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3,200 | abe4bbe4-6ddd-11ea-82a0-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-containing-5-0-moles-of-kcl-in-2-0-l-of-solut | 2.50 mol/L | start physical_unit 11 11 molarity mol/l qc_end physical_unit 11 11 8 9 mole qc_end physical_unit 6 6 13 14 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] KCl solution [IN] mol/L"}] | [{"type":"physical unit","value":"2.50 mol/L"}] | [{"type":"physical unit","value":"Mole [OF] KCl [=] \\pu{5.0 moles}"},{"type":"physical unit","value":"Volume [OF] KCl solution [=] \\pu{2.0 L}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution containing 5.0 moles of #KCl# in 2.0 L of solution?</h1> | null | 2.50 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Cocnetration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of Solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(5.0*"moles")/(2.0*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.5*mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Cocnetration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of Solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(5.0*"moles")/(2.0*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.5*mol*L^-1#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Cocnetration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of Solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(5.0*"moles")/(2.0*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#?? mol*L^-1#</mathjax></p></div>
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</article> | What is the molarity of a solution containing 5.0 moles of #KCl# in 2.0 L of solution? | null |
3,201 | abccf3e8-6ddd-11ea-855d-ccda262736ce | https://socratic.org/questions/how-would-you-write-a-chemical-equation-representing-the-third-ionization-energy | Li^2+(g) ->[\delta] Li^3+(g) + e- | start chemical_equation qc_end substance 13 13 qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Equation [OF] the chemical equation"}] | [{"type":"chemical equation","value":"Li^2+(g) ->[\\delta] Li^3+(g) + e-"}] | [{"type":"substance name","value":"Lithium"},{"type":"other","value":"The chemical equation representing the third ionization energy."}] | <h1 class="questionTitle" itemprop="name">How would you write a chemical equation representing the third ionization energy for lithium?</h1> | null | Li^2+(g) ->[\delta] Li^3+(g) + e- | <div class="answerDescription">
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<div class="markdown"><p>Clearly, the reaction as written would be endothermic.</p></div>
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<div class="markdown"><p><mathjax>#Li_((g))^(2+) + Delta rarr Li_((g))^(3+) + e^-#</mathjax></p>
<p>Someone might have measured this.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Clearly, the reaction as written would be endothermic.</p></div>
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<div class="markdown"><p><mathjax>#Li_((g))^(2+) + Delta rarr Li_((g))^(3+) + e^-#</mathjax></p>
<p>Someone might have measured this.</p></div>
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<div class="markdown"><p>Clearly, the reaction as written would be endothermic.</p></div>
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</article> | How would you write a chemical equation representing the third ionization energy for lithium? | null |
3,202 | aa14eea6-6ddd-11ea-a64e-ccda262736ce | https://socratic.org/questions/how-do-you-calculate-the-concentration-of-h-ions-in-a-0-27-m-naoh-solution | 3.70 × 10^(-14) M | start physical_unit 7 8 concentration mol/l qc_end physical_unit 13 14 11 12 molarity qc_end end | [{"type":"physical unit","value":"Concentration [OF] H+ ions [IN] M"}] | [{"type":"physical unit","value":"3.70 × 10^(-14) M"}] | [{"type":"physical unit","value":"Molarity [OF] NaoH solution [=] \\pu{0.27 M}"}] | <h1 class="questionTitle" itemprop="name">How do you calculate the concentration of #H^+# ions in a 0.27 M NaoH solution?</h1> | null | 3.70 × 10^(-14) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We gots the equilibrium <mathjax>#2H_2O(l)rightleftharpoonsH_3O^+ +HO^-#</mathjax></p>
<p>For which <mathjax>#K_w=[H_3O^+][HO^-]=10^(-14)#</mathjax> under standard conditions of temperature and pressure...and taking <mathjax>#log_10#</mathjax> of BOTH sides...</p>
<p><mathjax>#log_10K_w=log_10(10^-14)=underbrace(log_10[H_3O^+])_(-pH)+underbrace(log_10[HO^-])_(-pOH)#</mathjax></p>
<p>Now logarithms were used in the old days before electronic calculators given that they allow the multiplication of large or small numbers fairly easily..</p>
<p>But by definition of the logarithmic function, <mathjax>#log_10(10^-14)=-14#</mathjax>...and so our defining relationship...</p>
<p><mathjax>#pH+pOH=14#</mathjax></p>
<p>And we got <mathjax>#[HO^-]=0.27*mol*L^-1#</mathjax>...</p>
<p><mathjax>#pOH=-log_10(0.27)=-(-0.569)=0.569#</mathjax></p>
<p>And <mathjax>#pH=14-pOH=14-0.569=13.43#</mathjax> as required....</p>
<p>And finally, we take antilogarithms...<mathjax>#[H_3O^+]=10^(-13.43)=3.70xx10^-14*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#[H_3O^+]=10^(-13.43)*mol*L^-1=3.70xx10^-14*mol*L^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We gots the equilibrium <mathjax>#2H_2O(l)rightleftharpoonsH_3O^+ +HO^-#</mathjax></p>
<p>For which <mathjax>#K_w=[H_3O^+][HO^-]=10^(-14)#</mathjax> under standard conditions of temperature and pressure...and taking <mathjax>#log_10#</mathjax> of BOTH sides...</p>
<p><mathjax>#log_10K_w=log_10(10^-14)=underbrace(log_10[H_3O^+])_(-pH)+underbrace(log_10[HO^-])_(-pOH)#</mathjax></p>
<p>Now logarithms were used in the old days before electronic calculators given that they allow the multiplication of large or small numbers fairly easily..</p>
<p>But by definition of the logarithmic function, <mathjax>#log_10(10^-14)=-14#</mathjax>...and so our defining relationship...</p>
<p><mathjax>#pH+pOH=14#</mathjax></p>
<p>And we got <mathjax>#[HO^-]=0.27*mol*L^-1#</mathjax>...</p>
<p><mathjax>#pOH=-log_10(0.27)=-(-0.569)=0.569#</mathjax></p>
<p>And <mathjax>#pH=14-pOH=14-0.569=13.43#</mathjax> as required....</p>
<p>And finally, we take antilogarithms...<mathjax>#[H_3O^+]=10^(-13.43)=3.70xx10^-14*mol*L^-1#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How do you calculate the concentration of #H^+# ions in a 0.27 M NaoH solution?</h1>
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<div class="markdown"><p><mathjax>#[H_3O^+]=10^(-13.43)*mol*L^-1=3.70xx10^-14*mol*L^-1#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We gots the equilibrium <mathjax>#2H_2O(l)rightleftharpoonsH_3O^+ +HO^-#</mathjax></p>
<p>For which <mathjax>#K_w=[H_3O^+][HO^-]=10^(-14)#</mathjax> under standard conditions of temperature and pressure...and taking <mathjax>#log_10#</mathjax> of BOTH sides...</p>
<p><mathjax>#log_10K_w=log_10(10^-14)=underbrace(log_10[H_3O^+])_(-pH)+underbrace(log_10[HO^-])_(-pOH)#</mathjax></p>
<p>Now logarithms were used in the old days before electronic calculators given that they allow the multiplication of large or small numbers fairly easily..</p>
<p>But by definition of the logarithmic function, <mathjax>#log_10(10^-14)=-14#</mathjax>...and so our defining relationship...</p>
<p><mathjax>#pH+pOH=14#</mathjax></p>
<p>And we got <mathjax>#[HO^-]=0.27*mol*L^-1#</mathjax>...</p>
<p><mathjax>#pOH=-log_10(0.27)=-(-0.569)=0.569#</mathjax></p>
<p>And <mathjax>#pH=14-pOH=14-0.569=13.43#</mathjax> as required....</p>
<p>And finally, we take antilogarithms...<mathjax>#[H_3O^+]=10^(-13.43)=3.70xx10^-14*mol*L^-1#</mathjax></p></div>
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</article> | How do you calculate the concentration of #H^+# ions in a 0.27 M NaoH solution? | null |
3,203 | a864dd1a-6ddd-11ea-9022-ccda262736ce | https://socratic.org/questions/what-volume-of-a-2-5-m-stock-solution-of-acetic-acid-is-required-to-prepare-100- | 20 milliliters | start physical_unit 7 7 volume ml qc_end physical_unit 21 23 19 20 molarity qc_end physical_unit 21 23 15 16 volume qc_end end | [{"type":"physical unit","value":"Volume1 [OF] acetic acid stock solution [IN] milliliters"}] | [{"type":"physical unit","value":"20 milliliters"}] | [{"type":"physical unit","value":"Molarity1 [OF] acetic acid stock solution [=] \\pu{2.5 M}"},{"type":"physical unit","value":"Molarity2 [OF] acetic acid solution [=] \\pu{0.50 M}"},{"type":"physical unit","value":"Volume2 [OF] acetic acid solution [=] \\pu{100.0 milliliters}"}] | <h1 class="questionTitle" itemprop="name">What volume of a 2.5 M stock solution of acetic acid is required to prepare 100.0 milliliters of a .50 M acetic acid solution?</h1> | null | 20 milliliters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Whenever you are dealing with a dilution, (going from a higher concentration to a lower concentration), you can use the equation below:<br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/49DzrjFQxWgNFk9uoM5w_slide_28.jpg"/> </p>
<p>Let's determine our known and unknown variables:</p>
<p><mathjax>#color(blue)("Knowns:"#</mathjax></p>
<ul>
<li>Initial <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></li>
<li>Final Volume</li>
<li>Final <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></li>
</ul>
<p><mathjax>#color(orange)("Unknowns:"#</mathjax></p>
<ul>
<li>Initial Volume</li>
</ul>
<p>Next, we rearrange the equation to solve for <mathjax>#V_1#</mathjax> by dividing by <mathjax>#M_1#</mathjax> on both sides of the equation:</p>
<p><mathjax>#V_1=(M_2xxV_2)/M_1#</mathjax></p>
<p>Plug in the known values and solve for your unknown:<br/>
<mathjax>#V_1=(0.50cancel"M"xx100.0mL)/(2.5cancel"M")#</mathjax></p>
<p>Therefore,</p>
<p><mathjax>#V_1 = 20. mL#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div>
<div class="markdown"><p><mathjax>#20. "milliliters"#</mathjax> of acetic acid is required </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Whenever you are dealing with a dilution, (going from a higher concentration to a lower concentration), you can use the equation below:<br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/49DzrjFQxWgNFk9uoM5w_slide_28.jpg"/> </p>
<p>Let's determine our known and unknown variables:</p>
<p><mathjax>#color(blue)("Knowns:"#</mathjax></p>
<ul>
<li>Initial <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></li>
<li>Final Volume</li>
<li>Final <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></li>
</ul>
<p><mathjax>#color(orange)("Unknowns:"#</mathjax></p>
<ul>
<li>Initial Volume</li>
</ul>
<p>Next, we rearrange the equation to solve for <mathjax>#V_1#</mathjax> by dividing by <mathjax>#M_1#</mathjax> on both sides of the equation:</p>
<p><mathjax>#V_1=(M_2xxV_2)/M_1#</mathjax></p>
<p>Plug in the known values and solve for your unknown:<br/>
<mathjax>#V_1=(0.50cancel"M"xx100.0mL)/(2.5cancel"M")#</mathjax></p>
<p>Therefore,</p>
<p><mathjax>#V_1 = 20. mL#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What volume of a 2.5 M stock solution of acetic acid is required to prepare 100.0 milliliters of a .50 M acetic acid solution?</h1>
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<div class="markdown"><p><mathjax>#20. "milliliters"#</mathjax> of acetic acid is required </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Whenever you are dealing with a dilution, (going from a higher concentration to a lower concentration), you can use the equation below:<br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/49DzrjFQxWgNFk9uoM5w_slide_28.jpg"/> </p>
<p>Let's determine our known and unknown variables:</p>
<p><mathjax>#color(blue)("Knowns:"#</mathjax></p>
<ul>
<li>Initial <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></li>
<li>Final Volume</li>
<li>Final <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a></li>
</ul>
<p><mathjax>#color(orange)("Unknowns:"#</mathjax></p>
<ul>
<li>Initial Volume</li>
</ul>
<p>Next, we rearrange the equation to solve for <mathjax>#V_1#</mathjax> by dividing by <mathjax>#M_1#</mathjax> on both sides of the equation:</p>
<p><mathjax>#V_1=(M_2xxV_2)/M_1#</mathjax></p>
<p>Plug in the known values and solve for your unknown:<br/>
<mathjax>#V_1=(0.50cancel"M"xx100.0mL)/(2.5cancel"M")#</mathjax></p>
<p>Therefore,</p>
<p><mathjax>#V_1 = 20. mL#</mathjax></p></div>
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</article> | What volume of a 2.5 M stock solution of acetic acid is required to prepare 100.0 milliliters of a .50 M acetic acid solution? | null |
3,204 | aad5fbc8-6ddd-11ea-9455-ccda262736ce | https://socratic.org/questions/at-conditions-of-1-5-atm-of-pressure-and-15-0-c-temperature-a-gas-occupies-a-vol-1 | 65.3 mL | start physical_unit 25 27 volume ml qc_end physical_unit 12 12 17 18 volume qc_end physical_unit 12 12 8 9 temperature qc_end physical_unit 12 12 3 4 pressure qc_end physical_unit 25 27 32 33 temperature qc_end physical_unit 25 27 29 30 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the same gas [IN] mL"}] | [{"type":"physical unit","value":"65.3 mL"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{45.5 mL}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{15.0 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.5 atm}"},{"type":"physical unit","value":"Temperature2 [OF] the same gas [=] \\pu{30.0 ℃}"},{"type":"physical unit","value":"Pressure2 [OF] the same gas [=] \\pu{1.1 atm}"}] | <h1 class="questionTitle" itemprop="name">At conditions of 1.5 atm of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 1.1 atm and 30.0 °C? </h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>At conditions of 1.5 atm of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 1.1 atm and 30.0 °C? </p></div>
</h2>
</div>
</div> | 65.3 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>...that <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>...for a GIVEN molar quantity of gas....the which scenario pertains here. As usual we use <mathjax>#"absolute temperatures"#</mathjax>.</p>
<p>And so we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax></p>
<p><mathjax>#(1.5*atmxx45.5*mL)/(288.15*K)xx(303.15*K)/(1.1*atm)=65.3*mL#</mathjax>..</p>
<p>That volume has increased is consistent with the pressure drop, and temperature increase...the question should have specified that the gas was confined to a piston....with variable volume...</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"Well...what does the old combined gas equation say...?"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>...that <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>...for a GIVEN molar quantity of gas....the which scenario pertains here. As usual we use <mathjax>#"absolute temperatures"#</mathjax>.</p>
<p>And so we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax></p>
<p><mathjax>#(1.5*atmxx45.5*mL)/(288.15*K)xx(303.15*K)/(1.1*atm)=65.3*mL#</mathjax>..</p>
<p>That volume has increased is consistent with the pressure drop, and temperature increase...the question should have specified that the gas was confined to a piston....with variable volume...</p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">At conditions of 1.5 atm of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 1.1 atm and 30.0 °C? </h1>
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>At conditions of 1.5 atm of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 1.1 atm and 30.0 °C? </p></div>
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anor277
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Feb 27, 2018
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<div class="markdown"><p><mathjax>#"Well...what does the old combined gas equation say...?"#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>...that <mathjax>#(P_1V_1)/T_1=(P_2V_2)/T_2#</mathjax>...for a GIVEN molar quantity of gas....the which scenario pertains here. As usual we use <mathjax>#"absolute temperatures"#</mathjax>.</p>
<p>And so we solve for <mathjax>#V_2=(P_1V_1)/T_1xxT_2/P_2#</mathjax></p>
<p><mathjax>#(1.5*atmxx45.5*mL)/(288.15*K)xx(303.15*K)/(1.1*atm)=65.3*mL#</mathjax>..</p>
<p>That volume has increased is consistent with the pressure drop, and temperature increase...the question should have specified that the gas was confined to a piston....with variable volume...</p></div>
</div>
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Samrin M.
