br/>
Answer key tells me there is <mathjax>#\tt{2H_2O}#</mathjax> but that doesn't match with the <mathjax>#\tt{NO_3^(-)}#</mathjax> on the other end of the arror.</p>
<p>Plus, why add <mathjax>#\tt{3e^-}#</mathjax>? Bounded oxygen typically has a charge of -2, right? So three oxygens equals six electrons...?</p></div>
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<div class="markdown"><p><strong>And so elemental phosphorus <mathjax>#P(0)#</mathjax> is oxidized to phosphoric acid <mathjax>#P(V)#</mathjax>, a five electron oxidation....</strong></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#1/4P_4(s)+ 4H_2O rarr HPO_4^(2-)+7H^+ +5e^(-)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Charge and mass are balanced, so this is kosher. </p>
<p><strong>Nitrate <mathjax>#N(+V)#</mathjax> is REDUCED to <mathjax>#NO#</mathjax> <mathjax>#N(+II)#</mathjax>..</strong> . (Oxygen is slightly more electronegative than nitrogen, so it FORMALLY gets the electrons of the bonds. What is the oxidation number of oxygen in the REAL molecule, <mathjax>#OF_2#</mathjax>?)</p>
<p><mathjax>#NO_3^(-) + +4H^+ +3e^(-) rarr NO + 2H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>We add <mathjax>#3xx(i)+5xx(ii)#</mathjax> to get..</p>
<p><mathjax>#3/4P_4(s)+ 5NO_3^(-) +20H^+ +15e^(-)+12H_2O rarr 3HPO_4^(2-)+21H^+ +15e^(-)+5NO + 10H_2O#</mathjax></p>
<p>...and cancel away...</p>
<p><mathjax>#3/4P_4(s)+ 5NO_3^(-) +2H_2O rarr 3HPO_4^(2-) +5NO+1H^+#</mathjax></p>
<p>Is this balanced? Should it be? I ask because anybody can butcher the arifmetik. We know that if it is NOT balanced with respect to mass and charge, it cannot be accepted as representation of reality. </p>
<p>Oxygen is not involved in redox here...phosphorus is, and nitrogen ONLY.... And all I have really done is to assign oxidation states. </p></div>
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<div class="markdown"><p>Final balanced equation: <br/>
<mathjax>#3P_4(s)+20NO_3^(-)(aq)+8H^+(aq) +8H_2O(l)->20NO(g)+12H_2PO_4^(-)(aq)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Here we go:</p>
<p>In an acidic solution means I can strategically place hydrogen ion and water wherever in the equation necessary:</p>
<p><mathjax>#P_4(s)+16H_2O->4H_2PO_4^-)+20e^- )+24H^+#</mathjax></p>
<p><mathjax>#NO_3^-)+4H^(+)+ 3e^(- )->NO+2H_2O#</mathjax></p>
<p>Why add <mathjax>#3e^-#</mathjax>?<br/>
Because <mathjax>#N^(+5)#</mathjax> is reduced to <mathjax>#N^(+2)#</mathjax></p>
<p>The electrons on each side must be equal:</p>
<p><mathjax>#3(P_4(s)+16H_2O->4H_2PO_4^(-)+20e^(- )+24H^+)#</mathjax><br/>
<mathjax>#20(NO_3^-)+4H^(+)+ 3e^(- )->NO+2H_2O)#</mathjax></p>
<p><mathjax>#3P_4(s)+48H_2O->12H_2PO_4^-)+60e^- )+72H^+#</mathjax><br/>
<mathjax>#20NO_3^(-)+80H^(+)+ 60e^(- )->20NO+40H_2O#</mathjax></p>
<p>I am left with:<br/>
<mathjax>#3P_4(s)+20NO_3^(-)(aq)+8H^+(aq) +8H_2O(l)->20NO(g)+12H_2PO_4^(-)(aq)#</mathjax></p></div>
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</article> | Balance the following redox reaction in acidic solution? |
#\sf{P_4(s)+NO_3^(-)\toH_2PO_4^(-)(aq)+NO(g)}#
I have the oxidation, but not the reduction.
Answer key tells me there is #\tt{2H_2O}# but that doesn't match with the #\tt{NO_3^(-)}# on the other end of the arror.
Plus, why add #\tt{3e^-}#? Bounded oxygen typically has a charge of -2, right? So three oxygens equals six electrons...?
|
2,953 | ac963e07-6ddd-11ea-bf46-ccda262736ce | https://socratic.org/questions/a-tank-contains-n-2-at-1-0-atm-and-o-2-at-2-0-atm-helium-is-added-to-this-tank-u | 3.0 atm | start physical_unit 31 32 partial_pressure atm qc_end physical_unit 8 8 10 11 pressure qc_end physical_unit 3 3 5 6 pressure qc_end physical_unit 1 1 23 24 total_pressure qc_end end | [{"type":"physical unit","value":"Partial pressure2 [OF] the helium [IN] atm"}] | [{"type":"physical unit","value":"3.0 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] O2 [=] \\pu{2.0 atm}"},{"type":"physical unit","value":"Pressure1 [OF] N2 [=] \\pu{1.0 atm}"},{"type":"physical unit","value":"Total pressure2 [OF] tank [=] \\pu{6.0 atm}"}] | <h1 class="questionTitle" itemprop="name">A tank contains #N_2# at 1.0 atm and #O_2# at 2.0 atm. Helium is added to this tank until the total pressure is 6.0 atm. What is the partial pressure of the helium?</h1> | null | 3.0 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In a gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a gas is the same if that gas alone had occupied the container. The total pressure is the sum of the partial pressure. </p>
<p>We know that the total pressure is <mathjax>#6.0*atm#</mathjax>; i.e. <mathjax>#6.0*atm#</mathjax> <mathjax>#=#</mathjax> <mathjax>#p_(N_2)+p_(O_2)+p_(He)#</mathjax>.</p>
<p>Since we have been given the other partial pressures, calculation of <mathjax>#p_(He)#</mathjax> is trivial.</p></div>
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<div class="markdown"><p><mathjax>#p_(He)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.0*atm#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In a gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a gas is the same if that gas alone had occupied the container. The total pressure is the sum of the partial pressure. </p>
<p>We know that the total pressure is <mathjax>#6.0*atm#</mathjax>; i.e. <mathjax>#6.0*atm#</mathjax> <mathjax>#=#</mathjax> <mathjax>#p_(N_2)+p_(O_2)+p_(He)#</mathjax>.</p>
<p>Since we have been given the other partial pressures, calculation of <mathjax>#p_(He)#</mathjax> is trivial.</p></div>
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<h1 class="questionTitle" itemprop="name">A tank contains #N_2# at 1.0 atm and #O_2# at 2.0 atm. Helium is added to this tank until the total pressure is 6.0 atm. What is the partial pressure of the helium?</h1>
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<div class="markdown"><p><mathjax>#p_(He)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.0*atm#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>In a gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a gas is the same if that gas alone had occupied the container. The total pressure is the sum of the partial pressure. </p>
<p>We know that the total pressure is <mathjax>#6.0*atm#</mathjax>; i.e. <mathjax>#6.0*atm#</mathjax> <mathjax>#=#</mathjax> <mathjax>#p_(N_2)+p_(O_2)+p_(He)#</mathjax>.</p>
<p>Since we have been given the other partial pressures, calculation of <mathjax>#p_(He)#</mathjax> is trivial.</p></div>
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</article> | A tank contains #N_2# at 1.0 atm and #O_2# at 2.0 atm. Helium is added to this tank until the total pressure is 6.0 atm. What is the partial pressure of the helium? | null |
2,954 | abf314d0-6ddd-11ea-8542-ccda262736ce | https://socratic.org/questions/590b6efcb72cff7aa5fa3a57 | C2H4O | start chemical_formula qc_end chemical_equation 5 5 qc_end end | [{"type":"other","value":"Chemical Formula [OF] C4H8O2 [IN] empirical"}] | [{"type":"chemical equation","value":"C2H4O"}] | [{"type":"chemical equation","value":"C4H8O2"}] | <h1 class="questionTitle" itemprop="name">Given a #"molecular formula"# of #C_4H_8O_2#, what is the #"empirical formula"#?</h1> | null | C2H4O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And the <mathjax>#"empirical formula"#</mathjax> is the simplest whole number ratio that defines constituent atoms in a species, and thus...................</p>
<p><mathjax>#"empirical formula"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C_2H_4O#</mathjax></p>
<p>The molecular formula is always a whole number multiple of the empirical formula, i.e............</p>
<p><mathjax>#"molecular formula"=nxx{"empirical formula"}#</mathjax>.</p>
<p>To what does <mathjax>#n#</mathjax> equal here?</p></div>
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<div class="markdown"><p>Well the <mathjax>#"molecular formula"#</mathjax> is <mathjax>#C_4H_8O_2#</mathjax>.........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And the <mathjax>#"empirical formula"#</mathjax> is the simplest whole number ratio that defines constituent atoms in a species, and thus...................</p>
<p><mathjax>#"empirical formula"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C_2H_4O#</mathjax></p>
<p>The molecular formula is always a whole number multiple of the empirical formula, i.e............</p>
<p><mathjax>#"molecular formula"=nxx{"empirical formula"}#</mathjax>.</p>
<p>To what does <mathjax>#n#</mathjax> equal here?</p></div>
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<h1 class="questionTitle" itemprop="name">Given a #"molecular formula"# of #C_4H_8O_2#, what is the #"empirical formula"#?</h1>
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anor277
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<div class="markdown"><p>Well the <mathjax>#"molecular formula"#</mathjax> is <mathjax>#C_4H_8O_2#</mathjax>.........</p></div>
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<div class="markdown"><p>And the <mathjax>#"empirical formula"#</mathjax> is the simplest whole number ratio that defines constituent atoms in a species, and thus...................</p>
<p><mathjax>#"empirical formula"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#C_2H_4O#</mathjax></p>
<p>The molecular formula is always a whole number multiple of the empirical formula, i.e............</p>
<p><mathjax>#"molecular formula"=nxx{"empirical formula"}#</mathjax>.</p>
<p>To what does <mathjax>#n#</mathjax> equal here?</p></div>
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</article> | Given a #"molecular formula"# of #C_4H_8O_2#, what is the #"empirical formula"#? | null |
2,955 | a8489ca4-6ddd-11ea-b5a0-ccda262736ce | https://socratic.org/questions/how-much-heat-i-released-to-the-environment-as-245-g-of-steam-at-140-c-is-cooled | 765 kJ | start physical_unit 12 12 heat_energy kj qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 12 12 14 15 temperature qc_end physical_unit 12 12 19 20 temperature qc_end end | [{"type":"physical unit","value":"Released heat [OF] steam [IN] kJ"}] | [{"type":"physical unit","value":"765 kJ"}] | [{"type":"physical unit","value":"Mass [OF] steam [=] \\pu{245 g}"},{"type":"physical unit","value":"Temperature1 [OF] steam [=] \\pu{140 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] steam [=] \\pu{ -15 ℃}"}] | <h1 class="questionTitle" itemprop="name">How much heat i released to the environment as 245 g of steam at 140°C is cooled to -15°C?</h1> | null | 765 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are five separate heats involved in this problem:</p>
<ul>
<li><mathjax>#q_1#</mathjax> = heat required to cool the steam from 140 °C to 100 °C</li>
<li><mathjax>#q_2#</mathjax> = heat required to condense the steam to water at 100 °C</li>
<li><mathjax>#q_3#</mathjax> = heat required to cool the water from 100 °C to 0 °C</li>
<li><mathjax>#q_4#</mathjax> = heat required to freeze the water to ice at 0 °C</li>
<li><mathjax>#q_5#</mathjax> = heat required to cool the ice from 0 °C to -15 °C</li>
</ul>
<blockquote></blockquote>
<p><mathjax>#q = q_1 + q_2 + q_3 +q_4 + q_5#</mathjax></p>
<p><mathjax>#= mc_1ΔT_1 + mΔ_text(cond)H + mc_3ΔT_3 + mΔ_text(fus)H + mc_5ΔT_5#</mathjax></p>
<p>where</p>
<p><mathjax>#q_1, q_2, q_3, q_4,#</mathjax> and <mathjax>#q_5#</mathjax> are the heats involved in each step</p>
<p><mathjax>#m#</mathjax> is the mass of the sample</p>
<p><mathjax>#ΔT = T_"f" -T_"i"#</mathjax></p>
<p><mathjax>#c_1 = "the specific heat capacity of steam" = "2.010 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#c_3 = "the specific heat capacity of water" = "4.184 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#c_5 = "the specific heat capacity of ice" = "2.010 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#Δ_text(cond)H = "the enthalpy of condensation of steam" = "-2258 J·g"^"-1"#</mathjax></p>
<p><mathjax>#Δ_text(sol)H = "the enthalpy of solidification of water" = "-333.55 J·g"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_1#</mathjax></p>
<p><mathjax>#ΔT_1 = "100 °C - 140°C" = "-40 °C"#</mathjax></p>
<p><mathjax>#q_1 = mc_1ΔT_1 = 245 color(red)(cancel(color(black)("g"))) × 2.010 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-40") color(red)(cancel(color(black)("°C"))) = "-19 700 J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_2#</mathjax></p>
<p><mathjax>#q_2 = 245 color(red)(cancel(color(black)("g"))) × ("-2258")color(white)(l) "J"·color(red)(cancel(color(black)("g"^"-1"))) = "-553 200 kJ"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_3#</mathjax></p>
<p><mathjax>#ΔT = "0 °C - 100 °C" = "-100 °C"#</mathjax></p>
<p><mathjax>#q_3 = mcΔT = 245 color(red)(cancel(color(black)("g"))) × 4.184 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-100") color(red)(cancel(color(black)("°C"))) = "-102 500 J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_4#</mathjax></p>
<p><mathjax>#q_4 = 245 color(red)(cancel(color(black)("g"))) × ("-333.55")color(white)(l) "J"·color(red)(cancel(color(black)("g"^"-1"))) = "-81 720 kJ"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_5#</mathjax></p>
<p><mathjax>#ΔT_5 = "-15 °C - 0 °C" = "-15 °C"#</mathjax></p>
<p><mathjax>#q_5 = mcΔT = 245 color(red)(cancel(color(black)("g"))) × 2.010 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-15") color(red)(cancel(color(black)("°C"))) = "-7390 J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q = q_1 + q_2 + q_3 + q_4 + q_5#</mathjax><br/>
<mathjax>#= "-19 700 J" - "553 200 J" - "102 500 J" - "81 720 J" - "7390 J" = "-765 000 J" = "-765 kJ"#</mathjax></p>
<p>The process releases 765 kJ of heat into the environment.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><strong>Warning! Long answer!</strong> The process releases 765 kJ of heat into the environment.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are five separate heats involved in this problem:</p>
<ul>
<li><mathjax>#q_1#</mathjax> = heat required to cool the steam from 140 °C to 100 °C</li>
<li><mathjax>#q_2#</mathjax> = heat required to condense the steam to water at 100 °C</li>
<li><mathjax>#q_3#</mathjax> = heat required to cool the water from 100 °C to 0 °C</li>
<li><mathjax>#q_4#</mathjax> = heat required to freeze the water to ice at 0 °C</li>
<li><mathjax>#q_5#</mathjax> = heat required to cool the ice from 0 °C to -15 °C</li>
</ul>
<blockquote></blockquote>
<p><mathjax>#q = q_1 + q_2 + q_3 +q_4 + q_5#</mathjax></p>
<p><mathjax>#= mc_1ΔT_1 + mΔ_text(cond)H + mc_3ΔT_3 + mΔ_text(fus)H + mc_5ΔT_5#</mathjax></p>
<p>where</p>
<p><mathjax>#q_1, q_2, q_3, q_4,#</mathjax> and <mathjax>#q_5#</mathjax> are the heats involved in each step</p>
<p><mathjax>#m#</mathjax> is the mass of the sample</p>
<p><mathjax>#ΔT = T_"f" -T_"i"#</mathjax></p>
<p><mathjax>#c_1 = "the specific heat capacity of steam" = "2.010 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#c_3 = "the specific heat capacity of water" = "4.184 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#c_5 = "the specific heat capacity of ice" = "2.010 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#Δ_text(cond)H = "the enthalpy of condensation of steam" = "-2258 J·g"^"-1"#</mathjax></p>
<p><mathjax>#Δ_text(sol)H = "the enthalpy of solidification of water" = "-333.55 J·g"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_1#</mathjax></p>
<p><mathjax>#ΔT_1 = "100 °C - 140°C" = "-40 °C"#</mathjax></p>
<p><mathjax>#q_1 = mc_1ΔT_1 = 245 color(red)(cancel(color(black)("g"))) × 2.010 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-40") color(red)(cancel(color(black)("°C"))) = "-19 700 J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_2#</mathjax></p>
<p><mathjax>#q_2 = 245 color(red)(cancel(color(black)("g"))) × ("-2258")color(white)(l) "J"·color(red)(cancel(color(black)("g"^"-1"))) = "-553 200 kJ"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_3#</mathjax></p>
<p><mathjax>#ΔT = "0 °C - 100 °C" = "-100 °C"#</mathjax></p>
<p><mathjax>#q_3 = mcΔT = 245 color(red)(cancel(color(black)("g"))) × 4.184 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-100") color(red)(cancel(color(black)("°C"))) = "-102 500 J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_4#</mathjax></p>
<p><mathjax>#q_4 = 245 color(red)(cancel(color(black)("g"))) × ("-333.55")color(white)(l) "J"·color(red)(cancel(color(black)("g"^"-1"))) = "-81 720 kJ"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_5#</mathjax></p>
<p><mathjax>#ΔT_5 = "-15 °C - 0 °C" = "-15 °C"#</mathjax></p>
<p><mathjax>#q_5 = mcΔT = 245 color(red)(cancel(color(black)("g"))) × 2.010 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-15") color(red)(cancel(color(black)("°C"))) = "-7390 J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q = q_1 + q_2 + q_3 + q_4 + q_5#</mathjax><br/>
<mathjax>#= "-19 700 J" - "553 200 J" - "102 500 J" - "81 720 J" - "7390 J" = "-765 000 J" = "-765 kJ"#</mathjax></p>
<p>The process releases 765 kJ of heat into the environment.</p></div>
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<h1 class="questionTitle" itemprop="name">How much heat i released to the environment as 245 g of steam at 140°C is cooled to -15°C?</h1>
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<div class="markdown"><p><strong>Warning! Long answer!</strong> The process releases 765 kJ of heat into the environment.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>There are five separate heats involved in this problem:</p>
<ul>
<li><mathjax>#q_1#</mathjax> = heat required to cool the steam from 140 °C to 100 °C</li>
<li><mathjax>#q_2#</mathjax> = heat required to condense the steam to water at 100 °C</li>
<li><mathjax>#q_3#</mathjax> = heat required to cool the water from 100 °C to 0 °C</li>
<li><mathjax>#q_4#</mathjax> = heat required to freeze the water to ice at 0 °C</li>
<li><mathjax>#q_5#</mathjax> = heat required to cool the ice from 0 °C to -15 °C</li>
</ul>
<blockquote></blockquote>
<p><mathjax>#q = q_1 + q_2 + q_3 +q_4 + q_5#</mathjax></p>
<p><mathjax>#= mc_1ΔT_1 + mΔ_text(cond)H + mc_3ΔT_3 + mΔ_text(fus)H + mc_5ΔT_5#</mathjax></p>
<p>where</p>
<p><mathjax>#q_1, q_2, q_3, q_4,#</mathjax> and <mathjax>#q_5#</mathjax> are the heats involved in each step</p>
<p><mathjax>#m#</mathjax> is the mass of the sample</p>
<p><mathjax>#ΔT = T_"f" -T_"i"#</mathjax></p>
<p><mathjax>#c_1 = "the specific heat capacity of steam" = "2.010 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#c_3 = "the specific heat capacity of water" = "4.184 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#c_5 = "the specific heat capacity of ice" = "2.010 J·°C"^"-1""g"^"-1"#</mathjax></p>
<p><mathjax>#Δ_text(cond)H = "the enthalpy of condensation of steam" = "-2258 J·g"^"-1"#</mathjax></p>
<p><mathjax>#Δ_text(sol)H = "the enthalpy of solidification of water" = "-333.55 J·g"^"-1"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_1#</mathjax></p>
<p><mathjax>#ΔT_1 = "100 °C - 140°C" = "-40 °C"#</mathjax></p>
<p><mathjax>#q_1 = mc_1ΔT_1 = 245 color(red)(cancel(color(black)("g"))) × 2.010 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-40") color(red)(cancel(color(black)("°C"))) = "-19 700 J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_2#</mathjax></p>
<p><mathjax>#q_2 = 245 color(red)(cancel(color(black)("g"))) × ("-2258")color(white)(l) "J"·color(red)(cancel(color(black)("g"^"-1"))) = "-553 200 kJ"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_3#</mathjax></p>
<p><mathjax>#ΔT = "0 °C - 100 °C" = "-100 °C"#</mathjax></p>
<p><mathjax>#q_3 = mcΔT = 245 color(red)(cancel(color(black)("g"))) × 4.184 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-100") color(red)(cancel(color(black)("°C"))) = "-102 500 J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_4#</mathjax></p>
<p><mathjax>#q_4 = 245 color(red)(cancel(color(black)("g"))) × ("-333.55")color(white)(l) "J"·color(red)(cancel(color(black)("g"^"-1"))) = "-81 720 kJ"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#bbq_5#</mathjax></p>
<p><mathjax>#ΔT_5 = "-15 °C - 0 °C" = "-15 °C"#</mathjax></p>
<p><mathjax>#q_5 = mcΔT = 245 color(red)(cancel(color(black)("g"))) × 2.010 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × ("-15") color(red)(cancel(color(black)("°C"))) = "-7390 J"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#q = q_1 + q_2 + q_3 + q_4 + q_5#</mathjax><br/>
<mathjax>#= "-19 700 J" - "553 200 J" - "102 500 J" - "81 720 J" - "7390 J" = "-765 000 J" = "-765 kJ"#</mathjax></p>
<p>The process releases 765 kJ of heat into the environment.</p></div>
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</article> | How much heat i released to the environment as 245 g of steam at 140°C is cooled to -15°C? | null |
2,956 | a8f5a782-6ddd-11ea-931b-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-iron-in-feo | +2 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] iron"}] | [{"type":"physical unit","value":"+2"}] | [{"type":"chemical equation","value":"FeO"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of iron in #FeO#? </h1> | null | +2 | <div class="answerDescription">
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<div class="markdown"><p>The total of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> in a stable compound is zero. The more electronegative element takes on its typical oxidation state, which in the case of oxygen is -2. Therefore to give a total of zero, the oxidation state of iron would be +2.</p></div>
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<div class="markdown"><p>+2</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The total of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> in a stable compound is zero. The more electronegative element takes on its typical oxidation state, which in the case of oxygen is -2. Therefore to give a total of zero, the oxidation state of iron would be +2.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of iron in #FeO#? </h1>
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<div class="markdown"><p>+2</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The total of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> in a stable compound is zero. The more electronegative element takes on its typical oxidation state, which in the case of oxygen is -2. Therefore to give a total of zero, the oxidation state of iron would be +2.</p></div>
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</article> | What is the oxidation number of iron in #FeO#? | null |
2,957 | ab5478a2-6ddd-11ea-8670-ccda262736ce | https://socratic.org/questions/given-that-the-sulfate-iv-ion-so-2-2-is-converted-to-the-sulfate-vi-ion-so-4-2-i | Cr2O7^2- + 3 SO3^2- + 8 H+ -> 2 Cr^3+ + 3 SO4^2- + 4 H2O | start chemical_equation qc_end chemical_equation 5 5 qc_end chemical_equation 12 12 qc_end chemical_equation 27 27 qc_end chemical_equation 29 29 qc_end substance 17 17 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the redox reaction"}] | [{"type":"chemical equation","value":"Cr2O7^2- + 3 SO3^2- + 8 H+ -> 2 Cr^3+ + 3 SO4^2- + 4 H2O"}] | [{"type":"chemical equation","value":"SO2^2-"},{"type":"chemical equation","value":"SO4^2-"},{"type":"chemical equation","value":"Cr2O7^2-(aq)"},{"type":"chemical equation","value":"SO3^2-"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">Given that the sulfate(IV) ion, #SO_2^(-2)#, is converted to the sulfate(VI) ion, #SO_4^(-2)#, in the presence of water, deduce the balanced equation for the redox reaction between #Cr_2O_7^(-2)# (aq) and #SO_3^(-2)#?</h1> | null | Cr2O7^2- + 3 SO3^2- + 8 H+ -> 2 Cr^3+ + 3 SO4^2- + 4 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Oxidation half equation:"#</mathjax></p>
<p><mathjax>#SO_3^(2-) +H_2O rarr SO_4^(2-) + 2H^(+) + 2e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction half equation:"#</mathjax></p>
<p><mathjax>#Cr_2O_7^(2-) +14H^(+) + 6e^(-)rarr 2Cr^(3+) + 2H^(+) + 7H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Both equations are balanced with respect to mass and charge; as indeed they must be if they reflect chemical reality. So we simply cross-mulitply to eliminate the electrons, <mathjax>#3xx(i)+(ii):#</mathjax></p>
<p><mathjax>#Cr_2O_7^(2-)+3SO_3^(2-) +8H^(+)rarr 2Cr^(3+)+ 3SO_4^(2-) + 4H_2O#</mathjax></p>
<p>Given that the sulfate and sulfite ions are colourless, what would be the macroscopic colour change observed in this reaction?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#Cr_2O_7^(2-)+3SO_3^(2-) +8H^(+)rarr 2Cr^(3+)+ 3SO_4^(2-) + 4H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Oxidation half equation:"#</mathjax></p>
<p><mathjax>#SO_3^(2-) +H_2O rarr SO_4^(2-) + 2H^(+) + 2e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction half equation:"#</mathjax></p>
<p><mathjax>#Cr_2O_7^(2-) +14H^(+) + 6e^(-)rarr 2Cr^(3+) + 2H^(+) + 7H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Both equations are balanced with respect to mass and charge; as indeed they must be if they reflect chemical reality. So we simply cross-mulitply to eliminate the electrons, <mathjax>#3xx(i)+(ii):#</mathjax></p>
<p><mathjax>#Cr_2O_7^(2-)+3SO_3^(2-) +8H^(+)rarr 2Cr^(3+)+ 3SO_4^(2-) + 4H_2O#</mathjax></p>
<p>Given that the sulfate and sulfite ions are colourless, what would be the macroscopic colour change observed in this reaction?</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">Given that the sulfate(IV) ion, #SO_2^(-2)#, is converted to the sulfate(VI) ion, #SO_4^(-2)#, in the presence of water, deduce the balanced equation for the redox reaction between #Cr_2O_7^(-2)# (aq) and #SO_3^(-2)#?</h1>
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<div class="markdown"><p><mathjax>#Cr_2O_7^(2-)+3SO_3^(2-) +8H^(+)rarr 2Cr^(3+)+ 3SO_4^(2-) + 4H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Oxidation half equation:"#</mathjax></p>
<p><mathjax>#SO_3^(2-) +H_2O rarr SO_4^(2-) + 2H^(+) + 2e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p><mathjax>#"Reduction half equation:"#</mathjax></p>
<p><mathjax>#Cr_2O_7^(2-) +14H^(+) + 6e^(-)rarr 2Cr^(3+) + 2H^(+) + 7H_2O#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Both equations are balanced with respect to mass and charge; as indeed they must be if they reflect chemical reality. So we simply cross-mulitply to eliminate the electrons, <mathjax>#3xx(i)+(ii):#</mathjax></p>
<p><mathjax>#Cr_2O_7^(2-)+3SO_3^(2-) +8H^(+)rarr 2Cr^(3+)+ 3SO_4^(2-) + 4H_2O#</mathjax></p>
<p>Given that the sulfate and sulfite ions are colourless, what would be the macroscopic colour change observed in this reaction?</p></div>
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</article> | Given that the sulfate(IV) ion, #SO_2^(-2)#, is converted to the sulfate(VI) ion, #SO_4^(-2)#, in the presence of water, deduce the balanced equation for the redox reaction between #Cr_2O_7^(-2)# (aq) and #SO_3^(-2)#? | null |
2,958 | a8e86a2f-6ddd-11ea-be86-ccda262736ce | https://socratic.org/questions/a-45-0g-aqueous-solution-is-made-using-10-0g-of-ethanol-what-percentage-of-the-s-1 | 22.2% | start physical_unit 11 11 percent none qc_end physical_unit 3 4 1 2 mass qc_end physical_unit 11 11 8 9 mass qc_end end | [{"type":"physical unit","value":"Percentage [OF] ethanol in solution"}] | [{"type":"physical unit","value":"22.2%"}] | [{"type":"physical unit","value":"Mass [OF] aqueous solution [=] \\pu{45.0 g}"},{"type":"physical unit","value":"Mass [OF] ethanol [=] \\pu{10.0 g}"}] | <h1 class="questionTitle" itemprop="name">A 45.0g aqueous solution is made using 10.0g of ethanol. What percentage of the solution is ethanol?</h1> | null | 22.2% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your aim here is to find the mass of ethanol present in<mathjax>#"100 g"#</mathjax> of solution, which is essentially what you're referring to when talking about the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"% m/m"#</mathjax>.</p>
<p>You know that the total mass of the solution is equal to <mathjax>#"45.0 g"#</mathjax>. This mass includes the mass of water, your <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>, and the mass of ethanol, your <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>. </p>
<p>Now, this solution has a <strong>uniform composition</strong>, meaning that the solvent and the solute are <strong>evenly mixed</strong>. This implies that you can use this given composition to figure out how many grams of ethanol would be present in <mathjax>#"100 g"#</mathjax> of solution</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g solution"))) * overbrace("10.0 g ethanol"/(45.0color(red)(cancel(color(black)("g solution")))))^(color(purple)("known composition")) = "22.2 g ethanol"#</mathjax> </p>
</blockquote>
<p>Because <mathjax>#"100 g"#</mathjax> of solution contain <mathjax>#"22.2 g"#</mathjax> of ethanol, you can say that the solution's percent concentration by mass is equal to</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("% m/m" = "mass of ethanol in 100 g solution" = 22.2%)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#22.2%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your aim here is to find the mass of ethanol present in<mathjax>#"100 g"#</mathjax> of solution, which is essentially what you're referring to when talking about the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"% m/m"#</mathjax>.</p>
<p>You know that the total mass of the solution is equal to <mathjax>#"45.0 g"#</mathjax>. This mass includes the mass of water, your <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>, and the mass of ethanol, your <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>. </p>
<p>Now, this solution has a <strong>uniform composition</strong>, meaning that the solvent and the solute are <strong>evenly mixed</strong>. This implies that you can use this given composition to figure out how many grams of ethanol would be present in <mathjax>#"100 g"#</mathjax> of solution</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g solution"))) * overbrace("10.0 g ethanol"/(45.0color(red)(cancel(color(black)("g solution")))))^(color(purple)("known composition")) = "22.2 g ethanol"#</mathjax> </p>
</blockquote>
<p>Because <mathjax>#"100 g"#</mathjax> of solution contain <mathjax>#"22.2 g"#</mathjax> of ethanol, you can say that the solution's percent concentration by mass is equal to</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("% m/m" = "mass of ethanol in 100 g solution" = 22.2%)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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<h1 class="questionTitle" itemprop="name">A 45.0g aqueous solution is made using 10.0g of ethanol. What percentage of the solution is ethanol?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-09-16T00:04:23" itemprop="dateCreated">
Sep 16, 2016
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<div class="markdown"><p><mathjax>#22.2%#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your aim here is to find the mass of ethanol present in<mathjax>#"100 g"#</mathjax> of solution, which is essentially what you're referring to when talking about the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"% m/m"#</mathjax>.</p>
<p>You know that the total mass of the solution is equal to <mathjax>#"45.0 g"#</mathjax>. This mass includes the mass of water, your <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a></em>, and the mass of ethanol, your <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>. </p>
<p>Now, this solution has a <strong>uniform composition</strong>, meaning that the solvent and the solute are <strong>evenly mixed</strong>. This implies that you can use this given composition to figure out how many grams of ethanol would be present in <mathjax>#"100 g"#</mathjax> of solution</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g solution"))) * overbrace("10.0 g ethanol"/(45.0color(red)(cancel(color(black)("g solution")))))^(color(purple)("known composition")) = "22.2 g ethanol"#</mathjax> </p>
</blockquote>
<p>Because <mathjax>#"100 g"#</mathjax> of solution contain <mathjax>#"22.2 g"#</mathjax> of ethanol, you can say that the solution's percent concentration by mass is equal to</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("% m/m" = "mass of ethanol in 100 g solution" = 22.2%)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | A 45.0g aqueous solution is made using 10.0g of ethanol. What percentage of the solution is ethanol? | null |
2,959 | ab11bc01-6ddd-11ea-b8fb-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-sulfur-in-nahso4 | +6 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] sulfur"}] | [{"type":"physical unit","value":"+6"}] | [{"type":"chemical equation","value":"NaHSO4"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of sulfur in #"NaHSO"_4#?