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Stefan V.
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<div class="markdown"><p><mathjax>#"65.3 mL"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You'll be using the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a> since your question involves all three variables: </p>
<p><mathjax>#(P_1V_1)/T_1 = (P_2V_2)/T_2#</mathjax></p>
<p>If (<mathjax>#P =#</mathjax> pressure , <mathjax>#V =#</mathjax> volume, <mathjax>#T =#</mathjax> temperature), then we know in the equation that:</p>
<blockquote>
<ul>
<li><mathjax>#P_1 = "1.5 atm"#</mathjax></li>
<li><mathjax>#P_2 = "1.1 atm"#</mathjax></li>
<li><mathjax>#V_1 = "45.5 mL"#</mathjax></li>
<li><mathjax>#V_2 = ?#</mathjax></li>
<li><mathjax>#T_1 = 15^@"C"#</mathjax></li>
<li><mathjax>#T_2 = 30^@"C"#</mathjax></li>
</ul>
</blockquote>
<p>Before plugging in all values to solve for the volume, temperature MUST ALWAYS be in Kelvins, at least for gases. To convert Celsius to Kelvin, all you do is add <mathjax>#273#</mathjax>. So, </p>
<p><mathjax>#T_1 = "288 K "#</mathjax> and <mathjax>#" "T_2 = "303 K"#</mathjax> </p>
<p>Now we're ready to plug and chug!</p>
<p>Work for calculation:</p>
<p><mathjax>#(1.5*45.5)/288 = (1.1V_2)/303#</mathjax></p>
<p><mathjax>#0.24 = (1.1 V_2)/303#</mathjax></p>
<p><mathjax>#71.8 = 1.1 V_2#</mathjax></p>
<p><mathjax>#65.3 = V_2#</mathjax> </p>
<p>or </p>
<p><mathjax>#V_2 = 65.3#</mathjax></p>
<p>So that means the volume of the gas would be <em>about</em> <mathjax>#"65.3 mL"#</mathjax>.</p></div>
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</article> | At conditions of 1.5 atm of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 1.1 atm and 30.0 °C? |
At conditions of 1.5 atm of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 1.1 atm and 30.0 °C?
|
3,205 | aa16508d-6ddd-11ea-9c06-ccda262736ce | https://socratic.org/questions/coal-is-not-a-pure-chemical-compound-but-its-elemental-composition-can-be-approx | 1.81% | start physical_unit 26 28 mass_percent none qc_end chemical_equation 17 17 qc_end end | [{"type":"physical unit","value":"Percentage by mass [OF] sulfur in coal"}] | [{"type":"physical unit","value":"1.81%"}] | [{"type":"chemical equation","value":"C135H96NO9S"}] | <h1 class="questionTitle" itemprop="name">Coal is not a pure chemical compound, but its elemental composition can be approximated by the formula #C_135H_96NO_9S#. What is the approximate percentage by mass of sulfur in coal?</h1> | null | 1.81% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>1. Use the formula:</strong></p>
<p><strong>% by mass = (mass of part / mass of whole) x 100</strong></p>
<p><em>What to we need to know to use this formula?</em> </p>
<p>We need the mass of the part (<strong>sulfur</strong>) and the mass of the whole (<strong>coal</strong>). </p>
<p><em>What are we solving for?</em></p>
<p>We are solving for % by mass of sulfur in coal.</p>
<p><strong>2. To find the mass of the part (sulfur):</strong></p>
<p>Look at the molecular formula and see <strong>how many sulfur atoms</strong> there are.</p>
<p><mathjax>#C_135H_96NOS#</mathjax></p>
<p>There is just 1 sulfur (S) atom in the formula.</p>
<p>Since the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of sulfur is approximately 32.1 and there's only one sulfur atom here, <strong>the mass of the part is 32.1</strong>.</p>
<p><em>You can find information about atomic mass for any element on a periodic table.</em></p>
<p><strong>3. To find the mass of the whole (coal):</strong></p>
<p>First, count up the number of atoms of each element present in the formula.</p>
<p>C: 135 atoms<br/>
H: 96 atoms<br/>
N: 1 atom<br/>
O: 1 atom<br/>
S: 1 atom</p>
<p>Now multiply the # of atoms number by the atomic mass. </p>
<p>C: 135 atoms x 12.0<br/>
H: 96 atoms x 1.0<br/>
N: 1 atom x 14.0<br/>
O: 1 atom x 16.0<br/>
S: 1 atom x 32.1</p>
<p>C: 1620<br/>
H: 96<br/>
N: 14<br/>
O: 16<br/>
S: 32.1</p>
<p>Finally, add up your numbers to find the mass of the whole.</p>
<p><mathjax>#1620 + 96 + 14 + 16 + 32.1 = 1778.1#</mathjax></p>
<p><strong>1778.1 is the mass of the part.</strong></p>
<p><strong>4. Plug your answers into the formula and solve!</strong></p>
<p>% by mass <mathjax>#= (32.1 / 1778.1) * 100#</mathjax></p>
<p>% by mass <mathjax># = 1.80529779#</mathjax><br/>
% by mass <mathjax>#~~#</mathjax> 1.8</p>
<p>The approx. % by mass of sulfur in coal is 1.8%.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>1.8%</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>1. Use the formula:</strong></p>
<p><strong>% by mass = (mass of part / mass of whole) x 100</strong></p>
<p><em>What to we need to know to use this formula?</em> </p>
<p>We need the mass of the part (<strong>sulfur</strong>) and the mass of the whole (<strong>coal</strong>). </p>
<p><em>What are we solving for?</em></p>
<p>We are solving for % by mass of sulfur in coal.</p>
<p><strong>2. To find the mass of the part (sulfur):</strong></p>
<p>Look at the molecular formula and see <strong>how many sulfur atoms</strong> there are.</p>
<p><mathjax>#C_135H_96NOS#</mathjax></p>
<p>There is just 1 sulfur (S) atom in the formula.</p>
<p>Since the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of sulfur is approximately 32.1 and there's only one sulfur atom here, <strong>the mass of the part is 32.1</strong>.</p>
<p><em>You can find information about atomic mass for any element on a periodic table.</em></p>
<p><strong>3. To find the mass of the whole (coal):</strong></p>
<p>First, count up the number of atoms of each element present in the formula.</p>
<p>C: 135 atoms<br/>
H: 96 atoms<br/>
N: 1 atom<br/>
O: 1 atom<br/>
S: 1 atom</p>
<p>Now multiply the # of atoms number by the atomic mass. </p>
<p>C: 135 atoms x 12.0<br/>
H: 96 atoms x 1.0<br/>
N: 1 atom x 14.0<br/>
O: 1 atom x 16.0<br/>
S: 1 atom x 32.1</p>
<p>C: 1620<br/>
H: 96<br/>
N: 14<br/>
O: 16<br/>
S: 32.1</p>
<p>Finally, add up your numbers to find the mass of the whole.</p>
<p><mathjax>#1620 + 96 + 14 + 16 + 32.1 = 1778.1#</mathjax></p>
<p><strong>1778.1 is the mass of the part.</strong></p>
<p><strong>4. Plug your answers into the formula and solve!</strong></p>
<p>% by mass <mathjax>#= (32.1 / 1778.1) * 100#</mathjax></p>
<p>% by mass <mathjax># = 1.80529779#</mathjax><br/>
% by mass <mathjax>#~~#</mathjax> 1.8</p>
<p>The approx. % by mass of sulfur in coal is 1.8%.</p></div>
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<h1 class="questionTitle" itemprop="name">Coal is not a pure chemical compound, but its elemental composition can be approximated by the formula #C_135H_96NO_9S#. What is the approximate percentage by mass of sulfur in coal?</h1>
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Michelle
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<div class="markdown"><p>1.8%</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>1. Use the formula:</strong></p>
<p><strong>% by mass = (mass of part / mass of whole) x 100</strong></p>
<p><em>What to we need to know to use this formula?</em> </p>
<p>We need the mass of the part (<strong>sulfur</strong>) and the mass of the whole (<strong>coal</strong>). </p>
<p><em>What are we solving for?</em></p>
<p>We are solving for % by mass of sulfur in coal.</p>
<p><strong>2. To find the mass of the part (sulfur):</strong></p>
<p>Look at the molecular formula and see <strong>how many sulfur atoms</strong> there are.</p>
<p><mathjax>#C_135H_96NOS#</mathjax></p>
<p>There is just 1 sulfur (S) atom in the formula.</p>
<p>Since the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> of sulfur is approximately 32.1 and there's only one sulfur atom here, <strong>the mass of the part is 32.1</strong>.</p>
<p><em>You can find information about atomic mass for any element on a periodic table.</em></p>
<p><strong>3. To find the mass of the whole (coal):</strong></p>
<p>First, count up the number of atoms of each element present in the formula.</p>
<p>C: 135 atoms<br/>
H: 96 atoms<br/>
N: 1 atom<br/>
O: 1 atom<br/>
S: 1 atom</p>
<p>Now multiply the # of atoms number by the atomic mass. </p>
<p>C: 135 atoms x 12.0<br/>
H: 96 atoms x 1.0<br/>
N: 1 atom x 14.0<br/>
O: 1 atom x 16.0<br/>
S: 1 atom x 32.1</p>
<p>C: 1620<br/>
H: 96<br/>
N: 14<br/>
O: 16<br/>
S: 32.1</p>
<p>Finally, add up your numbers to find the mass of the whole.</p>
<p><mathjax>#1620 + 96 + 14 + 16 + 32.1 = 1778.1#</mathjax></p>
<p><strong>1778.1 is the mass of the part.</strong></p>
<p><strong>4. Plug your answers into the formula and solve!</strong></p>
<p>% by mass <mathjax>#= (32.1 / 1778.1) * 100#</mathjax></p>
<p>% by mass <mathjax># = 1.80529779#</mathjax><br/>
% by mass <mathjax>#~~#</mathjax> 1.8</p>
<p>The approx. % by mass of sulfur in coal is 1.8%.</p></div>
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</article> | Coal is not a pure chemical compound, but its elemental composition can be approximated by the formula #C_135H_96NO_9S#. What is the approximate percentage by mass of sulfur in coal? | null |
3,206 | aa1674d5-6ddd-11ea-93cd-ccda262736ce | https://socratic.org/questions/what-volume-in-ml-of-220-m-ethanol-solution-contains-2-90-x-10-3-mol-ethanol | 13.2 mL | start physical_unit 7 8 volume ml qc_end physical_unit 7 8 5 6 molarity qc_end physical_unit 7 7 10 13 mole qc_end end | [{"type":"physical unit","value":"Volume [OF] ethanol solution [IN] mL"}] | [{"type":"physical unit","value":"13.2 mL"}] | [{"type":"physical unit","value":"Molarity [OF] ethanol solution [=] \\pu{0.220 M}"},{"type":"physical unit","value":"Mole [OF] ethanol [=] \\pu{2.90 × 10^(-3) mol}"}] | <h1 class="questionTitle" itemprop="name">What volume (in mL) of #".220 M"# ethanol solution contains #2.90 xx 10^-3# #"mol"# ethanol?</h1> | null | 13.2 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax>#"Volume of EtOH" = 2.90 × 10^"-3" color(red)(cancel(color(black)("mol EtOH"))) × "1 L EtOH"/(0.220 color(red)(cancel(color(black)("mol EtOH")))) = "0.0132 L EtOH" = "13.2 mL EtOH"#</mathjax></p></div>
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<div class="markdown"><p>The answer is 13.2 mL.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><mathjax>#"Volume of EtOH" = 2.90 × 10^"-3" color(red)(cancel(color(black)("mol EtOH"))) × "1 L EtOH"/(0.220 color(red)(cancel(color(black)("mol EtOH")))) = "0.0132 L EtOH" = "13.2 mL EtOH"#</mathjax></p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What volume (in mL) of #".220 M"# ethanol solution contains #2.90 xx 10^-3# #"mol"# ethanol?</h1>
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<div class="markdown"><p>The answer is 13.2 mL.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p><mathjax>#"Volume of EtOH" = 2.90 × 10^"-3" color(red)(cancel(color(black)("mol EtOH"))) × "1 L EtOH"/(0.220 color(red)(cancel(color(black)("mol EtOH")))) = "0.0132 L EtOH" = "13.2 mL EtOH"#</mathjax></p></div>
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Stefan V.