</h1> | null | +6 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Sodium hydrogen sulfate</em>, <mathjax>#"NaHSO"_4#</mathjax>, is an <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compound</a> made up of sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and hydrogen sulfate anions, <mathjax>#"HSO"_4^(-)#</mathjax>. </p>
<p>The <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation state</a> of sulfur can thus be determined by looking at the oxidation states of the two other <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> that make up the hydrogen sulfate anion, hydrogen and oxygen. </p>
<p>Since you're dealing with a <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ion</a>, you know that the sum of the oxidation states of <strong>each individual atom</strong> that makes up the anion must add up to give the charge, which in this case is <mathjax>#(1-)#</mathjax>. </p>
<p>Oxygen will almost always have an oxidation state of <mathjax>#-2#</mathjax>, and this case is no exception. Likewise, hydrogen will almost always have an oxidation number equal to <mathjax>#+1#</mathjax>, as is the case here. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#overbrace(+1 xx 1)^(color(blue)("one H atom")) + overbrace(ON_"S")^(color(blue)("one S atom")) + overbrace(4 xx (-2))^(color(blue)("four O atoms")) = -1#</mathjax></p>
<p><mathjax>#ON_"S" = -1 - 1 + 8 = color(green)(+6)#</mathjax></p>
</blockquote>
<p>The oxidation state of sulfur in sodium hydrogen sulfate is <mathjax>#+6#</mathjax>. </p>
<p>
<iframe src="https://www.youtube.com/embed/81WdyqvLlVA?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div>
<div class="markdown"><p><mathjax>#+6#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Sodium hydrogen sulfate</em>, <mathjax>#"NaHSO"_4#</mathjax>, is an <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compound</a> made up of sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and hydrogen sulfate anions, <mathjax>#"HSO"_4^(-)#</mathjax>. </p>
<p>The <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation state</a> of sulfur can thus be determined by looking at the oxidation states of the two other <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> that make up the hydrogen sulfate anion, hydrogen and oxygen. </p>
<p>Since you're dealing with a <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ion</a>, you know that the sum of the oxidation states of <strong>each individual atom</strong> that makes up the anion must add up to give the charge, which in this case is <mathjax>#(1-)#</mathjax>. </p>
<p>Oxygen will almost always have an oxidation state of <mathjax>#-2#</mathjax>, and this case is no exception. Likewise, hydrogen will almost always have an oxidation number equal to <mathjax>#+1#</mathjax>, as is the case here. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#overbrace(+1 xx 1)^(color(blue)("one H atom")) + overbrace(ON_"S")^(color(blue)("one S atom")) + overbrace(4 xx (-2))^(color(blue)("four O atoms")) = -1#</mathjax></p>
<p><mathjax>#ON_"S" = -1 - 1 + 8 = color(green)(+6)#</mathjax></p>
</blockquote>
<p>The oxidation state of sulfur in sodium hydrogen sulfate is <mathjax>#+6#</mathjax>. </p>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of sulfur in #"NaHSO"_4#?
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Stefan V.
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<div class="markdown"><p><mathjax>#+6#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><em>Sodium hydrogen sulfate</em>, <mathjax>#"NaHSO"_4#</mathjax>, is an <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/ionic-compounds">ionic compound</a> made up of sodium cations, <mathjax>#"Na"^(+)#</mathjax>, and hydrogen sulfate anions, <mathjax>#"HSO"_4^(-)#</mathjax>. </p>
<p>The <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation state</a> of sulfur can thus be determined by looking at the oxidation states of the two other <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> that make up the hydrogen sulfate anion, hydrogen and oxygen. </p>
<p>Since you're dealing with a <a href="http://socratic.org/chemistry/ionic-bonds-and-formulas/polyatomic-ions">polyatomic ion</a>, you know that the sum of the oxidation states of <strong>each individual atom</strong> that makes up the anion must add up to give the charge, which in this case is <mathjax>#(1-)#</mathjax>. </p>
<p>Oxygen will almost always have an oxidation state of <mathjax>#-2#</mathjax>, and this case is no exception. Likewise, hydrogen will almost always have an oxidation number equal to <mathjax>#+1#</mathjax>, as is the case here. </p>
<p>This means that you have</p>
<blockquote>
<p><mathjax>#overbrace(+1 xx 1)^(color(blue)("one H atom")) + overbrace(ON_"S")^(color(blue)("one S atom")) + overbrace(4 xx (-2))^(color(blue)("four O atoms")) = -1#</mathjax></p>
<p><mathjax>#ON_"S" = -1 - 1 + 8 = color(green)(+6)#</mathjax></p>
</blockquote>
<p>The oxidation state of sulfur in sodium hydrogen sulfate is <mathjax>#+6#</mathjax>. </p>
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</article> | What is the oxidation number of sulfur in #"NaHSO"_4#?
| null |
2,960 | aa394560-6ddd-11ea-a54c-ccda262736ce | https://socratic.org/questions/if-a-0-5-l-container-of-juice-contains-15-g-of-sugar-what-is-the-concentration-o | 30.00 g/L | start physical_unit 11 11 concentration g/l qc_end physical_unit 6 6 2 3 volume qc_end physical_unit 11 11 8 9 mass qc_end end | [{"type":"physical unit","value":"Concentration [OF] sugar [IN] g/L"}] | [{"type":"physical unit","value":"30.00 g/L"}] | [{"type":"physical unit","value":"Volume [OF] juice [=] \\pu{0.5 L}"},{"type":"physical unit","value":"Mass [OF] sugar [=] \\pu{15 g}"}] | <h1 class="questionTitle" itemprop="name">If a 0.5 L container of juice contains 15 g of sugar, what is the concentration of sugar?</h1> | null | 30.00 g/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="mass"/"volume"=(15g)/(0.5L)=30gL^-1#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#30gL^-1#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="mass"/"volume"=(15g)/(0.5L)=30gL^-1#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If a 0.5 L container of juice contains 15 g of sugar, what is the concentration of sugar?</h1>
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Ben G.
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<div class="markdown"><p><mathjax>#30gL^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"="mass"/"volume"=(15g)/(0.5L)=30gL^-1#</mathjax></p></div>
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</article> | If a 0.5 L container of juice contains 15 g of sugar, what is the concentration of sugar? | null |
2,961 | abcbacfa-6ddd-11ea-8ffa-ccda262736ce | https://socratic.org/questions/if-8-40-kj-of-heat-is-needed-to-raise-the-temperature-of-a-sample-of-metal-from- | 25.20 kilojoules | start physical_unit 13 15 heat_energy kj qc_end physical_unit 13 15 1 2 heat_energy qc_end physical_unit 13 15 17 18 temperature qc_end physical_unit 13 15 20 21 temperature qc_end physical_unit 13 15 41 42 temperature qc_end physical_unit 13 15 44 45 temperature qc_end end | [{"type":"physical unit","value":"Required heat2 [OF] metal sample [IN] kilojoules"}] | [{"type":"physical unit","value":"25.20 kilojoules"}] | [{"type":"physical unit","value":"Needed heat1 [OF] metal sample [=] \\pu{8.40 kJ}"},{"type":"physical unit","value":"Temperature1 [OF] metal sample [=] \\pu{15 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] metal sample [=] \\pu{20 ℃}"},{"type":"physical unit","value":"Temperature3 [OF] metal sample [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature4 [OF] metal sample [=] \\pu{40 ℃}"}] | <h1 class="questionTitle" itemprop="name">If 8.40 kJ of heat is needed to raise the temperature of a sample of metal from 15 °C to 20 °C, how many kilojoules of heat will be required to raise the temperature of the same sample of metal from 25 °C to 40 °C?</h1> | null | 25.20 kilojoules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that because the sample of metal has <strong>the same mass</strong> in both cases, you can say that</p>
<blockquote>
<p><mathjax>#q_2 = (DeltaT_2)/(DeltaT_1) * q_1#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#q_1#</mathjax> is the amount of heat needed to raise the temperature of the sample by <mathjax>#DeltaT_1 = 20^@"C" - 15^@"C"#</mathjax></li>
<li><mathjax>#q_2#</mathjax> is the amount of heat needed to raise the temperature of the sample by <mathjax>#DeltaT_2 = 40^@"C" - 25^@"C"#</mathjax></li>
</ul>
</blockquote>
<p>This equation can be found by using the fact that the heat absorbed by the metal can be calculated using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(q = m * c * DeltaT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#m#</mathjax> is the <strong>mass</strong> of the sample</li>
<li><mathjax>#c#</mathjax> is the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of the metal</li>
</ul>
</blockquote>
<p>In your case, you can say that</p>
<blockquote>
<p><mathjax>#q_1 = m * c * DeltaT_1#</mathjax></p>
</blockquote>
<p>and </p>
<blockquote>
<p><mathjax>#q_2 = m * c * DeltaT_2#</mathjax></p>
</blockquote>
<p>Divide these two equations </p>
<blockquote>
<p><mathjax>#q_1/q_2 = (color(red)(cancel(color(black)(m * c))) * DeltaT_1)/(color(red)(cancel(color(black)(m * c))) * DeltaT_2)#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#q_2 = (DeltaT_2)/(DeltaT_1) * q_1#</mathjax></p>
<blockquote>
<p>This equation tells you that in order to increase the temperature of the metal by a factor <mathjax>#(DeltaT_2)/(DeltaT_1)#</mathjax> when the mass of the metal is <strong>constant</strong>, the amount of heat supplied must also <strong>increase</strong> by a factor of <mathjax>#(DeltaT_2)/(DeltaT_1)#</mathjax>.</p>
</blockquote>
</blockquote>
<p>So, plug in your values to find</p>
<blockquote>
<p><mathjax>#q_2 = ((40 - 25) color(red)(cancel(color(black)(""^@"C"))))/((20-15)color(red)(cancel(color(black)(""^@"C")))) * "8.40 kJ"#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(q_2 = "25 kJ")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"25 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that because the sample of metal has <strong>the same mass</strong> in both cases, you can say that</p>
<blockquote>
<p><mathjax>#q_2 = (DeltaT_2)/(DeltaT_1) * q_1#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#q_1#</mathjax> is the amount of heat needed to raise the temperature of the sample by <mathjax>#DeltaT_1 = 20^@"C" - 15^@"C"#</mathjax></li>
<li><mathjax>#q_2#</mathjax> is the amount of heat needed to raise the temperature of the sample by <mathjax>#DeltaT_2 = 40^@"C" - 25^@"C"#</mathjax></li>
</ul>
</blockquote>
<p>This equation can be found by using the fact that the heat absorbed by the metal can be calculated using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(q = m * c * DeltaT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#m#</mathjax> is the <strong>mass</strong> of the sample</li>
<li><mathjax>#c#</mathjax> is the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of the metal</li>
</ul>
</blockquote>
<p>In your case, you can say that</p>
<blockquote>
<p><mathjax>#q_1 = m * c * DeltaT_1#</mathjax></p>
</blockquote>
<p>and </p>
<blockquote>
<p><mathjax>#q_2 = m * c * DeltaT_2#</mathjax></p>
</blockquote>
<p>Divide these two equations </p>
<blockquote>
<p><mathjax>#q_1/q_2 = (color(red)(cancel(color(black)(m * c))) * DeltaT_1)/(color(red)(cancel(color(black)(m * c))) * DeltaT_2)#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#q_2 = (DeltaT_2)/(DeltaT_1) * q_1#</mathjax></p>
<blockquote>
<p>This equation tells you that in order to increase the temperature of the metal by a factor <mathjax>#(DeltaT_2)/(DeltaT_1)#</mathjax> when the mass of the metal is <strong>constant</strong>, the amount of heat supplied must also <strong>increase</strong> by a factor of <mathjax>#(DeltaT_2)/(DeltaT_1)#</mathjax>.</p>
</blockquote>
</blockquote>
<p>So, plug in your values to find</p>
<blockquote>
<p><mathjax>#q_2 = ((40 - 25) color(red)(cancel(color(black)(""^@"C"))))/((20-15)color(red)(cancel(color(black)(""^@"C")))) * "8.40 kJ"#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(q_2 = "25 kJ")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">If 8.40 kJ of heat is needed to raise the temperature of a sample of metal from 15 °C to 20 °C, how many kilojoules of heat will be required to raise the temperature of the same sample of metal from 25 °C to 40 °C?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2017-12-29T14:02:03" itemprop="dateCreated">
Dec 29, 2017
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<div class="markdown"><p><mathjax>#"25 kJ"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to realize that because the sample of metal has <strong>the same mass</strong> in both cases, you can say that</p>
<blockquote>
<p><mathjax>#q_2 = (DeltaT_2)/(DeltaT_1) * q_1#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#q_1#</mathjax> is the amount of heat needed to raise the temperature of the sample by <mathjax>#DeltaT_1 = 20^@"C" - 15^@"C"#</mathjax></li>
<li><mathjax>#q_2#</mathjax> is the amount of heat needed to raise the temperature of the sample by <mathjax>#DeltaT_2 = 40^@"C" - 25^@"C"#</mathjax></li>
</ul>
</blockquote>
<p>This equation can be found by using the fact that the heat absorbed by the metal can be calculated using the equation</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(q = m * c * DeltaT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#m#</mathjax> is the <strong>mass</strong> of the sample</li>
<li><mathjax>#c#</mathjax> is the <strong><a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a></strong> of the metal</li>
</ul>
</blockquote>
<p>In your case, you can say that</p>
<blockquote>
<p><mathjax>#q_1 = m * c * DeltaT_1#</mathjax></p>
</blockquote>
<p>and </p>
<blockquote>
<p><mathjax>#q_2 = m * c * DeltaT_2#</mathjax></p>
</blockquote>
<p>Divide these two equations </p>
<blockquote>
<p><mathjax>#q_1/q_2 = (color(red)(cancel(color(black)(m * c))) * DeltaT_1)/(color(red)(cancel(color(black)(m * c))) * DeltaT_2)#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#q_2 = (DeltaT_2)/(DeltaT_1) * q_1#</mathjax></p>
<blockquote>
<p>This equation tells you that in order to increase the temperature of the metal by a factor <mathjax>#(DeltaT_2)/(DeltaT_1)#</mathjax> when the mass of the metal is <strong>constant</strong>, the amount of heat supplied must also <strong>increase</strong> by a factor of <mathjax>#(DeltaT_2)/(DeltaT_1)#</mathjax>.</p>
</blockquote>
</blockquote>
<p>So, plug in your values to find</p>
<blockquote>
<p><mathjax>#q_2 = ((40 - 25) color(red)(cancel(color(black)(""^@"C"))))/((20-15)color(red)(cancel(color(black)(""^@"C")))) * "8.40 kJ"#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(q_2 = "25 kJ")))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | If 8.40 kJ of heat is needed to raise the temperature of a sample of metal from 15 °C to 20 °C, how many kilojoules of heat will be required to raise the temperature of the same sample of metal from 25 °C to 40 °C? | null |
2,962 | aa93963a-6ddd-11ea-9f6e-ccda262736ce | https://socratic.org/questions/lf-during-the-course-of-the-reaction-the-vessel-is-found-to-contain-7-00-mol-of- | 0.97 M | start physical_unit 4 5 reaction_quotient_q mol/l qc_end physical_unit 15 15 12 13 mole qc_end physical_unit 19 19 16 17 mole qc_end physical_unit 23 23 20 21 mole qc_end physical_unit 28 28 25 26 mole qc_end physical_unit 45 46 43 44 volume qc_end physical_unit 45 46 48 49 temperature qc_end chemical_equation 50 56 qc_end end | [{"type":"physical unit","value":"Reaction quotient Q [OF] the reaction [IN] M"}] | [{"type":"physical unit","value":"0.97 M"}] | [{"type":"physical unit","value":"Mole [OF] C [=] \\pu{7.00 mols}"},{"type":"physical unit","value":"Mole [OF] H2O [=] \\pu{14.0 mols}"},{"type":"physical unit","value":"Mole [OF] CO [=] \\pu{3.60 mols}"},{"type":"physical unit","value":"Mole [OF] H2 [=] \\pu{8.50 mols}"},{"type":"physical unit","value":"Volume [OF] reaction vessel [=] \\pu{2.25 L}"},{"type":"physical unit","value":"Temperature [OF] reaction vessel [=] \\pu{1100 K}"},{"type":"chemical equation","value":"C(s) + H2O(g) <=> CO(g) + H2(g)"}] | <h1 class="questionTitle" itemprop="name">During the course of the reaction, the vessel is found to contain #"7.00 mols"# of #C#, #"14.0 mols"# of #H_2O#, #"3.60 mols"# of #CO#, and #"8.50 mols"# of #H_2#. What is the reaction quotient #Q#? </h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The following reaction was carried out in a <mathjax>#"2.25 L"#</mathjax> reaction vessel at <mathjax>#"1100 K"#</mathjax>: </p>
<blockquote>
<p><mathjax>#C (s) + H_2O(g) rightleftharpoons CO (g) + H_2 (g)#</mathjax> </p>
</blockquote></div>
</h2>
</div>
</div> | 0.97 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equilibrium given to you looks like this </p>
<blockquote>
<p><mathjax>#"C"_ ((s)) + "H"_ 2"O"_ ((g)) rightleftharpoons "CO"_ ((g)) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>By definition, the <strong>reaction quotient</strong>, <mathjax>#Q_c#</mathjax>, tells you the ratio that exists between the <strong>concentrations</strong> of the products raised to the power of their respective stoichiometric coefficients and the <strong>concentrations</strong> of the reactants raised to the power of their respective stoichiometric coefficients.</p>
<p>Two important things to note here</p>
<blockquote>
<ul>
<li><em>the <strong>reaction quotient</strong> can be calculated by using the concentrations of the chemical species that take part in the reaction <strong>at any given moment</strong> in the course of the reaction</em></li>
<li><em>the concentration of <strong>solids</strong> and of <strong>pure liquids</strong> is excluded from the expression of the reaction quotient</em></li>
</ul>
</blockquote>
<p>In this case, the reaction quotient would take the form -- keep in mind that carbon, <mathjax>#"C"#</mathjax>, is in the <em>solid state</em> in this reaction</p>
<blockquote>
<p><mathjax>#Q_c = (["CO"] * ["H"_2])/(["H"_2"O"])#</mathjax></p>
</blockquote>
<p>Now, use the <em>volume</em> of the reaction vessel to calculate the <strong>concentrations</strong> of the chemical species that are of interest here</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/(V_"solution"[color(blue)("in liters")])color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#["H"_2"O"] = "14.0 moles"/"2.25 L" = "6.22 M"#</mathjax></p>
<p><mathjax>#["CO"] = "3.60 moles"/"2.25 L" = "1.6 M"#</mathjax></p>
<p><mathjax>#["H"_2] = "8.50 moles"/"2.25 L" = "3.78 M"#</mathjax></p>
</blockquote>
<p>Plug these values into the equation that gives you the reaction quotient to find</p>
<blockquote>
<p><mathjax>#Q_c = ("1.6 M" * 3.78 color(red)(cancel(color(black)("M"))))/(6.22color(red)(cancel(color(black)("M")))) = "0.972 M"#</mathjax></p>
</blockquote>
<p>The reaction quotient is usually expressed <em>without added units</em>, so your answer will be </p>
<blockquote>
<p><mathjax>#Q_c = color(green)(|bar(ul(color(white)(a/a)color(black)(0.972)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#Q_c = 0.972#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equilibrium given to you looks like this </p>
<blockquote>
<p><mathjax>#"C"_ ((s)) + "H"_ 2"O"_ ((g)) rightleftharpoons "CO"_ ((g)) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>By definition, the <strong>reaction quotient</strong>, <mathjax>#Q_c#</mathjax>, tells you the ratio that exists between the <strong>concentrations</strong> of the products raised to the power of their respective stoichiometric coefficients and the <strong>concentrations</strong> of the reactants raised to the power of their respective stoichiometric coefficients.</p>
<p>Two important things to note here</p>
<blockquote>
<ul>
<li><em>the <strong>reaction quotient</strong> can be calculated by using the concentrations of the chemical species that take part in the reaction <strong>at any given moment</strong> in the course of the reaction</em></li>
<li><em>the concentration of <strong>solids</strong> and of <strong>pure liquids</strong> is excluded from the expression of the reaction quotient</em></li>
</ul>
</blockquote>
<p>In this case, the reaction quotient would take the form -- keep in mind that carbon, <mathjax>#"C"#</mathjax>, is in the <em>solid state</em> in this reaction</p>
<blockquote>
<p><mathjax>#Q_c = (["CO"] * ["H"_2])/(["H"_2"O"])#</mathjax></p>
</blockquote>
<p>Now, use the <em>volume</em> of the reaction vessel to calculate the <strong>concentrations</strong> of the chemical species that are of interest here</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/(V_"solution"[color(blue)("in liters")])color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#["H"_2"O"] = "14.0 moles"/"2.25 L" = "6.22 M"#</mathjax></p>
<p><mathjax>#["CO"] = "3.60 moles"/"2.25 L" = "1.6 M"#</mathjax></p>
<p><mathjax>#["H"_2] = "8.50 moles"/"2.25 L" = "3.78 M"#</mathjax></p>
</blockquote>
<p>Plug these values into the equation that gives you the reaction quotient to find</p>
<blockquote>
<p><mathjax>#Q_c = ("1.6 M" * 3.78 color(red)(cancel(color(black)("M"))))/(6.22color(red)(cancel(color(black)("M")))) = "0.972 M"#</mathjax></p>
</blockquote>
<p>The reaction quotient is usually expressed <em>without added units</em>, so your answer will be </p>
<blockquote>
<p><mathjax>#Q_c = color(green)(|bar(ul(color(white)(a/a)color(black)(0.972)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">During the course of the reaction, the vessel is found to contain #"7.00 mols"# of #C#, #"14.0 mols"# of #H_2O#, #"3.60 mols"# of #CO#, and #"8.50 mols"# of #H_2#. What is the reaction quotient #Q#? </h1>
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<div class="markdown"><p>The following reaction was carried out in a <mathjax>#"2.25 L"#</mathjax> reaction vessel at <mathjax>#"1100 K"#</mathjax>: </p>
<blockquote>
<p><mathjax>#C (s) + H_2O(g) rightleftharpoons CO (g) + H_2 (g)#</mathjax> </p>
</blockquote></div>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-23T01:06:55" itemprop="dateCreated">
Jul 23, 2016
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<div class="markdown"><p><mathjax>#Q_c = 0.972#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equilibrium given to you looks like this </p>
<blockquote>
<p><mathjax>#"C"_ ((s)) + "H"_ 2"O"_ ((g)) rightleftharpoons "CO"_ ((g)) + "H"_ (2(g))#</mathjax></p>
</blockquote>
<p>By definition, the <strong>reaction quotient</strong>, <mathjax>#Q_c#</mathjax>, tells you the ratio that exists between the <strong>concentrations</strong> of the products raised to the power of their respective stoichiometric coefficients and the <strong>concentrations</strong> of the reactants raised to the power of their respective stoichiometric coefficients.</p>
<p>Two important things to note here</p>
<blockquote>
<ul>
<li><em>the <strong>reaction quotient</strong> can be calculated by using the concentrations of the chemical species that take part in the reaction <strong>at any given moment</strong> in the course of the reaction</em></li>
<li><em>the concentration of <strong>solids</strong> and of <strong>pure liquids</strong> is excluded from the expression of the reaction quotient</em></li>
</ul>
</blockquote>
<p>In this case, the reaction quotient would take the form -- keep in mind that carbon, <mathjax>#"C"#</mathjax>, is in the <em>solid state</em> in this reaction</p>
<blockquote>
<p><mathjax>#Q_c = (["CO"] * ["H"_2])/(["H"_2"O"])#</mathjax></p>
</blockquote>
<p>Now, use the <em>volume</em> of the reaction vessel to calculate the <strong>concentrations</strong> of the chemical species that are of interest here</p>
<blockquote>
<p><mathjax>#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/(V_"solution"[color(blue)("in liters")])color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>You will have</p>
<blockquote>
<p><mathjax>#["H"_2"O"] = "14.0 moles"/"2.25 L" = "6.22 M"#</mathjax></p>
<p><mathjax>#["CO"] = "3.60 moles"/"2.25 L" = "1.6 M"#</mathjax></p>
<p><mathjax>#["H"_2] = "8.50 moles"/"2.25 L" = "3.78 M"#</mathjax></p>
</blockquote>
<p>Plug these values into the equation that gives you the reaction quotient to find</p>
<blockquote>
<p><mathjax>#Q_c = ("1.6 M" * 3.78 color(red)(cancel(color(black)("M"))))/(6.22color(red)(cancel(color(black)("M")))) = "0.972 M"#</mathjax></p>
</blockquote>
<p>The reaction quotient is usually expressed <em>without added units</em>, so your answer will be </p>
<blockquote>
<p><mathjax>#Q_c = color(green)(|bar(ul(color(white)(a/a)color(black)(0.972)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | During the course of the reaction, the vessel is found to contain #"7.00 mols"# of #C#, #"14.0 mols"# of #H_2O#, #"3.60 mols"# of #CO#, and #"8.50 mols"# of #H_2#. What is the reaction quotient #Q#? |
The following reaction was carried out in a #"2.25 L"# reaction vessel at #"1100 K"#:
#C (s) + H_2O(g) rightleftharpoons CO (g) + H_2 (g)#
|
2,963 | aae2c9e4-6ddd-11ea-b787-ccda262736ce | https://socratic.org/questions/how-do-you-balance-cacl-2-cr-no-3-3-ca-no-3-2-crcl-3 | 3 CaCl2 + 2 Cr(NO3)3 -> 3 Ca(NO3)2 + 2 CrCl3 | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"3 CaCl2 + 2 Cr(NO3)3 -> 3 Ca(NO3)2 + 2 CrCl3"}] | [{"type":"chemical equation","value":"CaCl2 + Cr(NO3)3 -> Ca(NO3)2 + CrCl3"}] | <h1 class="questionTitle" itemprop="name">How do you balance #"CaCl"_2 + "Cr"("NO"_3)_3 -> "Ca"("NO"_3)_2 + "CrCl"_3#?</h1> | null | 3 CaCl2 + 2 Cr(NO3)3 -> 3 Ca(NO3)2 + 2 CrCl3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look for the element that appears the least and balance that first. (Just as when you solve simultaneous linear equation, the variable that appears least often is the easiest to eliminate.) In this case, it is either Ca or Cr. The equation given to us already has both balanced. </p>
<p>Now we try to balance Cl, while making sure that Ca and Cr remains balanced as well. We get:</p>
<p><mathjax>#"CaCl"_2 + 2/3"Cr"("NO"_3)_3 -> "Ca"("NO"_3)_2 + 2/3"CrCl"_3#</mathjax></p>
<p>To get integer coefficients, multiply everywhere by 3.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"3CaCl"_2 + 2"Cr"("NO"_3)_3 -> 3"Ca"("NO"_3)_2 + 2"CrCl"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look for the element that appears the least and balance that first. (Just as when you solve simultaneous linear equation, the variable that appears least often is the easiest to eliminate.) In this case, it is either Ca or Cr. The equation given to us already has both balanced. </p>
<p>Now we try to balance Cl, while making sure that Ca and Cr remains balanced as well. We get:</p>
<p><mathjax>#"CaCl"_2 + 2/3"Cr"("NO"_3)_3 -> "Ca"("NO"_3)_2 + 2/3"CrCl"_3#</mathjax></p>
<p>To get integer coefficients, multiply everywhere by 3.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you balance #"CaCl"_2 + "Cr"("NO"_3)_3 -> "Ca"("NO"_3)_2 + "CrCl"_3#?</h1>
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<div class="markdown"><p><mathjax>#"3CaCl"_2 + 2"Cr"("NO"_3)_3 -> 3"Ca"("NO"_3)_2 + 2"CrCl"_3#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Look for the element that appears the least and balance that first. (Just as when you solve simultaneous linear equation, the variable that appears least often is the easiest to eliminate.) In this case, it is either Ca or Cr. The equation given to us already has both balanced. </p>
<p>Now we try to balance Cl, while making sure that Ca and Cr remains balanced as well. We get:</p>
<p><mathjax>#"CaCl"_2 + 2/3"Cr"("NO"_3)_3 -> "Ca"("NO"_3)_2 + 2/3"CrCl"_3#</mathjax></p>
<p>To get integer coefficients, multiply everywhere by 3.</p></div>
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</article> | How do you balance #"CaCl"_2 + "Cr"("NO"_3)_3 -> "Ca"("NO"_3)_2 + "CrCl"_3#? | null |
2,964 | a91667fa-6ddd-11ea-9b98-ccda262736ce | https://socratic.org/questions/what-would-be-the-final-temperature-if-you-mixed-a-liter-of-32-c-water-with-4-li | 22.40 ℃ | start physical_unit 14 14 temperature °c qc_end physical_unit 14 14 16 17 volume qc_end physical_unit 14 14 12 13 temperature qc_end physical_unit 14 14 19 20 temperature qc_end end | [{"type":"physical unit","value":"Temperature3 [OF] water [IN] ℃"}] | [{"type":"physical unit","value":"22.40 ℃"}] | [{"type":"physical unit","value":"Volume1 [OF] water [=] \\pu{1 liter}"},{"type":"physical unit","value":"Volume2 [OF] water [=] \\pu{4 liters}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{32 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{20 ℃}"}] | <h1 class="questionTitle" itemprop="name">What would be the final temperature if you mixed a liter of 32°C water with 4 liters of 20°C water?</h1> | null | 22.40 ℃ | <div class="answerDescription">
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<div class="markdown"><p>Admitting that the calorific capacity doesn't change too much and since there is no change of state, the final temperature will be the weighted average:</p>
<p><mathjax>#(1xx32+4xx20)/5=112/5=22.4ºC#</mathjax></p></div>
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<div class="markdown"><p>Admitting that the calorific capacity doesn't change too much and since there is no change of state, the final temperature will be the weighted average:</p>
<p><mathjax>#(1xx32+4xx20)/5=112/5=22.4ºC#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What would be the final temperature if you mixed a liter of 32°C water with 4 liters of 20°C water?</h1>
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<div class="markdown"><p>Admitting that the calorific capacity doesn't change too much and since there is no change of state, the final temperature will be the weighted average:</p>
<p><mathjax>#(1xx32+4xx20)/5=112/5=22.4ºC#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#22.4^@"C"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the amount of heat <strong>lost</strong> by the warmer sample will be equal to the amount of heat <strong>gained</strong> by the colder sample. </p>
<p>Even without doing any calculations, you should be able to predict that the final temperature of the mixture will be <em>closer<strong> to <mathjax>#20^@"C"#</mathjax> than to <mathjax>#32^@"C"#</mathjax> because you have </strong>more</em>* warm water than room-temperature water. </p>
<p>The equation to use here is </p>
<blockquote>
<p><mathjax>#color(blue)(q = m * c * DeltaT)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#q#</mathjax> - the amount of heat gained / 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 water<br/>
<mathjax>#DeltaT#</mathjax> - the <em>change in temperature</em>, defined as the difference between the <strong>final temperature</strong> and the <strong>initial temperature</strong></p>
<p>Assuming that water's <em><a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a></em> is constant for both samples, the ratio of <em>volumes</em> will be equivalent to the ratios of <em>masses</em>. If you take <mathjax>#m_"warm"#</mathjax> to be the mass of the warmer sample, you will have</p>
<blockquote>
<p><mathjax>#m_"cold" = 4 * m_"warm"#</mathjax></p>
</blockquote>
<p>Since the heat lost by the warmer sample is <strong>equal</strong> to the heat gained by the colder sample, you will have</p>
<blockquote>
<p><mathjax>#-q_"warm" = q_"cold"#</mathjax></p>
</blockquote>
<p>Here the <em>minus sign</em> is used because <strong>heat lost</strong> carries a negative sign. </p>
<p>If you take <mathjax>#T_f#</mathjax> to be the final temperature of the mixture, you can say that</p>
<blockquote>
<p><mathjax>#-overbrace(color(red)(cancel(color(black)(m_"warm"))) * color(red)(cancel(color(black)(c))) * (T_f - 32^@"C"))^(color(blue)("heat lost by the warmer sample")) = overbrace(4 * color(red)(cancel(color(black)(m_"warm"))) * color(red)(cancel(color(black)(c))) * (T_f - 20^@"C"))^(color(purple)("heat gained by the colder sample"))#</mathjax></p>
</blockquote>
<p>This will get you </p>
<blockquote>
<p><mathjax>#-T_f + 32^@"C" = 4 * T_f - 80^@"C"#</mathjax></p>
</blockquote>
<p>Rearrange to solve for <mathjax>#T_f#</mathjax></p>
<blockquote>
<p><mathjax>#T_f = ((32 + 80)^@"C")/5 = color(green)(22.4^@"C")#</mathjax></p>
</blockquote>
<p>You <em>should</em> round this off to one <strong><a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig fig</a></strong>, since that's how many sig figs you have for the volumes of the two samples, but I'll leave it rounded to three sig figs.</p></div>
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</article> | What would be the final temperature if you mixed a liter of 32°C water with 4 liters of 20°C water? | null |
2,965 | a90441dd-6ddd-11ea-aaee-ccda262736ce | https://socratic.org/questions/56c0b4857c014946e01d59cf | C3H8 | start chemical_formula qc_end physical_unit 7 7 1 2 mole qc_end c_other Combusted qc_end physical_unit 15 15 12 13 mole qc_end physical_unit 20 20 17 18 mole qc_end end | [{"type":"other","value":"Chemical Formula [OF] this hydrocarbon [IN] empirical"}] | [{"type":"chemical equation","value":"C3H8"}] | [{"type":"physical unit","value":"Mole [OF] CxHy [=] \\pu{0.1 moles}"},{"type":"other","value":"combusted"},{"type":"physical unit","value":"Mole [OF] CO2 [=] \\pu{0.3 moles}"},{"type":"physical unit","value":"Mole [OF] H2O [=] \\pu{0.4 moles}"}] | <h1 class="questionTitle" itemprop="name">When #"0.1 mols"# of an unknown hydrocarbon #"C"_x"H"_y# was combusted, it produced #"0.3 mols"# of #"CO"_2# and #"0.4 mols"# of #"H"_2"O"#. What is the molecular formula of this hydrocarbon?</h1> | null | C3H8 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to recognize the fact that you're dealing with a <em>hydrocarbon</em>, which means that it will only contain <strong>carbon</strong> and <strong>hydrogen</strong>. </p>
<p>Now, the <em>complete combustion</em> of a hydrocarbon produces <em>carbon dioxide</em>, <mathjax>#"CO"_2#</mathjax>, and <em>water</em>, <mathjax>#"H"_2"O"#</mathjax>. </p>
<p>In your case, you know that when <mathjax>#0.1#</mathjax> moles of this hydrocarbon underwent combustion, the reaction produced <mathjax>#0.3#</mathjax> moles of carbon dioxide and <mathjax>#0.4#</mathjax> moles of water. </p>
<p>Now, another important thing to realize here is that <strong>all of the carbon</strong> that was initially a part of the hydrocarbon is now a part of the carbon dioxide, and <strong>all of the hydrogen</strong> that was a part of the hydrocarbon is now a part of the water. </p>
<p>Since <strong>one mole</strong> of carbon dioxide contains <mathjax>#1#</mathjax> <strong>mole</strong> of carbon, and <strong>one mole</strong> of water contains <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen, it follows that the hydrocarbon contained </p>
<blockquote>
<p><mathjax>#0.3color(red)(cancel(color(black)("moles CO"_2))) * "1 mole C"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.3 moles C"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#0.4color(red)(cancel(color(black)("moles H"_2"O"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.8 moles H"#</mathjax></p>
</blockquote>
<p>Now, let's say that your hydrocarbon has a <strong>molecular formula</strong> <mathjax>#"C"_x"H"_y#</mathjax>. You know that <strong>one mole</strong> of this hydrocarbon would produce</p>
<blockquote>
<ul>
<li><mathjax>#xcolor(white)(a)"moles of C"#</mathjax></li>
<li><mathjax>#ycolor(white)(a)"moles of H"#</mathjax></li>
</ul>
</blockquote>
<p>If this is the case, you can say that <mathjax>#0.1#</mathjax> moles of this hydrocarbon will produce</p>
<blockquote>
<p><mathjax>#(0.1 xx x)color(white)(a)"moles of C"#</mathjax><br/>
<mathjax>#(0.1 xx y)color(white)(a)"moles of H"#</mathjax></p>
</blockquote>
<p>But you already know how many moles of carbon and hydrogen were produced by the reaction, so you can say that</p>
<blockquote>
<p><mathjax>#0.1 xx x = 0.3 implies x = 0.3/0.1 = 3#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#0.1 xx y = 0.8 implies y = 0.8/0.1 = 8#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> of the hydrocarbon is </p>
<blockquote>
<p><mathjax>#color(green)("C"_3"H"_8) ->#</mathjax> <em>propane</em></p>
</blockquote>
<p><img alt="http://www.turbosquid.com/3d-models/propane-molecule-structure-3d-max/445795" src="https://useruploads.socratic.org/S0GoFbQyyX1jRB6k2a1Q_Propane-c-01.jpg1863f7bd-0838-4e23-b15f-7c3d98d904bfLarger.jpg"/> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"C"_3"H"_8#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to recognize the fact that you're dealing with a <em>hydrocarbon</em>, which means that it will only contain <strong>carbon</strong> and <strong>hydrogen</strong>. </p>
<p>Now, the <em>complete combustion</em> of a hydrocarbon produces <em>carbon dioxide</em>, <mathjax>#"CO"_2#</mathjax>, and <em>water</em>, <mathjax>#"H"_2"O"#</mathjax>. </p>
<p>In your case, you know that when <mathjax>#0.1#</mathjax> moles of this hydrocarbon underwent combustion, the reaction produced <mathjax>#0.3#</mathjax> moles of carbon dioxide and <mathjax>#0.4#</mathjax> moles of water. </p>
<p>Now, another important thing to realize here is that <strong>all of the carbon</strong> that was initially a part of the hydrocarbon is now a part of the carbon dioxide, and <strong>all of the hydrogen</strong> that was a part of the hydrocarbon is now a part of the water. </p>
<p>Since <strong>one mole</strong> of carbon dioxide contains <mathjax>#1#</mathjax> <strong>mole</strong> of carbon, and <strong>one mole</strong> of water contains <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen, it follows that the hydrocarbon contained </p>
<blockquote>
<p><mathjax>#0.3color(red)(cancel(color(black)("moles CO"_2))) * "1 mole C"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.3 moles C"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#0.4color(red)(cancel(color(black)("moles H"_2"O"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.8 moles H"#</mathjax></p>
</blockquote>
<p>Now, let's say that your hydrocarbon has a <strong>molecular formula</strong> <mathjax>#"C"_x"H"_y#</mathjax>. You know that <strong>one mole</strong> of this hydrocarbon would produce</p>
<blockquote>
<ul>
<li><mathjax>#xcolor(white)(a)"moles of C"#</mathjax></li>
<li><mathjax>#ycolor(white)(a)"moles of H"#</mathjax></li>
</ul>
</blockquote>
<p>If this is the case, you can say that <mathjax>#0.1#</mathjax> moles of this hydrocarbon will produce</p>
<blockquote>
<p><mathjax>#(0.1 xx x)color(white)(a)"moles of C"#</mathjax><br/>
<mathjax>#(0.1 xx y)color(white)(a)"moles of H"#</mathjax></p>
</blockquote>
<p>But you already know how many moles of carbon and hydrogen were produced by the reaction, so you can say that</p>
<blockquote>
<p><mathjax>#0.1 xx x = 0.3 implies x = 0.3/0.1 = 3#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#0.1 xx y = 0.8 implies y = 0.8/0.1 = 8#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> of the hydrocarbon is </p>
<blockquote>
<p><mathjax>#color(green)("C"_3"H"_8) ->#</mathjax> <em>propane</em></p>
</blockquote>
<p><img alt="http://www.turbosquid.com/3d-models/propane-molecule-structure-3d-max/445795" src="https://useruploads.socratic.org/S0GoFbQyyX1jRB6k2a1Q_Propane-c-01.jpg1863f7bd-0838-4e23-b15f-7c3d98d904bfLarger.jpg"/> </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">When #"0.1 mols"# of an unknown hydrocarbon #"C"_x"H"_y# was combusted, it produced #"0.3 mols"# of #"CO"_2# and #"0.4 mols"# of #"H"_2"O"#. What is the molecular formula of this hydrocarbon?</h1>
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Truong-Son N.