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<span class="dateCreated" datetime="2017-06-17T00:07:50" itemprop="dateCreated">
Jun 17, 2017
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<div class="markdown"><p><mathjax>#"13.2 mL"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution can be converted to the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3#</mathjax> <mathjax>#"mL"#</mathjax> </p>
</blockquote>
<p>of solution. In your case, a <mathjax>#"0.220-M"#</mathjax> solution of ethanol will contain <mathjax>#0.220#</mathjax> <strong>moles</strong> of ethanol for every <mathjax>#10^3#</mathjax> <mathjax>#"mL"#</mathjax> of solution. </p>
<p>This implies that <mathjax>#2.90 * 10^(-3)#</mathjax> <strong>moles</strong> of ethanol will be present in </p>
<blockquote>
<p><mathjax>#2.90 * color(blue)(cancel(color(black)(10^(-3))))color(red)(cancel(color(black)("moles ethanol"))) * overbrace((color(blue)(cancel(color(black)(10^3)))color(white)(.) "mL solution")/(0.220color(red)(cancel(color(black)("moles ethanol")))))^(color(blue)("= 0.220 M"))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax># = color(darkgreen)(ul(color(black)("13.2 mL solution")))#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p></div>
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</article> | What volume (in mL) of #".220 M"# ethanol solution contains #2.90 xx 10^-3# #"mol"# ethanol? | null |
3,207 | a8c09ef4-6ddd-11ea-8d46-ccda262736ce | https://socratic.org/questions/when-2-moles-of-hydrogen-is-heated-with-2-moles-of-iodine-2-96-moles-of-hydrogen | 4 | start physical_unit 25 29 equilibrium_constant_k none qc_end physical_unit 4 4 1 2 mole qc_end physical_unit 11 11 1 2 mole qc_end physical_unit 15 16 12 13 mole qc_end end | [{"type":"physical unit","value":"Equilibrium constant [OF] the formation of hydrogen iodide"}] | [{"type":"physical unit","value":"4"}] | [{"type":"physical unit","value":"Mole [OF] hydrogen [=] \\pu{2 moles}"},{"type":"physical unit","value":"Mole [OF] iodine [=] \\pu{2 moles}"},{"type":"physical unit","value":"Mole [OF] hydrogen iodide [=] \\pu{2.96 moles}"}] | <h1 class="questionTitle" itemprop="name">When 2 moles of hydrogen is heated with 2 moles of iodine,2.96 moles of hydrogen iodide is formed.what is the equilibrium constant for the formation of hydrogen iodide?</h1> | null | 4 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In this question, we aren't given the equilibrium concentrations of our reagents and products, we have to work it out ourselves using the ICE method. First, we must write out the balanced equation.</p>
<p><mathjax># color(white)(aaaaaaaaaaaaaaa)"H"_2 color(white)(aa)+ color(white)(aa)"I"_2 color(white)(aa)rightleftharpoons color(white)(aa)2"HI"#</mathjax></p>
<p>Initial moles: <mathjax>#color(white)(aaaaaz)2 color(white)(aaaaaaa)2 color(white)(aaaaaaaaa)0#</mathjax></p>
<p>Change in moles: <mathjax>#-1.48 color(white)(aa)-1.48 color(white)(aaa)+2.96#</mathjax></p>
<p>Equilibrium moles: <mathjax>#color(white)(a)0.53 color(white)(zacaa)0.53 color(white)(aaaaa)2.96#</mathjax></p>
<p>So we know how many moles we have of each at equilibrium. Since we aren't given the concentration, we must assume that the number of mols equals the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>.</p>
<p><mathjax>#"K"_"c" = ["HI"]^2/(["H"_2]["I"_2])=2.96^2/((1.48)(1.48))=4#</mathjax></p>
<p>The equilibrium constant has no units as they cancel out:</p>
<p><mathjax>#cancel("mol"^2)/(cancel(("mol")("mol")))#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"K"_"c" = 4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In this question, we aren't given the equilibrium concentrations of our reagents and products, we have to work it out ourselves using the ICE method. First, we must write out the balanced equation.</p>
<p><mathjax># color(white)(aaaaaaaaaaaaaaa)"H"_2 color(white)(aa)+ color(white)(aa)"I"_2 color(white)(aa)rightleftharpoons color(white)(aa)2"HI"#</mathjax></p>
<p>Initial moles: <mathjax>#color(white)(aaaaaz)2 color(white)(aaaaaaa)2 color(white)(aaaaaaaaa)0#</mathjax></p>
<p>Change in moles: <mathjax>#-1.48 color(white)(aa)-1.48 color(white)(aaa)+2.96#</mathjax></p>
<p>Equilibrium moles: <mathjax>#color(white)(a)0.53 color(white)(zacaa)0.53 color(white)(aaaaa)2.96#</mathjax></p>
<p>So we know how many moles we have of each at equilibrium. Since we aren't given the concentration, we must assume that the number of mols equals the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>.</p>
<p><mathjax>#"K"_"c" = ["HI"]^2/(["H"_2]["I"_2])=2.96^2/((1.48)(1.48))=4#</mathjax></p>
<p>The equilibrium constant has no units as they cancel out:</p>
<p><mathjax>#cancel("mol"^2)/(cancel(("mol")("mol")))#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">When 2 moles of hydrogen is heated with 2 moles of iodine,2.96 moles of hydrogen iodide is formed.what is the equilibrium constant for the formation of hydrogen iodide?</h1>
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Monzur R.
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<div class="markdown"><p><mathjax>#"K"_"c" = 4#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In this question, we aren't given the equilibrium concentrations of our reagents and products, we have to work it out ourselves using the ICE method. First, we must write out the balanced equation.</p>
<p><mathjax># color(white)(aaaaaaaaaaaaaaa)"H"_2 color(white)(aa)+ color(white)(aa)"I"_2 color(white)(aa)rightleftharpoons color(white)(aa)2"HI"#</mathjax></p>
<p>Initial moles: <mathjax>#color(white)(aaaaaz)2 color(white)(aaaaaaa)2 color(white)(aaaaaaaaa)0#</mathjax></p>
<p>Change in moles: <mathjax>#-1.48 color(white)(aa)-1.48 color(white)(aaa)+2.96#</mathjax></p>
<p>Equilibrium moles: <mathjax>#color(white)(a)0.53 color(white)(zacaa)0.53 color(white)(aaaaa)2.96#</mathjax></p>
<p>So we know how many moles we have of each at equilibrium. Since we aren't given the concentration, we must assume that the number of mols equals the <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a>.</p>
<p><mathjax>#"K"_"c" = ["HI"]^2/(["H"_2]["I"_2])=2.96^2/((1.48)(1.48))=4#</mathjax></p>
<p>The equilibrium constant has no units as they cancel out:</p>
<p><mathjax>#cancel("mol"^2)/(cancel(("mol")("mol")))#</mathjax></p></div>
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</article> | When 2 moles of hydrogen is heated with 2 moles of iodine,2.96 moles of hydrogen iodide is formed.what is the equilibrium constant for the formation of hydrogen iodide? | null |
3,208 | ac461c7e-6ddd-11ea-8340-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-one-mole-of-water | 18.00 grams | start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] water [IN] grams"}] | [{"type":"physical unit","value":"18.00 grams"}] | [{"type":"physical unit","value":"Mole [OF] water [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of one mole of water? </h1> | null | 18.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The mass of one mole of any atom/molecule is equal to its atomic molecular mass in grams. </p>
<p>The molecular formula of water is <mathjax>#H_2O#</mathjax></p>
<p>The atomic mass of <mathjax>#H#</mathjax> = 1</p>
<p>The atomic mass of <mathjax>#O#</mathjax> = 16</p>
<p>In a sense, the formula means <mathjax>#H#</mathjax> + <mathjax>#H#</mathjax> + <mathjax>#O#</mathjax></p>
<p>So... 1 + 1 + 16 = 18</p>
<p>Therefore, the mass of one mole of water = 18 Grams</p></div>
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<div class="markdown"><p>18 grams</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The mass of one mole of any atom/molecule is equal to its atomic molecular mass in grams. </p>
<p>The molecular formula of water is <mathjax>#H_2O#</mathjax></p>
<p>The atomic mass of <mathjax>#H#</mathjax> = 1</p>
<p>The atomic mass of <mathjax>#O#</mathjax> = 16</p>
<p>In a sense, the formula means <mathjax>#H#</mathjax> + <mathjax>#H#</mathjax> + <mathjax>#O#</mathjax></p>
<p>So... 1 + 1 + 16 = 18</p>
<p>Therefore, the mass of one mole of water = 18 Grams</p></div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">What is the mass of one mole of water? </h1>
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<div class="markdown"><p>18 grams</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The mass of one mole of any atom/molecule is equal to its atomic molecular mass in grams. </p>
<p>The molecular formula of water is <mathjax>#H_2O#</mathjax></p>
<p>The atomic mass of <mathjax>#H#</mathjax> = 1</p>
<p>The atomic mass of <mathjax>#O#</mathjax> = 16</p>
<p>In a sense, the formula means <mathjax>#H#</mathjax> + <mathjax>#H#</mathjax> + <mathjax>#O#</mathjax></p>
<p>So... 1 + 1 + 16 = 18</p>
<p>Therefore, the mass of one mole of water = 18 Grams</p></div>
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</article> | What is the mass of one mole of water? | null |
3,209 | ab11238c-6ddd-11ea-a87c-ccda262736ce | https://socratic.org/questions/what-volume-of-cl-2-gas-measured-at-684-torr-and-32-c-is-required-to-form-26-g-o | 6.18 L | start physical_unit 3 4 volume l qc_end physical_unit 3 4 7 8 pressure qc_end physical_unit 3 4 10 11 temperature qc_end physical_unit 19 19 16 17 mass qc_end end | [{"type":"physical unit","value":"Volume [OF] Cl2 gas [IN] L"}] | [{"type":"physical unit","value":"6.18 L"}] | [{"type":"physical unit","value":"Pressure [OF] Cl2 gas [=] \\pu{684 torr}"},{"type":"physical unit","value":"Temperature [OF] Cl2 gas [=] \\pu{32 ℃}"},{"type":"physical unit","value":"Mass [OF] NaCl [=] \\pu{26 g}"}] | <h1 class="questionTitle" itemprop="name">What volume of #Cl_2# gas, measured at 684 torr and 32 C, is required to form 26 g of #NaCl#?</h1> | null | 6.18 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Na(s)+1/2Cl_2(g) rarrNaCl(s)#</mathjax></p>
<p><mathjax>#"Moles of NaCl"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(26.0*g)/(58.44*g*mol^-1)=0.445*mol#</mathjax>.</p>
<p>And thus we need half an equiv of <mathjax>#Cl_2#</mathjax> gas, i.e. <mathjax>#0.222*mol#</mathjax>.</p>
<p>If we use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas law</a>, then <mathjax>#V=(nRT)/P#</mathjax>, and also we know that <mathjax>#760*mm*Hg-=1*atm#</mathjax>.</p>
<p>And thus <mathjax>#V=(0.222*molxx0.0821*L*atm*K^-1*mol^-1xx305*K)/((684*mm*Hg)/(760*mm*Hg*atm^-1))#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We require a volume of <mathjax>#~=6*L#</mathjax> <mathjax>#Cl_2#</mathjax> gas. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Na(s)+1/2Cl_2(g) rarrNaCl(s)#</mathjax></p>
<p><mathjax>#"Moles of NaCl"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(26.0*g)/(58.44*g*mol^-1)=0.445*mol#</mathjax>.</p>
<p>And thus we need half an equiv of <mathjax>#Cl_2#</mathjax> gas, i.e. <mathjax>#0.222*mol#</mathjax>.</p>
<p>If we use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas law</a>, then <mathjax>#V=(nRT)/P#</mathjax>, and also we know that <mathjax>#760*mm*Hg-=1*atm#</mathjax>.</p>
<p>And thus <mathjax>#V=(0.222*molxx0.0821*L*atm*K^-1*mol^-1xx305*K)/((684*mm*Hg)/(760*mm*Hg*atm^-1))#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What volume of #Cl_2# gas, measured at 684 torr and 32 C, is required to form 26 g of #NaCl#?</h1>
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anor277
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Dec 3, 2016
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<div class="markdown"><p>We require a volume of <mathjax>#~=6*L#</mathjax> <mathjax>#Cl_2#</mathjax> gas. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#Na(s)+1/2Cl_2(g) rarrNaCl(s)#</mathjax></p>
<p><mathjax>#"Moles of NaCl"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(26.0*g)/(58.44*g*mol^-1)=0.445*mol#</mathjax>.</p>
<p>And thus we need half an equiv of <mathjax>#Cl_2#</mathjax> gas, i.e. <mathjax>#0.222*mol#</mathjax>.</p>
<p>If we use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">Ideal Gas law</a>, then <mathjax>#V=(nRT)/P#</mathjax>, and also we know that <mathjax>#760*mm*Hg-=1*atm#</mathjax>.</p>
<p>And thus <mathjax>#V=(0.222*molxx0.0821*L*atm*K^-1*mol^-1xx305*K)/((684*mm*Hg)/(760*mm*Hg*atm^-1))#</mathjax></p></div>
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</article> | What volume of #Cl_2# gas, measured at 684 torr and 32 C, is required to form 26 g of #NaCl#? | null |
3,210 | ad0f1b02-6ddd-11ea-a45f-ccda262736ce | https://socratic.org/questions/the-volume-of-a-gas-is-250-0-ml-when-the-pressure-is-1-00-atm-at-the-same-temper | 3.13 atm | start physical_unit 27 28 pressure atm qc_end physical_unit 27 28 12 13 pressure qc_end physical_unit 27 28 6 7 volume qc_end physical_unit 27 28 30 31 volume qc_end c_other constant_temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"3.13 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.00 atm}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{250.0 mL}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{80 mL}"},{"type":"other","value":"ConstantTemperature"}] | <h1 class="questionTitle" itemprop="name">The volume of a gas is 250.0 mL when the pressure is 1.00 atm. At the same temperature, what is the pressure at which the volume of the gas is 80 mL?</h1> | null | 3.13 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to solve this problem we will use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a> which is the relationship between Pressure and Volume at a fixed temperature. </p>
<p>The equation is <mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#P_1 = 1atm#</mathjax><br/>
<mathjax>#V_1 = 250 mL#</mathjax><br/>
<mathjax>#P_2=?#</mathjax><br/>
<mathjax>#V_2=80mL#</mathjax></p>
<p><mathjax>#(1atm)(250mL)=P_2(80mL)#</mathjax></p>
<p><mathjax>#((1atm)(250cancel(mL)))/(80cancel(mL))=(P_2(cancel(80mL)))/(cancel(80mL))#</mathjax></p>
<p><mathjax>#3.125 atm = P_2#</mathjax></p>
<p>
<iframe src="https://www.youtube.com/embed/Hps5O-r6yDQ?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div class="markdown"><p>The new Pressure would be <mathjax>#3.125 atm#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to solve this problem we will use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a> which is the relationship between Pressure and Volume at a fixed temperature. </p>
<p>The equation is <mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#P_1 = 1atm#</mathjax><br/>
<mathjax>#V_1 = 250 mL#</mathjax><br/>
<mathjax>#P_2=?#</mathjax><br/>
<mathjax>#V_2=80mL#</mathjax></p>
<p><mathjax>#(1atm)(250mL)=P_2(80mL)#</mathjax></p>
<p><mathjax>#((1atm)(250cancel(mL)))/(80cancel(mL))=(P_2(cancel(80mL)))/(cancel(80mL))#</mathjax></p>
<p><mathjax>#3.125 atm = P_2#</mathjax></p>
<p>
<iframe src="https://www.youtube.com/embed/Hps5O-r6yDQ?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
</p></div>
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<h1 class="questionTitle" itemprop="name">The volume of a gas is 250.0 mL when the pressure is 1.00 atm. At the same temperature, what is the pressure at which the volume of the gas is 80 mL?</h1>
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<div class="markdown"><p>The new Pressure would be <mathjax>#3.125 atm#</mathjax></p></div>
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</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to solve this problem we will use <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a> which is the relationship between Pressure and Volume at a fixed temperature. </p>
<p>The equation is <mathjax>#P_1V_1=P_2V_2#</mathjax></p>
<p><mathjax>#P_1 = 1atm#</mathjax><br/>
<mathjax>#V_1 = 250 mL#</mathjax><br/>
<mathjax>#P_2=?#</mathjax><br/>
<mathjax>#V_2=80mL#</mathjax></p>
<p><mathjax>#(1atm)(250mL)=P_2(80mL)#</mathjax></p>
<p><mathjax>#((1atm)(250cancel(mL)))/(80cancel(mL))=(P_2(cancel(80mL)))/(cancel(80mL))#</mathjax></p>
<p><mathjax>#3.125 atm = P_2#</mathjax></p>
<p>
<iframe src="https://www.youtube.com/embed/Hps5O-r6yDQ?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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</article> | The volume of a gas is 250.0 mL when the pressure is 1.00 atm. At the same temperature, what is the pressure at which the volume of the gas is 80 mL? | null |
3,211 | aada5f6f-6ddd-11ea-805a-ccda262736ce | https://socratic.org/questions/find-the-number-of-waters-of-hydration-x-in-this-sample-of-epsom-salts | 7 | start physical_unit 4 7 number none qc_end chemical_equation 24 24 qc_end physical_unit 10 13 26 27 mass qc_end physical_unit 10 13 50 51 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Number [OF] waters of hydration (x)"}] | [{"type":"physical unit","value":"7"}] | [{"type":"chemical equation","value":"MgSO4.xH2O"},{"type":"physical unit","value":"Mass1 [OF] Epsom salts sample [=] \\pu{9.86 g}"},{"type":"physical unit","value":"Mass2 [OF] Epsom salts sample [=] \\pu{4.82 g}"},{"type":"other","value":"Complete dehydration."}] | <h1 class="questionTitle" itemprop="name"> Find the number of waters of hydration (x) in this sample of Epsom salts?</h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Epsom salts is a hydrated ionic compound with e formula MgSO4 · xH2O. A 9.86g sample of Epsom salts was heated to drive off the water of hydration. The mass of the sample after complete dehydration was 4.82g. </p></div>
</h2>
</div>
</div> | 7 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that heating the hydrate will drive off the <em>water of evaporation</em> and leave behind the <strong>anhydrous salt</strong>. </p>
<p>In your case, the anhydrous salt is magnesium sulfate, <mathjax>#"MgSO"_4#</mathjax>. Since you know that <strong>after</strong> complete dehydration the mass of the sample is equal to <mathjax>#"4.82 g"#</mathjax>, you can say that the hydrate contained</p>
<blockquote>
<p><mathjax>#"4.82 g " -> " MgSO"_4#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of the compound to convert this to <em>moles</em></p>
<blockquote>
<p><mathjax>#4.82 color(red)(cancel(color(black)("g"))) * "1 mole MgSO"_4/(120.37color(red)(cancel(color(black)("g")))) = "0.04004 moles MgSO"_4#</mathjax></p>
</blockquote>
<p>Now, the difference between the mass of the hydrate and the mass of the anhydrous salt will be equal to the mass of water of hydration.</p>
<p>You can thus say that the hydrate contained</p>
<blockquote>
<p><mathjax>#overbrace("9.86 g")^(color(blue)("mass of hydrate")) - overbrace("4.82 g")^(color(blue)("mass of anhydrous salt")) = overbrace("5.04 g")^(color(blue)("mass of water of hydration"))#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of water to convert this to moles</p>
<blockquote>
<p><mathjax>#5.04 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.2798 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>The number of waters of hydration is given by the number of moles of water present <strong>for every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of anhydrous salt. </p>
<p>In your case, you know that</p>
<blockquote>
<p><mathjax>#"0.04004 moles MgSO"_4 " " ->" " "0.2798 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>This means that <mathjax>#1#</mathjax> <strong>mole</strong> of magnesium sulfate will have</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("moles MgSO"_4))) * ("0.2798 moles H"_2"O")/(0.04004color(red)(cancel(color(black)("moles MgSO"_4)))) = 6.988 ~~ 7#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the chemical formula of the hydrate is</p>
<blockquote>
<p><mathjax>#"MgSO"_4 * 7"H"_2"O" ->#</mathjax> <em><strong>magnesium sulfate heptahydrate</strong></em></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"MgSO"_4 * 7"H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that heating the hydrate will drive off the <em>water of evaporation</em> and leave behind the <strong>anhydrous salt</strong>. </p>
<p>In your case, the anhydrous salt is magnesium sulfate, <mathjax>#"MgSO"_4#</mathjax>. Since you know that <strong>after</strong> complete dehydration the mass of the sample is equal to <mathjax>#"4.82 g"#</mathjax>, you can say that the hydrate contained</p>
<blockquote>
<p><mathjax>#"4.82 g " -> " MgSO"_4#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of the compound to convert this to <em>moles</em></p>
<blockquote>
<p><mathjax>#4.82 color(red)(cancel(color(black)("g"))) * "1 mole MgSO"_4/(120.37color(red)(cancel(color(black)("g")))) = "0.04004 moles MgSO"_4#</mathjax></p>
</blockquote>
<p>Now, the difference between the mass of the hydrate and the mass of the anhydrous salt will be equal to the mass of water of hydration.</p>
<p>You can thus say that the hydrate contained</p>
<blockquote>
<p><mathjax>#overbrace("9.86 g")^(color(blue)("mass of hydrate")) - overbrace("4.82 g")^(color(blue)("mass of anhydrous salt")) = overbrace("5.04 g")^(color(blue)("mass of water of hydration"))#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of water to convert this to moles</p>
<blockquote>
<p><mathjax>#5.04 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.2798 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>The number of waters of hydration is given by the number of moles of water present <strong>for every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of anhydrous salt. </p>
<p>In your case, you know that</p>
<blockquote>
<p><mathjax>#"0.04004 moles MgSO"_4 " " ->" " "0.2798 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>This means that <mathjax>#1#</mathjax> <strong>mole</strong> of magnesium sulfate will have</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("moles MgSO"_4))) * ("0.2798 moles H"_2"O")/(0.04004color(red)(cancel(color(black)("moles MgSO"_4)))) = 6.988 ~~ 7#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the chemical formula of the hydrate is</p>
<blockquote>
<p><mathjax>#"MgSO"_4 * 7"H"_2"O" ->#</mathjax> <em><strong>magnesium sulfate heptahydrate</strong></em></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name"> Find the number of waters of hydration (x) in this sample of Epsom salts?</h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>Epsom salts is a hydrated ionic compound with e formula MgSO4 · xH2O. A 9.86g sample of Epsom salts was heated to drive off the water of hydration. The mass of the sample after complete dehydration was 4.82g. </p></div>
</h2>
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<a class="topContributorPic" href="/users/stefan-zdre"><img alt="" class="" src="https://profilepictures.socratic.org/LrguokJzR9yQlbiWbCvr_proba_1.jpg" title=""/></a>
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Stefan V.
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<span class="dateCreated" datetime="2017-06-17T01:25:05" itemprop="dateCreated">
Jun 17, 2017
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<div class="markdown"><p><mathjax>#"MgSO"_4 * 7"H"_2"O"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that heating the hydrate will drive off the <em>water of evaporation</em> and leave behind the <strong>anhydrous salt</strong>. </p>
<p>In your case, the anhydrous salt is magnesium sulfate, <mathjax>#"MgSO"_4#</mathjax>. Since you know that <strong>after</strong> complete dehydration the mass of the sample is equal to <mathjax>#"4.82 g"#</mathjax>, you can say that the hydrate contained</p>
<blockquote>
<p><mathjax>#"4.82 g " -> " MgSO"_4#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of the compound to convert this to <em>moles</em></p>
<blockquote>
<p><mathjax>#4.82 color(red)(cancel(color(black)("g"))) * "1 mole MgSO"_4/(120.37color(red)(cancel(color(black)("g")))) = "0.04004 moles MgSO"_4#</mathjax></p>
</blockquote>
<p>Now, the difference between the mass of the hydrate and the mass of the anhydrous salt will be equal to the mass of water of hydration.</p>
<p>You can thus say that the hydrate contained</p>
<blockquote>
<p><mathjax>#overbrace("9.86 g")^(color(blue)("mass of hydrate")) - overbrace("4.82 g")^(color(blue)("mass of anhydrous salt")) = overbrace("5.04 g")^(color(blue)("mass of water of hydration"))#</mathjax></p>
</blockquote>
<p>Use the <strong>molar mass</strong> of water to convert this to moles</p>
<blockquote>
<p><mathjax>#5.04 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.2798 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>The number of waters of hydration is given by the number of moles of water present <strong>for every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of anhydrous salt. </p>
<p>In your case, you know that</p>
<blockquote>
<p><mathjax>#"0.04004 moles MgSO"_4 " " ->" " "0.2798 moles H"_2"O"#</mathjax></p>
</blockquote>
<p>This means that <mathjax>#1#</mathjax> <strong>mole</strong> of magnesium sulfate will have</p>
<blockquote>
<p><mathjax>#1 color(red)(cancel(color(black)("moles MgSO"_4))) * ("0.2798 moles H"_2"O")/(0.04004color(red)(cancel(color(black)("moles MgSO"_4)))) = 6.988 ~~ 7#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the chemical formula of the hydrate is</p>
<blockquote>
<p><mathjax>#"MgSO"_4 * 7"H"_2"O" ->#</mathjax> <em><strong>magnesium sulfate heptahydrate</strong></em></p>
</blockquote></div>
</div>
</div>
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<a href="https://socratic.org/answers/440828" itemprop="url">Answer link</a>
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</article> | Find the number of waters of hydration (x) in this sample of Epsom salts? |
Epsom salts is a hydrated ionic compound with e formula MgSO4 · xH2O. A 9.86g sample of Epsom salts was heated to drive off the water of hydration. The mass of the sample after complete dehydration was 4.82g.
|
3,212 | a8f68fe8-6ddd-11ea-8feb-ccda262736ce | https://socratic.org/questions/how-many-mmols-in-naoh-are-used-in-an-experiment-5ml-of-6-m-naoh-is-used-along-w | 0.8 mmoles | start physical_unit 4 4 mole mmol qc_end physical_unit 25 25 10 11 volume qc_end physical_unit 4 4 13 14 molarity qc_end physical_unit 22 23 20 21 volume qc_end physical_unit 25 25 40 41 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mole [OF] NaOH [IN] mmoles"}] | [{"type":"physical unit","value":"0.8 mmoles"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{5 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{6 M}"},{"type":"physical unit","value":"Volume [OF] distilled water [=] \\pu{300 mL}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{7.92 mL}"},{"type":"other","value":"The solution of the NaOH was titrated into a vinegar solution."}] | <h1 class="questionTitle" itemprop="name">How many mmols in NaOH are used? In an experiment, 5mL of 6 M NaOH is used along with 300 ml distilled water. The solution of the NaOH was titrated into a vinegar solution and it was found that 7.92 mL of NaOH was used. </h1> | <div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>My teacher wrote this down on the board: 6M NaOH/305 ml X 5ml =.1 M NaOH</p></div>
</h2>
</div>
</div> | 0.8 mmoles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you're <strong>diluting</strong> the stock solution to a total volume of </p>
<blockquote>
<p><mathjax>#overbrace("5 mL")^(color(blue)("the volume of the stock solution")) + overbrace("300 mL")^(color(blue)("the volume of water")) = "305 mL"#</mathjax></p>
</blockquote>
<p>The thing to keep in mind about dilutions is that the ratio that exists between the volume of the diluted solution and the volume of the stock solution</p>
<blockquote>
<p><mathjax>#V_"diluted"/V_"stock"#</mathjax></p>
</blockquote>
<p>and the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution</p>
<blockquote>
<p><mathjax>#c_"stock"/c_"diluted"#</mathjax></p>
</blockquote>
<p>must be <strong>equal</strong>. You can thus say that you have--here <mathjax>#"DF"#</mathjax> is the <strong>dilution factor</strong></p>
<blockquote>
<p><mathjax>#"DF" = c_"stock"/c_"diluted" = V_"diluted"/V_"stock"#</mathjax></p>
</blockquote>
<p>In your case, the dilution factor is </p>
<blockquote>
<p><mathjax>#"DF" = (305 color(red)(cancel(color(black)("mL"))))/(5color(red)(cancel(color(black)("mL")))) = color(blue)(61)#</mathjax></p>
</blockquote>
<p>which means that the concentration of the diluted sodium hydroxide solution is </p>
<blockquote>
<p><mathjax>#c_"diluted" = c_"stock"/color(blue)("DF")#</mathjax></p>
<p><mathjax>#c_"diluted" = "6 M"/color(blue)(61) = "0.0984 M"#</mathjax></p>
</blockquote>
<p>Now all you have to do is to use the volume of the solution to find the number of <em>millimoles</em> of sodium hydroxide present in the sample</p>
<blockquote>
<p><mathjax>#7.92 color(red)(cancel(color(black)("mL solution"))) * (0.0984 * color(blue)(cancel(color(black)(10^3)))color(white)(.)"mmoles NaOH")/(color(blue)(cancel(color(black)(10^3)))color(red)(cancel(color(black)("mL solution")))) = color(darkgreen)(ul(color(black)("0.8 mmoles NaOH")))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>.</p>
<p>Notice that this is exactly what your teacher calculated by doing</p>
<blockquote>
<p><mathjax>#"6 M"/"305 mL" * "5 mL" ~~ "0.1 M"#</mathjax></p>
</blockquote>