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<div class="markdown"><p><mathjax>#"C"_3"H"_8#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The trick here is to recognize the fact that you're dealing with a <em>hydrocarbon</em>, which means that it will only contain <strong>carbon</strong> and <strong>hydrogen</strong>. </p>
<p>Now, the <em>complete combustion</em> of a hydrocarbon produces <em>carbon dioxide</em>, <mathjax>#"CO"_2#</mathjax>, and <em>water</em>, <mathjax>#"H"_2"O"#</mathjax>. </p>
<p>In your case, you know that when <mathjax>#0.1#</mathjax> moles of this hydrocarbon underwent combustion, the reaction produced <mathjax>#0.3#</mathjax> moles of carbon dioxide and <mathjax>#0.4#</mathjax> moles of water. </p>
<p>Now, another important thing to realize here is that <strong>all of the carbon</strong> that was initially a part of the hydrocarbon is now a part of the carbon dioxide, and <strong>all of the hydrogen</strong> that was a part of the hydrocarbon is now a part of the water. </p>
<p>Since <strong>one mole</strong> of carbon dioxide contains <mathjax>#1#</mathjax> <strong>mole</strong> of carbon, and <strong>one mole</strong> of water contains <mathjax>#2#</mathjax> <strong>moles</strong> of hydrogen, it follows that the hydrocarbon contained </p>
<blockquote>
<p><mathjax>#0.3color(red)(cancel(color(black)("moles CO"_2))) * "1 mole C"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.3 moles C"#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#0.4color(red)(cancel(color(black)("moles H"_2"O"))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "0.8 moles H"#</mathjax></p>
</blockquote>
<p>Now, let's say that your hydrocarbon has a <strong>molecular formula</strong> <mathjax>#"C"_x"H"_y#</mathjax>. You know that <strong>one mole</strong> of this hydrocarbon would produce</p>
<blockquote>
<ul>
<li><mathjax>#xcolor(white)(a)"moles of C"#</mathjax></li>
<li><mathjax>#ycolor(white)(a)"moles of H"#</mathjax></li>
</ul>
</blockquote>
<p>If this is the case, you can say that <mathjax>#0.1#</mathjax> moles of this hydrocarbon will produce</p>
<blockquote>
<p><mathjax>#(0.1 xx x)color(white)(a)"moles of C"#</mathjax><br/>
<mathjax>#(0.1 xx y)color(white)(a)"moles of H"#</mathjax></p>
</blockquote>
<p>But you already know how many moles of carbon and hydrogen were produced by the reaction, so you can say that</p>
<blockquote>
<p><mathjax>#0.1 xx x = 0.3 implies x = 0.3/0.1 = 3#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#0.1 xx y = 0.8 implies y = 0.8/0.1 = 8#</mathjax></p>
</blockquote>
<p>Therefore, the <strong>molecular formula</strong> of the hydrocarbon is </p>
<blockquote>
<p><mathjax>#color(green)("C"_3"H"_8) ->#</mathjax> <em>propane</em></p>
</blockquote>
<p><img alt="http://www.turbosquid.com/3d-models/propane-molecule-structure-3d-max/445795" src="https://useruploads.socratic.org/S0GoFbQyyX1jRB6k2a1Q_Propane-c-01.jpg1863f7bd-0838-4e23-b15f-7c3d98d904bfLarger.jpg"/> </p></div>
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</article> | When #"0.1 mols"# of an unknown hydrocarbon #"C"_x"H"_y# was combusted, it produced #"0.3 mols"# of #"CO"_2# and #"0.4 mols"# of #"H"_2"O"#. What is the molecular formula of this hydrocarbon? | null |
2,966 | a8a8a78c-6ddd-11ea-9f2d-ccda262736ce | https://socratic.org/questions/how-many-grams-of-copper-are-required-to-replace-5-moles-of-silver-nitrate-which | 15.89 grams | start physical_unit 4 4 mass g qc_end physical_unit 12 13 9 10 mole qc_end c_other OTHER qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] copper [IN] grams"}] | [{"type":"physical unit","value":"15.89 grams"}] | [{"type":"physical unit","value":"Mole [OF] silver nitrate [=] \\pu{0.5 moles}"},{"type":"other","value":"Silver nitrate is dissolved in water."},{"type":"other","value":"Assuming a copper (II) product."}] | <h1 class="questionTitle" itemprop="name">How many grams of copper are required to replace .5 moles of silver nitrate (which is dissolved in water), assuming a copper (II) product?</h1> | null | 15.89 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The equation for the reaction is</p>
<p><mathjax>#"2Ag"^+("aq") + "Cu(s)" → "2Ag(s)" + "Cu"^(2+)("aq")#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1</strong>. Calculate the moles of <mathjax>#"Cu"#</mathjax> required.</p>
<p><mathjax>#0.5 color(red)(cancel(color(black)("mol Ag"^+))) × "1 mol Cu"/(2 color(red)(cancel(color(black)("mol Ag"^+)))) = "0.25 mol Cu"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2</strong>. Calculate the mass of <mathjax>#"Cu"#</mathjax>.</p>
<p><mathjax>#0.25 color(red)(cancel(color(black)("mol Cu"))) × "63.55 g Cu"/(1 color(red)(cancel(color(black)("mol Cu")))) = "16 g Cu"#</mathjax></p></div>
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<div class="markdown"><p>You need 16 g of copper.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The equation for the reaction is</p>
<p><mathjax>#"2Ag"^+("aq") + "Cu(s)" → "2Ag(s)" + "Cu"^(2+)("aq")#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1</strong>. Calculate the moles of <mathjax>#"Cu"#</mathjax> required.</p>
<p><mathjax>#0.5 color(red)(cancel(color(black)("mol Ag"^+))) × "1 mol Cu"/(2 color(red)(cancel(color(black)("mol Ag"^+)))) = "0.25 mol Cu"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2</strong>. Calculate the mass of <mathjax>#"Cu"#</mathjax>.</p>
<p><mathjax>#0.25 color(red)(cancel(color(black)("mol Cu"))) × "63.55 g Cu"/(1 color(red)(cancel(color(black)("mol Cu")))) = "16 g Cu"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of copper are required to replace .5 moles of silver nitrate (which is dissolved in water), assuming a copper (II) product?</h1>
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<div class="markdown"><p>You need 16 g of copper.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>The equation for the reaction is</p>
<p><mathjax>#"2Ag"^+("aq") + "Cu(s)" → "2Ag(s)" + "Cu"^(2+)("aq")#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 1</strong>. Calculate the moles of <mathjax>#"Cu"#</mathjax> required.</p>
<p><mathjax>#0.5 color(red)(cancel(color(black)("mol Ag"^+))) × "1 mol Cu"/(2 color(red)(cancel(color(black)("mol Ag"^+)))) = "0.25 mol Cu"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2</strong>. Calculate the mass of <mathjax>#"Cu"#</mathjax>.</p>
<p><mathjax>#0.25 color(red)(cancel(color(black)("mol Cu"))) × "63.55 g Cu"/(1 color(red)(cancel(color(black)("mol Cu")))) = "16 g Cu"#</mathjax></p></div>
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</article> | How many grams of copper are required to replace .5 moles of silver nitrate (which is dissolved in water), assuming a copper (II) product? | null |
2,967 | ac5c4146-6ddd-11ea-907a-ccda262736ce | https://socratic.org/questions/in-the-reaction-2h-2o-2h-2-o-2-if-you-start-with-2-mol-of-water-how-many-moles-o | 2.00 moles | start physical_unit 22 23 mole mol qc_end chemical_equation 3 9 qc_end physical_unit 17 17 14 15 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] hydrogen gas [IN] moles"}] | [{"type":"physical unit","value":"2.00 moles"}] | [{"type":"chemical equation","value":"2 H2O -> 2 H2 + O2"},{"type":"physical unit","value":"Mole [OF] water [=] \\pu{2 mol}"}] | <h1 class="questionTitle" itemprop="name">In the reaction #2H_2O -> 2H_2 + O_2#, if you start with 2 mol of water, how many moles of hydrogen gas are produced?</h1> | null | 2.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a fairly simple example of a titration. We use titrations to calculate how many (grams, moles, litres, molecules, etc.) of one compound/element we would have if we are given a specific amount(grams, moles, litres, molecules, etc.) of another compound/element. To do a titration you must have a balanced equation. </p>
<p>The equation for this example looks like this:</p>
<p>2.0 mol <mathjax>#H_2O#</mathjax> (2.0 mol <mathjax>#H_2#</mathjax>/2.0 mol <mathjax>#H_2O#</mathjax>) =2.0 mol <mathjax>#H_2#</mathjax></p>
<p>You use the coefficients (big numbers in front of a compound/element in the balanced equation) to get the numbers for the equation. In this question the coefficient 2 in front of <mathjax>#H_2#</mathjax> in your balanced equation becomes the 2.0 we multiply our original number by. </p>
<p>Essentially what you are doing in this equation is taking the number the equation gives you and converting it to moles through unit conversion. This example was already in moles so we skipped that. Then you do another unit conversion from moles of your original compound to moles of the compound you want(using the relationship between the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> found in the balanced equation; 2=2)</p>
<p>I hope this helps!</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p>You would end up with 2 moles of Hydrogen gas.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a fairly simple example of a titration. We use titrations to calculate how many (grams, moles, litres, molecules, etc.) of one compound/element we would have if we are given a specific amount(grams, moles, litres, molecules, etc.) of another compound/element. To do a titration you must have a balanced equation. </p>
<p>The equation for this example looks like this:</p>
<p>2.0 mol <mathjax>#H_2O#</mathjax> (2.0 mol <mathjax>#H_2#</mathjax>/2.0 mol <mathjax>#H_2O#</mathjax>) =2.0 mol <mathjax>#H_2#</mathjax></p>
<p>You use the coefficients (big numbers in front of a compound/element in the balanced equation) to get the numbers for the equation. In this question the coefficient 2 in front of <mathjax>#H_2#</mathjax> in your balanced equation becomes the 2.0 we multiply our original number by. </p>
<p>Essentially what you are doing in this equation is taking the number the equation gives you and converting it to moles through unit conversion. This example was already in moles so we skipped that. Then you do another unit conversion from moles of your original compound to moles of the compound you want(using the relationship between the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> found in the balanced equation; 2=2)</p>
<p>I hope this helps!</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">In the reaction #2H_2O -> 2H_2 + O_2#, if you start with 2 mol of water, how many moles of hydrogen gas are produced?</h1>
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<div class="markdown"><p>You would end up with 2 moles of Hydrogen gas.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a fairly simple example of a titration. We use titrations to calculate how many (grams, moles, litres, molecules, etc.) of one compound/element we would have if we are given a specific amount(grams, moles, litres, molecules, etc.) of another compound/element. To do a titration you must have a balanced equation. </p>
<p>The equation for this example looks like this:</p>
<p>2.0 mol <mathjax>#H_2O#</mathjax> (2.0 mol <mathjax>#H_2#</mathjax>/2.0 mol <mathjax>#H_2O#</mathjax>) =2.0 mol <mathjax>#H_2#</mathjax></p>
<p>You use the coefficients (big numbers in front of a compound/element in the balanced equation) to get the numbers for the equation. In this question the coefficient 2 in front of <mathjax>#H_2#</mathjax> in your balanced equation becomes the 2.0 we multiply our original number by. </p>
<p>Essentially what you are doing in this equation is taking the number the equation gives you and converting it to moles through unit conversion. This example was already in moles so we skipped that. Then you do another unit conversion from moles of your original compound to moles of the compound you want(using the relationship between the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> found in the balanced equation; 2=2)</p>
<p>I hope this helps!</p></div>
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</article> | In the reaction #2H_2O -> 2H_2 + O_2#, if you start with 2 mol of water, how many moles of hydrogen gas are produced? | null |
2,968 | ac106155-6ddd-11ea-ba17-ccda262736ce | https://socratic.org/questions/583e72ef7c014963e4b9fee0 | 0.71 M | start physical_unit 19 19 concentration mol/l qc_end physical_unit 6 6 11 11 ph qc_end physical_unit 6 6 13 14 temperature qc_end end | [{"type":"physical unit","value":"Concentration [OF] NaF [IN] M"}] | [{"type":"physical unit","value":"0.71 M"}] | [{"type":"physical unit","value":"pH [OF] the solution [=] \\pu{8.51}"},{"type":"physical unit","value":"Temperature [OF] the solution [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">If you want to prepare a solution that has #"pH" = 8.51# at #25^@ "C"#, what initial concentration of #"NaF"# would you need to achieve?</h1> | null | 0.71 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#sf(F^-)#</mathjax> is the co - base of <mathjax>#sf(HF)#</mathjax>:</p>
<p><mathjax>#sf(F^(-)+H_2OrightleftharpoonsHF+OH^-)#</mathjax></p>
<p>You need to look up <mathjax>#sf(pK_b)#</mathjax> for the <mathjax>#sf(F^-)#</mathjax> ion which = 10.82.</p>
<p>From the ICE table you get this expression:</p>
<p><mathjax>#sf(pOH=1/2(pK_a-log[F^-])#</mathjax>`</p>
<p><mathjax>#sf(pOH=14-pH=14-8.51=5.49)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(5.49=1/2(10.82-log[F^-])#</mathjax></p>
<p><mathjax>#sf(1/2log[F^-]=-0.08)#</mathjax></p>
<p><mathjax>#sf(log[F^-]=-0.16)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf([F^-]=0.69color(white)(x)"mol/l")#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#sf([HF]=0.69color(white)(x)"mol/l")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#sf(F^-)#</mathjax> is the co - base of <mathjax>#sf(HF)#</mathjax>:</p>
<p><mathjax>#sf(F^(-)+H_2OrightleftharpoonsHF+OH^-)#</mathjax></p>
<p>You need to look up <mathjax>#sf(pK_b)#</mathjax> for the <mathjax>#sf(F^-)#</mathjax> ion which = 10.82.</p>
<p>From the ICE table you get this expression:</p>
<p><mathjax>#sf(pOH=1/2(pK_a-log[F^-])#</mathjax>`</p>
<p><mathjax>#sf(pOH=14-pH=14-8.51=5.49)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(5.49=1/2(10.82-log[F^-])#</mathjax></p>
<p><mathjax>#sf(1/2log[F^-]=-0.08)#</mathjax></p>
<p><mathjax>#sf(log[F^-]=-0.16)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf([F^-]=0.69color(white)(x)"mol/l")#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">If you want to prepare a solution that has #"pH" = 8.51# at #25^@ "C"#, what initial concentration of #"NaF"# would you need to achieve?</h1>
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<div class="markdown"><p><mathjax>#sf([HF]=0.69color(white)(x)"mol/l")#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#sf(F^-)#</mathjax> is the co - base of <mathjax>#sf(HF)#</mathjax>:</p>
<p><mathjax>#sf(F^(-)+H_2OrightleftharpoonsHF+OH^-)#</mathjax></p>
<p>You need to look up <mathjax>#sf(pK_b)#</mathjax> for the <mathjax>#sf(F^-)#</mathjax> ion which = 10.82.</p>
<p>From the ICE table you get this expression:</p>
<p><mathjax>#sf(pOH=1/2(pK_a-log[F^-])#</mathjax>`</p>
<p><mathjax>#sf(pOH=14-pH=14-8.51=5.49)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(5.49=1/2(10.82-log[F^-])#</mathjax></p>
<p><mathjax>#sf(1/2log[F^-]=-0.08)#</mathjax></p>
<p><mathjax>#sf(log[F^-]=-0.16)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf([F^-]=0.69color(white)(x)"mol/l")#</mathjax></p></div>
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<div class="markdown"><p>I am also getting approximately <mathjax>#"0.70 M"#</mathjax>. Specifically I get <mathjax>#"0.708 M"#</mathjax>. You might want to try doing the problem backwards and work from the solved concentration to check that the <mathjax>#"pH"#</mathjax> still turns out to be close to <mathjax>#8.51#</mathjax>.</p>
<hr/>
<p>As I mentioned earlier, you were given the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> which allows you to find the equilibrium concentration of <mathjax>#"H"^(+)#</mathjax>. Note the main equilibrium in this problem is:</p>
<blockquote>
<blockquote>
<p><mathjax>#"F"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HF"(aq) + "OH"^(-)(aq)#</mathjax></p>
</blockquote>
<p><mathjax>#"I"" "c" M"" "" "" "-" "" "" ""0 M"" "" ""0 M"#</mathjax> <br/>
<mathjax>#"C"" "-x " ""M"" "-" "" "+x " M"" "+x" M"#</mathjax> <br/>
<mathjax>#"E"" "(c - x) "M"" "-" "" "" "x" M"" "" "x" M"#</mathjax> </p>
</blockquote>
<p>where we let <mathjax>#c#</mathjax> be the initial concentration of <mathjax>#"F"^(-)#</mathjax>, which is equal to that of <mathjax>#"NaF"#</mathjax> by their 1:1 <a href="https://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a>.</p>
<p>You should have the <mathjax>#K_a#</mathjax> in your book, but I shall use the <mathjax>#"pKa"#</mathjax> of <mathjax>#3.17#</mathjax> for <mathjax>#"HF"#</mathjax>, which corresponds to a <mathjax>#"pKb"#</mathjax> of <mathjax>#10.83#</mathjax> for <mathjax>#"F"^(-)#</mathjax>, since <mathjax>#"pKw" = "pKa" + "pKb" = 14#</mathjax> at <mathjax>#25^@ "C"#</mathjax>.</p>
<p>Therefore, <mathjax>#K_b = 10^(-"pKb") = 10^(-10.83) = 1.479xx10^(-11)#</mathjax>. So, the <mathjax>#K_a#</mathjax> in your book should be around <mathjax>#6.671xx10^(-4)#</mathjax>. Now, using the definition of <mathjax>#K_b#</mathjax>:</p>
<blockquote>
<p><mathjax>#K_b = (["HF"]["OH"^(-)])/(["F"^(-)]) = x^2/(c - x)#</mathjax></p>
</blockquote>
<p>Since we were given that <mathjax>#"pH" = 8.51#</mathjax>, we know that <mathjax>#"pOH" = 14 - "pH" = 5.49#</mathjax> at <mathjax>#25^@ "C"#</mathjax>. Therefore:</p>
<blockquote>
<p><mathjax>#["OH"^(-)]#</mathjax> at equilibrium is <mathjax>#10^(-"pOH") = 3.236xx10^(-6)#</mathjax> <mathjax>#"M" = x#</mathjax>. </p>
</blockquote>
<p>This means we already solved for <mathjax>#x#</mathjax>. So, we can plug it back into <mathjax>#K_b#</mathjax>:</p>
<blockquote>
<p><mathjax>#K_b = (3.236xx10^(-6) "M")^2/(c - 3.236xx10^(-6) "M") = 1.479xx10^(-11)#</mathjax></p>
</blockquote>
<p>By the small <mathjax>#K_b#</mathjax> approximation, since <mathjax>#K_b "<<" 10^(-5)#</mathjax>, we can say that the true answer allows us to neglect the <mathjax>#-3.236xx10^(-6) "M"#</mathjax> term that is being subtracted in the denominator with minimal error.</p>
<p>As a result, our evaluation simplifies to (where <mathjax>#K_b#</mathjax> implicitly has units of <mathjax>#"M"#</mathjax> since it had units of <mathjax>#"M"^2/"M"#</mathjax> from its definition):</p>
<blockquote>
<p><mathjax>#(1.479xx10^(-11) "M")c = (3.326xx10^(-6) "M")^2#</mathjax></p>
<p><mathjax>#=> color(blue)(c = ["F"^(-)] = ["NaF"] = "0.708 M")#</mathjax></p>
</blockquote></div>
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</article> | If you want to prepare a solution that has #"pH" = 8.51# at #25^@ "C"#, what initial concentration of #"NaF"# would you need to achieve? | null |
2,969 | a8735ffa-6ddd-11ea-8d51-ccda262736ce | https://socratic.org/questions/what-is-the-standard-emf-of-a-galvanic-cell-made-of-a-sn-electrode-in-a-1-0-m-sn | +0.47 V | start physical_unit 6 8 standard_emf v qc_end physical_unit 18 19 16 17 molarity qc_end physical_unit 28 29 16 17 molarity qc_end c_other OTHER qc_end physical_unit 7 8 31 32 temperature qc_end end | [{"type":"physical unit","value":"Standard emf [OF] a galvanic cell [IN] V"}] | [{"type":"physical unit","value":"+0.47 V"}] | [{"type":"physical unit","value":"Molarity [OF] Sn(NO3)2 solution [=] \\pu{1.0 M}"},{"type":"physical unit","value":"Molarity [OF] Cu(NO3)2 solution [=] \\pu{1.0 M}"},{"type":"other","value":"a galvanic cell made of a Sn electrode in Sn(NO3)2 solution and a Cu electrode in Cu(NO3)2 solution."},{"type":"physical unit","value":"Temperature [OF] galvanic cell [=] \\pu{25 ℃}"}] | <h1 class="questionTitle" itemprop="name">What is the standard emf of a galvanic cell made of a Sn electrode in a 1.0 M #Sn(NO_3)_2# solution and a Cu electrode in a 1.0 M #Cu(NO_3)_2# solution at 25°C? </h1> | null | +0.47 V | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You need to look up standard electrode potentials and list them -ve to +ve:</p>
<p><mathjax>#" "E^@("V")#</mathjax><br/>
<mathjax>#stackrel(color(white)(xxxxxxxxxxxxxxxxxx))(color(red)(larr))#</mathjax></p>
<p><mathjax>#Sn_((aq))^(2+)+2erightleftharpoonsSn_((s))" "-0.13#</mathjax></p>
<p><mathjax>#Cu_((aq))^(2+)+2erightleftharpoonsCu_((s))" "+0.34#</mathjax><br/>
<mathjax>#stackrel(color(white)(xxxxxxxxxxxxxxxxxx))(color(blue)(rarr))#</mathjax></p>
<p>You can see that the more +ve 1/2 cell will take in the electrons so the 1/2 cell reactions proceed in the direction indicated by the arrows.</p>
<p>Emf is an experimentally measured quantity and must always have a +ve value.</p>
<p>If you try to measure the emf of a cell and get a -ve reading, it means you have connected the voltmeter to the wrong terminals.</p>
<p>So to get the emf of the cell, always subtract the <strong>least</strong> positive potential from the <strong>most</strong> positive potential <mathjax>#rArr#</mathjax></p>
<p><mathjax>#E_(cell)^@=+0.34-(-0.13)=+0.47"V"#</mathjax></p>
<p>I have adopted the convention which is used in the UK.</p>
<p>I understand other conventions would reverse the sign of the <mathjax>#Sn^(2+)"/"Sn#</mathjax> 1/2 cell then add.</p>
<p>Other conventions I have seen write the 1/2 cells in the other direction and reverse the voltage.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#E_(cell)^@=+0.47"V"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You need to look up standard electrode potentials and list them -ve to +ve:</p>
<p><mathjax>#" "E^@("V")#</mathjax><br/>
<mathjax>#stackrel(color(white)(xxxxxxxxxxxxxxxxxx))(color(red)(larr))#</mathjax></p>
<p><mathjax>#Sn_((aq))^(2+)+2erightleftharpoonsSn_((s))" "-0.13#</mathjax></p>
<p><mathjax>#Cu_((aq))^(2+)+2erightleftharpoonsCu_((s))" "+0.34#</mathjax><br/>
<mathjax>#stackrel(color(white)(xxxxxxxxxxxxxxxxxx))(color(blue)(rarr))#</mathjax></p>
<p>You can see that the more +ve 1/2 cell will take in the electrons so the 1/2 cell reactions proceed in the direction indicated by the arrows.</p>
<p>Emf is an experimentally measured quantity and must always have a +ve value.</p>
<p>If you try to measure the emf of a cell and get a -ve reading, it means you have connected the voltmeter to the wrong terminals.</p>
<p>So to get the emf of the cell, always subtract the <strong>least</strong> positive potential from the <strong>most</strong> positive potential <mathjax>#rArr#</mathjax></p>
<p><mathjax>#E_(cell)^@=+0.34-(-0.13)=+0.47"V"#</mathjax></p>
<p>I have adopted the convention which is used in the UK.</p>
<p>I understand other conventions would reverse the sign of the <mathjax>#Sn^(2+)"/"Sn#</mathjax> 1/2 cell then add.</p>
<p>Other conventions I have seen write the 1/2 cells in the other direction and reverse the voltage.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the standard emf of a galvanic cell made of a Sn electrode in a 1.0 M #Sn(NO_3)_2# solution and a Cu electrode in a 1.0 M #Cu(NO_3)_2# solution at 25°C? </h1>
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<div class="markdown"><p><mathjax>#E_(cell)^@=+0.47"V"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You need to look up standard electrode potentials and list them -ve to +ve:</p>
<p><mathjax>#" "E^@("V")#</mathjax><br/>
<mathjax>#stackrel(color(white)(xxxxxxxxxxxxxxxxxx))(color(red)(larr))#</mathjax></p>
<p><mathjax>#Sn_((aq))^(2+)+2erightleftharpoonsSn_((s))" "-0.13#</mathjax></p>
<p><mathjax>#Cu_((aq))^(2+)+2erightleftharpoonsCu_((s))" "+0.34#</mathjax><br/>
<mathjax>#stackrel(color(white)(xxxxxxxxxxxxxxxxxx))(color(blue)(rarr))#</mathjax></p>
<p>You can see that the more +ve 1/2 cell will take in the electrons so the 1/2 cell reactions proceed in the direction indicated by the arrows.</p>
<p>Emf is an experimentally measured quantity and must always have a +ve value.</p>
<p>If you try to measure the emf of a cell and get a -ve reading, it means you have connected the voltmeter to the wrong terminals.</p>
<p>So to get the emf of the cell, always subtract the <strong>least</strong> positive potential from the <strong>most</strong> positive potential <mathjax>#rArr#</mathjax></p>
<p><mathjax>#E_(cell)^@=+0.34-(-0.13)=+0.47"V"#</mathjax></p>
<p>I have adopted the convention which is used in the UK.</p>
<p>I understand other conventions would reverse the sign of the <mathjax>#Sn^(2+)"/"Sn#</mathjax> 1/2 cell then add.</p>
<p>Other conventions I have seen write the 1/2 cells in the other direction and reverse the voltage.</p></div>
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</article> | What is the standard emf of a galvanic cell made of a Sn electrode in a 1.0 M #Sn(NO_3)_2# solution and a Cu electrode in a 1.0 M #Cu(NO_3)_2# solution at 25°C? | null |
2,970 | ab51b912-6ddd-11ea-87b1-ccda262736ce | https://socratic.org/questions/how-many-grams-of-cacl-2-would-be-required-to-produce-a-3-5-m-solution-with-a-vo | 600 grams | start physical_unit 4 4 mass g qc_end physical_unit 13 13 18 19 volume qc_end physical_unit 4 4 11 12 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] CaCl2 [IN] grams"}] | [{"type":"physical unit","value":"600 grams"}] | [{"type":"physical unit","value":"Volume [OF] CaCl2 solution [=] \\pu{2.0 L}"},{"type":"physical unit","value":"Molarity [OF] CaCl2 solution [=] \\pu{3.5 M}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #CaCl_2# would be required to produce a 3.5 M solution with a volume of 2.0 L?</h1> | null | 600 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"=("moles of solute")/("liters of solution")#</mathjax></p>
<p>A <mathjax>#"3.5 M CaCl"_2"#</mathjax> solution contains <mathjax>#"3.5 moles"#</mathjax> <mathjax>#"CaCl"_2"/L solution"#</mathjax>.</p>
<p>We need to convert moles <mathjax>#"CaCl"_2#</mathjax> to grams. We do this by multiplying moles <mathjax>#"CaCl"_2#</mathjax> by its molar mass <mathjax>#("110.978 g/mol")#</mathjax>. </p>
<p><mathjax>#3.5color(red)cancel(color(black)("mol CaCl"_2))xx(110.978"g CaCl"_2)/(1color(red)cancel(color(black)("mol CaCl"_2)))="300 g CaCl"_2"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p>
<p>Since <mathjax>#"300 g CaCl"_2"#</mathjax> are needed to make <mathjax>#"1 L"#</mathjax> of a <mathjax>#"3.5 M"#</mathjax> solution, <mathjax>#"600 g CaCl"_2#</mathjax> are needed to make <mathjax>#"2 L"#</mathjax> of a <mathjax>#"3.5 M"#</mathjax> solution.</p></div>
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<div class="markdown"><p><mathjax>#"600 g CaCl"_2#</mathjax> would be required to make <mathjax>#"2 L"#</mathjax> of a <mathjax>#"3.5 M"#</mathjax> solution.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Molarity"=("moles of solute")/("liters of solution")#</mathjax></p>
<p>A <mathjax>#"3.5 M CaCl"_2"#</mathjax> solution contains <mathjax>#"3.5 moles"#</mathjax> <mathjax>#"CaCl"_2"/L solution"#</mathjax>.</p>
<p>We need to convert moles <mathjax>#"CaCl"_2#</mathjax> to grams. We do this by multiplying moles <mathjax>#"CaCl"_2#</mathjax> by its molar mass <mathjax>#("110.978 g/mol")#</mathjax>. </p>
<p><mathjax>#3.5color(red)cancel(color(black)("mol CaCl"_2))xx(110.978"g CaCl"_2)/(1color(red)cancel(color(black)("mol CaCl"_2)))="300 g CaCl"_2"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p>
<p>Since <mathjax>#"300 g CaCl"_2"#</mathjax> are needed to make <mathjax>#"1 L"#</mathjax> of a <mathjax>#"3.5 M"#</mathjax> solution, <mathjax>#"600 g CaCl"_2#</mathjax> are needed to make <mathjax>#"2 L"#</mathjax> of a <mathjax>#"3.5 M"#</mathjax> solution.</p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of #CaCl_2# would be required to produce a 3.5 M solution with a volume of 2.0 L?</h1>
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<div class="markdown"><p><mathjax>#"600 g CaCl"_2#</mathjax> would be required to make <mathjax>#"2 L"#</mathjax> of a <mathjax>#"3.5 M"#</mathjax> solution.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Molarity"=("moles of solute")/("liters of solution")#</mathjax></p>