<p>because this is actually the concentration of the stock solution divided by the dilution factor</p>
<blockquote>
<p><mathjax>#"6 M"/("305 mL"/"5 mL") = "6 M"/"305 mL" * "5 mL"#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"0.8 mmoles NaOH"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you're <strong>diluting</strong> the stock solution to a total volume of </p>
<blockquote>
<p><mathjax>#overbrace("5 mL")^(color(blue)("the volume of the stock solution")) + overbrace("300 mL")^(color(blue)("the volume of water")) = "305 mL"#</mathjax></p>
</blockquote>
<p>The thing to keep in mind about dilutions is that the ratio that exists between the volume of the diluted solution and the volume of the stock solution</p>
<blockquote>
<p><mathjax>#V_"diluted"/V_"stock"#</mathjax></p>
</blockquote>
<p>and the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution</p>
<blockquote>
<p><mathjax>#c_"stock"/c_"diluted"#</mathjax></p>
</blockquote>
<p>must be <strong>equal</strong>. You can thus say that you have--here <mathjax>#"DF"#</mathjax> is the <strong>dilution factor</strong></p>
<blockquote>
<p><mathjax>#"DF" = c_"stock"/c_"diluted" = V_"diluted"/V_"stock"#</mathjax></p>
</blockquote>
<p>In your case, the dilution factor is </p>
<blockquote>
<p><mathjax>#"DF" = (305 color(red)(cancel(color(black)("mL"))))/(5color(red)(cancel(color(black)("mL")))) = color(blue)(61)#</mathjax></p>
</blockquote>
<p>which means that the concentration of the diluted sodium hydroxide solution is </p>
<blockquote>
<p><mathjax>#c_"diluted" = c_"stock"/color(blue)("DF")#</mathjax></p>
<p><mathjax>#c_"diluted" = "6 M"/color(blue)(61) = "0.0984 M"#</mathjax></p>
</blockquote>
<p>Now all you have to do is to use the volume of the solution to find the number of <em>millimoles</em> of sodium hydroxide present in the sample</p>
<blockquote>
<p><mathjax>#7.92 color(red)(cancel(color(black)("mL solution"))) * (0.0984 * color(blue)(cancel(color(black)(10^3)))color(white)(.)"mmoles NaOH")/(color(blue)(cancel(color(black)(10^3)))color(red)(cancel(color(black)("mL solution")))) = color(darkgreen)(ul(color(black)("0.8 mmoles NaOH")))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>.</p>
<p>Notice that this is exactly what your teacher calculated by doing</p>
<blockquote>
<p><mathjax>#"6 M"/"305 mL" * "5 mL" ~~ "0.1 M"#</mathjax></p>
</blockquote>
<p>because this is actually the concentration of the stock solution divided by the dilution factor</p>
<blockquote>
<p><mathjax>#"6 M"/("305 mL"/"5 mL") = "6 M"/"305 mL" * "5 mL"#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How many mmols in NaOH are used? In an experiment, 5mL of 6 M NaOH is used along with 300 ml distilled water. The solution of the NaOH was titrated into a vinegar solution and it was found that 7.92 mL of NaOH was used. </h1>
<div class="questionDetailsContainer">
<div class="collapsedQuestionDetails">
<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>My teacher wrote this down on the board: 6M NaOH/305 ml X 5ml =.1 M NaOH</p></div>
</h2>
</div>
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<div class="markdown"><p><mathjax>#"0.8 mmoles NaOH"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that you're <strong>diluting</strong> the stock solution to a total volume of </p>
<blockquote>
<p><mathjax>#overbrace("5 mL")^(color(blue)("the volume of the stock solution")) + overbrace("300 mL")^(color(blue)("the volume of water")) = "305 mL"#</mathjax></p>
</blockquote>
<p>The thing to keep in mind about dilutions is that the ratio that exists between the volume of the diluted solution and the volume of the stock solution</p>
<blockquote>
<p><mathjax>#V_"diluted"/V_"stock"#</mathjax></p>
</blockquote>
<p>and the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution</p>
<blockquote>
<p><mathjax>#c_"stock"/c_"diluted"#</mathjax></p>
</blockquote>
<p>must be <strong>equal</strong>. You can thus say that you have--here <mathjax>#"DF"#</mathjax> is the <strong>dilution factor</strong></p>
<blockquote>
<p><mathjax>#"DF" = c_"stock"/c_"diluted" = V_"diluted"/V_"stock"#</mathjax></p>
</blockquote>
<p>In your case, the dilution factor is </p>
<blockquote>
<p><mathjax>#"DF" = (305 color(red)(cancel(color(black)("mL"))))/(5color(red)(cancel(color(black)("mL")))) = color(blue)(61)#</mathjax></p>
</blockquote>
<p>which means that the concentration of the diluted sodium hydroxide solution is </p>
<blockquote>
<p><mathjax>#c_"diluted" = c_"stock"/color(blue)("DF")#</mathjax></p>
<p><mathjax>#c_"diluted" = "6 M"/color(blue)(61) = "0.0984 M"#</mathjax></p>
</blockquote>
<p>Now all you have to do is to use the volume of the solution to find the number of <em>millimoles</em> of sodium hydroxide present in the sample</p>
<blockquote>
<p><mathjax>#7.92 color(red)(cancel(color(black)("mL solution"))) * (0.0984 * color(blue)(cancel(color(black)(10^3)))color(white)(.)"mmoles NaOH")/(color(blue)(cancel(color(black)(10^3)))color(red)(cancel(color(black)("mL solution")))) = color(darkgreen)(ul(color(black)("0.8 mmoles NaOH")))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to one <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figure</a></strong>.</p>
<p>Notice that this is exactly what your teacher calculated by doing</p>
<blockquote>
<p><mathjax>#"6 M"/"305 mL" * "5 mL" ~~ "0.1 M"#</mathjax></p>
</blockquote>
<p>because this is actually the concentration of the stock solution divided by the dilution factor</p>
<blockquote>
<p><mathjax>#"6 M"/("305 mL"/"5 mL") = "6 M"/"305 mL" * "5 mL"#</mathjax></p>
</blockquote></div>
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</article> | How many mmols in NaOH are used? In an experiment, 5mL of 6 M NaOH is used along with 300 ml distilled water. The solution of the NaOH was titrated into a vinegar solution and it was found that 7.92 mL of NaOH was used. |
My teacher wrote this down on the board: 6M NaOH/305 ml X 5ml =.1 M NaOH
|
3,213 | ad24d3a4-6ddd-11ea-a4d9-ccda262736ce | https://socratic.org/questions/how-many-grams-of-alcl-3-can-be-produced-by-reacting-80-0-grams-of-cl-2-with-an- | 100.27 grams | start physical_unit 4 4 mass g qc_end physical_unit 13 13 10 11 mass qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Mass [OF] AlCl3 [IN] grams"}] | [{"type":"physical unit","value":"100.27 grams"}] | [{"type":"physical unit","value":"Mass [OF] Cl2 [=] \\pu{80.0 grams}"},{"type":"other","value":"Unlimited supply of Al."}] | <h1 class="questionTitle" itemprop="name">How many grams of #AlCl_3# can be produced by reacting 80.0 grams of #Cl_2# with an unlimited supply of #Al#?</h1> | null | 100.27 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your starting point here will be the <strong>balanced</strong> chemical equation for this <strong><a href="http://socratic.org/chemistry/chemical-reactions/synthesis-reactions">synthesis reaction</a></strong>.</p>
<p>Aluminium metal, <mathjax>#"Al"#</mathjax>, will react with chlorine gas, <mathjax>#"Cl"_2#</mathjax>, to produce aluminium chloride, <mathjax>#"AlCl"_3#</mathjax>, according to the balanced chemical equation</p>
<blockquote>
<p><mathjax>#2"Al"_text((s]) + color(red)(3)"Cl"_text(2(g]) -> color(blue)(2)"AlCl"_text(3(s])#</mathjax></p>
</blockquote>
<p>Now, you know that aluminium is <em>in excess</em>, which implies that the reaction will <strong>completely consume</strong> the chlorine gas. </p>
<p>Notice that you have a <mathjax>#color(red)(3):color(blue)(2)#</mathjax> mole ratio between chlorine gas and aluminium chloride. This tells you that for every three moles of chlorine gas that take part in the reaction, you get two moles of aluminium chloride. </p>
<p>Use the <strong>molar mass</strong> of chlorine gas to determine how many <em>moles</em> you have in that <mathjax>#"80.0-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#80.0 color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("g")))) = "1.128 moles Cl"_2#</mathjax></p>
</blockquote>
<p>Next, use the aforementioned mole ratio to determine how many <em>moles</em> of aluminium chloride will be produced by this many moles of chlorine gas</p>
<blockquote>
<p><mathjax>#1.128color(red)(cancel(color(black)("moles Cl"_2))) * (color(blue)(2)color(white)(a)"moles AlCl"_3)/(color(red)(3)color(red)(cancel(color(black)("moles Cl"_2)))) = "0.7520 moles AlCl"_3#</mathjax></p>
</blockquote>
<p>Finally, to find the mass of aluminium chloride that would contain that many moles, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#0.7520color(red)(cancel(color(black)("moles AlCl"_3))) * "133.34 g"/(1color(red)(cancel(color(black)("mole AlCl"_3)))) = "100.27 g"#</mathjax></p>
</blockquote>
<p>Rounded to three **<a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the mass of chlorine gas, the answer will be</p>
<blockquote>
<p><mathjax>#m_(AlCl_3) = color(green)(|bar(ul(color(white)(a/a)"100 g"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"100 g AlCl"_3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your starting point here will be the <strong>balanced</strong> chemical equation for this <strong><a href="http://socratic.org/chemistry/chemical-reactions/synthesis-reactions">synthesis reaction</a></strong>.</p>
<p>Aluminium metal, <mathjax>#"Al"#</mathjax>, will react with chlorine gas, <mathjax>#"Cl"_2#</mathjax>, to produce aluminium chloride, <mathjax>#"AlCl"_3#</mathjax>, according to the balanced chemical equation</p>
<blockquote>
<p><mathjax>#2"Al"_text((s]) + color(red)(3)"Cl"_text(2(g]) -> color(blue)(2)"AlCl"_text(3(s])#</mathjax></p>
</blockquote>
<p>Now, you know that aluminium is <em>in excess</em>, which implies that the reaction will <strong>completely consume</strong> the chlorine gas. </p>
<p>Notice that you have a <mathjax>#color(red)(3):color(blue)(2)#</mathjax> mole ratio between chlorine gas and aluminium chloride. This tells you that for every three moles of chlorine gas that take part in the reaction, you get two moles of aluminium chloride. </p>
<p>Use the <strong>molar mass</strong> of chlorine gas to determine how many <em>moles</em> you have in that <mathjax>#"80.0-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#80.0 color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("g")))) = "1.128 moles Cl"_2#</mathjax></p>
</blockquote>
<p>Next, use the aforementioned mole ratio to determine how many <em>moles</em> of aluminium chloride will be produced by this many moles of chlorine gas</p>
<blockquote>
<p><mathjax>#1.128color(red)(cancel(color(black)("moles Cl"_2))) * (color(blue)(2)color(white)(a)"moles AlCl"_3)/(color(red)(3)color(red)(cancel(color(black)("moles Cl"_2)))) = "0.7520 moles AlCl"_3#</mathjax></p>
</blockquote>
<p>Finally, to find the mass of aluminium chloride that would contain that many moles, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#0.7520color(red)(cancel(color(black)("moles AlCl"_3))) * "133.34 g"/(1color(red)(cancel(color(black)("mole AlCl"_3)))) = "100.27 g"#</mathjax></p>
</blockquote>
<p>Rounded to three **<a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the mass of chlorine gas, the answer will be</p>
<blockquote>
<p><mathjax>#m_(AlCl_3) = color(green)(|bar(ul(color(white)(a/a)"100 g"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">How many grams of #AlCl_3# can be produced by reacting 80.0 grams of #Cl_2# with an unlimited supply of #Al#?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"100 g AlCl"_3#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your starting point here will be the <strong>balanced</strong> chemical equation for this <strong><a href="http://socratic.org/chemistry/chemical-reactions/synthesis-reactions">synthesis reaction</a></strong>.</p>
<p>Aluminium metal, <mathjax>#"Al"#</mathjax>, will react with chlorine gas, <mathjax>#"Cl"_2#</mathjax>, to produce aluminium chloride, <mathjax>#"AlCl"_3#</mathjax>, according to the balanced chemical equation</p>
<blockquote>
<p><mathjax>#2"Al"_text((s]) + color(red)(3)"Cl"_text(2(g]) -> color(blue)(2)"AlCl"_text(3(s])#</mathjax></p>
</blockquote>
<p>Now, you know that aluminium is <em>in excess</em>, which implies that the reaction will <strong>completely consume</strong> the chlorine gas. </p>
<p>Notice that you have a <mathjax>#color(red)(3):color(blue)(2)#</mathjax> mole ratio between chlorine gas and aluminium chloride. This tells you that for every three moles of chlorine gas that take part in the reaction, you get two moles of aluminium chloride. </p>
<p>Use the <strong>molar mass</strong> of chlorine gas to determine how many <em>moles</em> you have in that <mathjax>#"80.0-g"#</mathjax> sample</p>
<blockquote>
<p><mathjax>#80.0 color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("g")))) = "1.128 moles Cl"_2#</mathjax></p>
</blockquote>
<p>Next, use the aforementioned mole ratio to determine how many <em>moles</em> of aluminium chloride will be produced by this many moles of chlorine gas</p>
<blockquote>
<p><mathjax>#1.128color(red)(cancel(color(black)("moles Cl"_2))) * (color(blue)(2)color(white)(a)"moles AlCl"_3)/(color(red)(3)color(red)(cancel(color(black)("moles Cl"_2)))) = "0.7520 moles AlCl"_3#</mathjax></p>
</blockquote>
<p>Finally, to find the mass of aluminium chloride that would contain that many moles, use the compound's <strong>molar mass</strong></p>
<blockquote>
<p><mathjax>#0.7520color(red)(cancel(color(black)("moles AlCl"_3))) * "133.34 g"/(1color(red)(cancel(color(black)("mole AlCl"_3)))) = "100.27 g"#</mathjax></p>
</blockquote>
<p>Rounded to three **<a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the mass of chlorine gas, the answer will be</p>
<blockquote>
<p><mathjax>#m_(AlCl_3) = color(green)(|bar(ul(color(white)(a/a)"100 g"color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | How many grams of #AlCl_3# can be produced by reacting 80.0 grams of #Cl_2# with an unlimited supply of #Al#? | null |
3,214 | ac1ebf5c-6ddd-11ea-b41b-ccda262736ce | https://socratic.org/questions/calculate-the-number-of-moles-of-cl-ions-in-1-75-l-of-1-0-x-10-3-m-alcl-3-how-wo | 5.25 moles | start physical_unit 6 7 mole mol qc_end physical_unit 16 16 12 15 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] Cl- ions [IN] moles"}] | [{"type":"physical unit","value":"5.25 moles"}] | [{"type":"physical unit","value":"Volume [OF] AlCl3 solution [=] \\pu{1.75 L}"},{"type":"physical unit","value":"Molarity [OF] AlCl3 solution [=] \\pu{1.0 × 10^(-3) M}"}] | <h1 class="questionTitle" itemprop="name">Calculate the number of moles of #Cl^(-)# ions in 1.75 L of #1.0 xx 10^-3# M #AlCl_3#.