<p>A <mathjax>#"3.5 M CaCl"_2"#</mathjax> solution contains <mathjax>#"3.5 moles"#</mathjax> <mathjax>#"CaCl"_2"/L solution"#</mathjax>.</p>
<p>We need to convert moles <mathjax>#"CaCl"_2#</mathjax> to grams. We do this by multiplying moles <mathjax>#"CaCl"_2#</mathjax> by its molar mass <mathjax>#("110.978 g/mol")#</mathjax>. </p>
<p><mathjax>#3.5color(red)cancel(color(black)("mol CaCl"_2))xx(110.978"g CaCl"_2)/(1color(red)cancel(color(black)("mol CaCl"_2)))="300 g CaCl"_2"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p>
<p>Since <mathjax>#"300 g CaCl"_2"#</mathjax> are needed to make <mathjax>#"1 L"#</mathjax> of a <mathjax>#"3.5 M"#</mathjax> solution, <mathjax>#"600 g CaCl"_2#</mathjax> are needed to make <mathjax>#"2 L"#</mathjax> of a <mathjax>#"3.5 M"#</mathjax> solution.</p></div>
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</article> | How many grams of #CaCl_2# would be required to produce a 3.5 M solution with a volume of 2.0 L? | null |
2,971 | ad07ef07-6ddd-11ea-87ad-ccda262736ce | https://socratic.org/questions/a-compound-has-the-empirical-formula-of-ch20-and-a-molar-mass-of-60-g-mole-what- | C2H4O2 | start chemical_formula qc_end physical_unit 1 1 13 14 molar_mass qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"C2H4O2"}] | [{"type":"physical unit","value":"Molar mass [OF] the compound [=] \\pu{60 g/mole}"},{"type":"other","value":"A compound has the empirical formula of CH2O."}] | <h1 class="questionTitle" itemprop="name">A compound has the empirical formula of #"CH"_2"O"# and a molar mass of 60 g/mole. What is the molecular formula? </h1> | null | C2H4O2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Empirical formula <mathjax>#"= CH"_2"O"#</mathjax><br/>
Empirical formula mass <mathjax>#"= 12g/mol + 2 g/mol + 16 g/mol = 30 g/mol"#</mathjax></p>
<blockquote>
<p><mathjax>#"Molecular formula mass"/"Empirical formula mass" = "60 g/mol"/"30 g/mol" = 2#</mathjax></p>
</blockquote>
<p>Molecular formula <mathjax>#= 2 × "CH"_2"O" = "C"_2"H"_4"O"_2#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"C"_2"H"_4"O"_2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Empirical formula <mathjax>#"= CH"_2"O"#</mathjax><br/>
Empirical formula mass <mathjax>#"= 12g/mol + 2 g/mol + 16 g/mol = 30 g/mol"#</mathjax></p>
<blockquote>
<p><mathjax>#"Molecular formula mass"/"Empirical formula mass" = "60 g/mol"/"30 g/mol" = 2#</mathjax></p>
</blockquote>
<p>Molecular formula <mathjax>#= 2 × "CH"_2"O" = "C"_2"H"_4"O"_2#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">A compound has the empirical formula of #"CH"_2"O"# and a molar mass of 60 g/mole. What is the molecular formula? </h1>
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<div class="markdown"><p><mathjax>#"C"_2"H"_4"O"_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Empirical formula <mathjax>#"= CH"_2"O"#</mathjax><br/>
Empirical formula mass <mathjax>#"= 12g/mol + 2 g/mol + 16 g/mol = 30 g/mol"#</mathjax></p>
<blockquote>
<p><mathjax>#"Molecular formula mass"/"Empirical formula mass" = "60 g/mol"/"30 g/mol" = 2#</mathjax></p>
</blockquote>
<p>Molecular formula <mathjax>#= 2 × "CH"_2"O" = "C"_2"H"_4"O"_2#</mathjax></p></div>
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</article> | A compound has the empirical formula of #"CH"_2"O"# and a molar mass of 60 g/mole. What is the molecular formula? | null |
2,972 | a8c5b666-6ddd-11ea-b010-ccda262736ce | https://socratic.org/questions/what-mass-of-solution-containing-6-50-sodium-sulfate-by-mass-contains-1-50-g-sod | 23.08 g | start physical_unit 3 3 mass g qc_end physical_unit 6 7 11 12 mass qc_end end | [{"type":"physical unit","value":"Mass [OF] solution [IN] g"}] | [{"type":"physical unit","value":"23.08 g"}] | [{"type":"physical unit","value":"Percent by mass [OF] sodium sulfate in the solution [=] \\pu{6.50%}"},{"type":"physical unit","value":"Mass [OF] sodium sulfate [=] \\pu{1.50 g}"}] | <h1 class="questionTitle" itemprop="name">What mass of solution containing 6.50% sodium sulfate, by mass contains 1.50 g sodium sulfate? </h1> | null | 23.08 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Mass of solute"/"Mass of solution"xx100%=6.5%#</mathjax>.</p>
<p>Thus <mathjax># "mass of solute"/(0.065)="mass of solution"#</mathjax> <mathjax>#=??g#</mathjax>.</p></div>
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<div class="markdown"><p>About <mathjax>#23#</mathjax> <mathjax>#g#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Mass of solute"/"Mass of solution"xx100%=6.5%#</mathjax>.</p>
<p>Thus <mathjax># "mass of solute"/(0.065)="mass of solution"#</mathjax> <mathjax>#=??g#</mathjax>.</p></div>
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anor277
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<span class="dateCreated" datetime="2016-07-17T07:00:46" itemprop="dateCreated">
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<div class="markdown"><p>About <mathjax>#23#</mathjax> <mathjax>#g#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Mass of solute"/"Mass of solution"xx100%=6.5%#</mathjax>.</p>
<p>Thus <mathjax># "mass of solute"/(0.065)="mass of solution"#</mathjax> <mathjax>#=??g#</mathjax>.</p></div>
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</article> | What mass of solution containing 6.50% sodium sulfate, by mass contains 1.50 g sodium sulfate? | null |
2,973 | a99c34a4-6ddd-11ea-a84c-ccda262736ce | https://socratic.org/questions/an-ammonia-solution-is-6-00-m-and-the-density-is-0-950-g-ml-what-is-the-mass-mas-2 | 10.76% | start physical_unit 2 2 mass_percent none qc_end physical_unit 1 2 4 5 molarity qc_end physical_unit 1 2 10 11 density qc_end physical_unit 19 19 20 21 molar_mass qc_end end | [{"type":"physical unit","value":"Mass/mass percent concentration [OF] NH3 in the solution"}] | [{"type":"physical unit","value":"10.76%"}] | [{"type":"physical unit","value":"Molarity [OF] ammonia solution [=] \\pu{6.00 M}"},{"type":"physical unit","value":"Density [OF] ammonia solution [=] \\pu{0.950 g/mL}"},{"type":"physical unit","value":"Molar mass [OF] NH3 [=] \\pu{17.03 g/mol}"}] | <h1 class="questionTitle" itemprop="name">An ammonia solution is 6.00 M and the density is 0.950 g mL. What is the mass/mass percent concentration of #NH_3# (17.03 g/mol)? </h1> | null | 10.76% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to pick a sample <em>volume</em> of this solution and calculate</p>
<blockquote>
<ul>
<li><em>the <strong>total mass</strong> of this sample</em></li>
<li><em>the <strong>mass of ammonia</strong> it contains</em></li>
</ul>
</blockquote>
<p>To make the calculations easier, let's pick a <mathjax>#"1000 mL"#</mathjax> sample. This solution is said to have a <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of <mathjax>#"0.950 g mL"^(-1)#</mathjax>, which means that <strong>every milliliter</strong> of solution has a mass of <mathjax>#"0.950 g"#</mathjax>. </p>
<p>The sample we've picked will have a mass of </p>
<blockquote>
<p><mathjax>#1000 color(red)(cancel(color(black)("mL"))) * "0.950 g"/(1color(red)(cancel(color(black)("mL")))) = "950 g"#</mathjax></p>
</blockquote>
<p>Now, you know that this solution has a <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of <mathjax>#"6.00 M"#</mathjax>, which basically means that <strong>every liter</strong>, which is the equivalent of <mathjax>#"1000 mL"#</mathjax>, will contain <mathjax>#6.00#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>Since we've picked a sample of <mathjax>#"1000 mL"#</mathjax>, you can say that it will contain <mathjax>#6.00#</mathjax> <strong>moles</strong> of ammonia. To convert this to <em>grams</em>, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#6.00 color(red)(cancel(color(black)("moles NH"_3))) * "17.03 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = "102.18 g"#</mathjax></p>
</blockquote>
<p>Now, the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"%m/m"#</mathjax>, tells you how many grams of ammonia, the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>, you get <strong>per</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>You already know how many grams of ammonia you have in <mathjax>#"950 g"#</mathjax> of solution, so use this known composition to figure out how many grams of ammonia you'd get in <mathjax>#"100 g"#</mathjax> of solution</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g solution"))) * "102.18 g NH"_3/(950color(red)(cancel(color(black)("g solution")))) = "10.76 g NH"_3#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the solution's mass by mass percent concentration is</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("% m/m" = 10.8%"NH"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#"10.8 % NH"_3#</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 pick a sample <em>volume</em> of this solution and calculate</p>
<blockquote>
<ul>
<li><em>the <strong>total mass</strong> of this sample</em></li>
<li><em>the <strong>mass of ammonia</strong> it contains</em></li>
</ul>
</blockquote>
<p>To make the calculations easier, let's pick a <mathjax>#"1000 mL"#</mathjax> sample. This solution is said to have a <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of <mathjax>#"0.950 g mL"^(-1)#</mathjax>, which means that <strong>every milliliter</strong> of solution has a mass of <mathjax>#"0.950 g"#</mathjax>. </p>
<p>The sample we've picked will have a mass of </p>
<blockquote>
<p><mathjax>#1000 color(red)(cancel(color(black)("mL"))) * "0.950 g"/(1color(red)(cancel(color(black)("mL")))) = "950 g"#</mathjax></p>
</blockquote>
<p>Now, you know that this solution has a <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of <mathjax>#"6.00 M"#</mathjax>, which basically means that <strong>every liter</strong>, which is the equivalent of <mathjax>#"1000 mL"#</mathjax>, will contain <mathjax>#6.00#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>Since we've picked a sample of <mathjax>#"1000 mL"#</mathjax>, you can say that it will contain <mathjax>#6.00#</mathjax> <strong>moles</strong> of ammonia. To convert this to <em>grams</em>, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#6.00 color(red)(cancel(color(black)("moles NH"_3))) * "17.03 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = "102.18 g"#</mathjax></p>
</blockquote>
<p>Now, the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"%m/m"#</mathjax>, tells you how many grams of ammonia, the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>, you get <strong>per</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>You already know how many grams of ammonia you have in <mathjax>#"950 g"#</mathjax> of solution, so use this known composition to figure out how many grams of ammonia you'd get in <mathjax>#"100 g"#</mathjax> of solution</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g solution"))) * "102.18 g NH"_3/(950color(red)(cancel(color(black)("g solution")))) = "10.76 g NH"_3#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the solution's mass by mass percent concentration is</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("% m/m" = 10.8%"NH"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">An ammonia solution is 6.00 M and the density is 0.950 g mL. What is the mass/mass percent concentration of #NH_3# (17.03 g/mol)? </h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-10-01T00:04:11" itemprop="dateCreated">
Oct 1, 2016
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<div class="markdown"><p><mathjax>#"10.8 % NH"_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you need to pick a sample <em>volume</em> of this solution and calculate</p>
<blockquote>
<ul>
<li><em>the <strong>total mass</strong> of this sample</em></li>
<li><em>the <strong>mass of ammonia</strong> it contains</em></li>
</ul>
</blockquote>
<p>To make the calculations easier, let's pick a <mathjax>#"1000 mL"#</mathjax> sample. This solution is said to have a <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of <mathjax>#"0.950 g mL"^(-1)#</mathjax>, which means that <strong>every milliliter</strong> of solution has a mass of <mathjax>#"0.950 g"#</mathjax>. </p>
<p>The sample we've picked will have a mass of </p>
<blockquote>
<p><mathjax>#1000 color(red)(cancel(color(black)("mL"))) * "0.950 g"/(1color(red)(cancel(color(black)("mL")))) = "950 g"#</mathjax></p>
</blockquote>
<p>Now, you know that this solution has a <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of <mathjax>#"6.00 M"#</mathjax>, which basically means that <strong>every liter</strong>, which is the equivalent of <mathjax>#"1000 mL"#</mathjax>, will contain <mathjax>#6.00#</mathjax> <strong>moles</strong> of ammonia. </p>
<p>Since we've picked a sample of <mathjax>#"1000 mL"#</mathjax>, you can say that it will contain <mathjax>#6.00#</mathjax> <strong>moles</strong> of ammonia. To convert this to <em>grams</em>, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#6.00 color(red)(cancel(color(black)("moles NH"_3))) * "17.03 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = "102.18 g"#</mathjax></p>
</blockquote>
<p>Now, the solution's <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> by mass</strong>, <mathjax>#"%m/m"#</mathjax>, tells you how many grams of ammonia, the <em><a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>, you get <strong>per</strong> <mathjax>#"100 g"#</mathjax> <strong>of solution</strong>. </p>
<p>You already know how many grams of ammonia you have in <mathjax>#"950 g"#</mathjax> of solution, so use this known composition to figure out how many grams of ammonia you'd get in <mathjax>#"100 g"#</mathjax> of solution</p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g solution"))) * "102.18 g NH"_3/(950color(red)(cancel(color(black)("g solution")))) = "10.76 g NH"_3#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the solution's mass by mass percent concentration is</p>
<blockquote>
<p><mathjax>#color(green)(bar(ul(|color(white)(a/a)color(black)("% m/m" = 10.8%"NH"_3)color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | An ammonia solution is 6.00 M and the density is 0.950 g mL. What is the mass/mass percent concentration of #NH_3# (17.03 g/mol)? | null |
2,974 | abcb10e8-6ddd-11ea-8eae-ccda262736ce | https://socratic.org/questions/5665f25011ef6b4355e84a23 | 4 Cr2O7^2- + 26 H+ + 3 S2O3^2- -> 8 Cr^3+ + 6 SO4^2- + 13 H2O(l) | start chemical_equation qc_end chemical_equation 10 10 qc_end chemical_equation 14 14 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the oxidation"}] | [{"type":"chemical equation","value":"4 Cr2O7^2- + 26 H+ + 3 S2O3^2- -> 8 Cr^3+ + 6 SO4^2- + 13 H2O(l)"}] | [{"type":"chemical equation","value":"S2O3^2-"},{"type":"chemical equation","value":"Cr2O7^2-"}] | <h1 class="questionTitle" itemprop="name">How do we quantitatively represent the oxidation of sodium thiosulfate, #S_2O_3^(2-)#, by dichromate ion, #Cr_2O_7^(2-)#? </h1> | null | 4 Cr2O7^2- + 26 H+ + 3 S2O3^2- -> 8 Cr^3+ + 6 SO4^2- + 13 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Cr(VI)#</mathjax> is reduced to <mathjax>#Cr(III)#</mathjax>:</p>
<p><mathjax>#Cr_2O_7^(2-) +14H^(+)+ 6e^(-)rarr 2Cr^(3+) + 7H_2O(l)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Thiolsulfate is oxidized to sulfate:</p>
<p><mathjax>#S_2O_3^(2-) + 5H_2O(l) rarr 2SO_4^(2-) + 10H^(+) + 8e^-#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Both equations (I think) are balanced with respect to mass and charge (as they must be!). </p>
<p>So I cross multiply to give the overall redox equation (I wish to remove the electrons!) and cancel common reagents:</p>
<p><mathjax>#4xx(i)+3xx(ii)#</mathjax>:</p>
<p><mathjax>#4Cr_2O_7^(2-) +26H^(+)+ 3S_2O_3^(2-) rarr 8Cr^(3+) + 6SO_4^(2-) + 13H_2O(l)#</mathjax></p>
<p>So at the end of your titration, you should have <mathjax>#Cr^(3+)#</mathjax>, and sulfate ions; acidium ions will be in excess with these reactions. What should signal the endpoint in this redox reaction?</p>
<p>Note that in thiosulfate, I have always found it useful to regard the terminal sulfur as the precise analogue of oxygen, which has the <mathjax>#-II#</mathjax> oxidation state, whereas the central sulfur has a <mathjax>#VI^+#</mathjax> oxidation state, as in sulfate. The average sulfur oxidation state in thiosulfate is still <mathjax>#II^+#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>This is a somewhat complicated redox equation. You should end with <mathjax>#Cr^(3+)#</mathjax> and sulfate ions. </p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Cr(VI)#</mathjax> is reduced to <mathjax>#Cr(III)#</mathjax>:</p>
<p><mathjax>#Cr_2O_7^(2-) +14H^(+)+ 6e^(-)rarr 2Cr^(3+) + 7H_2O(l)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Thiolsulfate is oxidized to sulfate:</p>
<p><mathjax>#S_2O_3^(2-) + 5H_2O(l) rarr 2SO_4^(2-) + 10H^(+) + 8e^-#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Both equations (I think) are balanced with respect to mass and charge (as they must be!). </p>
<p>So I cross multiply to give the overall redox equation (I wish to remove the electrons!) and cancel common reagents:</p>
<p><mathjax>#4xx(i)+3xx(ii)#</mathjax>:</p>
<p><mathjax>#4Cr_2O_7^(2-) +26H^(+)+ 3S_2O_3^(2-) rarr 8Cr^(3+) + 6SO_4^(2-) + 13H_2O(l)#</mathjax></p>
<p>So at the end of your titration, you should have <mathjax>#Cr^(3+)#</mathjax>, and sulfate ions; acidium ions will be in excess with these reactions. What should signal the endpoint in this redox reaction?</p>
<p>Note that in thiosulfate, I have always found it useful to regard the terminal sulfur as the precise analogue of oxygen, which has the <mathjax>#-II#</mathjax> oxidation state, whereas the central sulfur has a <mathjax>#VI^+#</mathjax> oxidation state, as in sulfate. The average sulfur oxidation state in thiosulfate is still <mathjax>#II^+#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">How do we quantitatively represent the oxidation of sodium thiosulfate, #S_2O_3^(2-)#, by dichromate ion, #Cr_2O_7^(2-)#? </h1>
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anor277
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<div class="markdown"><p>This is a somewhat complicated redox equation. You should end with <mathjax>#Cr^(3+)#</mathjax> and sulfate ions. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Cr(VI)#</mathjax> is reduced to <mathjax>#Cr(III)#</mathjax>:</p>
<p><mathjax>#Cr_2O_7^(2-) +14H^(+)+ 6e^(-)rarr 2Cr^(3+) + 7H_2O(l)#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>Thiolsulfate is oxidized to sulfate:</p>
<p><mathjax>#S_2O_3^(2-) + 5H_2O(l) rarr 2SO_4^(2-) + 10H^(+) + 8e^-#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>Both equations (I think) are balanced with respect to mass and charge (as they must be!). </p>
<p>So I cross multiply to give the overall redox equation (I wish to remove the electrons!) and cancel common reagents:</p>
<p><mathjax>#4xx(i)+3xx(ii)#</mathjax>:</p>
<p><mathjax>#4Cr_2O_7^(2-) +26H^(+)+ 3S_2O_3^(2-) rarr 8Cr^(3+) + 6SO_4^(2-) + 13H_2O(l)#</mathjax></p>
<p>So at the end of your titration, you should have <mathjax>#Cr^(3+)#</mathjax>, and sulfate ions; acidium ions will be in excess with these reactions. What should signal the endpoint in this redox reaction?</p>
<p>Note that in thiosulfate, I have always found it useful to regard the terminal sulfur as the precise analogue of oxygen, which has the <mathjax>#-II#</mathjax> oxidation state, whereas the central sulfur has a <mathjax>#VI^+#</mathjax> oxidation state, as in sulfate. The average sulfur oxidation state in thiosulfate is still <mathjax>#II^+#</mathjax>.</p></div>
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</article> | How do we quantitatively represent the oxidation of sodium thiosulfate, #S_2O_3^(2-)#, by dichromate ion, #Cr_2O_7^(2-)#? | null |
2,975 | ac872040-6ddd-11ea-a0cd-ccda262736ce | https://socratic.org/questions/how-would-you-balance-na-s-cl2-nacl-s | 2 Na(s) + Cl2 -> 2 NaCl(s) | 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 Na(s) + Cl2 -> 2 NaCl(s)"}] | [{"type":"chemical equation","value":"Na(s) + Cl2 -> NaCl(s)"}] | <h1 class="questionTitle" itemprop="name">How would you balance Na(s) + Cl2 → NaCl(s)?</h1> | null | 2 Na(s) + Cl2 -> 2 NaCl(s) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>How to Balance the Equation <mathjax>#"Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"NaCl"#</mathjax>.</strong></p>
<p>Since there are two chlorine atoms on the left side, there needs to be two chlorine atoms on the right side. So I put a coefficient of two in front of the formula <mathjax>#"NaCl"#</mathjax>.</p>
<p><mathjax>#"Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NaCl"#</mathjax></p>
<p>Now there are two sodium and two chlorine atoms on the right side. However, the left side has only one sodium atom. So I put a coefficient of two in front of the formula <mathjax>#"Na"#</mathjax>.</p>
<p><mathjax>#"2Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NaCl(s)"#</mathjax></p>
<p>Now there are two sodium and two chlorine atoms on both sides of the equation and it is balanced.</p>
<p>Notice that I only changed the amount of each substance by changing the coefficients. I did not change the formulas.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The balanced equation is <mathjax>#"2Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NaCl(s)"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>How to Balance the Equation <mathjax>#"Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"NaCl"#</mathjax>.</strong></p>
<p>Since there are two chlorine atoms on the left side, there needs to be two chlorine atoms on the right side. So I put a coefficient of two in front of the formula <mathjax>#"NaCl"#</mathjax>.</p>
<p><mathjax>#"Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NaCl"#</mathjax></p>
<p>Now there are two sodium and two chlorine atoms on the right side. However, the left side has only one sodium atom. So I put a coefficient of two in front of the formula <mathjax>#"Na"#</mathjax>.</p>
<p><mathjax>#"2Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NaCl(s)"#</mathjax></p>
<p>Now there are two sodium and two chlorine atoms on both sides of the equation and it is balanced.</p>
<p>Notice that I only changed the amount of each substance by changing the coefficients. I did not change the formulas.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance Na(s) + Cl2 → NaCl(s)?</h1>
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<div class="markdown"><p>The balanced equation is <mathjax>#"2Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NaCl(s)"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>How to Balance the Equation <mathjax>#"Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"NaCl"#</mathjax>.</strong></p>
<p>Since there are two chlorine atoms on the left side, there needs to be two chlorine atoms on the right side. So I put a coefficient of two in front of the formula <mathjax>#"NaCl"#</mathjax>.</p>
<p><mathjax>#"Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NaCl"#</mathjax></p>
<p>Now there are two sodium and two chlorine atoms on the right side. However, the left side has only one sodium atom. So I put a coefficient of two in front of the formula <mathjax>#"Na"#</mathjax>.</p>
<p><mathjax>#"2Na(s)"+"Cl"_2("g")"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NaCl(s)"#</mathjax></p>
<p>Now there are two sodium and two chlorine atoms on both sides of the equation and it is balanced.</p>
<p>Notice that I only changed the amount of each substance by changing the coefficients. I did not change the formulas.</p></div>
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</article> | How would you balance Na(s) + Cl2 → NaCl(s)? | null |
2,976 | ac4523a8-6ddd-11ea-a884-ccda262736ce | https://socratic.org/questions/56d9cb8811ef6b7783c29316 | Pb(NO3)2(aq) + Na2CO3(s) -> 2 NaNO3(aq) + PbCO3(s) | start chemical_equation qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"Pb(NO3)2(aq) + Na2CO3(s) -> 2 NaNO3(aq) + PbCO3(s)"}] | [{"type":"substance name","value":"Lead(II) nitrate aqueous solution"},{"type":"substance name","value":"Sodium carbonate solution"}] | <h1 class="questionTitle" itemprop="name">What occurs when an aqueous solution of #"lead(II) nitrate"# is mixed with a solution of #"sodium carbonate"#?</h1> | null | Pb(NO3)2(aq) + Na2CO3(s) -> 2 NaNO3(aq) + PbCO3(s) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a partner exchange or metathesis reaction. You know that solubilities follow a hierarchy. The salts of the alkali metals are SOLUBLE as are nitrate salts; carbonate salts are generally insoluble. How do you know? You commit the solubility tables to memory. Lead salts are generally insoluble (save for nitrates!). So what would occur in this reaction:</p>
<p><mathjax>#Pb(NO_3)_2(aq) + Na_2SO_4(aq) rarr ??#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#Pb(NO_3)_2(aq) + Na_2CO_3(s) rarr 2NaNO_3(aq) + PbCO_3(s) darr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>This is a partner exchange or metathesis reaction. You know that solubilities follow a hierarchy. The salts of the alkali metals are SOLUBLE as are nitrate salts; carbonate salts are generally insoluble. How do you know? You commit the solubility tables to memory. Lead salts are generally insoluble (save for nitrates!). So what would occur in this reaction:</p>
<p><mathjax>#Pb(NO_3)_2(aq) + Na_2SO_4(aq) rarr ??#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What occurs when an aqueous solution of #"lead(II) nitrate"# is mixed with a solution of #"sodium carbonate"#?</h1>
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anor277
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<div class="markdown"><p><mathjax>#Pb(NO_3)_2(aq) + Na_2CO_3(s) rarr 2NaNO_3(aq) + PbCO_3(s) darr#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>This is a partner exchange or metathesis reaction. You know that solubilities follow a hierarchy. The salts of the alkali metals are SOLUBLE as are nitrate salts; carbonate salts are generally insoluble. How do you know? You commit the solubility tables to memory. Lead salts are generally insoluble (save for nitrates!). So what would occur in this reaction:</p>
<p><mathjax>#Pb(NO_3)_2(aq) + Na_2SO_4(aq) rarr ??#</mathjax></p></div>
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</article> | What occurs when an aqueous solution of #"lead(II) nitrate"# is mixed with a solution of #"sodium carbonate"#? | null |
2,977 | aad476e7-6ddd-11ea-87bc-ccda262736ce | https://socratic.org/questions/what-volume-will-a-300-0-ml-sample-of-a-gas-at-stp-occupy-when-the-pressure-is-d | 150.00 mL | start physical_unit 9 9 volume ml qc_end physical_unit 9 9 4 5 volume qc_end c_other STP qc_end c_other OTHER qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] mL"}] | [{"type":"physical unit","value":"150.00 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas sample [=] \\pu{300.0 mL}"},{"type":"other","value":"STP"},{"type":"other","value":"The pressure is doubled."},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">What volume will a 300.0 mL sample of a gas at STP occupy when the pressure is doubled at constant temperature? </h1> | null | 150.00 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When <em>number of moles</em> and <em>temperature</em> are kept <strong>constant</strong>, pressure and volume have an <strong>inverse relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>. </p>
<p>In simple terms, when pressure <strong>increases</strong>, volume <em>decreases</em>, and when pressure <strong>decreases</strong>, volume <em>Increases</em>. </p>