How would you solve this?</h1> | null | 5.25 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, calculate how many moles of aluminium chloride would be needed in order to get a <mathjax>#1.0 * 10^(-3)#</mathjax> <mathjax>#"M"#</mathjax> solution in <mathjax>#"1.75 L"#</mathjax> of solution. </p>
<p>As you know, <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> tells you the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in <mathjax>#"1 L"#</mathjax> of solution. This means that <mathjax>#"1 L"#</mathjax> of <mathjax>#1.0 * 10^(-3)#</mathjax> <mathjax>#"M"#</mathjax> aluminium chloride solution will contain <mathjax>#1.0 * 10^(-3)#</mathjax> <strong>moles</strong> of aluminium chloride, the solute.</p>
<p>In your case, the solution will contain</p>
<blockquote>
<p><mathjax>#1.75 color(red)(cancel(color(black)("L solution"))) * overbrace((1.0 * 10^(-3)color(white)(.)"moles AlCl"_3)/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)(=1.0 * 10^(-3)"M AlCl"_3)) = 1.75 * 10^(-3)#</mathjax> <mathjax>#"moles AlCl"_3#</mathjax></p>
</blockquote>
<p>So, you know that you must dissolve <mathjax>#1.75 * 10^(-3)#</mathjax> <strong>moles</strong> of aluminium chloride in enough water to make the volume of the solution equal to <mathjax>#"1.75 L"#</mathjax>.</p>
<p>Now, aluminium chloride is a <strong>soluble</strong> ionic compound, which implies that it dissociates completely in aqueous solution to produce aluminium actions and chloride anions</p>
<blockquote>
<p><mathjax>#"AlCl"_ (color(red)(3)(aq)) -> "Al"_ ((aq))^(3+) + color(red)(3)"Cl"^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium chloride dissolved in water produces <mathjax>#color(red)(3)#</mathjax> <strong>moles</strong> of chloride anions.</p>
<p>This implies that your solution will contain</p>
<blockquote>
<p><mathjax>#1.75 * 10^(-3) color(red)(cancel(color(black)("moles AlCl"_3))) * (color(red)(3)color(white)(.)"moles Cl"^(-))/(1color(red)(cancel(color(black)("mole AlCl"_3))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = color(darkgreen)(ul(color(black)(5.3 * 10^(-3)color(white)(.)"moles Cl"^(-))))#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The answer must be rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#5.3 * 10^(-3)"moles Cl"^(-)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, calculate how many moles of aluminium chloride would be needed in order to get a <mathjax>#1.0 * 10^(-3)#</mathjax> <mathjax>#"M"#</mathjax> solution in <mathjax>#"1.75 L"#</mathjax> of solution. </p>
<p>As you know, <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> tells you the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in <mathjax>#"1 L"#</mathjax> of solution. This means that <mathjax>#"1 L"#</mathjax> of <mathjax>#1.0 * 10^(-3)#</mathjax> <mathjax>#"M"#</mathjax> aluminium chloride solution will contain <mathjax>#1.0 * 10^(-3)#</mathjax> <strong>moles</strong> of aluminium chloride, the solute.</p>
<p>In your case, the solution will contain</p>
<blockquote>
<p><mathjax>#1.75 color(red)(cancel(color(black)("L solution"))) * overbrace((1.0 * 10^(-3)color(white)(.)"moles AlCl"_3)/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)(=1.0 * 10^(-3)"M AlCl"_3)) = 1.75 * 10^(-3)#</mathjax> <mathjax>#"moles AlCl"_3#</mathjax></p>
</blockquote>
<p>So, you know that you must dissolve <mathjax>#1.75 * 10^(-3)#</mathjax> <strong>moles</strong> of aluminium chloride in enough water to make the volume of the solution equal to <mathjax>#"1.75 L"#</mathjax>.</p>
<p>Now, aluminium chloride is a <strong>soluble</strong> ionic compound, which implies that it dissociates completely in aqueous solution to produce aluminium actions and chloride anions</p>
<blockquote>
<p><mathjax>#"AlCl"_ (color(red)(3)(aq)) -> "Al"_ ((aq))^(3+) + color(red)(3)"Cl"^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium chloride dissolved in water produces <mathjax>#color(red)(3)#</mathjax> <strong>moles</strong> of chloride anions.</p>
<p>This implies that your solution will contain</p>
<blockquote>
<p><mathjax>#1.75 * 10^(-3) color(red)(cancel(color(black)("moles AlCl"_3))) * (color(red)(3)color(white)(.)"moles Cl"^(-))/(1color(red)(cancel(color(black)("mole AlCl"_3))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = color(darkgreen)(ul(color(black)(5.3 * 10^(-3)color(white)(.)"moles Cl"^(-))))#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The answer must be rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Calculate the number of moles of #Cl^(-)# ions in 1.75 L of #1.0 xx 10^-3# M #AlCl_3#.
How would you solve this?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2017-05-09T15:42:52" itemprop="dateCreated">
May 9, 2017
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<div class="markdown"><p><mathjax>#5.3 * 10^(-3)"moles Cl"^(-)#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>For starters, calculate how many moles of aluminium chloride would be needed in order to get a <mathjax>#1.0 * 10^(-3)#</mathjax> <mathjax>#"M"#</mathjax> solution in <mathjax>#"1.75 L"#</mathjax> of solution. </p>
<p>As you know, <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> tells you the number of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> present in <mathjax>#"1 L"#</mathjax> of solution. This means that <mathjax>#"1 L"#</mathjax> of <mathjax>#1.0 * 10^(-3)#</mathjax> <mathjax>#"M"#</mathjax> aluminium chloride solution will contain <mathjax>#1.0 * 10^(-3)#</mathjax> <strong>moles</strong> of aluminium chloride, the solute.</p>
<p>In your case, the solution will contain</p>
<blockquote>
<p><mathjax>#1.75 color(red)(cancel(color(black)("L solution"))) * overbrace((1.0 * 10^(-3)color(white)(.)"moles AlCl"_3)/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)(=1.0 * 10^(-3)"M AlCl"_3)) = 1.75 * 10^(-3)#</mathjax> <mathjax>#"moles AlCl"_3#</mathjax></p>
</blockquote>
<p>So, you know that you must dissolve <mathjax>#1.75 * 10^(-3)#</mathjax> <strong>moles</strong> of aluminium chloride in enough water to make the volume of the solution equal to <mathjax>#"1.75 L"#</mathjax>.</p>
<p>Now, aluminium chloride is a <strong>soluble</strong> ionic compound, which implies that it dissociates completely in aqueous solution to produce aluminium actions and chloride anions</p>
<blockquote>
<p><mathjax>#"AlCl"_ (color(red)(3)(aq)) -> "Al"_ ((aq))^(3+) + color(red)(3)"Cl"^(-)#</mathjax></p>
</blockquote>
<p>Notice that <strong>every</strong> <mathjax>#1#</mathjax> <strong>mole</strong> of aluminium chloride dissolved in water produces <mathjax>#color(red)(3)#</mathjax> <strong>moles</strong> of chloride anions.</p>
<p>This implies that your solution will contain</p>
<blockquote>
<p><mathjax>#1.75 * 10^(-3) color(red)(cancel(color(black)("moles AlCl"_3))) * (color(red)(3)color(white)(.)"moles Cl"^(-))/(1color(red)(cancel(color(black)("mole AlCl"_3))))#</mathjax></p>
<blockquote>
<blockquote>
<blockquote>
<p><mathjax># = color(darkgreen)(ul(color(black)(5.3 * 10^(-3)color(white)(.)"moles Cl"^(-))))#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
</blockquote>
<p>The answer must be rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
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</article> | Calculate the number of moles of #Cl^(-)# ions in 1.75 L of #1.0 xx 10^-3# M #AlCl_3#.
How would you solve this? | null |
3,215 | ac7df88a-6ddd-11ea-a77d-ccda262736ce | https://socratic.org/questions/what-pressure-in-atmospheres-is-required-to-compress-1-00-l-of-gas-at-760-mm-hg- | 20.00 atmospheres | start physical_unit 11 11 pressure atm qc_end physical_unit 11 11 8 9 volume qc_end physical_unit 11 11 13 14 pressure qc_end physical_unit 11 11 20 21 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] atmospheres"}] | [{"type":"physical unit","value":"20.00 atmospheres"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{1.00 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{760 mmHg}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{50.0 mL}"}] | <h1 class="questionTitle" itemprop="name">What pressure (in atmospheres) is required to compress 1.00 L of gas at 760. mm Hg pressure to a volume of 50.0 mL?</h1> | null | 20.00 atmospheres | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax>, now <mathjax>#P_1=1*atm#</mathjax>, and so we solve for <mathjax>#P_2#</mathjax>:</p>
<p><mathjax>#P_2=(P_1V_1)/V_2=(1.00*cancelLxx1*atm)/(0.050*cancelL)=??atm#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"20 atmospheres........"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax>, now <mathjax>#P_1=1*atm#</mathjax>, and so we solve for <mathjax>#P_2#</mathjax>:</p>
<p><mathjax>#P_2=(P_1V_1)/V_2=(1.00*cancelLxx1*atm)/(0.050*cancelL)=??atm#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What pressure (in atmospheres) is required to compress 1.00 L of gas at 760. mm Hg pressure to a volume of 50.0 mL?</h1>
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anor277
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<div class="markdown"><p><mathjax>#"20 atmospheres........"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#P_1V_1=P_2V_2#</mathjax>, now <mathjax>#P_1=1*atm#</mathjax>, and so we solve for <mathjax>#P_2#</mathjax>:</p>
<p><mathjax>#P_2=(P_1V_1)/V_2=(1.00*cancelLxx1*atm)/(0.050*cancelL)=??atm#</mathjax></p></div>
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</article> | What pressure (in atmospheres) is required to compress 1.00 L of gas at 760. mm Hg pressure to a volume of 50.0 mL? | null |
3,216 | a8deedec-6ddd-11ea-ba25-ccda262736ce | https://socratic.org/questions/5984b3bd7c01497466fde781 | 583.24 psi | start physical_unit 7 7 pressure psi qc_end physical_unit 7 7 15 16 volume qc_end physical_unit 7 7 24 25 volume qc_end physical_unit 7 7 9 10 pressure qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] psi"}] | [{"type":"physical unit","value":"583.24 psi"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{6.53 L}"},{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{12.45 L}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1112 psi}"}] | <h1 class="questionTitle" itemprop="name">What is the new pressure of a gas under #1112*"psi"# with a volume of #6.53*L# that is expanded to a volume of #12.45*L#?</h1> | null | 583.24 psi | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given the proportionality, we could use any units of pressure, and volume, i.e. <mathjax>#"psi"#</mathjax>, or <mathjax>#"pints"#</mathjax>, or <mathjax>#"gallons"#</mathjax> as Boyle, working in the 17th century, undoubtedly did. </p>
<p>And so <mathjax>#P_2=(P_1V_1)/V_2=(1112*"psi"xx6.53*L)/(12.45*L)=??"psi"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Well <mathjax>#P_1V_1=P_2V_2#</mathjax> according to old <mathjax>#"Boyle's Law"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given the proportionality, we could use any units of pressure, and volume, i.e. <mathjax>#"psi"#</mathjax>, or <mathjax>#"pints"#</mathjax>, or <mathjax>#"gallons"#</mathjax> as Boyle, working in the 17th century, undoubtedly did. </p>
<p>And so <mathjax>#P_2=(P_1V_1)/V_2=(1112*"psi"xx6.53*L)/(12.45*L)=??"psi"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the new pressure of a gas under #1112*"psi"# with a volume of #6.53*L# that is expanded to a volume of #12.45*L#?</h1>
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<div class="markdown"><p>Well <mathjax>#P_1V_1=P_2V_2#</mathjax> according to old <mathjax>#"Boyle's Law"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given the proportionality, we could use any units of pressure, and volume, i.e. <mathjax>#"psi"#</mathjax>, or <mathjax>#"pints"#</mathjax>, or <mathjax>#"gallons"#</mathjax> as Boyle, working in the 17th century, undoubtedly did. </p>
<p>And so <mathjax>#P_2=(P_1V_1)/V_2=(1112*"psi"xx6.53*L)/(12.45*L)=??"psi"#</mathjax></p></div>
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</article> | What is the new pressure of a gas under #1112*"psi"# with a volume of #6.53*L# that is expanded to a volume of #12.45*L#? | null |
3,217 | ac0316e6-6ddd-11ea-9b2d-ccda262736ce | https://socratic.org/questions/592d8a0e11ef6b28431c1d95 | 6.26 mL | start physical_unit 19 20 volume ml qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Evolved volume [OF] dihydrogen gas [IN] mL"}] | [{"type":"physical unit","value":"6.26 mL"}] | [{"type":"physical unit","value":"Volume [OF] hydrochloric acid solution [=] \\pu{55 mL}"},{"type":"other","value":"Excess of magnesium turnings."}] | <h1 class="questionTitle" itemprop="name">A #55*mL# volume of hydrochloric acid was poured on to an excess of magnesium turnings. What volume of dihydrogen gas evolved?</h1> | null | 6.26 mL | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus we interrogate the redox reaction........</p>
<p><mathjax>#Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>And thus the moles of magnesium is equivalent to the moles of dihydrogen gas if there is molar equivalence.</p>
<p>Now I happen to know that if <mathjax>#rho_"HCl"=1.18*g*mol^-1#</mathjax>, then the concentration of <mathjax>#HCl#</mathjax> is <mathjax>#10.17*mol*L^-1#</mathjax> (and such data SHOULD have been supplied with the question).</p>
<p>And so we have a molar quantity of <mathjax>#55xx10^-3*Lxx10.17*mol*L^-1=0.559*mol#</mathjax> with respect to <mathjax>#HCl#</mathjax>.</p>
<p><mathjax>#"Moles of magnesium"=(0.865*g)/(24.305*g*mol^-1)=0.0356*mol#</mathjax>.</p>
<p>Magnesium metal is in excess, and we do not have a handle on the volume of dihydrogen evolved. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well, one mole of gas at <mathjax>#"STP"#</mathjax> occupies <mathjax>#22.4*L#</mathjax>.......We do not know the concentration of hydrochloric acid.......</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus we interrogate the redox reaction........</p>
<p><mathjax>#Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>And thus the moles of magnesium is equivalent to the moles of dihydrogen gas if there is molar equivalence.</p>