<p><img alt="https://saylordotorg.github.io/text_introductory-chemistry/s10-03-gas-laws.html" src="https://useruploads.socratic.org/pg9wYkkUTK6710cOskG2_72ab7ad8a1fdc1f4fdadd93d52db9294.jpg"/> </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(P_1V_1 = P_2V_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> - the pressure and volume at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> - the pressure and volume at a final state</p>
<p>So, if pressure <strong>doubles</strong>, what would you expect to happen to the sample's volume? </p>
<p>Well, the only way the product between the pressure and the volume would remain constant is if the volume would be <strong>halved</strong>. </p>
<p>Now, you don't need to know the <em>actual</em> values of the two pressures, all you need to know is the relationship that exists between them. </p>
<blockquote>
<p><mathjax>#P_2 = 2 * P_1#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1#</mathjax></p>
<p><mathjax>#V_2 = color(red)(cancel(color(black)(P_1)))/(2 * color(red)(cancel(color(black)(P_1)))) * V_1 = 1/2 * V_1#</mathjax></p>
<p><mathjax>#V_2 = 1/2 * "300.0 mL" = color(green)("150.0 mL")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"150.0 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When <em>number of moles</em> and <em>temperature</em> are kept <strong>constant</strong>, pressure and volume have an <strong>inverse relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>. </p>
<p>In simple terms, when pressure <strong>increases</strong>, volume <em>decreases</em>, and when pressure <strong>decreases</strong>, volume <em>Increases</em>. </p>
<p><img alt="https://saylordotorg.github.io/text_introductory-chemistry/s10-03-gas-laws.html" src="https://useruploads.socratic.org/pg9wYkkUTK6710cOskG2_72ab7ad8a1fdc1f4fdadd93d52db9294.jpg"/> </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(P_1V_1 = P_2V_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> - the pressure and volume at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> - the pressure and volume at a final state</p>
<p>So, if pressure <strong>doubles</strong>, what would you expect to happen to the sample's volume? </p>
<p>Well, the only way the product between the pressure and the volume would remain constant is if the volume would be <strong>halved</strong>. </p>
<p>Now, you don't need to know the <em>actual</em> values of the two pressures, all you need to know is the relationship that exists between them. </p>
<blockquote>
<p><mathjax>#P_2 = 2 * P_1#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1#</mathjax></p>
<p><mathjax>#V_2 = color(red)(cancel(color(black)(P_1)))/(2 * color(red)(cancel(color(black)(P_1)))) * V_1 = 1/2 * V_1#</mathjax></p>
<p><mathjax>#V_2 = 1/2 * "300.0 mL" = color(green)("150.0 mL")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What volume will a 300.0 mL sample of a gas at STP occupy when the pressure is doubled at constant temperature? </h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-11-25T10:52:04" itemprop="dateCreated">
Nov 25, 2015
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<div class="markdown"><p><mathjax>#"150.0 mL"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>When <em>number of moles</em> and <em>temperature</em> are kept <strong>constant</strong>, pressure and volume have an <strong>inverse relationship</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>. </p>
<p>In simple terms, when pressure <strong>increases</strong>, volume <em>decreases</em>, and when pressure <strong>decreases</strong>, volume <em>Increases</em>. </p>
<p><img alt="https://saylordotorg.github.io/text_introductory-chemistry/s10-03-gas-laws.html" src="https://useruploads.socratic.org/pg9wYkkUTK6710cOskG2_72ab7ad8a1fdc1f4fdadd93d52db9294.jpg"/> </p>
<p>Mathematically, this is written as </p>
<blockquote>
<p><mathjax>#color(blue)(P_1V_1 = P_2V_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax> - the pressure and volume at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax> - the pressure and volume at a final state</p>
<p>So, if pressure <strong>doubles</strong>, what would you expect to happen to the sample's volume? </p>
<p>Well, the only way the product between the pressure and the volume would remain constant is if the volume would be <strong>halved</strong>. </p>
<p>Now, you don't need to know the <em>actual</em> values of the two pressures, all you need to know is the relationship that exists between them. </p>
<blockquote>
<p><mathjax>#P_2 = 2 * P_1#</mathjax></p>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1#</mathjax></p>
<p><mathjax>#V_2 = color(red)(cancel(color(black)(P_1)))/(2 * color(red)(cancel(color(black)(P_1)))) * V_1 = 1/2 * V_1#</mathjax></p>
<p><mathjax>#V_2 = 1/2 * "300.0 mL" = color(green)("150.0 mL")#</mathjax></p>
</blockquote></div>
</div>
</div>
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</article> | What volume will a 300.0 mL sample of a gas at STP occupy when the pressure is doubled at constant temperature? | null |
2,978 | a98f1999-6ddd-11ea-80a8-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-state-of-chromium-in-crcl3 | +3 | start physical_unit 6 6 oxidation_state none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation state [OF] chromium"}] | [{"type":"physical unit","value":"+3"}] | [{"type":"chemical equation","value":"CrCl3"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation state of chromium in CrCl3?</h1> | null | +3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In this compound, <mathjax>#CrCl_3#</mathjax>, it has a <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of <mathjax>#0#</mathjax>. </p>
<p>Chlorine is more electronegative than chromium, and so it will possess its usual state of <mathjax>#-1#</mathjax>. There are three chlorine atoms in total, and so their total charge will be <mathjax>#-1*3=-3#</mathjax>.</p>
<p>Let <mathjax>#x#</mathjax> be the oxidation number of chromium. Then we have,</p>
<p><mathjax>#x-3=0#</mathjax></p>
<p><mathjax>#x=3#</mathjax></p>
<p>And so, chromium would have an oxidation state of <mathjax>#+3#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#+3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In this compound, <mathjax>#CrCl_3#</mathjax>, it has a <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of <mathjax>#0#</mathjax>. </p>
<p>Chlorine is more electronegative than chromium, and so it will possess its usual state of <mathjax>#-1#</mathjax>. There are three chlorine atoms in total, and so their total charge will be <mathjax>#-1*3=-3#</mathjax>.</p>
<p>Let <mathjax>#x#</mathjax> be the oxidation number of chromium. Then we have,</p>
<p><mathjax>#x-3=0#</mathjax></p>
<p><mathjax>#x=3#</mathjax></p>
<p>And so, chromium would have an oxidation state of <mathjax>#+3#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation state of chromium in CrCl3?</h1>
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<div class="markdown"><p><mathjax>#+3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In this compound, <mathjax>#CrCl_3#</mathjax>, it has a <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of <mathjax>#0#</mathjax>. </p>
<p>Chlorine is more electronegative than chromium, and so it will possess its usual state of <mathjax>#-1#</mathjax>. There are three chlorine atoms in total, and so their total charge will be <mathjax>#-1*3=-3#</mathjax>.</p>
<p>Let <mathjax>#x#</mathjax> be the oxidation number of chromium. Then we have,</p>
<p><mathjax>#x-3=0#</mathjax></p>
<p><mathjax>#x=3#</mathjax></p>
<p>And so, chromium would have an oxidation state of <mathjax>#+3#</mathjax>.</p></div>
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</article> | What is the oxidation state of chromium in CrCl3? | null |
2,979 | ac3abee2-6ddd-11ea-abb8-ccda262736ce | https://socratic.org/questions/in-the-neutralization-of-sulfuric-acid-with-sodium-hydroxide-the-theoretical-yie | 72.00% | start physical_unit 36 37 percent_yield none qc_end physical_unit 4 5 13 14 mass qc_end physical_unit 36 37 19 20 theoretical_yield qc_end physical_unit 36 37 28 29 actual_yield qc_end substance 7 8 qc_end end | [{"type":"physical unit","value":"Percentage yield [OF] this synthesis"}] | [{"type":"physical unit","value":"72.00%"}] | [{"type":"physical unit","value":"Mass [OF] sulfuric acid [=] \\pu{6.9 g}"},{"type":"physical unit","value":"Theoretical yield [OF] this synthesis [=] \\pu{10 g}"},{"type":"physical unit","value":"Actual yield [OF] this synthesis [=] \\pu{7.2 g}"},{"type":"substance name","value":"Sodium hydroxide"}] | <h1 class="questionTitle" itemprop="name">In the neutralization of sulfuric acid with sodium hydroxide, the theoretical yield from 6.9g of sulfuric acid is 10g. In a synthesis, the actual yield is 7.2g. What is the percentage yield for this synthesis? </h1> | null | 72.00% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus for this example <mathjax>#%"Yield"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.2*g)/(10*g)xx100%#</mathjax></p>
<p>And we don't even need a calculator.........</p>
<p>Most of the time, we write a stoichiometric equation, that shows the molar equivalence between reactant and product, so that we can work out the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> etc.......</p>
<p><mathjax>#H_2SO_4(aq) + 2NaOH(aq) rarr Na_2SO_4(aq) + 2H_2O#</mathjax>.</p>
<p>At any rate, you have not really quoted starting conditions here. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#%"Yield"="Mass recovered"/"Mass expected"xx100%#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus for this example <mathjax>#%"Yield"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.2*g)/(10*g)xx100%#</mathjax></p>
<p>And we don't even need a calculator.........</p>
<p>Most of the time, we write a stoichiometric equation, that shows the molar equivalence between reactant and product, so that we can work out the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> etc.......</p>
<p><mathjax>#H_2SO_4(aq) + 2NaOH(aq) rarr Na_2SO_4(aq) + 2H_2O#</mathjax>.</p>
<p>At any rate, you have not really quoted starting conditions here. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">In the neutralization of sulfuric acid with sodium hydroxide, the theoretical yield from 6.9g of sulfuric acid is 10g. In a synthesis, the actual yield is 7.2g. What is the percentage yield for this synthesis? </h1>
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<div class="markdown"><p><mathjax>#%"Yield"="Mass recovered"/"Mass expected"xx100%#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus for this example <mathjax>#%"Yield"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(7.2*g)/(10*g)xx100%#</mathjax></p>
<p>And we don't even need a calculator.........</p>
<p>Most of the time, we write a stoichiometric equation, that shows the molar equivalence between reactant and product, so that we can work out the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> etc.......</p>
<p><mathjax>#H_2SO_4(aq) + 2NaOH(aq) rarr Na_2SO_4(aq) + 2H_2O#</mathjax>.</p>
<p>At any rate, you have not really quoted starting conditions here. </p></div>
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</article> | In the neutralization of sulfuric acid with sodium hydroxide, the theoretical yield from 6.9g of sulfuric acid is 10g. In a synthesis, the actual yield is 7.2g. What is the percentage yield for this synthesis? | null |
2,980 | aad42711-6ddd-11ea-bd05-ccda262736ce | https://socratic.org/questions/0-314-mol-of-a-diatomic-molecule-has-a-mass-of-22-26g-how-would-you-identify-the | Cl2 | start chemical_formula qc_end physical_unit 4 5 0 1 mole qc_end physical_unit 4 5 10 11 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the molecule [IN] default"}] | [{"type":"chemical equation","value":"Cl2"}] | [{"type":"physical unit","value":"Mole [OF] the diatomic molecule [=] \\pu{0.314 mol}"},{"type":"physical unit","value":"Mass [OF] the diatomic molecule [=] \\pu{22.26 g}"}] | <h1 class="questionTitle" itemprop="name">0.314 mol of a diatomic molecule has a mass of 22.26g. How would you identify the molecule?</h1> | null | Cl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus <mathjax>#"mass"/"molar quantity"=(22.26*g)/(0.314*mol^-1)#</mathjax></p>
<p><mathjax>#-=70.9*g*mol^-1#</mathjax> WITH RESPECT TO <mathjax>#X_2#</mathjax>, and thus the molar mass, with respect to <mathjax>#X#</mathjax> was <mathjax>#35.45*g*mol^-1#</mathjax>, and clearly <mathjax>#X=Cl#</mathjax>, i.e. <mathjax>#X_2-=Cl_2#</mathjax>.</p>
<p>As a tip, which I would expect A2 students, and certainly 1st year undergraduates, to know, <mathjax>#"ALL of the elemental gases"#</mathjax> (save the Noble Gases), <mathjax>#"dihydrogen, dinitrogen, dioxygen, dihalogen,"#</mathjax> are BINUCLEAR, i.e. <mathjax>#H_2#</mathjax>, <mathjax>#N_2#</mathjax>, <mathjax>#X_2#</mathjax> etc.</p>
<p>And of course <mathjax>#Br_2#</mathjax> is a room temperature liquid, and <mathjax>#I_2#</mathjax> is a room temperature solid. </p></div>
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<div class="markdown"><p>We take the quotient <mathjax>#"mass"/"molar quantity"#</mathjax>............, and we indentify the molecule as <mathjax>#Cl_2#</mathjax>........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus <mathjax>#"mass"/"molar quantity"=(22.26*g)/(0.314*mol^-1)#</mathjax></p>
<p><mathjax>#-=70.9*g*mol^-1#</mathjax> WITH RESPECT TO <mathjax>#X_2#</mathjax>, and thus the molar mass, with respect to <mathjax>#X#</mathjax> was <mathjax>#35.45*g*mol^-1#</mathjax>, and clearly <mathjax>#X=Cl#</mathjax>, i.e. <mathjax>#X_2-=Cl_2#</mathjax>.</p>
<p>As a tip, which I would expect A2 students, and certainly 1st year undergraduates, to know, <mathjax>#"ALL of the elemental gases"#</mathjax> (save the Noble Gases), <mathjax>#"dihydrogen, dinitrogen, dioxygen, dihalogen,"#</mathjax> are BINUCLEAR, i.e. <mathjax>#H_2#</mathjax>, <mathjax>#N_2#</mathjax>, <mathjax>#X_2#</mathjax> etc.</p>
<p>And of course <mathjax>#Br_2#</mathjax> is a room temperature liquid, and <mathjax>#I_2#</mathjax> is a room temperature solid. </p></div>
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<h1 class="questionTitle" itemprop="name">0.314 mol of a diatomic molecule has a mass of 22.26g. How would you identify the molecule?</h1>
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anor277
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<div class="markdown"><p>We take the quotient <mathjax>#"mass"/"molar quantity"#</mathjax>............, and we indentify the molecule as <mathjax>#Cl_2#</mathjax>........</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus <mathjax>#"mass"/"molar quantity"=(22.26*g)/(0.314*mol^-1)#</mathjax></p>
<p><mathjax>#-=70.9*g*mol^-1#</mathjax> WITH RESPECT TO <mathjax>#X_2#</mathjax>, and thus the molar mass, with respect to <mathjax>#X#</mathjax> was <mathjax>#35.45*g*mol^-1#</mathjax>, and clearly <mathjax>#X=Cl#</mathjax>, i.e. <mathjax>#X_2-=Cl_2#</mathjax>.</p>
<p>As a tip, which I would expect A2 students, and certainly 1st year undergraduates, to know, <mathjax>#"ALL of the elemental gases"#</mathjax> (save the Noble Gases), <mathjax>#"dihydrogen, dinitrogen, dioxygen, dihalogen,"#</mathjax> are BINUCLEAR, i.e. <mathjax>#H_2#</mathjax>, <mathjax>#N_2#</mathjax>, <mathjax>#X_2#</mathjax> etc.</p>
<p>And of course <mathjax>#Br_2#</mathjax> is a room temperature liquid, and <mathjax>#I_2#</mathjax> is a room temperature solid. </p></div>
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</article> | 0.314 mol of a diatomic molecule has a mass of 22.26g. How would you identify the molecule? | null |
2,981 | acb1fab6-6ddd-11ea-9b1c-ccda262736ce | https://socratic.org/questions/nitrogen-and-hydrogen-gas-react-to-form-ammonia-gas-at-a-certain-temperature-and | 2.4 L | start physical_unit 38 38 volume l qc_end physical_unit 18 18 15 16 volume qc_end physical_unit 24 24 21 22 volume qc_end c_other OTHER qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume [OF] NH3 [IN] L"}] | [{"type":"physical unit","value":"2.4 L"}] | [{"type":"physical unit","value":"Volume [OF] N2 [=] \\pu{1.2 L}"},{"type":"physical unit","value":"Volume [OF] H2 [=] \\pu{3.6 L}"},{"type":"other","value":"Nitrogen and hydrogen gas react to form ammonia gas."},{"type":"other","value":"At a certain temperature and pressure."}] | <h1 class="questionTitle" itemprop="name">Nitrogen and hydrogen gas react to form ammonia gas. At a certain temperature and pressure 1.2 L of N2 reacts with 3.6 L of H2. If all of both gases are consumed what is the final volume of NH3 produced?</h1> | null | 2.4 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here is the balanced equation</p>
<p><mathjax>#N_2+3H_2= 2NH_3#</mathjax></p>
<p>At STP, 1 mole has a volume of 22.4 Liters. To do a dimensional analysis conversion from Liters to moles:</p>
<p><mathjax>#(N" Liters")/1(1" mole")/(22.4" Liters") = N/22.4" moles"#</mathjax></p>
<p>Please observe that Liters and moles vary directly, therefore, we can simply multiply each of the coefficients (that represent moles) by <mathjax>#1.2" L"#</mathjax> to obtain the volumes:</p>
<p><mathjax>#(1.2" L")N_2+(3.6" L")H_2= (2.4" L")NH_3#</mathjax></p>
<p>It does not really matter what number of moles that we have as long as we maintain the <mathjax>#1:3:2#</mathjax> relationship.</p></div>
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<div class="markdown"><p>The final volume is <mathjax>#2.4" L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Here is the balanced equation</p>
<p><mathjax>#N_2+3H_2= 2NH_3#</mathjax></p>
<p>At STP, 1 mole has a volume of 22.4 Liters. To do a dimensional analysis conversion from Liters to moles:</p>
<p><mathjax>#(N" Liters")/1(1" mole")/(22.4" Liters") = N/22.4" moles"#</mathjax></p>
<p>Please observe that Liters and moles vary directly, therefore, we can simply multiply each of the coefficients (that represent moles) by <mathjax>#1.2" L"#</mathjax> to obtain the volumes:</p>
<p><mathjax>#(1.2" L")N_2+(3.6" L")H_2= (2.4" L")NH_3#</mathjax></p>
<p>It does not really matter what number of moles that we have as long as we maintain the <mathjax>#1:3:2#</mathjax> relationship.</p></div>
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<h1 class="questionTitle" itemprop="name">Nitrogen and hydrogen gas react to form ammonia gas. At a certain temperature and pressure 1.2 L of N2 reacts with 3.6 L of H2. If all of both gases are consumed what is the final volume of NH3 produced?</h1>
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<div class="markdown"><p>The final volume is <mathjax>#2.4" L"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Here is the balanced equation</p>
<p><mathjax>#N_2+3H_2= 2NH_3#</mathjax></p>
<p>At STP, 1 mole has a volume of 22.4 Liters. To do a dimensional analysis conversion from Liters to moles:</p>
<p><mathjax>#(N" Liters")/1(1" mole")/(22.4" Liters") = N/22.4" moles"#</mathjax></p>
<p>Please observe that Liters and moles vary directly, therefore, we can simply multiply each of the coefficients (that represent moles) by <mathjax>#1.2" L"#</mathjax> to obtain the volumes:</p>
<p><mathjax>#(1.2" L")N_2+(3.6" L")H_2= (2.4" L")NH_3#</mathjax></p>
<p>It does not really matter what number of moles that we have as long as we maintain the <mathjax>#1:3:2#</mathjax> relationship.</p></div>
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</article> | Nitrogen and hydrogen gas react to form ammonia gas. At a certain temperature and pressure 1.2 L of N2 reacts with 3.6 L of H2. If all of both gases are consumed what is the final volume of NH3 produced? | null |
2,982 | a8ff5f90-6ddd-11ea-bc73-ccda262736ce | https://socratic.org/questions/what-is-the-equilibrium-constant-kc-for-the-reaction-at-this-temperature | 2.25 × 10^3 | start physical_unit 7 8 equilibrium_constant_k none qc_end chemical_equation 24 29 qc_end physical_unit 7 8 22 23 temperature qc_end physical_unit 40 40 42 43 concentration qc_end physical_unit 45 45 47 48 concentration qc_end physical_unit 26 26 55 56 equilibrium_concentration qc_end end | [{"type":"physical unit","value":"equilibrium concentration Kc [OF] the reaction"}] | [{"type":"physical unit","value":"2.25 × 10^3"}] | [{"type":"chemical equation","value":"H2(g) + I2(g) <=> 2 HI(g)"},{"type":"physical unit","value":"Temperature [OF] the reaction [=] \\pu{753 ℃}"},{"type":"physical unit","value":"concentration [OF] [H2] [=] \\pu{3.10 M}"},{"type":"physical unit","value":"concentration [OF] [I2] [=] \\pu{2.55 M}"},{"type":"physical unit","value":"Equilibrium concentration [OF] I2 [=] \\pu{0.0200 M}"}] | <h1 class="questionTitle" itemprop="name">What is the equilibrium constant, #K_c#, for the reaction at this temperature?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The following reaction was performed in a sealed vessel at 753 ∘C :<br/>
H2(g)+I2(g)⇌2HI(g)</p>
<p>Initially, only H2 and I2 were present at concentrations of [H2]=3.10M and [I2]=2.55M. The equilibrium concentration of I2 is 0.0200 M .</p></div>
</h2>
</div>
</div> | 2.25 × 10^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that when the following reaction takes place at <mathjax>#753^@"C"#</mathjax></p>
<blockquote>
<p><mathjax>#"H"_ (2(g)) + "I"_ (2(g)) rightleftharpoons color(red)(2)"HI"_ ((g))#</mathjax></p>
</blockquote>
<p>the <strong>initial concentrations</strong> of the two reactants are</p>
<blockquote>
<p><mathjax>#["H"_ 2]_ 0 = "3.10 M" " "#</mathjax> and <mathjax>#" "["I"_ 2]_ 0 = "2.55 M"#</mathjax></p>
</blockquote>
<p>and the <strong>equilibrium concentration</strong> of iodine gas is </p>
<blockquote>
<p><mathjax>#["I"_2] = "0.0200 M"#</mathjax></p>
</blockquote>
<p>This tells you that the concentration of iodine gas <strong>decreased</strong> by </p>
<blockquote>
<p><mathjax>#| "0.0200 M " - " 2.55 M"| = "2.53 M"#</mathjax></p>
</blockquote>
<p>In other words, the reaction <strong>consumed</strong> <mathjax>#"2.53 M"#</mathjax> of iodine gas. Since iodine gas and hydrogen gas react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, you can say that the reaction also consumed <mathjax>#"2.53 M"#</mathjax> of hydrogen gas. </p>
<p>This means that the <strong>equilibrium concentration</strong> of hydrogen gas will be</p>
<blockquote>
<p><mathjax>#["H"_ 2] = ["H"_ 2]_ 0 - "2.53 M"#</mathjax></p>
<p><mathjax>#["H"_ 2] = "3.10 M" - "2.53 M"#</mathjax></p>
<p><mathjax>#["H"_ 2] = "0.57 M"#</mathjax></p>
</blockquote>
<p>Now, notice that hydrogen iodide, the product of the reaction, is produced in a <mathjax>#color(red)(2):1#</mathjax> <strong>mole ratio</strong> with iodine gas and hydrogen gas. </p>
<p>This means that for every <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen gas and <mathjax>#1#</mathjax> <strong>mole</strong> of iodine gas that the reaction consumes, you get <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of hydrogen iodide. </p>
<p>Therefore, the <strong>equilibrium concentration</strong> of hydrogen iodide will be</p>
<blockquote>
<p><mathjax>#["HI"] = color(red)(2) * "2.53 M"#</mathjax></p>
<p><mathjax>#["HI"] = "5.06 M"#</mathjax></p>
</blockquote>
<p>By definition, the <strong>equilibrium constant</strong> that describes this equilibrium is equal to</p>
<blockquote>
<p><mathjax>#K_c = (["HI"]^color(red)(2))/(["H"_2] * ["I"_2])#</mathjax></p>
</blockquote>
<p>Plug in your values to find--I'll leave the calculation <em>without added units</em>!</p>
<blockquote>
<p><mathjax>#K_c = (5.06)^color(red)(2)/(0.57 * 0.0200) = color(darkgreen)(ul(color(black)(2.25 * 10^(3))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>So, does this value make sense?</p>
<p>Notice that the concentrations of the two reactants <strong>decrease significantly</strong> as the reaction proceeds. This tells you that at this temperature, the equilibrium <strong>lies to the right</strong>, i.e. the <em>forward reaction</em> is favored. </p>
<p>Therefore, you should expect to find </p>
<blockquote>
<p><mathjax>#K_c > 1#</mathjax></p>
</blockquote>
<p>which is exactly what happens here. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#K_c = 2.25 * 10^(3)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that when the following reaction takes place at <mathjax>#753^@"C"#</mathjax></p>
<blockquote>
<p><mathjax>#"H"_ (2(g)) + "I"_ (2(g)) rightleftharpoons color(red)(2)"HI"_ ((g))#</mathjax></p>
</blockquote>
<p>the <strong>initial concentrations</strong> of the two reactants are</p>
<blockquote>
<p><mathjax>#["H"_ 2]_ 0 = "3.10 M" " "#</mathjax> and <mathjax>#" "["I"_ 2]_ 0 = "2.55 M"#</mathjax></p>
</blockquote>
<p>and the <strong>equilibrium concentration</strong> of iodine gas is </p>
<blockquote>
<p><mathjax>#["I"_2] = "0.0200 M"#</mathjax></p>
</blockquote>
<p>This tells you that the concentration of iodine gas <strong>decreased</strong> by </p>
<blockquote>
<p><mathjax>#| "0.0200 M " - " 2.55 M"| = "2.53 M"#</mathjax></p>
</blockquote>
<p>In other words, the reaction <strong>consumed</strong> <mathjax>#"2.53 M"#</mathjax> of iodine gas. Since iodine gas and hydrogen gas react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, you can say that the reaction also consumed <mathjax>#"2.53 M"#</mathjax> of hydrogen gas. </p>
<p>This means that the <strong>equilibrium concentration</strong> of hydrogen gas will be</p>
<blockquote>
<p><mathjax>#["H"_ 2] = ["H"_ 2]_ 0 - "2.53 M"#</mathjax></p>
<p><mathjax>#["H"_ 2] = "3.10 M" - "2.53 M"#</mathjax></p>
<p><mathjax>#["H"_ 2] = "0.57 M"#</mathjax></p>
</blockquote>
<p>Now, notice that hydrogen iodide, the product of the reaction, is produced in a <mathjax>#color(red)(2):1#</mathjax> <strong>mole ratio</strong> with iodine gas and hydrogen gas. </p>
<p>This means that for every <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen gas and <mathjax>#1#</mathjax> <strong>mole</strong> of iodine gas that the reaction consumes, you get <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of hydrogen iodide. </p>
<p>Therefore, the <strong>equilibrium concentration</strong> of hydrogen iodide will be</p>
<blockquote>
<p><mathjax>#["HI"] = color(red)(2) * "2.53 M"#</mathjax></p>
<p><mathjax>#["HI"] = "5.06 M"#</mathjax></p>
</blockquote>
<p>By definition, the <strong>equilibrium constant</strong> that describes this equilibrium is equal to</p>
<blockquote>
<p><mathjax>#K_c = (["HI"]^color(red)(2))/(["H"_2] * ["I"_2])#</mathjax></p>
</blockquote>
<p>Plug in your values to find--I'll leave the calculation <em>without added units</em>!</p>
<blockquote>
<p><mathjax>#K_c = (5.06)^color(red)(2)/(0.57 * 0.0200) = color(darkgreen)(ul(color(black)(2.25 * 10^(3))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>So, does this value make sense?</p>
<p>Notice that the concentrations of the two reactants <strong>decrease significantly</strong> as the reaction proceeds. This tells you that at this temperature, the equilibrium <strong>lies to the right</strong>, i.e. the <em>forward reaction</em> is favored. </p>
<p>Therefore, you should expect to find </p>
<blockquote>
<p><mathjax>#K_c > 1#</mathjax></p>
</blockquote>
<p>which is exactly what happens here. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the equilibrium constant, #K_c#, for the reaction at this temperature?</h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>The following reaction was performed in a sealed vessel at 753 ∘C :<br/>
H2(g)+I2(g)⇌2HI(g)</p>
<p>Initially, only H2 and I2 were present at concentrations of [H2]=3.10M and [I2]=2.55M. The equilibrium concentration of I2 is 0.0200 M .</p></div>
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Stefan V.
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<span class="dateCreated" datetime="2018-03-04T00:15:05" itemprop="dateCreated">
Mar 4, 2018
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<div class="markdown"><p><mathjax>#K_c = 2.25 * 10^(3)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You know that when the following reaction takes place at <mathjax>#753^@"C"#</mathjax></p>
<blockquote>
<p><mathjax>#"H"_ (2(g)) + "I"_ (2(g)) rightleftharpoons color(red)(2)"HI"_ ((g))#</mathjax></p>
</blockquote>
<p>the <strong>initial concentrations</strong> of the two reactants are</p>
<blockquote>
<p><mathjax>#["H"_ 2]_ 0 = "3.10 M" " "#</mathjax> and <mathjax>#" "["I"_ 2]_ 0 = "2.55 M"#</mathjax></p>
</blockquote>
<p>and the <strong>equilibrium concentration</strong> of iodine gas is </p>
<blockquote>
<p><mathjax>#["I"_2] = "0.0200 M"#</mathjax></p>
</blockquote>
<p>This tells you that the concentration of iodine gas <strong>decreased</strong> by </p>
<blockquote>
<p><mathjax>#| "0.0200 M " - " 2.55 M"| = "2.53 M"#</mathjax></p>
</blockquote>
<p>In other words, the reaction <strong>consumed</strong> <mathjax>#"2.53 M"#</mathjax> of iodine gas. Since iodine gas and hydrogen gas react in a <mathjax>#1:1#</mathjax> <strong>mole ratio</strong>, you can say that the reaction also consumed <mathjax>#"2.53 M"#</mathjax> of hydrogen gas. </p>
<p>This means that the <strong>equilibrium concentration</strong> of hydrogen gas will be</p>
<blockquote>
<p><mathjax>#["H"_ 2] = ["H"_ 2]_ 0 - "2.53 M"#</mathjax></p>
<p><mathjax>#["H"_ 2] = "3.10 M" - "2.53 M"#</mathjax></p>
<p><mathjax>#["H"_ 2] = "0.57 M"#</mathjax></p>
</blockquote>
<p>Now, notice that hydrogen iodide, the product of the reaction, is produced in a <mathjax>#color(red)(2):1#</mathjax> <strong>mole ratio</strong> with iodine gas and hydrogen gas. </p>
<p>This means that for every <mathjax>#1#</mathjax> <strong>mole</strong> of hydrogen gas and <mathjax>#1#</mathjax> <strong>mole</strong> of iodine gas that the reaction consumes, you get <mathjax>#color(red)(2)#</mathjax> <strong>moles</strong> of hydrogen iodide. </p>
<p>Therefore, the <strong>equilibrium concentration</strong> of hydrogen iodide will be</p>
<blockquote>
<p><mathjax>#["HI"] = color(red)(2) * "2.53 M"#</mathjax></p>
<p><mathjax>#["HI"] = "5.06 M"#</mathjax></p>
</blockquote>
<p>By definition, the <strong>equilibrium constant</strong> that describes this equilibrium is equal to</p>
<blockquote>
<p><mathjax>#K_c = (["HI"]^color(red)(2))/(["H"_2] * ["I"_2])#</mathjax></p>
</blockquote>
<p>Plug in your values to find--I'll leave the calculation <em>without added units</em>!</p>
<blockquote>
<p><mathjax>#K_c = (5.06)^color(red)(2)/(0.57 * 0.0200) = color(darkgreen)(ul(color(black)(2.25 * 10^(3))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<p>So, does this value make sense?</p>
<p>Notice that the concentrations of the two reactants <strong>decrease significantly</strong> as the reaction proceeds. This tells you that at this temperature, the equilibrium <strong>lies to the right</strong>, i.e. the <em>forward reaction</em> is favored. </p>
<p>Therefore, you should expect to find </p>
<blockquote>
<p><mathjax>#K_c > 1#</mathjax></p>
</blockquote>
<p>which is exactly what happens here. </p></div>
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</article> | What is the equilibrium constant, #K_c#, for the reaction at this temperature? |
The following reaction was performed in a sealed vessel at 753 ∘C :
H2(g)+I2(g)⇌2HI(g)
Initially, only H2 and I2 were present at concentrations of [H2]=3.10M and [I2]=2.55M. The equilibrium concentration of I2 is 0.0200 M .
|
2,983 | aa9591dc-6ddd-11ea-a5f7-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-solution-made-by-mixing-100-0-ml-of-0-10-m-hno-3-50-0-ml-of- | 1.10 | start physical_unit 6 6 ph none qc_end physical_unit 6 6 10 11 volume qc_end physical_unit 15 15 13 14 molarity qc_end physical_unit 6 6 16 17 volume qc_end physical_unit 21 21 19 20 molarity qc_end physical_unit 26 26 10 11 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"1.10"}] | [{"type":"physical unit","value":"Volume [OF] HNO3 solution [=] \\pu{100.0 mL}"},{"type":"physical unit","value":"Molarity [OF] HNO3 solution [=] \\pu{0.10 M}"},{"type":"physical unit","value":"Volume [OF] HCl solution [=] \\pu{50.0 mL}"},{"type":"physical unit","value":"Molarity [OF] HCl solution [=] \\pu{0.20 M}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{100.0 mL}"},{"type":"other","value":"Assume that the volumes are additive."}] | <h1 class="questionTitle" itemprop="name">What is the pH of a solution made by mixing 100.0 mL of 0.10 M #HNO_3#, 50.0 mL of 0.20 M #HCl# and 100.0 mL of water? Assume that the volumes are additive.</h1> | null | 1.10 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing first, calculate the total volume of the resulting solution</p>
<blockquote>
<p><mathjax>#V_"total" = "100.0 mL + 50.0 mL + 100.0 mL"#</mathjax></p>
<p><mathjax>#V_"total" = "250.0 mL"#</mathjax></p>
</blockquote>
<p>Now, you are dealing with two <strong>strong acids</strong> that ionize completely in aqueous solution. Both nitric acid and hydrochloric acid produce hydronium cations in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>, so you know that</p>
<blockquote>
<p><mathjax>#["H"_ 3"O"^(+)]_ ("coming from HNO"_ 3) = ["HNO"_3]#</mathjax></p>
<p><mathjax>#["H"_ 3"O"^(+)]_ ("coming from HCl") = ["HCl"]#</mathjax></p>
</blockquote>
<p>As you know, <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> is defined as the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution. For the nitric acid solution, you have</p>
<blockquote>
<p><mathjax>#"100.0 mL" = (10^3color(white)(.)"mL")/color(blue)(10) => n_( "H"_3"O"^(+)) = "0.10 moles"/color(blue)(10) = "0.010 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
<p>For hydrochloric acid, you have</p>
<blockquote>
<p><mathjax>#"50.0 mL"= (10^3color(white)(.)"mL")/color(blue)(20) implies n_ ("H"_ 3"O"^(+)) = "0.20 moles"/color(blue)(20) = "0.010 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
<p>The <strong>total number of moles</strong> of hydronium cations delivered by the two acids in the resulting solution will be</p>
<blockquote>
<p><mathjax>#n_ ("H"_ 3"O"^(+)) = "0.010 moles + 0.010 moles"#</mathjax></p>
<p><mathjax>#n_ ("H"_ 3"O"^(+)) = "0.020 moles"#</mathjax></p>
</blockquote>
<p>The <strong>concentration</strong> of the hydronium cations in the resulting solution will be </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = "0.020 moles"/(250.0 * 10^(-3)"L") = "0.080 M"#</mathjax></p>
</blockquote>
<p>As you know, you have</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>This will give you </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = - log(0.080) = 1.10)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <em>decimal places</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the molarities of the two acids. </p>
<p>As a fun fact, a mixture of nitric acid and hydrochloric acid is called <strong><a href="https://en.wikipedia.org/wiki/Aqua_regia" rel="nofollow">aqua regia</a></strong>. Ideally, aqua regia contains nitric acid and hydrochloric acid in a <mathjax>#1:3#</mathjax> mole ratio, not in a <mathjax>#1:1#</mathjax> mole ratio like you have here. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"pH" = 1.10#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing first, calculate the total volume of the resulting solution</p>
<blockquote>
<p><mathjax>#V_"total" = "100.0 mL + 50.0 mL + 100.0 mL"#</mathjax></p>
<p><mathjax>#V_"total" = "250.0 mL"#</mathjax></p>
</blockquote>
<p>Now, you are dealing with two <strong>strong acids</strong> that ionize completely in aqueous solution. Both nitric acid and hydrochloric acid produce hydronium cations in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>, so you know that</p>
<blockquote>
<p><mathjax>#["H"_ 3"O"^(+)]_ ("coming from HNO"_ 3) = ["HNO"_3]#</mathjax></p>
<p><mathjax>#["H"_ 3"O"^(+)]_ ("coming from HCl") = ["HCl"]#</mathjax></p>
</blockquote>
<p>As you know, <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> is defined as the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution. For the nitric acid solution, you have</p>
<blockquote>
<p><mathjax>#"100.0 mL" = (10^3color(white)(.)"mL")/color(blue)(10) => n_( "H"_3"O"^(+)) = "0.10 moles"/color(blue)(10) = "0.010 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
<p>For hydrochloric acid, you have</p>
<blockquote>
<p><mathjax>#"50.0 mL"= (10^3color(white)(.)"mL")/color(blue)(20) implies n_ ("H"_ 3"O"^(+)) = "0.20 moles"/color(blue)(20) = "0.010 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
<p>The <strong>total number of moles</strong> of hydronium cations delivered by the two acids in the resulting solution will be</p>
<blockquote>
<p><mathjax>#n_ ("H"_ 3"O"^(+)) = "0.010 moles + 0.010 moles"#</mathjax></p>
<p><mathjax>#n_ ("H"_ 3"O"^(+)) = "0.020 moles"#</mathjax></p>
</blockquote>
<p>The <strong>concentration</strong> of the hydronium cations in the resulting solution will be </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = "0.020 moles"/(250.0 * 10^(-3)"L") = "0.080 M"#</mathjax></p>
</blockquote>
<p>As you know, you have</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>This will give you </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = - log(0.080) = 1.10)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <em>decimal places</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the molarities of the two acids. </p>
<p>As a fun fact, a mixture of nitric acid and hydrochloric acid is called <strong><a href="https://en.wikipedia.org/wiki/Aqua_regia" rel="nofollow">aqua regia</a></strong>. Ideally, aqua regia contains nitric acid and hydrochloric acid in a <mathjax>#1:3#</mathjax> mole ratio, not in a <mathjax>#1:1#</mathjax> mole ratio like you have here. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the pH of a solution made by mixing 100.0 mL of 0.10 M #HNO_3#, 50.0 mL of 0.20 M #HCl# and 100.0 mL of water? Assume that the volumes are additive.</h1>
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Stefan V.
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Apr 23, 2017
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<div class="markdown"><p><mathjax>#"pH" = 1.10#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First thing first, calculate the total volume of the resulting solution</p>
<blockquote>
<p><mathjax>#V_"total" = "100.0 mL + 50.0 mL + 100.0 mL"#</mathjax></p>
<p><mathjax>#V_"total" = "250.0 mL"#</mathjax></p>
</blockquote>
<p>Now, you are dealing with two <strong>strong acids</strong> that ionize completely in aqueous solution. Both nitric acid and hydrochloric acid produce hydronium cations in <mathjax>#1:1#</mathjax> <strong><a href="https://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratios</a></strong>, so you know that</p>
<blockquote>
<p><mathjax>#["H"_ 3"O"^(+)]_ ("coming from HNO"_ 3) = ["HNO"_3]#</mathjax></p>
<p><mathjax>#["H"_ 3"O"^(+)]_ ("coming from HCl") = ["HCl"]#</mathjax></p>
</blockquote>
<p>As you know, <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> is defined as the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution. For the nitric acid solution, you have</p>
<blockquote>
<p><mathjax>#"100.0 mL" = (10^3color(white)(.)"mL")/color(blue)(10) => n_( "H"_3"O"^(+)) = "0.10 moles"/color(blue)(10) = "0.010 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
<p>For hydrochloric acid, you have</p>
<blockquote>
<p><mathjax>#"50.0 mL"= (10^3color(white)(.)"mL")/color(blue)(20) implies n_ ("H"_ 3"O"^(+)) = "0.20 moles"/color(blue)(20) = "0.010 moles H"_3"O"^(+)#</mathjax></p>
</blockquote>
<p>The <strong>total number of moles</strong> of hydronium cations delivered by the two acids in the resulting solution will be</p>
<blockquote>
<p><mathjax>#n_ ("H"_ 3"O"^(+)) = "0.010 moles + 0.010 moles"#</mathjax></p>
<p><mathjax>#n_ ("H"_ 3"O"^(+)) = "0.020 moles"#</mathjax></p>
</blockquote>
<p>The <strong>concentration</strong> of the hydronium cations in the resulting solution will be </p>
<blockquote>
<p><mathjax>#["H"_3"O"^(+)] = "0.020 moles"/(250.0 * 10^(-3)"L") = "0.080 M"#</mathjax></p>
</blockquote>
<p>As you know, you have</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#</mathjax></p>
</blockquote>
<p>This will give you </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("pH" = - log(0.080) = 1.10)))#</mathjax></p>
</blockquote>
<p>The answer is rounded to two <em>decimal places</em>, the number of <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong> you have for the molarities of the two acids. </p>
<p>As a fun fact, a mixture of nitric acid and hydrochloric acid is called <strong><a href="https://en.wikipedia.org/wiki/Aqua_regia" rel="nofollow">aqua regia</a></strong>. Ideally, aqua regia contains nitric acid and hydrochloric acid in a <mathjax>#1:3#</mathjax> mole ratio, not in a <mathjax>#1:1#</mathjax> mole ratio like you have here. </p></div>
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</article> | What is the pH of a solution made by mixing 100.0 mL of 0.10 M #HNO_3#, 50.0 mL of 0.20 M #HCl# and 100.0 mL of water? Assume that the volumes are additive. | null |
2,984 | ab4e54c6-6ddd-11ea-8d7e-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-state-of-s-in-s-2o-3-2 | +2 | start physical_unit 6 6 oxidation_state none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation state [OF] S"}] | [{"type":"physical unit","value":"+2"}] | [{"type":"chemical equation","value":"S2O3^2-"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation state of S in #S_2O_3^-2#?</h1> | null | +2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can see that the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of the molecule is <mathjax>#-2#</mathjax>.<br/>
To start of we find the oxidation state of oxygen, in the molecule.<br/>
Oxygen has a oxidation state of <mathjax>#-2#</mathjax> since we have 3 oxygens this would give a total oxidation state of <mathjax>#-6#</mathjax> for oxygen. But since the net charge of the molecule is <mathjax>#-2#</mathjax> we have to all 2 the the total charge of oxygen, thereby giving <mathjax>#-4#</mathjax>. Now we can see that the sulfur must have a charge of <mathjax>#+2#</mathjax>. </p></div>
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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>We can see that the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of the molecule is <mathjax>#-2#</mathjax>.<br/>
To start of we find the oxidation state of oxygen, in the molecule.<br/>
Oxygen has a oxidation state of <mathjax>#-2#</mathjax> since we have 3 oxygens this would give a total oxidation state of <mathjax>#-6#</mathjax> for oxygen. But since the net charge of the molecule is <mathjax>#-2#</mathjax> we have to all 2 the the total charge of oxygen, thereby giving <mathjax>#-4#</mathjax>. Now we can see that the sulfur must have a charge of <mathjax>#+2#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation state of S in #S_2O_3^-2#?</h1>
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Shulgin
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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>We can see that the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/net-charge">net charge</a> of the molecule is <mathjax>#-2#</mathjax>.<br/>
To start of we find the oxidation state of oxygen, in the molecule.<br/>
Oxygen has a oxidation state of <mathjax>#-2#</mathjax> since we have 3 oxygens this would give a total oxidation state of <mathjax>#-6#</mathjax> for oxygen. But since the net charge of the molecule is <mathjax>#-2#</mathjax> we have to all 2 the the total charge of oxygen, thereby giving <mathjax>#-4#</mathjax>. Now we can see that the sulfur must have a charge of <mathjax>#+2#</mathjax>. </p></div>
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anor277
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<div class="markdown"><p>Formally we gots <mathjax>#stackrel(VI+)S#</mathjax> and <mathjax>#stackrel(-II)S#</mathjax>...</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>And thus <mathjax>#S_"average oxidation state"=(VI+(-II))/2=+II#</mathjax>.</p>
<p><mathjax>#"Thiosulfate ion"#</mathjax> is an interesting customer in terms of sulfur oxidation state. If we look at sulfate, <mathjax>#SO_4^(2-)#</mathjax>, CLEARLY we got <mathjax>#S^(VI+)#</mathjax> and <mathjax>#4xxO^(-II)#</mathjax>..and as usual, the weighted sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a>, <mathjax>#6-8=-2#</mathjax>, i.e. the charge on the ion.</p>
<p>In <mathjax>#"thiosulfate"#</mathjax>, <mathjax>#S_2O_3^(2-)#</mathjax>, I like to think that ONE of the oxygen atoms of sulfate has BEEN REPLACED by one sulfur as sulfide. And thus the central sulfur is <mathjax>#+VI#</mathjax>, and the terminal sulfur is <mathjax>#S^(-II)#</mathjax>, i.e. its oxidation state is PRECISELY the same as its Group 16 congener, oxygen. Of course, this is a formalism, but so is the whole concept of oxidation state and oxidation number.</p>
<p>Claro? </p></div>
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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>We got the thiosulfate ion <mathjax>#S_2O_3^(2-)#</mathjax>.</p>
<p>Since oxygen is more electronegative than sulfur, then oxygen will have its usual <mathjax>#-2#</mathjax> oxidation state. There are three oxygen atoms, and so the total charge of the oxygens is <mathjax>#-2*3=-6#</mathjax>.</p>
<p>Let <mathjax>#x#</mathjax> be the total charge of two sulfur atoms.</p>
<p>We got:</p>
<p><mathjax>#x-6=-2#</mathjax></p>
<p><mathjax>#x=-2+6#</mathjax></p>
<p><mathjax>#x=4#</mathjax></p>
<p>So, the sum of the <a href="https://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of the sulfur atoms is <mathjax>#+4#</mathjax>. If we consider both sulfur atoms to have the same oxidation number, then each sulfur will have an oxidation number of <mathjax>#(+4)/2=+2#</mathjax>.</p></div>
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</article> | What is the oxidation state of S in #S_2O_3^-2#? | null |
2,985 | ad0b51c1-6ddd-11ea-961d-ccda262736ce | https://socratic.org/questions/getting-chemical-equations-from-word-equations | CuO(s) + H2(g) -> Cu(s) + H2O(l) | start chemical_equation qc_end substance 6 8 qc_end substance 15 15 qc_end substance 16 17 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"CuO(s) + H2(g) -> Cu(s) + H2O(l)"}] | [{"type":"substance name","value":"Copper II Oxide"},{"type":"substance name","value":"Hydrogen"},{"type":"substance name","value":"Copper metal"}] | <h1 class="questionTitle" itemprop="name">Getting chemical equations from word equations?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>I understand the basic concepts, but I don't know whether to use H2 for hydrogen or a single H for example (and for all BrNCLHOF problems)<br/>
Right now I am stuck on:<br/>
<em>"Copper II Oxide is heated in the presence of Hydrogen. Copper metal and one other product (figure out its identity) are formed"</em></p></div>
</h2>
</div>
</div> | CuO(s) + H2(g) -> Cu(s) + H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So to answer your question of <mathjax>#H_2#</mathjax> vs. <mathjax>#H#</mathjax>, I'd say always use <mathjax>#H_2#</mathjax> when considering reactions. This is just since hydrogen gas is diatomic in its standard state (i.e. most thermodynamically favorable state), and hence when you are reacting something with hydrogen gas, it will more often than not be <mathjax>#H_2#</mathjax>. </p>
<p>One notable exception to this would be if you were considering acid-base reactions. In these types of reactions, you'll see the <mathjax>#H^+#</mathjax> ion a lot, and this would obviously be just a single <mathjax>#H#</mathjax>. </p>
<p>Hope that helped :) </p></div>
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<div class="markdown"><p>Equation:<br/>
<mathjax>#CuO(s) + H_2(g) -> Cu(s) + H_2O(l)#</mathjax></p>
<p>So your other product would be <mathjax>#H_2O#</mathjax>, or water. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So to answer your question of <mathjax>#H_2#</mathjax> vs. <mathjax>#H#</mathjax>, I'd say always use <mathjax>#H_2#</mathjax> when considering reactions. This is just since hydrogen gas is diatomic in its standard state (i.e. most thermodynamically favorable state), and hence when you are reacting something with hydrogen gas, it will more often than not be <mathjax>#H_2#</mathjax>. </p>
<p>One notable exception to this would be if you were considering acid-base reactions. In these types of reactions, you'll see the <mathjax>#H^+#</mathjax> ion a lot, and this would obviously be just a single <mathjax>#H#</mathjax>. </p>
<p>Hope that helped :) </p></div>
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<div class="markdown"><p>I understand the basic concepts, but I don't know whether to use H2 for hydrogen or a single H for example (and for all BrNCLHOF problems)<br/>
Right now I am stuck on:<br/>
<em>"Copper II Oxide is heated in the presence of Hydrogen. Copper metal and one other product (figure out its identity) are formed"</em></p></div>
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Darshan Senthil
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<span class="dateCreated" datetime="2016-12-07T03:25:41" itemprop="dateCreated">
Dec 7, 2016
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<div class="markdown"><p>Equation:<br/>
<mathjax>#CuO(s) + H_2(g) -> Cu(s) + H_2O(l)#</mathjax></p>
<p>So your other product would be <mathjax>#H_2O#</mathjax>, or water. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So to answer your question of <mathjax>#H_2#</mathjax> vs. <mathjax>#H#</mathjax>, I'd say always use <mathjax>#H_2#</mathjax> when considering reactions. This is just since hydrogen gas is diatomic in its standard state (i.e. most thermodynamically favorable state), and hence when you are reacting something with hydrogen gas, it will more often than not be <mathjax>#H_2#</mathjax>. </p>
<p>One notable exception to this would be if you were considering acid-base reactions. In these types of reactions, you'll see the <mathjax>#H^+#</mathjax> ion a lot, and this would obviously be just a single <mathjax>#H#</mathjax>. </p>
<p>Hope that helped :) </p></div>
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</article> | Getting chemical equations from word equations? |
I understand the basic concepts, but I don't know whether to use H2 for hydrogen or a single H for example (and for all BrNCLHOF problems)
Right now I am stuck on:
"Copper II Oxide is heated in the presence of Hydrogen. Copper metal and one other product (figure out its identity) are formed"
|
2,986 | ac18c5f4-6ddd-11ea-8a0c-ccda262736ce | https://socratic.org/questions/what-is-the-concentration-of-a-solution-that-with-a-volume-of-660-ml-that-contai | 50.61 g/L | start physical_unit 21 21 concentration g/l qc_end physical_unit 6 6 12 13 volume qc_end physical_unit 21 21 16 17 mass qc_end end | [{"type":"physical unit","value":"Concentration [OF] Al(C2H3O2)3 solution [IN] g/L"}] | [{"type":"physical unit","value":"50.61 g/L"}] | [{"type":"physical unit","value":"Volume [OF] Al(C2H3O2)3 solution [=] \\pu{660 mL}"},{"type":"physical unit","value":"Mass [OF] Al(C2H3O2)3 [=] \\pu{33.4 grams}"}] | <h1 class="questionTitle" itemprop="name">What is the concentration of a solution that with a volume of 660 mL that contains 33.4 grams of aluminum acetate #Al(C_2H_3O_2)_3#?</h1> | null | 50.61 g/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>One formula for concentration is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) "Concentration" = "mass of solute"/"volume of solution"color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#"Concentration" = "33.4 g"/"0.660 L" = "50.6 g/L"#</mathjax></p></div>
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<div class="markdown"><p>The concentration is 50.6 g/L.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>One formula for concentration is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) "Concentration" = "mass of solute"/"volume of solution"color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#"Concentration" = "33.4 g"/"0.660 L" = "50.6 g/L"#</mathjax></p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the concentration of a solution that with a volume of 660 mL that contains 33.4 grams of aluminum acetate #Al(C_2H_3O_2)_3#?</h1>
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Ernest Z.
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<div class="markdown"><p>The concentration is 50.6 g/L.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p>One formula for concentration is</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(white)(a/a) "Concentration" = "mass of solute"/"volume of solution"color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p><mathjax>#"Concentration" = "33.4 g"/"0.660 L" = "50.6 g/L"#</mathjax></p></div>
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</article> | What is the concentration of a solution that with a volume of 660 mL that contains 33.4 grams of aluminum acetate #Al(C_2H_3O_2)_3#? | null |
2,987 | ab026564-6ddd-11ea-ac6e-ccda262736ce | https://socratic.org/questions/what-is-the-formula-for-magnesium-carbonate | MgCO3 | start chemical_formula qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Formula [OF] magnesium carbonate [IN] default"}] | [{"type":"chemical equation","value":"MgCO3"}] | [{"type":"substance name","value":"Magnesium carbonate"}] | <h1 class="questionTitle" itemprop="name">What is the formula for magnesium carbonate?</h1> | null | MgCO3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Magnesium is an alkaline earth metal, and forms a <mathjax>#Mg^(2+)#</mathjax> ion....</p>
<p>And carbonate ion, <mathjax>#CO_3^(2-)#</mathjax>, has an equal an opposite charge. And thus the ionic species is <mathjax>#MgCO_3#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#MgCO_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Magnesium is an alkaline earth metal, and forms a <mathjax>#Mg^(2+)#</mathjax> ion....</p>
<p>And carbonate ion, <mathjax>#CO_3^(2-)#</mathjax>, has an equal an opposite charge. And thus the ionic species is <mathjax>#MgCO_3#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the formula for magnesium carbonate?</h1>
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anor277
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<div class="markdown"><p><mathjax>#MgCO_3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Magnesium is an alkaline earth metal, and forms a <mathjax>#Mg^(2+)#</mathjax> ion....</p>
<p>And carbonate ion, <mathjax>#CO_3^(2-)#</mathjax>, has an equal an opposite charge. And thus the ionic species is <mathjax>#MgCO_3#</mathjax>.</p></div>
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</article> | What is the formula for magnesium carbonate? | null |
2,988 | ab81cfb0-6ddd-11ea-a952-ccda262736ce | https://socratic.org/questions/what-is-the-molar-mass-of-an-ideal-gas-if-a-0-622-g-sample-of-this-gas-occupies- | 51 g/mol | start physical_unit 7 8 molar_mass g/mol qc_end physical_unit 13 16 11 12 mass qc_end physical_unit 13 16 21 22 volume qc_end physical_unit 13 16 24 25 temperature qc_end physical_unit 13 16 27 28 pressure qc_end end | [{"type":"physical unit","value":"Molar mass [OF] the ideal gas [IN] g/mol"}] | [{"type":"physical unit","value":"51 g/mol"}] | [{"type":"physical unit","value":"Mass [OF] this gas sample [=] \\pu{0.622 g}"},{"type":"physical unit","value":"Volume [OF] this gas sample [=] \\pu{300 mL}"},{"type":"physical unit","value":"Temperature [OF] this gas sample [=] \\pu{35 ℃}"},{"type":"physical unit","value":"Pressure [OF] this gas sample [=] \\pu{789 mmHg}"}] | <h1 class="questionTitle" itemprop="name"> What is the molar mass of an ideal gas if a 0.622 g sample of this gas occupies a volume of 300 mL at 35 °C and 789 mm Hg?</h1> | null | 51 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Whenever you see molar masses in gas law questions, more often than not <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation <mathjax>#PV = nRT#</mathjax> to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:</p>
<p><mathjax>#M = (dRT)/P#</mathjax></p>
<p>M = molar mass (g/mol)<br/>
d = density (g/L)<br/>
R = Ideal Gas Constant (<mathjax>#~~0.0821 (atm*L)/(mol*K)#</mathjax>) <br/>
T = Temperature (In Kelvin) <br/>
P = Pressure (atm)</p>
<p>As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> in which the identity of the gas plays a role in your calculations. Just something to take note of. </p>
<p><strong>Back to the problem:</strong> Now, looking back at what we're given, we will need to make some <a href="https://socratic.org/chemistry/measurement-in-chemistry/unit-conversions">unit conversions</a> to ensure everything matches the dimensions required by the equation:</p>
<p><mathjax>#T = 35^oC + 273.15 =#</mathjax> 308.15 K<br/>
<mathjax>#V = 300 mL * (1000 mL)/(1 L) =#</mathjax> 0.300 L<br/>
<mathjax>#P = 789 mm Hg * (1 atm)/(760 mm Hg) =#</mathjax> 1.038 atm</p>
<p>So, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:</p>
<p><mathjax>#d = (0.622 g)/(0.300 L) =#</mathjax> 2.073 g/L</p>
<p>Now, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.</p>
<p><mathjax>#M = (dRT)/P = ((2.073)(0.0821)(308.15))/(1.038) =#</mathjax> 51 g/mol</p>
<p><em>Rounded to 2 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></em></p>
<p>Now if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol). </p>
<p>Hope that helped :) </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>51 g/mol</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Whenever you see molar masses in gas law questions, more often than not <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation <mathjax>#PV = nRT#</mathjax> to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:</p>
<p><mathjax>#M = (dRT)/P#</mathjax></p>
<p>M = molar mass (g/mol)<br/>
d = density (g/L)<br/>
R = Ideal Gas Constant (<mathjax>#~~0.0821 (atm*L)/(mol*K)#</mathjax>) <br/>
T = Temperature (In Kelvin) <br/>
P = Pressure (atm)</p>
<p>As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> in which the identity of the gas plays a role in your calculations. Just something to take note of. </p>
<p><strong>Back to the problem:</strong> Now, looking back at what we're given, we will need to make some <a href="https://socratic.org/chemistry/measurement-in-chemistry/unit-conversions">unit conversions</a> to ensure everything matches the dimensions required by the equation:</p>
<p><mathjax>#T = 35^oC + 273.15 =#</mathjax> 308.15 K<br/>
<mathjax>#V = 300 mL * (1000 mL)/(1 L) =#</mathjax> 0.300 L<br/>
<mathjax>#P = 789 mm Hg * (1 atm)/(760 mm Hg) =#</mathjax> 1.038 atm</p>
<p>So, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:</p>
<p><mathjax>#d = (0.622 g)/(0.300 L) =#</mathjax> 2.073 g/L</p>
<p>Now, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.</p>
<p><mathjax>#M = (dRT)/P = ((2.073)(0.0821)(308.15))/(1.038) =#</mathjax> 51 g/mol</p>
<p><em>Rounded to 2 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></em></p>
<p>Now if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol). </p>
<p>Hope that helped :) </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> What is the molar mass of an ideal gas if a 0.622 g sample of this gas occupies a volume of 300 mL at 35 °C and 789 mm Hg?</h1>
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Darshan Senthil