<p>Now I happen to know that if <mathjax>#rho_"HCl"=1.18*g*mol^-1#</mathjax>, then the concentration of <mathjax>#HCl#</mathjax> is <mathjax>#10.17*mol*L^-1#</mathjax> (and such data SHOULD have been supplied with the question).</p>
<p>And so we have a molar quantity of <mathjax>#55xx10^-3*Lxx10.17*mol*L^-1=0.559*mol#</mathjax> with respect to <mathjax>#HCl#</mathjax>.</p>
<p><mathjax>#"Moles of magnesium"=(0.865*g)/(24.305*g*mol^-1)=0.0356*mol#</mathjax>.</p>
<p>Magnesium metal is in excess, and we do not have a handle on the volume of dihydrogen evolved. </p></div>
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<h1 class="questionTitle" itemprop="name">A #55*mL# volume of hydrochloric acid was poured on to an excess of magnesium turnings. What volume of dihydrogen gas evolved?</h1>
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<div class="markdown"><p>Well, one mole of gas at <mathjax>#"STP"#</mathjax> occupies <mathjax>#22.4*L#</mathjax>.......We do not know the concentration of hydrochloric acid.......</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>And thus we interrogate the redox reaction........</p>
<p><mathjax>#Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr#</mathjax></p>
<p>And thus the moles of magnesium is equivalent to the moles of dihydrogen gas if there is molar equivalence.</p>
<p>Now I happen to know that if <mathjax>#rho_"HCl"=1.18*g*mol^-1#</mathjax>, then the concentration of <mathjax>#HCl#</mathjax> is <mathjax>#10.17*mol*L^-1#</mathjax> (and such data SHOULD have been supplied with the question).</p>
<p>And so we have a molar quantity of <mathjax>#55xx10^-3*Lxx10.17*mol*L^-1=0.559*mol#</mathjax> with respect to <mathjax>#HCl#</mathjax>.</p>
<p><mathjax>#"Moles of magnesium"=(0.865*g)/(24.305*g*mol^-1)=0.0356*mol#</mathjax>.</p>
<p>Magnesium metal is in excess, and we do not have a handle on the volume of dihydrogen evolved. </p></div>
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</article> | A #55*mL# volume of hydrochloric acid was poured on to an excess of magnesium turnings. What volume of dihydrogen gas evolved? | null |
3,218 | a86d4192-6ddd-11ea-b836-ccda262736ce | https://socratic.org/questions/590f5a0a11ef6b54efc4de3d | C2H2 | start chemical_formula qc_end physical_unit 20 20 17 18 mass qc_end physical_unit 25 25 22 23 mass qc_end physical_unit 31 32 27 28 mass qc_end physical_unit 31 32 37 38 volume qc_end physical_unit 31 32 40 41 temperature qc_end c_other OTHER qc_end end | [{"type":"other","value":"Chemical Formula [OF] a material [IN] molecular"}] | [{"type":"chemical equation","value":"C2H2"}] | [{"type":"physical unit","value":"Mass [OF] CO2(g) [=] \\pu{3.38 g}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{0.69 g}"},{"type":"physical unit","value":"Mass [OF] this gas [=] \\pu{11.6 g}"},{"type":"physical unit","value":"Volume [OF] this gas [=] \\pu{10.0 L}"},{"type":"physical unit","value":"Temperature [OF] this gas [=] \\pu{0 ℃}"},{"type":"other","value":"Complete combustion."}] | <h1 class="questionTitle" itemprop="name">What is the molecular formula of a material for which complete combustion of a unknown mass gives #3.38*g# of #CO_2(g)#, and #0.69*g# of water. An #11.6*g# mass of this gas gives a volume of #10.0*L# at #0# #""^@C# of #10*L#?</h1> | null | C2H2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can only interrogate the molecular formula on the basis of the Ideal Gas equation. The combustion data that you gave us were non-kosher (neither the mass of the combusted gas, nor the water content of the combustions were included).</p>
<p>If <mathjax>#PV=nRT#</mathjax>, then <mathjax>#1*atmxx10L=(11.6*g)/"Molar mass"xx0.0821(L*atm)/(K*mol)xx298*K#</mathjax></p>
<p>OR</p>
<p><mathjax>#"Molar mass"=(11.6*g)/(1*atmxx10*L)xxxx0.0821(L*atm)/(K*mol)xx298*K=28.4*g*mol^-1#</mathjax>.</p>
<p>Now this molecular mass is consistent with that of <mathjax>#"ethylene"#</mathjax>, <mathjax>#H_2C=CH_2#</mathjax> whose molecular mass is <mathjax>#28.0*g*mol^-1#</mathjax>. </p>
<p>Anyway, I don't think you have included all of the data for this question.</p>
<p>Had we calculated as molecular mass of <mathjax>#26*g*mol^-1#</mathjax>, or <mathjax>#30*g*mol^-1#</mathjax>, what would be the likely identity of the gas under the given scenario?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>You probably have......<mathjax>#"ethylene"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We can only interrogate the molecular formula on the basis of the Ideal Gas equation. The combustion data that you gave us were non-kosher (neither the mass of the combusted gas, nor the water content of the combustions were included).</p>
<p>If <mathjax>#PV=nRT#</mathjax>, then <mathjax>#1*atmxx10L=(11.6*g)/"Molar mass"xx0.0821(L*atm)/(K*mol)xx298*K#</mathjax></p>
<p>OR</p>
<p><mathjax>#"Molar mass"=(11.6*g)/(1*atmxx10*L)xxxx0.0821(L*atm)/(K*mol)xx298*K=28.4*g*mol^-1#</mathjax>.</p>
<p>Now this molecular mass is consistent with that of <mathjax>#"ethylene"#</mathjax>, <mathjax>#H_2C=CH_2#</mathjax> whose molecular mass is <mathjax>#28.0*g*mol^-1#</mathjax>. </p>
<p>Anyway, I don't think you have included all of the data for this question.</p>
<p>Had we calculated as molecular mass of <mathjax>#26*g*mol^-1#</mathjax>, or <mathjax>#30*g*mol^-1#</mathjax>, what would be the likely identity of the gas under the given scenario?</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molecular formula of a material for which complete combustion of a unknown mass gives #3.38*g# of #CO_2(g)#, and #0.69*g# of water. An #11.6*g# mass of this gas gives a volume of #10.0*L# at #0# #""^@C# of #10*L#?</h1>
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anor277
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<div class="markdown"><p>You probably have......<mathjax>#"ethylene"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We can only interrogate the molecular formula on the basis of the Ideal Gas equation. The combustion data that you gave us were non-kosher (neither the mass of the combusted gas, nor the water content of the combustions were included).</p>
<p>If <mathjax>#PV=nRT#</mathjax>, then <mathjax>#1*atmxx10L=(11.6*g)/"Molar mass"xx0.0821(L*atm)/(K*mol)xx298*K#</mathjax></p>
<p>OR</p>
<p><mathjax>#"Molar mass"=(11.6*g)/(1*atmxx10*L)xxxx0.0821(L*atm)/(K*mol)xx298*K=28.4*g*mol^-1#</mathjax>.</p>
<p>Now this molecular mass is consistent with that of <mathjax>#"ethylene"#</mathjax>, <mathjax>#H_2C=CH_2#</mathjax> whose molecular mass is <mathjax>#28.0*g*mol^-1#</mathjax>. </p>
<p>Anyway, I don't think you have included all of the data for this question.</p>
<p>Had we calculated as molecular mass of <mathjax>#26*g*mol^-1#</mathjax>, or <mathjax>#30*g*mol^-1#</mathjax>, what would be the likely identity of the gas under the given scenario?</p></div>
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<div class="markdown"><p>The molecular formula is <mathjax>#"C"_2"H"_2#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><blockquote></blockquote>
<p>This is an empirical formula/molecular formula problem.</p>
<blockquote></blockquote>
<p><strong>(a)</strong> Calculate the moles of <mathjax>#"C"#</mathjax>.</p>
<p><mathjax>#"Moles of C" = 3.38 color(red)(cancel(color(black)("g CO"_2))) × ("1 mol CO"_2)/(44.01 color(red)(cancel(color(black)("g CO"_2)))) × "12.01 g C"/(1 "mol C") = "0.076 80 mol CO"_2#</mathjax></p>
<blockquote></blockquote>
<p><strong>(b)</strong> Calculate the moles of <mathjax>#"H"#</mathjax>.</p>
<p>I assume that you meant 0.69 g of water.</p>
<p><mathjax>#"Moles of H" = 0.69 color(red)(cancel(color(black)("g H"_2"O"))) × (1 color(red)(cancel(color(black)("mol H"_2"O"))))/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) × "2 mol H"/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "0.0766 mol H"#</mathjax></p>
<blockquote></blockquote>
<p><strong>(c)</strong> Find the molar ratios.</p>
<p>From here on, I like to summarize the calculations in a table.</p>
<p><mathjax>#bb("Element"color(white)(Xll) "Moles"color(white)(m) "Ratio" color(white)(m)"Integers")#</mathjax><br/>
<mathjax>#color(white)(mm)"C" color(white)(mmmll)0.07680color(white)(X)1.003color(white)(mmml)1#</mathjax><br/>
<mathjax>#color(white)(mm)"H" color(white)(mmmll)0.0766 color(white)(mll)1 color(white)(Xmmmml)1#</mathjax></p>
<p>The empirical formula is <mathjax>#"CH"#</mathjax>.</p>
<blockquote></blockquote>
<p><strong>(d)</strong> Calculate the molar mass of the gas</p>
<p>You don’t give the pressure or temperature of the gas, so I will assume STP (1 bar and 0 °C).</p>
<p><mathjax>#pV = nRT = m/MRT#</mathjax></p>
<p><mathjax>#M = (mRT)/(pV) = ("11.6 g" × "0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar"))) × 10 color(red)(cancel(color(black)("L")))) = "26.3 g/mol"#</mathjax></p>
<blockquote></blockquote>
<p><strong>(d)</strong> Calculate the formula of the gas</p>
<p>The empirical formula mass of <mathjax>#"CH"#</mathjax> is 13.02 u.</p>
<p>The molecular mass of the gas is 26.3 u.</p>
<p>The molecular mass must be an integral multiple of the empirical formula mass.</p>
<p><mathjax>#"MM"/"EFM" = (26.3 color(red)(cancel(color(black)("u"))))/(13.02 color(red)(cancel(color(black)("u")))) = 2.02 ≈ 2#</mathjax></p>
<p>The molecular formula must be twice the empirical formula.</p>
<p><mathjax>#"MF" = ("EF")_2 = ("CH")_2 = "C"_2"H"_2#</mathjax></p></div>
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</article> | What is the molecular formula of a material for which complete combustion of a unknown mass gives #3.38*g# of #CO_2(g)#, and #0.69*g# of water. An #11.6*g# mass of this gas gives a volume of #10.0*L# at #0# #""^@C# of #10*L#? | null |
3,219 | a9022774-6ddd-11ea-aaae-ccda262736ce | https://socratic.org/questions/31-5-grams-of-an-unknown-substance-is-heated-to-102-4-c-and-then-placed-into-a-c | 0.38 J/(g * ℃) | start physical_unit 42 44 specific_heat j/(°c_·_g) qc_end physical_unit 42 44 0 1 mass qc_end physical_unit 42 44 9 10 temperature qc_end physical_unit 21 21 18 19 mass qc_end physical_unit 21 21 23 24 temperature qc_end physical_unit 21 21 34 35 temperature qc_end end | [{"type":"physical unit","value":"Specific heat [OF] the unknown substance [IN] J/(g * ℃)"}] | [{"type":"physical unit","value":"0.38 J/(g * ℃)"}] | [{"type":"physical unit","value":"Mass [OF] the unknown substance [=] \\pu{31.5 grams}"},{"type":"physical unit","value":"Temperature [OF] the unknown substance [=] \\pu{102.4 ℃}"},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{103.5 grams}"},{"type":"physical unit","value":"Temperature1 [OF] water [=] \\pu{24.5 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] water [=] \\pu{32.5 ℃}"}] | <h1 class="questionTitle" itemprop="name">31.5 grams of an unknown substance is heated to 102.4 C and then placed into a calorimeter containing 103.5 grams of water at 24.5 C. If the final temperature reached in the calorimeter is 32.5 C, what is the specific heat of the unknown substance?</h1> | null | 0.38 J/(g * ℃) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since heat gained by substance equals heat lost by water, and heat transferred is given by<br/>
<mathjax>#H = m*c*Delta (theta)#</mathjax><br/>
where, m is the mass of the substance,<br/>
c is its <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity<br/>
And <mathjax>#Delta (theta)#</mathjax> is change in temperature,<br/>
<mathjax>#31.5*c*(102.4-32.5)=103.5*1*(32.5-24.5)#</mathjax><br/>
Then,<br/>
<mathjax>#c=(103.5*8)/(31.5*69.9)#</mathjax><br/>
<mathjax>#=828/2201.85#</mathjax><br/>
<mathjax>#=0.38#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#0.38 C g^-1 (degree C)^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since heat gained by substance equals heat lost by water, and heat transferred is given by<br/>
<mathjax>#H = m*c*Delta (theta)#</mathjax><br/>
where, m is the mass of the substance,<br/>
c is its <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity<br/>
And <mathjax>#Delta (theta)#</mathjax> is change in temperature,<br/>
<mathjax>#31.5*c*(102.4-32.5)=103.5*1*(32.5-24.5)#</mathjax><br/>
Then,<br/>
<mathjax>#c=(103.5*8)/(31.5*69.9)#</mathjax><br/>
<mathjax>#=828/2201.85#</mathjax><br/>
<mathjax>#=0.38#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">31.5 grams of an unknown substance is heated to 102.4 C and then placed into a calorimeter containing 103.5 grams of water at 24.5 C. If the final temperature reached in the calorimeter is 32.5 C, what is the specific heat of the unknown substance?</h1>
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<div class="markdown"><p><mathjax>#0.38 C g^-1 (degree C)^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since heat gained by substance equals heat lost by water, and heat transferred is given by<br/>
<mathjax>#H = m*c*Delta (theta)#</mathjax><br/>
where, m is the mass of the substance,<br/>
c is its <a href="https://socratic.org/chemistry/thermochemistry/specific-heat">specific heat</a> capacity<br/>
And <mathjax>#Delta (theta)#</mathjax> is change in temperature,<br/>
<mathjax>#31.5*c*(102.4-32.5)=103.5*1*(32.5-24.5)#</mathjax><br/>
Then,<br/>
<mathjax>#c=(103.5*8)/(31.5*69.9)#</mathjax><br/>
<mathjax>#=828/2201.85#</mathjax><br/>
<mathjax>#=0.38#</mathjax></p></div>
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</article> | 31.5 grams of an unknown substance is heated to 102.4 C and then placed into a calorimeter containing 103.5 grams of water at 24.5 C. If the final temperature reached in the calorimeter is 32.5 C, what is the specific heat of the unknown substance? | null |