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<div class="markdown"><p>51 g/mol</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Whenever you see molar masses in gas law questions, more often than not <a href="https://socratic.org/chemistry/measurement-in-chemistry/density">density</a> will be involved. This question is no different. To solve this, however, we will first need to play with the combined ideal gas equation <mathjax>#PV = nRT#</mathjax> to make it work for density and molar mass. The derivation is simple but for the sake of time and space, I will skip it. Hence, just take my word for it that you will end up with the equation:</p>
<p><mathjax>#M = (dRT)/P#</mathjax></p>
<p>M = molar mass (g/mol)<br/>
d = density (g/L)<br/>
R = Ideal Gas Constant (<mathjax>#~~0.0821 (atm*L)/(mol*K)#</mathjax>) <br/>
T = Temperature (In Kelvin) <br/>
P = Pressure (atm)</p>
<p>As an aside, note that because calculations with this equation involve molar mass, this is the only variation of the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> in which the identity of the gas plays a role in your calculations. Just something to take note of. </p>
<p><strong>Back to the problem:</strong> Now, looking back at what we're given, we will need to make some <a href="https://socratic.org/chemistry/measurement-in-chemistry/unit-conversions">unit conversions</a> to ensure everything matches the dimensions required by the equation:</p>
<p><mathjax>#T = 35^oC + 273.15 =#</mathjax> 308.15 K<br/>
<mathjax>#V = 300 mL * (1000 mL)/(1 L) =#</mathjax> 0.300 L<br/>
<mathjax>#P = 789 mm Hg * (1 atm)/(760 mm Hg) =#</mathjax> 1.038 atm</p>
<p>So, we have almost everything we need to simply plug into the equation. The last thing we need is density. How do we find density? Notice we're given the mass of the sample (0.622 g). All we need to do is divide this by volume, and we have density:</p>
<p><mathjax>#d = (0.622 g)/(0.300 L) =#</mathjax> 2.073 g/L</p>
<p>Now, we can plug in everything. When you punch the numbers into your calculator, however, make sure you use the stored values you got from the actual conversions, and not the rounded ones. This will help you ensure accuracy.</p>
<p><mathjax>#M = (dRT)/P = ((2.073)(0.0821)(308.15))/(1.038) =#</mathjax> 51 g/mol</p>
<p><em>Rounded to 2 <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></em></p>
<p>Now if you were asked to identify which element this is based on your calculation, your best bet would probably be Vandium (molar mass 50.94 g/mol). </p>
<p>Hope that helped :) </p></div>
</div>
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<a href="https://socratic.org/answers/348081" itemprop="url">Answer link</a>
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</article> | What is the molar mass of an ideal gas if a 0.622 g sample of this gas occupies a volume of 300 mL at 35 °C and 789 mm Hg? | null |
2,989 | acfbc4d8-6ddd-11ea-a823-ccda262736ce | https://socratic.org/questions/what-is-the-voltage-of-a-galvanic-cell-made-with-silver-and-nickel | +1.05 V | start physical_unit 6 7 voltage v qc_end substance 10 10 qc_end substance 12 12 qc_end end | [{"type":"physical unit","value":"Voltage [OF] the galvanic cell [IN] V"}] | [{"type":"physical unit","value":"+1.05 V"}] | [{"type":"substance name","value":"Silver"},{"type":"substance name","value":"Nickel"}] | <h1 class="questionTitle" itemprop="name">What is the voltage of a galvanic cell made with silver and nickel?</h1> | null | +1.05 V | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>List the standard electrode potentials from -ve to +ve:</p>
<p><mathjax>#" "E^@("V")#</mathjax></p>
<p><mathjax>#stackrelcolor(blue)(leftarrow)color(white)(xxxxxxxxxxxxxx)#</mathjax><br/>
<mathjax>#Ni_((aq))^(2+)+2erightleftharpoonsNi_((s))" "-0.25#</mathjax></p>
<p><mathjax>#Ag_((aq))^++erightleftharpoonsAg_((s))" "+0.8#</mathjax></p>
<p><mathjax>#stackrelcolor(red)(rightarrow)color(white)(xxxxxxxxxxxxxx)#</mathjax><br/>
You can see that the <mathjax>#"Ag"^+"/""Ag"#</mathjax> 1/2 cell has the most +ve <mathjax>#E^@#</mathjax> value so this will take in electrons and move left to right.</p>
<p>The <mathjax>#"Ni"^(2+)"/""Ni"#</mathjax> 1/2 cell will, therefore, take in these electrons and move right to left.</p>
<p>The voltage of the cell is an empirically measured quantity so must always have a +ve value. So to calculate <mathjax>#E_(cell)^@#</mathjax> subtract the least +ve <mathjax>#E^@#</mathjax> from the most +ve <mathjax>#E^@#</mathjax> value<mathjax>#rArr#</mathjax></p>
<p><mathjax>#E_(cell)^@=+0.8-(-0.25)=+1.05"V"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#E_(cell)^@=+1.05"V"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>List the standard electrode potentials from -ve to +ve:</p>
<p><mathjax>#" "E^@("V")#</mathjax></p>
<p><mathjax>#stackrelcolor(blue)(leftarrow)color(white)(xxxxxxxxxxxxxx)#</mathjax><br/>
<mathjax>#Ni_((aq))^(2+)+2erightleftharpoonsNi_((s))" "-0.25#</mathjax></p>
<p><mathjax>#Ag_((aq))^++erightleftharpoonsAg_((s))" "+0.8#</mathjax></p>
<p><mathjax>#stackrelcolor(red)(rightarrow)color(white)(xxxxxxxxxxxxxx)#</mathjax><br/>
You can see that the <mathjax>#"Ag"^+"/""Ag"#</mathjax> 1/2 cell has the most +ve <mathjax>#E^@#</mathjax> value so this will take in electrons and move left to right.</p>
<p>The <mathjax>#"Ni"^(2+)"/""Ni"#</mathjax> 1/2 cell will, therefore, take in these electrons and move right to left.</p>
<p>The voltage of the cell is an empirically measured quantity so must always have a +ve value. So to calculate <mathjax>#E_(cell)^@#</mathjax> subtract the least +ve <mathjax>#E^@#</mathjax> from the most +ve <mathjax>#E^@#</mathjax> value<mathjax>#rArr#</mathjax></p>
<p><mathjax>#E_(cell)^@=+0.8-(-0.25)=+1.05"V"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the voltage of a galvanic cell made with silver and nickel?</h1>
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<div class="markdown"><p><mathjax>#E_(cell)^@=+1.05"V"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>List the standard electrode potentials from -ve to +ve:</p>
<p><mathjax>#" "E^@("V")#</mathjax></p>
<p><mathjax>#stackrelcolor(blue)(leftarrow)color(white)(xxxxxxxxxxxxxx)#</mathjax><br/>
<mathjax>#Ni_((aq))^(2+)+2erightleftharpoonsNi_((s))" "-0.25#</mathjax></p>
<p><mathjax>#Ag_((aq))^++erightleftharpoonsAg_((s))" "+0.8#</mathjax></p>
<p><mathjax>#stackrelcolor(red)(rightarrow)color(white)(xxxxxxxxxxxxxx)#</mathjax><br/>
You can see that the <mathjax>#"Ag"^+"/""Ag"#</mathjax> 1/2 cell has the most +ve <mathjax>#E^@#</mathjax> value so this will take in electrons and move left to right.</p>
<p>The <mathjax>#"Ni"^(2+)"/""Ni"#</mathjax> 1/2 cell will, therefore, take in these electrons and move right to left.</p>
<p>The voltage of the cell is an empirically measured quantity so must always have a +ve value. So to calculate <mathjax>#E_(cell)^@#</mathjax> subtract the least +ve <mathjax>#E^@#</mathjax> from the most +ve <mathjax>#E^@#</mathjax> value<mathjax>#rArr#</mathjax></p>
<p><mathjax>#E_(cell)^@=+0.8-(-0.25)=+1.05"V"#</mathjax></p></div>
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</article> | What is the voltage of a galvanic cell made with silver and nickel? | null |
2,990 | abfb50a3-6ddd-11ea-b9a4-ccda262736ce | https://socratic.org/questions/what-is-the-coefficient-for-o-2-when-the-equation-for-the-combustion-of-c-5h-12- | 8 | start physical_unit 5 5 coefficient none qc_end chemical_equation 13 13 qc_end chemical_equation 15 15 qc_end chemical_equation 17 17 qc_end end | [{"type":"physical unit","value":"Coefficient [OF] O2"}] | [{"type":"physical unit","value":"8"}] | [{"type":"chemical equation","value":"C5H12"},{"type":"chemical equation","value":"CO2"},{"type":"chemical equation","value":"H2O"}] | <h1 class="questionTitle" itemprop="name">What is the coefficient for #O_2# when the equation for the combustion of #C_5H_12# to #CO_2# and #H_2O# is balanced?</h1> | null | 8 | <div class="answerDescription">
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<div class="markdown"><p>Is the equation above balanced? How do you know? Hexanes (<mathjax>#C_6H_14#</mathjax>) is also a common fuel. How would you represent its combustion? How is energy transferred in the above reaction?</p></div>
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<div class="markdown"><p><mathjax>#C_5H_(12(l)) + 8O_(2(g)) rarr 5CO_(2(g)) + 6H_2O_((g))#</mathjax></p></div>
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<div class="markdown"><p>Is the equation above balanced? How do you know? Hexanes (<mathjax>#C_6H_14#</mathjax>) is also a common fuel. How would you represent its combustion? How is energy transferred in the above reaction?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the coefficient for #O_2# when the equation for the combustion of #C_5H_12# to #CO_2# and #H_2O# is balanced?</h1>
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<div class="markdown"><p><mathjax>#C_5H_(12(l)) + 8O_(2(g)) rarr 5CO_(2(g)) + 6H_2O_((g))#</mathjax></p></div>
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<div class="markdown"><p>Is the equation above balanced? How do you know? Hexanes (<mathjax>#C_6H_14#</mathjax>) is also a common fuel. How would you represent its combustion? How is energy transferred in the above reaction?</p></div>
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</article> | What is the coefficient for #O_2# when the equation for the combustion of #C_5H_12# to #CO_2# and #H_2O# is balanced? | null |
2,991 | ab04da9b-6ddd-11ea-a0b9-ccda262736ce | https://socratic.org/questions/how-many-grams-of-nacl-should-be-weighed-to-prepare-1-l-of-20-ppm-solution-of-na | 0.05 grams | start physical_unit 4 4 mass g qc_end physical_unit 15 15 10 11 volume qc_end physical_unit 17 17 13 14 molarity qc_end end | [{"type":"physical unit","value":"Weight [OF] NaCl [IN] grams"}] | [{"type":"physical unit","value":"0.05 grams"}] | [{"type":"physical unit","value":"Volume [OF] Na+ solution [=] \\pu{1 L}"},{"type":"physical unit","value":"Molarity [OF] Na+ solution [=] \\pu{20 ppm}"}] | <h1 class="questionTitle" itemprop="name">How many grams of #NaCl# should be weighed to prepare 1 L of 20 ppm solution of #Na^+#?</h1> | null | 0.05 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to determine how much sodium chloride must be dissolved in <strong>1 L</strong> of water, start from the definition of <em>parts per million</em>, <strong>ppm</strong>. </p>
<p>A concentration of <strong>1 ppm</strong> is equivalent to 1 part <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, in your case sodium chloride, for every <strong>1 million</strong> parts <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, in your case water. </p>
<p>To get a solution's concentration in ppm, you multiply the ratio that exists between the mass of the solute and the mass of the water by <strong>1 million</strong>, or <mathjax>#10^6#</mathjax>. </p>
<p>This is the exact same approach you use when calculating <em>percentage</em>, the only difference being the fact that you need to multiply the ratio by 1 million, instead of by 100. </p>
<p>So, you can safely assume the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of water to be equal to <mathjax>#"1 g/mL"#</mathjax>. This would make the mass of water equal to </p>
<p><mathjax>#1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1 g"/(1cancel("mL")) = "1000 g"#</mathjax></p>
<p>Sodium chloride will dissociate completely in aqueous solution to form sodium cations, <mathjax>#Na^(+)#</mathjax>, and chloride anions, <mathjax>#Cl^(-)#</mathjax>. </p>
<p><mathjax>#NaCl_((aq)) -> Na_((aq))^(+) + Cl_((aq))^(-)#</mathjax></p>
<p>The important thing to notice here is that you have a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between sodium chloride and sodium cations. This mole ratio will help you determine the mass of sodium chloride you need to dissolve in order to get this particular solution.</p>
<p>The concentration of the target solution (in ppm) will be </p>
<p><mathjax>#"ppm" = m_"solute"/m_"water" * 10^6#</mathjax></p>
<p>Plug in your values and solve for <mathjax>#m_"solute"#</mathjax>.</p>
<p><mathjax>#m_"solute" = ("ppm" * m_"water")/10^6#</mathjax></p>
<p><mathjax>#m_"solute" = (20 * "1000 g")/10^6 = "0.02 g"#</mathjax></p>
<p>This means that your solution must contain <strong>0.02 g</strong> of sodium cations, <mathjax>#Na^(+)#</mathjax>. Use sodium's molar mass to determine how many moles of sodium cations would be present </p>
<p><mathjax>#0.02cancel("g") * "1 mole"/(23.0cancel("g")) = "0.000870 moles"#</mathjax> <mathjax>#Na^(+)#</mathjax></p>
<p>The aforementioned mole ratio tells you that you must add the <em>same number of moles</em> of sodium chloride to the solution. </p>
<p><mathjax>#0.000870cancel("moles"Na^(+)) * ("1 mole"NaCl)/(1cancel("mole"Na^(+))) = "0.000870 moles"#</mathjax> <mathjax>#NaCl#</mathjax> </p>
<p>Now use sodium chloride's molar mass to determine how much you need</p>
<p><mathjax>#0.000870cancel("moles") * "58.44 g"/(1cancel("mole")) = color(green)("0.05 g NaCl")#</mathjax></p>
<p>So, if you dissolve <strong>0.05 g</strong> of sodium chloride in <strong>1 L</strong> of water you'll get a 20-ppm <mathjax>#Na^(+)#</mathjax> solution. </p></div>
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<div class="markdown"><p>You need <strong>0.05 g</strong> of sodium chloride.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to determine how much sodium chloride must be dissolved in <strong>1 L</strong> of water, start from the definition of <em>parts per million</em>, <strong>ppm</strong>. </p>
<p>A concentration of <strong>1 ppm</strong> is equivalent to 1 part <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, in your case sodium chloride, for every <strong>1 million</strong> parts <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, in your case water. </p>
<p>To get a solution's concentration in ppm, you multiply the ratio that exists between the mass of the solute and the mass of the water by <strong>1 million</strong>, or <mathjax>#10^6#</mathjax>. </p>
<p>This is the exact same approach you use when calculating <em>percentage</em>, the only difference being the fact that you need to multiply the ratio by 1 million, instead of by 100. </p>
<p>So, you can safely assume the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of water to be equal to <mathjax>#"1 g/mL"#</mathjax>. This would make the mass of water equal to </p>
<p><mathjax>#1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1 g"/(1cancel("mL")) = "1000 g"#</mathjax></p>
<p>Sodium chloride will dissociate completely in aqueous solution to form sodium cations, <mathjax>#Na^(+)#</mathjax>, and chloride anions, <mathjax>#Cl^(-)#</mathjax>. </p>
<p><mathjax>#NaCl_((aq)) -> Na_((aq))^(+) + Cl_((aq))^(-)#</mathjax></p>
<p>The important thing to notice here is that you have a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between sodium chloride and sodium cations. This mole ratio will help you determine the mass of sodium chloride you need to dissolve in order to get this particular solution.</p>
<p>The concentration of the target solution (in ppm) will be </p>
<p><mathjax>#"ppm" = m_"solute"/m_"water" * 10^6#</mathjax></p>
<p>Plug in your values and solve for <mathjax>#m_"solute"#</mathjax>.</p>
<p><mathjax>#m_"solute" = ("ppm" * m_"water")/10^6#</mathjax></p>
<p><mathjax>#m_"solute" = (20 * "1000 g")/10^6 = "0.02 g"#</mathjax></p>
<p>This means that your solution must contain <strong>0.02 g</strong> of sodium cations, <mathjax>#Na^(+)#</mathjax>. Use sodium's molar mass to determine how many moles of sodium cations would be present </p>
<p><mathjax>#0.02cancel("g") * "1 mole"/(23.0cancel("g")) = "0.000870 moles"#</mathjax> <mathjax>#Na^(+)#</mathjax></p>
<p>The aforementioned mole ratio tells you that you must add the <em>same number of moles</em> of sodium chloride to the solution. </p>
<p><mathjax>#0.000870cancel("moles"Na^(+)) * ("1 mole"NaCl)/(1cancel("mole"Na^(+))) = "0.000870 moles"#</mathjax> <mathjax>#NaCl#</mathjax> </p>
<p>Now use sodium chloride's molar mass to determine how much you need</p>
<p><mathjax>#0.000870cancel("moles") * "58.44 g"/(1cancel("mole")) = color(green)("0.05 g NaCl")#</mathjax></p>
<p>So, if you dissolve <strong>0.05 g</strong> of sodium chloride in <strong>1 L</strong> of water you'll get a 20-ppm <mathjax>#Na^(+)#</mathjax> solution. </p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of #NaCl# should be weighed to prepare 1 L of 20 ppm solution of #Na^+#?</h1>
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<div class="markdown"><p>You need <strong>0.05 g</strong> of sodium chloride.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to determine how much sodium chloride must be dissolved in <strong>1 L</strong> of water, start from the definition of <em>parts per million</em>, <strong>ppm</strong>. </p>
<p>A concentration of <strong>1 ppm</strong> is equivalent to 1 part <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, in your case sodium chloride, for every <strong>1 million</strong> parts <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solvent">solvent</a>, in your case water. </p>
<p>To get a solution's concentration in ppm, you multiply the ratio that exists between the mass of the solute and the mass of the water by <strong>1 million</strong>, or <mathjax>#10^6#</mathjax>. </p>
<p>This is the exact same approach you use when calculating <em>percentage</em>, the only difference being the fact that you need to multiply the ratio by 1 million, instead of by 100. </p>
<p>So, you can safely assume the <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> of water to be equal to <mathjax>#"1 g/mL"#</mathjax>. This would make the mass of water equal to </p>
<p><mathjax>#1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1 g"/(1cancel("mL")) = "1000 g"#</mathjax></p>
<p>Sodium chloride will dissociate completely in aqueous solution to form sodium cations, <mathjax>#Na^(+)#</mathjax>, and chloride anions, <mathjax>#Cl^(-)#</mathjax>. </p>
<p><mathjax>#NaCl_((aq)) -> Na_((aq))^(+) + Cl_((aq))^(-)#</mathjax></p>
<p>The important thing to notice here is that you have a <mathjax>#1:1#</mathjax> <a href="http://socratic.org/chemistry/stoichiometry/mole-ratios">mole ratio</a> between sodium chloride and sodium cations. This mole ratio will help you determine the mass of sodium chloride you need to dissolve in order to get this particular solution.</p>
<p>The concentration of the target solution (in ppm) will be </p>
<p><mathjax>#"ppm" = m_"solute"/m_"water" * 10^6#</mathjax></p>
<p>Plug in your values and solve for <mathjax>#m_"solute"#</mathjax>.</p>
<p><mathjax>#m_"solute" = ("ppm" * m_"water")/10^6#</mathjax></p>
<p><mathjax>#m_"solute" = (20 * "1000 g")/10^6 = "0.02 g"#</mathjax></p>
<p>This means that your solution must contain <strong>0.02 g</strong> of sodium cations, <mathjax>#Na^(+)#</mathjax>. Use sodium's molar mass to determine how many moles of sodium cations would be present </p>
<p><mathjax>#0.02cancel("g") * "1 mole"/(23.0cancel("g")) = "0.000870 moles"#</mathjax> <mathjax>#Na^(+)#</mathjax></p>
<p>The aforementioned mole ratio tells you that you must add the <em>same number of moles</em> of sodium chloride to the solution. </p>
<p><mathjax>#0.000870cancel("moles"Na^(+)) * ("1 mole"NaCl)/(1cancel("mole"Na^(+))) = "0.000870 moles"#</mathjax> <mathjax>#NaCl#</mathjax> </p>
<p>Now use sodium chloride's molar mass to determine how much you need</p>
<p><mathjax>#0.000870cancel("moles") * "58.44 g"/(1cancel("mole")) = color(green)("0.05 g NaCl")#</mathjax></p>
<p>So, if you dissolve <strong>0.05 g</strong> of sodium chloride in <strong>1 L</strong> of water you'll get a 20-ppm <mathjax>#Na^(+)#</mathjax> solution. </p></div>
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</article> | How many grams of #NaCl# should be weighed to prepare 1 L of 20 ppm solution of #Na^+#? | null |
2,992 | aa895dde-6ddd-11ea-bfe1-ccda262736ce | https://socratic.org/questions/at-stp-if-4-20l-of-o-2-reacts-with-n-2h-4-how-many-liters-of-water-vapor-will-be | 8.40 liters | start physical_unit 14 15 volume l qc_end physical_unit 6 6 3 4 volume qc_end chemical_equation 9 9 qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] water vapor [IN] liters"}] | [{"type":"physical unit","value":"8.40 liters"}] | [{"type":"physical unit","value":"Volume [OF] O2 [=] \\pu{4.20 L}"},{"type":"chemical equation","value":"N2H4"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">At STP, if 4.20L of #O_2# reacts with #N_2H_4#, how many liters of water vapor will be produced?</h1> | null | 8.40 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I apologize in advance because this explanation is very long!</p>
<p>First, we need the entire chemical reaction in order to arrive at the answer:</p>
<p><mathjax>#O_2#</mathjax> <mathjax>#+#</mathjax> <mathjax>#N_2H_4#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2H_2O#</mathjax> <mathjax>#+#</mathjax> <mathjax>#N_2#</mathjax></p>
<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, <br/>
<mathjax>#PxxV=nxxRxxT#</mathjax></p>
<p>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 <mathjax>#(Lxxatm)/(molxxK)#</mathjax>), 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 water. 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 <mathjax>#(Lxxatm)/(molxxK)#</mathjax>.</p>
<p>The only problem is that we are given the volume of oxygen, so we have to convert the given volume of oxygen to moles of oxygen. Then we must use the balanced equation to go from moles of oxygen to moles of water, which will allow us to go to volume of water using the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>. </p>
<p>Rearrange the equation to solve for moles of <mathjax>#O_2#</mathjax>:</p>
<p>n <mathjax>#=#</mathjax> <mathjax>#(PxxV)/(RxxT)#</mathjax></p>
<p>n = <mathjax>#(1 (cancel "atm") xx4.20(cancel"L"))/(0.0821(cancel"L"x cancel"atm")/(molxxcancel"K")xx273cancel"K")#</mathjax> = 0.1874 mol <mathjax>#O_2#</mathjax></p>
<p>Moles of <mathjax>#2H_2O#</mathjax> = (0.1874 (<mathjax>#cancel ("mol#</mathjax><mathjax>#O_2#</mathjax>)<mathjax># #</mathjax><mathjax>#xx#</mathjax> 2 mol <mathjax>#2H_2O#</mathjax>)/(<mathjax>#1 (cancel "mol" #</mathjax><mathjax>#O_2#</mathjax>)) = 0.3748 moles <mathjax>#H_2O#</mathjax></p>
<p>Rearrange the equation to solve for volume of <mathjax>#H_2O#</mathjax>:<br/>
<mathjax>#(nxxRxxT)/P#</mathjax> <mathjax>#=#</mathjax><mathjax>#V#</mathjax></p>
<p><mathjax>#(0.3748 cancel"mol" xx0.0821(L xx cancel"atm")/(cancel"mol" xxcancel"K")xx273(cancel"K"))/(1cancel"atm")
#</mathjax>=<mathjax># #</mathjax>8.40 L</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>8.40 L of <mathjax>#H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I apologize in advance because this explanation is very long!</p>
<p>First, we need the entire chemical reaction in order to arrive at the answer:</p>
<p><mathjax>#O_2#</mathjax> <mathjax>#+#</mathjax> <mathjax>#N_2H_4#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2H_2O#</mathjax> <mathjax>#+#</mathjax> <mathjax>#N_2#</mathjax></p>
<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, <br/>
<mathjax>#PxxV=nxxRxxT#</mathjax></p>
<p>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 <mathjax>#(Lxxatm)/(molxxK)#</mathjax>), 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 water. 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 <mathjax>#(Lxxatm)/(molxxK)#</mathjax>.</p>
<p>The only problem is that we are given the volume of oxygen, so we have to convert the given volume of oxygen to moles of oxygen. Then we must use the balanced equation to go from moles of oxygen to moles of water, which will allow us to go to volume of water using the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>. </p>
<p>Rearrange the equation to solve for moles of <mathjax>#O_2#</mathjax>:</p>
<p>n <mathjax>#=#</mathjax> <mathjax>#(PxxV)/(RxxT)#</mathjax></p>
<p>n = <mathjax>#(1 (cancel "atm") xx4.20(cancel"L"))/(0.0821(cancel"L"x cancel"atm")/(molxxcancel"K")xx273cancel"K")#</mathjax> = 0.1874 mol <mathjax>#O_2#</mathjax></p>
<p>Moles of <mathjax>#2H_2O#</mathjax> = (0.1874 (<mathjax>#cancel ("mol#</mathjax><mathjax>#O_2#</mathjax>)<mathjax># #</mathjax><mathjax>#xx#</mathjax> 2 mol <mathjax>#2H_2O#</mathjax>)/(<mathjax>#1 (cancel "mol" #</mathjax><mathjax>#O_2#</mathjax>)) = 0.3748 moles <mathjax>#H_2O#</mathjax></p>
<p>Rearrange the equation to solve for volume of <mathjax>#H_2O#</mathjax>:<br/>
<mathjax>#(nxxRxxT)/P#</mathjax> <mathjax>#=#</mathjax><mathjax>#V#</mathjax></p>
<p><mathjax>#(0.3748 cancel"mol" xx0.0821(L xx cancel"atm")/(cancel"mol" xxcancel"K")xx273(cancel"K"))/(1cancel"atm")
#</mathjax>=<mathjax># #</mathjax>8.40 L</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">At STP, if 4.20L of #O_2# reacts with #N_2H_4#, how many liters of water vapor will be produced?</h1>
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Jun 6, 2016
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<div class="markdown"><p>8.40 L of <mathjax>#H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I apologize in advance because this explanation is very long!</p>
<p>First, we need the entire chemical reaction in order to arrive at the answer:</p>
<p><mathjax>#O_2#</mathjax> <mathjax>#+#</mathjax> <mathjax>#N_2H_4#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#2H_2O#</mathjax> <mathjax>#+#</mathjax> <mathjax>#N_2#</mathjax></p>
<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, <br/>
<mathjax>#PxxV=nxxRxxT#</mathjax></p>
<p>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 <mathjax>#(Lxxatm)/(molxxK)#</mathjax>), 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 water. 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 <mathjax>#(Lxxatm)/(molxxK)#</mathjax>.</p>
<p>The only problem is that we are given the volume of oxygen, so we have to convert the given volume of oxygen to moles of oxygen. Then we must use the balanced equation to go from moles of oxygen to moles of water, which will allow us to go to volume of water using the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>. </p>
<p>Rearrange the equation to solve for moles of <mathjax>#O_2#</mathjax>:</p>
<p>n <mathjax>#=#</mathjax> <mathjax>#(PxxV)/(RxxT)#</mathjax></p>
<p>n = <mathjax>#(1 (cancel "atm") xx4.20(cancel"L"))/(0.0821(cancel"L"x cancel"atm")/(molxxcancel"K")xx273cancel"K")#</mathjax> = 0.1874 mol <mathjax>#O_2#</mathjax></p>
<p>Moles of <mathjax>#2H_2O#</mathjax> = (0.1874 (<mathjax>#cancel ("mol#</mathjax><mathjax>#O_2#</mathjax>)<mathjax># #</mathjax><mathjax>#xx#</mathjax> 2 mol <mathjax>#2H_2O#</mathjax>)/(<mathjax>#1 (cancel "mol" #</mathjax><mathjax>#O_2#</mathjax>)) = 0.3748 moles <mathjax>#H_2O#</mathjax></p>
<p>Rearrange the equation to solve for volume of <mathjax>#H_2O#</mathjax>:<br/>
<mathjax>#(nxxRxxT)/P#</mathjax> <mathjax>#=#</mathjax><mathjax>#V#</mathjax></p>
<p><mathjax>#(0.3748 cancel"mol" xx0.0821(L xx cancel"atm")/(cancel"mol" xxcancel"K")xx273(cancel"K"))/(1cancel"atm")
#</mathjax>=<mathjax># #</mathjax>8.40 L</p></div>
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</article> | At STP, if 4.20L of #O_2# reacts with #N_2H_4#, how many liters of water vapor will be produced? | null |
2,993 | a9c6d088-6ddd-11ea-9b9f-ccda262736ce | https://socratic.org/questions/a-mixture-of-four-gases-exerts-a-total-pressure-of-860-mm-hg-gases-a-and-b-each- | 310.00 mmHg | start physical_unit 30 31 pressure mmhg qc_end physical_unit 1 1 10 11 total_pressure qc_end physical_unit 20 21 23 24 pressure qc_end end | [{"type":"physical unit","value":"Exerted pressure [OF] gas D [IN] mmHg"}] | [{"type":"physical unit","value":"310.00 mmHg"}] | [{"type":"physical unit","value":"Total pressure [OF] the mixture [=] \\pu{860 mmHg}"},{"type":"physical unit","value":"Exerted pressure [OF] gas A [=] \\pu{220 mmHg}"},{"type":"physical unit","value":"Exerted pressure [OF] gas B [=] \\pu{220 mmHg}"},{"type":"physical unit","value":"Exerted pressure [OF] gas C [=] \\pu{110 mmHg}"}] | <h1 class="questionTitle" itemprop="name">A mixture of four gases exerts a total pressure of 860 mmHg. Gases A and B each exert 220 mmHg. Gas C exerts 110 mmHg. What pressure is exerted by gas D?</h1> | null | 310.00 mmHg | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here in order to figure out the pressure exerted by gas <mathjax>#"D"#</mathjax> is use <strong>Dalton's Law of Partial Pressures</strong>.</p>
<p>In essence, <em>Dalton's Law of Partial Pressures</em> states that the <strong>total pressure</strong> exerted by a mixture of gases will be equal to the sum of the <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a></em> each constituent of the mixture exerts <strong>in the same volume</strong> and under the <strong>same conditions</strong> for pressure and temperature. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(blue)(P_"total" = sum_i P_icolor(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><img alt="http://ch301.cm.utexas.edu/gases/#mixtures/mixtures-all.php" src="https://useruploads.socratic.org/kqG6DPRZitTaEjdru4Pw_partial-pressures.svg"/> </p>
<p>In your case, four gases labeled <mathjax>#"A"#</mathjax>, <mathjax>#"B"#</mathjax>, <mathjax>#"C"#</mathjax>, and <mathjax>#"D"#</mathjax> are placed in a container. The pressure of the gaseous mixture is said to be equal to <mathjax>#"860 mmHg"#</mathjax>. </p>
<p>This means that the partial pressures of the four gases <strong>must add up</strong> to give <mathjax>#"860 mmHg"#</mathjax>. </p>
<p>You will thus have </p>
<blockquote>
<p><mathjax>#P_"total" = P_A + P_B + P_C + P_D#</mathjax></p>
</blockquote>
<p>This means that the partial pressure of gas <mathjax>#"D"#</mathjax> will be equal to </p>
<blockquote>
<p><mathjax>#P_D = "860 mmHg" - (#</mathjax> <mathjax>#overbrace("220 mmHg")^(color(blue)(P_A)) + overbrace("220 mmHg")^(color(red)(P_B)) + overbrace("110 mmHg")^(color(brown)(P_C))#</mathjax> <mathjax>#)#</mathjax></p>
<p><mathjax>#P_D = "860 mmHg" - "550 mm Hg" = color(green)(|bar(ul(color(white)(a/a)"310 mmHg"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"310 mmHg"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here in order to figure out the pressure exerted by gas <mathjax>#"D"#</mathjax> is use <strong>Dalton's Law of Partial Pressures</strong>.</p>
<p>In essence, <em>Dalton's Law of Partial Pressures</em> states that the <strong>total pressure</strong> exerted by a mixture of gases will be equal to the sum of the <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a></em> each constituent of the mixture exerts <strong>in the same volume</strong> and under the <strong>same conditions</strong> for pressure and temperature. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(blue)(P_"total" = sum_i P_icolor(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><img alt="http://ch301.cm.utexas.edu/gases/#mixtures/mixtures-all.php" src="https://useruploads.socratic.org/kqG6DPRZitTaEjdru4Pw_partial-pressures.svg"/> </p>
<p>In your case, four gases labeled <mathjax>#"A"#</mathjax>, <mathjax>#"B"#</mathjax>, <mathjax>#"C"#</mathjax>, and <mathjax>#"D"#</mathjax> are placed in a container. The pressure of the gaseous mixture is said to be equal to <mathjax>#"860 mmHg"#</mathjax>. </p>
<p>This means that the partial pressures of the four gases <strong>must add up</strong> to give <mathjax>#"860 mmHg"#</mathjax>. </p>
<p>You will thus have </p>
<blockquote>
<p><mathjax>#P_"total" = P_A + P_B + P_C + P_D#</mathjax></p>
</blockquote>
<p>This means that the partial pressure of gas <mathjax>#"D"#</mathjax> will be equal to </p>
<blockquote>
<p><mathjax>#P_D = "860 mmHg" - (#</mathjax> <mathjax>#overbrace("220 mmHg")^(color(blue)(P_A)) + overbrace("220 mmHg")^(color(red)(P_B)) + overbrace("110 mmHg")^(color(brown)(P_C))#</mathjax> <mathjax>#)#</mathjax></p>
<p><mathjax>#P_D = "860 mmHg" - "550 mm Hg" = color(green)(|bar(ul(color(white)(a/a)"310 mmHg"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">A mixture of four gases exerts a total pressure of 860 mmHg. Gases A and B each exert 220 mmHg. Gas C exerts 110 mmHg. What pressure is exerted by gas D?</h1>
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<div class="markdown"><p><mathjax>#"310 mmHg"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>All you have to do here in order to figure out the pressure exerted by gas <mathjax>#"D"#</mathjax> is use <strong>Dalton's Law of Partial Pressures</strong>.</p>
<p>In essence, <em>Dalton's Law of Partial Pressures</em> states that the <strong>total pressure</strong> exerted by a mixture of gases will be equal to the sum of the <em><a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a></em> each constituent of the mixture exerts <strong>in the same volume</strong> and under the <strong>same conditions</strong> for pressure and temperature. </p>
<blockquote>
<p><mathjax>#color(blue)(|bar(ul(color(blue)(P_"total" = sum_i P_icolor(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p><img alt="http://ch301.cm.utexas.edu/gases/#mixtures/mixtures-all.php" src="https://useruploads.socratic.org/kqG6DPRZitTaEjdru4Pw_partial-pressures.svg"/> </p>
<p>In your case, four gases labeled <mathjax>#"A"#</mathjax>, <mathjax>#"B"#</mathjax>, <mathjax>#"C"#</mathjax>, and <mathjax>#"D"#</mathjax> are placed in a container. The pressure of the gaseous mixture is said to be equal to <mathjax>#"860 mmHg"#</mathjax>. </p>
<p>This means that the partial pressures of the four gases <strong>must add up</strong> to give <mathjax>#"860 mmHg"#</mathjax>. </p>
<p>You will thus have </p>
<blockquote>
<p><mathjax>#P_"total" = P_A + P_B + P_C + P_D#</mathjax></p>
</blockquote>
<p>This means that the partial pressure of gas <mathjax>#"D"#</mathjax> will be equal to </p>
<blockquote>
<p><mathjax>#P_D = "860 mmHg" - (#</mathjax> <mathjax>#overbrace("220 mmHg")^(color(blue)(P_A)) + overbrace("220 mmHg")^(color(red)(P_B)) + overbrace("110 mmHg")^(color(brown)(P_C))#</mathjax> <mathjax>#)#</mathjax></p>
<p><mathjax>#P_D = "860 mmHg" - "550 mm Hg" = color(green)(|bar(ul(color(white)(a/a)"310 mmHg"color(white)(a/a)|)))#</mathjax></p>
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</article> | A mixture of four gases exerts a total pressure of 860 mmHg. Gases A and B each exert 220 mmHg. Gas C exerts 110 mmHg. What pressure is exerted by gas D? | null |
2,994 | a86c3094-6ddd-11ea-8c52-ccda262736ce | https://socratic.org/questions/what-volume-does-0-0250-mole-h-2-occupy-at-0-821-atm-pressure-and-300-k | 0.75 L | start physical_unit 5 5 volume l qc_end physical_unit 5 5 3 4 mole qc_end physical_unit 5 5 8 9 pressure qc_end physical_unit 5 5 12 13 temperature qc_end end | [{"type":"physical unit","value":"Volume [OF] H2 [IN] L"}] | [{"type":"physical unit","value":"0.75 L"}] | [{"type":"physical unit","value":"Mole [OF] H2 [=] \\pu{0.0250 moles}"},{"type":"physical unit","value":"Pressure [OF] H2 [=] \\pu{0.821 atm}"},{"type":"physical unit","value":"Temperature [OF] H2 [=] \\pu{300 K}"}] | <h1 class="questionTitle" itemprop="name">What volume does 0.0250 mole #H_2# occupy at 0.821 atm pressure and 300 K?</h1> | null | 0.75 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Hydrogen behaves like an ideal gas.</p>
<p><mathjax>#P*V=n*R*T#</mathjax>, where</p>
<p><mathjax>#P = "the pressure" = 0.821 color(red)(cancel(color(black)("atm")))*(1.01325*10^5 "Pa")/(1 color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#P = "83 190" color(red)(cancel(color(black)("Pa"))) * "1N/m"^2/(1 color(red)(cancel(color(black)("Pa")))) = "83 190 N/m"^2#</mathjax></p>
<p><mathjax>#T= "the temperature"#</mathjax></p>
<p><mathjax>#n = "number of moles"#</mathjax></p>
<p><mathjax>#R= "ideal gas constant" = 8.314 color(red)(cancel(color(black)("J")))//("K·mol") * "1 N·m"/(1 color(red)(cancel(color(black)("J")))) = "8.314 N·m/(K·mol)"#</mathjax></p>
<p><mathjax>#V = "the volume"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#V=(n*R*T)/P#</mathjax></p>
<p><mathjax>#V=((0.250 color(red)(cancel(color(black)("mol"))) * 8.314 color(red)(cancel(color(black)("N")))*"m")//(color(red)(cancel(color(black)("K·mol")))) · 300 color(red)(cancel(color(black)("K"))))/("83 190" color(red)(cancel(color(black)("N")))//"m"^2)#</mathjax></p>
<p><mathjax>#V = 7.50*10^-4 " m"^3 = "0.750 dm"^3#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The volume is <mathjax>#"0.750 dm"^3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Hydrogen behaves like an ideal gas.</p>
<p><mathjax>#P*V=n*R*T#</mathjax>, where</p>
<p><mathjax>#P = "the pressure" = 0.821 color(red)(cancel(color(black)("atm")))*(1.01325*10^5 "Pa")/(1 color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#P = "83 190" color(red)(cancel(color(black)("Pa"))) * "1N/m"^2/(1 color(red)(cancel(color(black)("Pa")))) = "83 190 N/m"^2#</mathjax></p>
<p><mathjax>#T= "the temperature"#</mathjax></p>
<p><mathjax>#n = "number of moles"#</mathjax></p>
<p><mathjax>#R= "ideal gas constant" = 8.314 color(red)(cancel(color(black)("J")))//("K·mol") * "1 N·m"/(1 color(red)(cancel(color(black)("J")))) = "8.314 N·m/(K·mol)"#</mathjax></p>
<p><mathjax>#V = "the volume"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#V=(n*R*T)/P#</mathjax></p>
<p><mathjax>#V=((0.250 color(red)(cancel(color(black)("mol"))) * 8.314 color(red)(cancel(color(black)("N")))*"m")//(color(red)(cancel(color(black)("K·mol")))) · 300 color(red)(cancel(color(black)("K"))))/("83 190" color(red)(cancel(color(black)("N")))//"m"^2)#</mathjax></p>
<p><mathjax>#V = 7.50*10^-4 " m"^3 = "0.750 dm"^3#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What volume does 0.0250 mole #H_2# occupy at 0.821 atm pressure and 300 K?</h1>
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<div class="markdown"><p>The volume is <mathjax>#"0.750 dm"^3#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>Hydrogen behaves like an ideal gas.</p>
<p><mathjax>#P*V=n*R*T#</mathjax>, where</p>
<p><mathjax>#P = "the pressure" = 0.821 color(red)(cancel(color(black)("atm")))*(1.01325*10^5 "Pa")/(1 color(red)(cancel(color(black)("atm"))))#</mathjax></p>
<p><mathjax>#P = "83 190" color(red)(cancel(color(black)("Pa"))) * "1N/m"^2/(1 color(red)(cancel(color(black)("Pa")))) = "83 190 N/m"^2#</mathjax></p>
<p><mathjax>#T= "the temperature"#</mathjax></p>
<p><mathjax>#n = "number of moles"#</mathjax></p>
<p><mathjax>#R= "ideal gas constant" = 8.314 color(red)(cancel(color(black)("J")))//("K·mol") * "1 N·m"/(1 color(red)(cancel(color(black)("J")))) = "8.314 N·m/(K·mol)"#</mathjax></p>
<p><mathjax>#V = "the volume"#</mathjax></p>
<blockquote></blockquote>
<p><mathjax>#V=(n*R*T)/P#</mathjax></p>
<p><mathjax>#V=((0.250 color(red)(cancel(color(black)("mol"))) * 8.314 color(red)(cancel(color(black)("N")))*"m")//(color(red)(cancel(color(black)("K·mol")))) · 300 color(red)(cancel(color(black)("K"))))/("83 190" color(red)(cancel(color(black)("N")))//"m"^2)#</mathjax></p>
<p><mathjax>#V = 7.50*10^-4 " m"^3 = "0.750 dm"^3#</mathjax></p></div>
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<div class="markdown"><p>The volume will be <mathjax>#"0.7 L"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>.</p>
<p><mathjax>#PV=nRT#</mathjax>, where <mathjax>#n#</mathjax> is moles, and <mathjax>#R#</mathjax> is the gas constant.</p>
<p><strong>Given/Known</strong><br/>
<mathjax>#P="0.821 atm"#</mathjax><br/>
<mathjax>#n="0.0250 mol"#</mathjax><br/>
<mathjax>#R="0.082057338 L atm K"^(-1) "mol"^(-1)#</mathjax><br/>
<mathjax>#T="300 K"#</mathjax></p>
<p><strong>Unknown</strong></p>
<p><mathjax>#V#</mathjax></p>
<p><strong>Equation</strong></p>
<p><mathjax>#PV=nRT#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#V#</mathjax> and solve.</p>
<p><mathjax>#V=(nRT)/P#</mathjax></p>
<p><mathjax>#V=((0.0250cancel"mol"xx0.08205733" L" cancel"atm" cancel("K"^(-1)) cancel("mol"^(-1)) xx 300cancel"K"))/(0.821cancel"atm")="0.7 L"#</mathjax> (rounded to one significant figure due to 300 K)</p></div>
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</article> | What volume does 0.0250 mole #H_2# occupy at 0.821 atm pressure and 300 K? | null |
2,995 | acb6eec0-6ddd-11ea-8c2b-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-0-0067-m-koh-solution | 11.83 | start physical_unit 8 9 ph none qc_end physical_unit 8 9 6 7 molarity qc_end end | [{"type":"physical unit","value":"pH [OF] KOH solution"}] | [{"type":"physical unit","value":"11.83"}] | [{"type":"physical unit","value":"Molarity [OF] KOH solution [=] \\pu{0.0067 M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a 0.0067 M KOH solution? </h1> | null | 11.83 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium hydroxide fully ionizes when dissolved in water according to the following equation:</p>
<p><mathjax>#KOH -> K^+ + OH^-#</mathjax></p>
<p>Now, let's find the relationship between <mathjax>#KOH#</mathjax> and <mathjax>#OH^-#</mathjax></p>
<p><mathjax>#" " 1 " : "1 " ratio "=> [KOH] =[OH^-]= 6.7xx10^-3M#</mathjax></p>
<p>In any aqueous solution, <mathjax># [H_3O^+] #</mathjax>and <mathjax>#[OH^-]#</mathjax> must satisfy the following condition:</p>
<p><mathjax>#[H_3O^+] [OH^-]= K_w#</mathjax></p>
<p><mathjax>#[H_3O^+] = K_w/([OH^-])#</mathjax></p>
<p><mathjax>#[H_3O^+] = (1.0xx10^-14)/(6.7xx10^-3)#</mathjax></p>
<p><mathjax>#[H_3O^+] = 1.5xx10^-12 M#</mathjax></p>
<p>Now, after finding the concentration of the hydronium ion, the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution is determined:</p>
<p><mathjax>#pH = -log[H_3O^+] #</mathjax></p>
<p><mathjax>#pH = - log[1.5xx10^-12]#</mathjax></p>
<p><mathjax>#pH=11.83#</mathjax></p>
<p>Other Method<br/>
Find the pOH using the concentration of the hydroxide ion, then use the formula <mathjax>#" "pH + POH = 14 #</mathjax> to find the pH.</p></div>
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<div class="markdown"><p><mathjax>#pH=11.83#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium hydroxide fully ionizes when dissolved in water according to the following equation:</p>
<p><mathjax>#KOH -> K^+ + OH^-#</mathjax></p>
<p>Now, let's find the relationship between <mathjax>#KOH#</mathjax> and <mathjax>#OH^-#</mathjax></p>
<p><mathjax>#" " 1 " : "1 " ratio "=> [KOH] =[OH^-]= 6.7xx10^-3M#</mathjax></p>
<p>In any aqueous solution, <mathjax># [H_3O^+] #</mathjax>and <mathjax>#[OH^-]#</mathjax> must satisfy the following condition:</p>
<p><mathjax>#[H_3O^+] [OH^-]= K_w#</mathjax></p>
<p><mathjax>#[H_3O^+] = K_w/([OH^-])#</mathjax></p>
<p><mathjax>#[H_3O^+] = (1.0xx10^-14)/(6.7xx10^-3)#</mathjax></p>
<p><mathjax>#[H_3O^+] = 1.5xx10^-12 M#</mathjax></p>
<p>Now, after finding the concentration of the hydronium ion, the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution is determined:</p>
<p><mathjax>#pH = -log[H_3O^+] #</mathjax></p>
<p><mathjax>#pH = - log[1.5xx10^-12]#</mathjax></p>
<p><mathjax>#pH=11.83#</mathjax></p>
<p>Other Method<br/>
Find the pOH using the concentration of the hydroxide ion, then use the formula <mathjax>#" "pH + POH = 14 #</mathjax> to find the pH.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a 0.0067 M KOH solution? </h1>
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<div class="markdown"><p><mathjax>#pH=11.83#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Potassium hydroxide fully ionizes when dissolved in water according to the following equation:</p>
<p><mathjax>#KOH -> K^+ + OH^-#</mathjax></p>
<p>Now, let's find the relationship between <mathjax>#KOH#</mathjax> and <mathjax>#OH^-#</mathjax></p>
<p><mathjax>#" " 1 " : "1 " ratio "=> [KOH] =[OH^-]= 6.7xx10^-3M#</mathjax></p>
<p>In any aqueous solution, <mathjax># [H_3O^+] #</mathjax>and <mathjax>#[OH^-]#</mathjax> must satisfy the following condition:</p>
<p><mathjax>#[H_3O^+] [OH^-]= K_w#</mathjax></p>
<p><mathjax>#[H_3O^+] = K_w/([OH^-])#</mathjax></p>
<p><mathjax>#[H_3O^+] = (1.0xx10^-14)/(6.7xx10^-3)#</mathjax></p>
<p><mathjax>#[H_3O^+] = 1.5xx10^-12 M#</mathjax></p>
<p>Now, after finding the concentration of the hydronium ion, the <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of the solution is determined:</p>
<p><mathjax>#pH = -log[H_3O^+] #</mathjax></p>
<p><mathjax>#pH = - log[1.5xx10^-12]#</mathjax></p>
<p><mathjax>#pH=11.83#</mathjax></p>
<p>Other Method<br/>
Find the pOH using the concentration of the hydroxide ion, then use the formula <mathjax>#" "pH + POH = 14 #</mathjax> to find the pH.</p></div>
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</article> | What is the pH of a 0.0067 M KOH solution? | null |
2,996 | a8f5b6b6-6ddd-11ea-8681-ccda262736ce | https://socratic.org/questions/1-for-a-given-amount-type-and-temperature-of-gas-when-its-pressure-is-increased- | 1.2 cubic meters | start physical_unit 9 9 volume m^3 qc_end physical_unit 9 9 31 33 volume qc_end physical_unit 9 9 16 17 pressure qc_end physical_unit 9 9 19 20 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] cubic meters"}] | [{"type":"physical unit","value":"1.2 cubic meters"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{3 cubic meters}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{200 kPa}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{500 kPa}"}] | <h1 class="questionTitle" itemprop="name">1: For a given amount, type, and temperature of gas, when its pressure is increased from 200 kPa to 500 kPa, what will its volume be if its initial volume is 3 cubic meters? </h1> | null | 1.2 cubic meters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P*V=n*R*T " " #</mathjax>(Ideal gas law)</p>
<p>Assuming that the amount of gas and the temperature are fixed, the product of the pressure and volume is constant at any given moment. Using this, you can find the simple equation:</p>
<p>For <mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax> as the initial pressure and volume and <mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax> as the pressure and volume at any given moment.</p>
<p><mathjax>#P_1 xx V_1 = P_2 xx V_2#</mathjax></p>
<p>Writing the values, we get:</p>
<p><mathjax>#"3 m"^3 xx "200 kPa" = V_2 xx "500 kPa"#</mathjax></p>
<p>So </p>
<p><mathjax>#"600 kPa m"^3 = V_2 xx "500 kPa"#</mathjax></p>
<p><mathjax>#V_2 = "600 kPa m"^3/"500 kPa" = 6/5 quad "m"^3 = "1.2 m"^3#</mathjax></p>
<p>Since we only played with numbers, the volume is still as cubic meters. </p></div>
</div>
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<div class="answerSummary">
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<div class="markdown"><p><mathjax>#"1.2 m"^3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P*V=n*R*T " " #</mathjax>(Ideal gas law)</p>
<p>Assuming that the amount of gas and the temperature are fixed, the product of the pressure and volume is constant at any given moment. Using this, you can find the simple equation:</p>
<p>For <mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax> as the initial pressure and volume and <mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax> as the pressure and volume at any given moment.</p>
<p><mathjax>#P_1 xx V_1 = P_2 xx V_2#</mathjax></p>
<p>Writing the values, we get:</p>
<p><mathjax>#"3 m"^3 xx "200 kPa" = V_2 xx "500 kPa"#</mathjax></p>
<p>So </p>
<p><mathjax>#"600 kPa m"^3 = V_2 xx "500 kPa"#</mathjax></p>
<p><mathjax>#V_2 = "600 kPa m"^3/"500 kPa" = 6/5 quad "m"^3 = "1.2 m"^3#</mathjax></p>
<p>Since we only played with numbers, the volume is still as cubic meters. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">1: For a given amount, type, and temperature of gas, when its pressure is increased from 200 kPa to 500 kPa, what will its volume be if its initial volume is 3 cubic meters? </h1>
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<div class="markdown"><p><mathjax>#"1.2 m"^3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P*V=n*R*T " " #</mathjax>(Ideal gas law)</p>
<p>Assuming that the amount of gas and the temperature are fixed, the product of the pressure and volume is constant at any given moment. Using this, you can find the simple equation:</p>
<p>For <mathjax>#P_1#</mathjax> and <mathjax>#V_1#</mathjax> as the initial pressure and volume and <mathjax>#P_2#</mathjax> and <mathjax>#V_2#</mathjax> as the pressure and volume at any given moment.</p>
<p><mathjax>#P_1 xx V_1 = P_2 xx V_2#</mathjax></p>
<p>Writing the values, we get:</p>
<p><mathjax>#"3 m"^3 xx "200 kPa" = V_2 xx "500 kPa"#</mathjax></p>
<p>So </p>
<p><mathjax>#"600 kPa m"^3 = V_2 xx "500 kPa"#</mathjax></p>
<p><mathjax>#V_2 = "600 kPa m"^3/"500 kPa" = 6/5 quad "m"^3 = "1.2 m"^3#</mathjax></p>
<p>Since we only played with numbers, the volume is still as cubic meters. </p></div>
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</article> | 1: For a given amount, type, and temperature of gas, when its pressure is increased from 200 kPa to 500 kPa, what will its volume be if its initial volume is 3 cubic meters? | null |
2,997 | ab116e02-6ddd-11ea-95a3-ccda262736ce | https://socratic.org/questions/hydrogen-sulfide-decomposes-according-to-the-following-reaction-for-which-kc-9-3 | 0.01 mol/L | start physical_unit 23 23 equilibrium_concentration mol/l qc_end physical_unit 7 7 12 14 equilibrium_constant_k qc_end physical_unit 7 7 16 18 temperature qc_end physical_unit 20 20 27 28 mole qc_end physical_unit 37 37 35 36 volume qc_end chemical_equation 19 25 qc_end end | [{"type":"physical unit","value":"Equilibrium concentration [OF] H2(g) [IN] mol/L"}] | [{"type":"physical unit","value":"0.01 mol/L"}] | [{"type":"physical unit","value":"Kc [OF] the reaction [=] \\pu{9.30 × 10^(-8)}"},{"type":"physical unit","value":"Temperature1 [OF] the reaction [=] \\pu{700 degrees Celsius}"},{"type":"physical unit","value":"Mole [OF] H2S [=] \\pu{0.29 moles}"},{"type":"physical unit","value":"Volume [OF] container [=] \\pu{3.0 L}"},{"type":"chemical equation","value":"2 H2S(g) -> 2 H2(g) + S2(g)"}] | <h1 class="questionTitle" itemprop="name">Hydrogen sulfide decomposes according to the following reaction, for which Kc=9.30E-8 at 700 degrees Celsius. 2 H2S(g) --> 2 H2(g) + S2(g) If 0.29 moles of H2S is placed in a 3.0-L container, What is the equilibrium concentration of H2(g) at 700 degrees Celsius?</h1> | null | 0.01 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the initial concentrations</strong></p>
<p><mathjax>#["H"_2"S"]_0 = "0.29 mol"/"3.0 L" = "0.0967 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Write the balanced equation and set up an ICE table.</strong></p>
<p><mathjax>#color(white)(mmmmmmm)"2H"_2"S"color(white)(m) ⇌color(white)(m) "2H"_2 + "S"_2#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(m) "0.096 67" color(white)(mmmm)0 color(white)(mm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mm) "-"2xcolor(white)(mmmm) "+2"xcolor(white)(m) "+"x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1": "0.096 67 - 2"xcolor(white)(lmm) 2xcolor(white)(ml) x#</mathjax></p>
<blockquote></blockquote>
<p><strong>Write the <mathjax>#K_"c"#</mathjax> expression and solve for <mathjax>#x#</mathjax></strong>.</p>
<p><mathjax>#K_"c" = (["H"_2]^2["S"_2])/["H"_2"S"]^2 = 9.30 × 10^"-8"#</mathjax></p>
<p><mathjax>#((2x)^2×x)/(("0.096 67" – 2x)^2) = 9.30 × 10^"-8"#</mathjax></p>
<p><mathjax>#0.09667/(9.30 ×10^"-8") = 1 × 10^"-6" > 400#</mathjax>. ∴<mathjax>#x ≪ "0.099 67"#</mathjax></p>
<p>∴ <mathjax>#(4x^3)/"0.096 67"^2 = 9.30 × 10^"-8"#</mathjax></p>
<p><mathjax>#4x^3 = "0.096 67"^2 × 9.30 × 10^"-8" = 8.691 × 10^"-10"#</mathjax></p>
<p><mathjax>#x^3 = 2.173 × 10^"-10"#</mathjax></p>
<p><mathjax>#x = 6.012 × 10^"-4"#</mathjax></p>
<p><mathjax>#["H"_2] = 2xcolor(white)(l) "mol/L" = 2 × 6.012 × 10^"-4"color(white)(l) "mol/L" = "0.0012 mol/L"#</mathjax></p>
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<div class="answerSummary">
<div>
<div class="markdown"><p>The equilibrium concentration of <mathjax>#"H"_2#</mathjax> is 0.012 mol/L.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the initial concentrations</strong></p>
<p><mathjax>#["H"_2"S"]_0 = "0.29 mol"/"3.0 L" = "0.0967 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Write the balanced equation and set up an ICE table.</strong></p>
<p><mathjax>#color(white)(mmmmmmm)"2H"_2"S"color(white)(m) ⇌color(white)(m) "2H"_2 + "S"_2#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(m) "0.096 67" color(white)(mmmm)0 color(white)(mm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mm) "-"2xcolor(white)(mmmm) "+2"xcolor(white)(m) "+"x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1": "0.096 67 - 2"xcolor(white)(lmm) 2xcolor(white)(ml) x#</mathjax></p>
<blockquote></blockquote>
<p><strong>Write the <mathjax>#K_"c"#</mathjax> expression and solve for <mathjax>#x#</mathjax></strong>.</p>
<p><mathjax>#K_"c" = (["H"_2]^2["S"_2])/["H"_2"S"]^2 = 9.30 × 10^"-8"#</mathjax></p>
<p><mathjax>#((2x)^2×x)/(("0.096 67" – 2x)^2) = 9.30 × 10^"-8"#</mathjax></p>
<p><mathjax>#0.09667/(9.30 ×10^"-8") = 1 × 10^"-6" > 400#</mathjax>. ∴<mathjax>#x ≪ "0.099 67"#</mathjax></p>
<p>∴ <mathjax>#(4x^3)/"0.096 67"^2 = 9.30 × 10^"-8"#</mathjax></p>
<p><mathjax>#4x^3 = "0.096 67"^2 × 9.30 × 10^"-8" = 8.691 × 10^"-10"#</mathjax></p>
<p><mathjax>#x^3 = 2.173 × 10^"-10"#</mathjax></p>
<p><mathjax>#x = 6.012 × 10^"-4"#</mathjax></p>
<p><mathjax>#["H"_2] = 2xcolor(white)(l) "mol/L" = 2 × 6.012 × 10^"-4"color(white)(l) "mol/L" = "0.0012 mol/L"#</mathjax></p>
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<h1 class="questionTitle" itemprop="name">Hydrogen sulfide decomposes according to the following reaction, for which Kc=9.30E-8 at 700 degrees Celsius. 2 H2S(g) --> 2 H2(g) + S2(g) If 0.29 moles of H2S is placed in a 3.0-L container, What is the equilibrium concentration of H2(g) at 700 degrees Celsius?</h1>
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<div class="markdown"><p>The equilibrium concentration of <mathjax>#"H"_2#</mathjax> is 0.012 mol/L.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1. Calculate the initial concentrations</strong></p>
<p><mathjax>#["H"_2"S"]_0 = "0.29 mol"/"3.0 L" = "0.0967 mol/L"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2. Write the balanced equation and set up an ICE table.</strong></p>
<p><mathjax>#color(white)(mmmmmmm)"2H"_2"S"color(white)(m) ⇌color(white)(m) "2H"_2 + "S"_2#</mathjax><br/>
<mathjax>#"I/mol·L"^"-1":color(white)(m) "0.096 67" color(white)(mmmm)0 color(white)(mm)0#</mathjax><br/>
<mathjax>#"C/mol·L"^"-1":color(white)(mm) "-"2xcolor(white)(mmmm) "+2"xcolor(white)(m) "+"x#</mathjax><br/>
<mathjax>#"E/mol·L"^"-1": "0.096 67 - 2"xcolor(white)(lmm) 2xcolor(white)(ml) x#</mathjax></p>
<blockquote></blockquote>
<p><strong>Write the <mathjax>#K_"c"#</mathjax> expression and solve for <mathjax>#x#</mathjax></strong>.</p>
<p><mathjax>#K_"c" = (["H"_2]^2["S"_2])/["H"_2"S"]^2 = 9.30 × 10^"-8"#</mathjax></p>
<p><mathjax>#((2x)^2×x)/(("0.096 67" – 2x)^2) = 9.30 × 10^"-8"#</mathjax></p>
<p><mathjax>#0.09667/(9.30 ×10^"-8") = 1 × 10^"-6" > 400#</mathjax>. ∴<mathjax>#x ≪ "0.099 67"#</mathjax></p>
<p>∴ <mathjax>#(4x^3)/"0.096 67"^2 = 9.30 × 10^"-8"#</mathjax></p>
<p><mathjax>#4x^3 = "0.096 67"^2 × 9.30 × 10^"-8" = 8.691 × 10^"-10"#</mathjax></p>
<p><mathjax>#x^3 = 2.173 × 10^"-10"#</mathjax></p>
<p><mathjax>#x = 6.012 × 10^"-4"#</mathjax></p>
<p><mathjax>#["H"_2] = 2xcolor(white)(l) "mol/L" = 2 × 6.012 × 10^"-4"color(white)(l) "mol/L" = "0.0012 mol/L"#</mathjax></p>
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</article> | Hydrogen sulfide decomposes according to the following reaction, for which Kc=9.30E-8 at 700 degrees Celsius. 2 H2S(g) --> 2 H2(g) + S2(g) If 0.29 moles of H2S is placed in a 3.0-L container, What is the equilibrium concentration of H2(g) at 700 degrees Celsius? | null |
2,998 | a9611dd1-6ddd-11ea-8a84-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-oxygen-in-of-2 | +2 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] oxygen"}] | [{"type":"physical unit","value":"+2"}] | [{"type":"chemical equation","value":"OF2"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of oxygen in #OF_2#?</h1> | null | +2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxygen always has an oxidation number of <mathjax>#-2#</mathjax> unless in peroxides, like <mathjax>#H_2O_2#</mathjax>, where it has an oxidation of <mathjax>#-1#</mathjax>.</p>
<p>When <mathjax>#O#</mathjax> is attached to an even more electronegative element (<mathjax>#F#</mathjax>), it has a positive oxidation number.</p>
<p>You can also check this because <mathjax>#F#</mathjax> normally has an oxidation of <mathjax>#-1#</mathjax> in its compounds, so </p>
<p><mathjax>#(O)+2(F)=0#</mathjax></p>
<p><mathjax>#(+2)+2(-1)=0#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#+2#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxygen always has an oxidation number of <mathjax>#-2#</mathjax> unless in peroxides, like <mathjax>#H_2O_2#</mathjax>, where it has an oxidation of <mathjax>#-1#</mathjax>.</p>
<p>When <mathjax>#O#</mathjax> is attached to an even more electronegative element (<mathjax>#F#</mathjax>), it has a positive oxidation number.</p>
<p>You can also check this because <mathjax>#F#</mathjax> normally has an oxidation of <mathjax>#-1#</mathjax> in its compounds, so </p>
<p><mathjax>#(O)+2(F)=0#</mathjax></p>
<p><mathjax>#(+2)+2(-1)=0#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of oxygen in #OF_2#?</h1>
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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>Oxygen always has an oxidation number of <mathjax>#-2#</mathjax> unless in peroxides, like <mathjax>#H_2O_2#</mathjax>, where it has an oxidation of <mathjax>#-1#</mathjax>.</p>
<p>When <mathjax>#O#</mathjax> is attached to an even more electronegative element (<mathjax>#F#</mathjax>), it has a positive oxidation number.</p>
<p>You can also check this because <mathjax>#F#</mathjax> normally has an oxidation of <mathjax>#-1#</mathjax> in its compounds, so </p>
<p><mathjax>#(O)+2(F)=0#</mathjax></p>
<p><mathjax>#(+2)+2(-1)=0#</mathjax></p></div>
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</article> | What is the oxidation number of oxygen in #OF_2#? | null |
2,999 | ac0699ca-6ddd-11ea-9693-ccda262736ce | https://socratic.org/questions/what-is-the-molar-mass-of-a-gas-if-a-40-0-gram-sample-of-the-gas-occupies-11-2-l | 80.02 g/mol | start physical_unit 14 15 molar_mass g/mol qc_end physical_unit 7 7 10 11 mass qc_end physical_unit 7 7 17 18 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Molar mass [OF] the gas [IN] g/mol"}] | [{"type":"physical unit","value":"80.02 g/mol"}] | [{"type":"physical unit","value":"Mass [OF] gas sample [=] \\pu{40.0 gram}"},{"type":"physical unit","value":"Volume [OF] gas sample [=] \\pu{11.2 liters}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the molar mass of a gas if a 40.0-gram sample of the gas occupies 11.2 liters of space at STP?</h1> | null | 80.02 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this solution one would use the variation of the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> <br/>
<mathjax>#PM = dRT#</mathjax><br/>
<mathjax>#P = Pressure#</mathjax> in <mathjax>#atm#</mathjax><br/>
<mathjax>#M = Molar Mass#</mathjax> in <mathjax>#g/(mol)#</mathjax> <br/>
<mathjax>#d = Density#</mathjax> in <mathjax>#g/L#</mathjax><br/>
<mathjax>#R = 0.0821 (atmL)/(molK)#</mathjax> the gas law constant<br/>
<mathjax>#T = Temperature#</mathjax> <mathjax>#K#</mathjax></p>
<p>The <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> can be found by dividing the mass by the volume. </p>
<p><mathjax>#d = (mass)/(volume) = g/L#</mathjax></p>
<p><mathjax>#mass = 40.0 g#</mathjax><br/>
<mathjax>#volume = 11.2 L#</mathjax></p>
<p><mathjax>#d = (40.0g)/(11.2 L)#</mathjax><br/>
<mathjax>#d = 3.57 g/L#</mathjax></p>
<p>The STP values (standard temperature and pressure) are 1 atm and 273 K.</p>
<p><mathjax>#P = 1 atm#</mathjax><br/>
<mathjax>#M = ?#</mathjax><br/>
<mathjax>#d = 3.57 g/L#</mathjax><br/>
<mathjax>#R = 0.0821 (atmL)/(molK)#</mathjax><br/>
<mathjax>#T = 273K#</mathjax></p>
<p><mathjax>#PM=dRT#</mathjax><br/>
<mathjax>#(1 atm)(M)=(3.57g/L)(0.0821 (atmL)/(molK))(273K)#</mathjax> <br/>
<mathjax>#(M)=((3.57g/cancel(L))(0.0821 (cancel(atm)cancel(L))/(molcancel(K))(273cancelK)))/(1 cancel(atm))#</mathjax><br/>
<mathjax>#M = 80.0 g/(mol)#</mathjax><br/>
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<div class="markdown"><p><mathjax>#M = 80.0 g/(mol)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this solution one would use the variation of the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> <br/>
<mathjax>#PM = dRT#</mathjax><br/>
<mathjax>#P = Pressure#</mathjax> in <mathjax>#atm#</mathjax><br/>
<mathjax>#M = Molar Mass#</mathjax> in <mathjax>#g/(mol)#</mathjax> <br/>
<mathjax>#d = Density#</mathjax> in <mathjax>#g/L#</mathjax><br/>
<mathjax>#R = 0.0821 (atmL)/(molK)#</mathjax> the gas law constant<br/>
<mathjax>#T = Temperature#</mathjax> <mathjax>#K#</mathjax></p>
<p>The <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> can be found by dividing the mass by the volume. </p>
<p><mathjax>#d = (mass)/(volume) = g/L#</mathjax></p>
<p><mathjax>#mass = 40.0 g#</mathjax><br/>
<mathjax>#volume = 11.2 L#</mathjax></p>
<p><mathjax>#d = (40.0g)/(11.2 L)#</mathjax><br/>
<mathjax>#d = 3.57 g/L#</mathjax></p>
<p>The STP values (standard temperature and pressure) are 1 atm and 273 K.</p>
<p><mathjax>#P = 1 atm#</mathjax><br/>
<mathjax>#M = ?#</mathjax><br/>
<mathjax>#d = 3.57 g/L#</mathjax><br/>
<mathjax>#R = 0.0821 (atmL)/(molK)#</mathjax><br/>
<mathjax>#T = 273K#</mathjax></p>
<p><mathjax>#PM=dRT#</mathjax><br/>
<mathjax>#(1 atm)(M)=(3.57g/L)(0.0821 (atmL)/(molK))(273K)#</mathjax> <br/>
<mathjax>#(M)=((3.57g/cancel(L))(0.0821 (cancel(atm)cancel(L))/(molcancel(K))(273cancelK)))/(1 cancel(atm))#</mathjax><br/>
<mathjax>#M = 80.0 g/(mol)#</mathjax><br/>
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<h1 class="questionTitle" itemprop="name">What is the molar mass of a gas if a 40.0-gram sample of the gas occupies 11.2 liters of space at STP?</h1>
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<div class="markdown"><p><mathjax>#M = 80.0 g/(mol)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For this solution one would use the variation of the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas Law</a> <br/>
<mathjax>#PM = dRT#</mathjax><br/>
<mathjax>#P = Pressure#</mathjax> in <mathjax>#atm#</mathjax><br/>
<mathjax>#M = Molar Mass#</mathjax> in <mathjax>#g/(mol)#</mathjax> <br/>
<mathjax>#d = Density#</mathjax> in <mathjax>#g/L#</mathjax><br/>
<mathjax>#R = 0.0821 (atmL)/(molK)#</mathjax> the gas law constant<br/>
<mathjax>#T = Temperature#</mathjax> <mathjax>#K#</mathjax></p>
<p>The <a href="http://socratic.org/chemistry/measurement-in-chemistry/density">density</a> can be found by dividing the mass by the volume. </p>
<p><mathjax>#d = (mass)/(volume) = g/L#</mathjax></p>
<p><mathjax>#mass = 40.0 g#</mathjax><br/>
<mathjax>#volume = 11.2 L#</mathjax></p>
<p><mathjax>#d = (40.0g)/(11.2 L)#</mathjax><br/>
<mathjax>#d = 3.57 g/L#</mathjax></p>
<p>The STP values (standard temperature and pressure) are 1 atm and 273 K.</p>
<p><mathjax>#P = 1 atm#</mathjax><br/>
<mathjax>#M = ?#</mathjax><br/>
<mathjax>#d = 3.57 g/L#</mathjax><br/>
<mathjax>#R = 0.0821 (atmL)/(molK)#</mathjax><br/>
<mathjax>#T = 273K#</mathjax></p>
<p><mathjax>#PM=dRT#</mathjax><br/>
<mathjax>#(1 atm)(M)=(3.57g/L)(0.0821 (atmL)/(molK))(273K)#</mathjax> <br/>
<mathjax>#(M)=((3.57g/cancel(L))(0.0821 (cancel(atm)cancel(L))/(molcancel(K))(273cancelK)))/(1 cancel(atm))#</mathjax><br/>
<mathjax>#M = 80.0 g/(mol)#</mathjax><br/>
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</article> | What is the molar mass of a gas if a 40.0-gram sample of the gas occupies 11.2 liters of space at STP? | null |
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