3,220 | a8efdd62-6ddd-11ea-a563-ccda262736ce | https://socratic.org/questions/5842ec9911ef6b057e97abd0 | 1.45 × 10^25 | start physical_unit 5 5 number none qc_end physical_unit 12 13 8 9 mole qc_end end | [{"type":"physical unit","value":"Number [OF] electrons"}] | [{"type":"physical unit","value":"1.45 × 10^25"}] | [{"type":"physical unit","value":"Mole [OF] dioxygen gas [=] \\pu{1.5 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the number of electrons in a #1.5*mol# quantity of dioxygen gas?</h1> | null | 1.45 × 10^25 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The oxygen <mathjax>#"atom"#</mathjax> has <mathjax>#8#</mathjax> electrons. And thus if we have <mathjax>#1.5*mol#</mathjax> <mathjax>#O_2#</mathjax> molecules, there are <mathjax>#N_Axx1.5xx2xx8=24xxN_A" electrons"#</mathjax>, where <mathjax>#N_A="Avogadro's Number"-=6.022xx10^23*mol^-1#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>In <mathjax>#1.5*mol#</mathjax> of <mathjax>#"dioxygen gas"#</mathjax>?</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The oxygen <mathjax>#"atom"#</mathjax> has <mathjax>#8#</mathjax> electrons. And thus if we have <mathjax>#1.5*mol#</mathjax> <mathjax>#O_2#</mathjax> molecules, there are <mathjax>#N_Axx1.5xx2xx8=24xxN_A" electrons"#</mathjax>, where <mathjax>#N_A="Avogadro's Number"-=6.022xx10^23*mol^-1#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the number of electrons in a #1.5*mol# quantity of dioxygen gas?</h1>
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anor277
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<div class="markdown"><p>In <mathjax>#1.5*mol#</mathjax> of <mathjax>#"dioxygen gas"#</mathjax>?</p></div>
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<div class="markdown"><p>The oxygen <mathjax>#"atom"#</mathjax> has <mathjax>#8#</mathjax> electrons. And thus if we have <mathjax>#1.5*mol#</mathjax> <mathjax>#O_2#</mathjax> molecules, there are <mathjax>#N_Axx1.5xx2xx8=24xxN_A" electrons"#</mathjax>, where <mathjax>#N_A="Avogadro's Number"-=6.022xx10^23*mol^-1#</mathjax>.</p></div>
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</article> | What is the number of electrons in a #1.5*mol# quantity of dioxygen gas? | null |
3,221 | abfc13fe-6ddd-11ea-8858-ccda262736ce | https://socratic.org/questions/a-10-0-ml-sample-of-h-2so-4-solution-from-an-automobile-battery-requires-32-75-m | 1.64 M | start physical_unit 5 6 molarity mol/l qc_end physical_unit 3 6 1 2 volume qc_end physical_unit 6 6 12 13 volume qc_end physical_unit 17 17 15 16 molarity qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Molarity [OF] H2SO4 solution [IN] M"}] | [{"type":"physical unit","value":"1.64 M"}] | [{"type":"physical unit","value":"Volume [OF] H2SO4 solution sample [=] \\pu{10.0 mL}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{32.75 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{1.0 M}"},{"type":"other","value":"Complete reaction."}] | <h1 class="questionTitle" itemprop="name">A 10.0 mL sample of #H_2SO_4# solution from an automobile battery requires 32.75 mL of M NaOH for a complete reaction. What is the molarity of the #H_2SO_4# solution?</h1> | null | 1.64 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since both <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are a strong acid/base, no <mathjax>#K_a//K_b#</mathjax> calculations are required. In a reaction between sulfuric acid and sodium hydroxide, two mols of base are required to fully neutralize one mol of acid.</p>
<p>Also, for the purposes of this question, I will assume that the molarity of <mathjax>#NaOH_{(aq)}=1.0\ mols#</mathjax>, since this information was accidentaly skipped in the question. (Or defaults to 1.0 mols/L anyways, as Michael suggested)</p>
<p>Step 1) Write the equation.</p>
<p><mathjax>#mols\ H_2SO_{4(aq)} * 2= mols\ NaOH_{(aq)}#</mathjax></p>
<p><mathjax>#10.0\ mL * {1\ L}/{1000\ mL} * {2\ mols\ NaOH_{(aq)}}/{1\ mol\ H_2SO_{4(aq)}} * x = 32.75\ mL * {1\ L}/{1000\ mL}*1.0\ {mols}/{L}#</mathjax></p>
<p><mathjax>#0.010\ L * {2\ mols\ NaOH_{(aq)}}/{1\ mol\ H_2SO_{4(aq)}} * x = 0.03275\ L * 1.0\ {mols}/{L}#</mathjax></p>
<p>Step 2) Isolate the variable being solved for. (<mathjax>#x#</mathjax>)</p>
<p><mathjax>#x={0.03275\ L}/{0.010\ L} * {1\ mol\ H_2SO_{4(aq)}}/{2\ mols\ NaOH_{(aq)}} * 1.0\ {mols}/{L}#</mathjax></p>
<p>After that, solve, and you have your answer of <mathjax>#1.6375\ {mols}/{L}#</mathjax>, or properly rounded, <mathjax>#1.64\ {mols}/{L}#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#1.64\ {mols}/{L}#</mathjax>, assuming that the molarity of <mathjax>#NaOH_{(aq)}#</mathjax> is <mathjax>#1.0\ {mols}/{L}#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since both <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are a strong acid/base, no <mathjax>#K_a//K_b#</mathjax> calculations are required. In a reaction between sulfuric acid and sodium hydroxide, two mols of base are required to fully neutralize one mol of acid.</p>
<p>Also, for the purposes of this question, I will assume that the molarity of <mathjax>#NaOH_{(aq)}=1.0\ mols#</mathjax>, since this information was accidentaly skipped in the question. (Or defaults to 1.0 mols/L anyways, as Michael suggested)</p>
<p>Step 1) Write the equation.</p>
<p><mathjax>#mols\ H_2SO_{4(aq)} * 2= mols\ NaOH_{(aq)}#</mathjax></p>
<p><mathjax>#10.0\ mL * {1\ L}/{1000\ mL} * {2\ mols\ NaOH_{(aq)}}/{1\ mol\ H_2SO_{4(aq)}} * x = 32.75\ mL * {1\ L}/{1000\ mL}*1.0\ {mols}/{L}#</mathjax></p>
<p><mathjax>#0.010\ L * {2\ mols\ NaOH_{(aq)}}/{1\ mol\ H_2SO_{4(aq)}} * x = 0.03275\ L * 1.0\ {mols}/{L}#</mathjax></p>
<p>Step 2) Isolate the variable being solved for. (<mathjax>#x#</mathjax>)</p>
<p><mathjax>#x={0.03275\ L}/{0.010\ L} * {1\ mol\ H_2SO_{4(aq)}}/{2\ mols\ NaOH_{(aq)}} * 1.0\ {mols}/{L}#</mathjax></p>
<p>After that, solve, and you have your answer of <mathjax>#1.6375\ {mols}/{L}#</mathjax>, or properly rounded, <mathjax>#1.64\ {mols}/{L}#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">A 10.0 mL sample of #H_2SO_4# solution from an automobile battery requires 32.75 mL of M NaOH for a complete reaction. What is the molarity of the #H_2SO_4# solution?</h1>
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<div class="markdown"><p><mathjax>#1.64\ {mols}/{L}#</mathjax>, assuming that the molarity of <mathjax>#NaOH_{(aq)}#</mathjax> is <mathjax>#1.0\ {mols}/{L}#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since both <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solutions">solutions</a> are a strong acid/base, no <mathjax>#K_a//K_b#</mathjax> calculations are required. In a reaction between sulfuric acid and sodium hydroxide, two mols of base are required to fully neutralize one mol of acid.</p>
<p>Also, for the purposes of this question, I will assume that the molarity of <mathjax>#NaOH_{(aq)}=1.0\ mols#</mathjax>, since this information was accidentaly skipped in the question. (Or defaults to 1.0 mols/L anyways, as Michael suggested)</p>
<p>Step 1) Write the equation.</p>
<p><mathjax>#mols\ H_2SO_{4(aq)} * 2= mols\ NaOH_{(aq)}#</mathjax></p>
<p><mathjax>#10.0\ mL * {1\ L}/{1000\ mL} * {2\ mols\ NaOH_{(aq)}}/{1\ mol\ H_2SO_{4(aq)}} * x = 32.75\ mL * {1\ L}/{1000\ mL}*1.0\ {mols}/{L}#</mathjax></p>
<p><mathjax>#0.010\ L * {2\ mols\ NaOH_{(aq)}}/{1\ mol\ H_2SO_{4(aq)}} * x = 0.03275\ L * 1.0\ {mols}/{L}#</mathjax></p>
<p>Step 2) Isolate the variable being solved for. (<mathjax>#x#</mathjax>)</p>
<p><mathjax>#x={0.03275\ L}/{0.010\ L} * {1\ mol\ H_2SO_{4(aq)}}/{2\ mols\ NaOH_{(aq)}} * 1.0\ {mols}/{L}#</mathjax></p>
<p>After that, solve, and you have your answer of <mathjax>#1.6375\ {mols}/{L}#</mathjax>, or properly rounded, <mathjax>#1.64\ {mols}/{L}#</mathjax>.</p></div>
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<div class="markdown"><p>The <a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of the <mathjax>#"H"_2"SO"_4#</mathjax> is 1.64 mol/L.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p><strong>Step 1.</strong> Write the balanced chemical equation for the reaction</p>
<p><mathjax>#"H"_2"SO"_4 + "2NaOH" → "Na"_2"SO"_4 + "2H"_2"O"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 2.</strong> Calculate the moles of <mathjax>#"NaOH"#</mathjax></p>
<p>I assume that the molarity of the <mathjax>#"NaOH"#</mathjax> is 1.00 mol/L.</p>
<p><mathjax>#"Moles of NaOH" = "0.032 75" color(red)(cancel(color(black)("L NaOH"))) × "1.00 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.032 75 mol NaOH"#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 3.</strong> Calculate the moles of <mathjax>#"H"_2"SO"_4#</mathjax>.</p>
<p><mathjax>#"Moles of H"_2"SO"_4 = "0.032 75" color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol H"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaOH")))) = "0.016 375 mol H"_2"SO"_4#</mathjax></p>
<blockquote></blockquote>
<p><strong>Step 4.</strong> Calculate the molarity of the <mathjax>#"H"_2"SO"_4#</mathjax></p>
<p><mathjax>#"Molarity" = "moles"/"litres" = "0.016 375 mol"/"0.0100 L" = "1.64 mol/L"#</mathjax></p></div>
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</article> | A 10.0 mL sample of #H_2SO_4# solution from an automobile battery requires 32.75 mL of M NaOH for a complete reaction. What is the molarity of the #H_2SO_4# solution? | null |
3,222 | a9e0b0cc-6ddd-11ea-a1be-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-the-masonic-acid-solution-h-2c-3h-2o-4-2naoh-na-2c-3h-2o | 8 × 10^(-2) M | start physical_unit 5 8 molarity mol/l qc_end chemical_equation 9 17 qc_end physical_unit 31 32 29 30 molarity qc_end physical_unit 31 32 34 35 volume qc_end physical_unit 8 8 45 46 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Molarity [OF] the masonic acid solution [IN] M"}] | [{"type":"physical unit","value":"8 × 10^(-2) M"}] | [{"type":"chemical equation","value":"H2C3H2O4 + 2 NaOH -> Na2C3H2O4 + 2 H2O"},{"type":"physical unit","value":"Molarity [OF] NaOH solution [=] \\pu{0.1000 M}"},{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{20.76 mL}"},{"type":"physical unit","value":"Volume [OF] the malonic acid solution [=] \\pu{13.15 mL}"},{"type":"other","value":"Neutralize completely."}] | <h1 class="questionTitle" itemprop="name">What is the molarity of the masonic acid solution?
#H_2C_3H_2O_4 + 2NaOH -> Na_2C_3H_2O_4 + 2H_2O#</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>A solution of malonic acid, <mathjax>#H_2C_3H_2O_4#</mathjax>, was standardized by titration with 0.1000 M <mathjax>#NaOH#</mathjax> solution. If 20.76 mL of the <mathjax>#NaOH#</mathjax> solution were required to neutralize completely 13.15 mL of the malonic acid...</p></div>
</h2>
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</div> | 8 × 10^(-2) M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of NaOH"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.1000*mol*L^-1xx20.76xx10^-3L=2.076xx10^-3*mol#</mathjax></p>
<p><mathjax>#"Concentration of malonic acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1/2xx2.076xx10^-3*mol)/(13.15xx10^-3*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol*L^-1#</mathjax></p>
<p>Note that I had to use <mathjax>#1/2xx2.076xx10^-3*mol" with respect to malonic acid"#</mathjax> as malonic acid is a diacid, <mathjax>#HO(O=C)CH_2C(=O)OH#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#"Concentration of malonic acid "~=#</mathjax> <mathjax>#8xx10^-2*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of NaOH"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.1000*mol*L^-1xx20.76xx10^-3L=2.076xx10^-3*mol#</mathjax></p>
<p><mathjax>#"Concentration of malonic acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1/2xx2.076xx10^-3*mol)/(13.15xx10^-3*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol*L^-1#</mathjax></p>
<p>Note that I had to use <mathjax>#1/2xx2.076xx10^-3*mol" with respect to malonic acid"#</mathjax> as malonic acid is a diacid, <mathjax>#HO(O=C)CH_2C(=O)OH#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of the masonic acid solution?
#H_2C_3H_2O_4 + 2NaOH -> Na_2C_3H_2O_4 + 2H_2O#</h1>
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<div class="markdown"><p>A solution of malonic acid, <mathjax>#H_2C_3H_2O_4#</mathjax>, was standardized by titration with 0.1000 M <mathjax>#NaOH#</mathjax> solution. If 20.76 mL of the <mathjax>#NaOH#</mathjax> solution were required to neutralize completely 13.15 mL of the malonic acid...</p></div>
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<div class="markdown"><p><mathjax>#"Concentration of malonic acid "~=#</mathjax> <mathjax>#8xx10^-2*mol*L^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Moles of NaOH"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.1000*mol*L^-1xx20.76xx10^-3L=2.076xx10^-3*mol#</mathjax></p>
<p><mathjax>#"Concentration of malonic acid"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1/2xx2.076xx10^-3*mol)/(13.15xx10^-3*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??mol*L^-1#</mathjax></p>
<p>Note that I had to use <mathjax>#1/2xx2.076xx10^-3*mol" with respect to malonic acid"#</mathjax> as malonic acid is a diacid, <mathjax>#HO(O=C)CH_2C(=O)OH#</mathjax>.</p></div>
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</article> | What is the molarity of the masonic acid solution?
#H_2C_3H_2O_4 + 2NaOH -> Na_2C_3H_2O_4 + 2H_2O# |
A solution of malonic acid, #H_2C_3H_2O_4#, was standardized by titration with 0.1000 M #NaOH# solution. If 20.76 mL of the #NaOH# solution were required to neutralize completely 13.15 mL of the malonic acid...
|
3,223 | aad1dcf5-6ddd-11ea-a513-ccda262736ce | https://socratic.org/questions/if-a-0-614-g-sample-of-a-gas-maintains-a-pressure-of-238-mm-hg-when-contained-in | 44.0 g/mol | start physical_unit 26 27 molecular_weight g/mol qc_end physical_unit 7 7 2 3 mass qc_end physical_unit 7 7 12 13 pressure qc_end physical_unit 7 7 22 23 temperature qc_end physical_unit 20 20 18 19 volume qc_end end | [{"type":"physical unit","value":"Molecular weight [OF] the gases [IN] g/mol"}] | [{"type":"physical unit","value":"44.0 g/mol"}] | [{"type":"physical unit","value":"Mass [OF] gas sample [=] \\pu{0.614 g}"},{"type":"physical unit","value":"Pressure [OF] gas sample [=] \\pu{238 mmHg}"},{"type":"physical unit","value":"Temperature [OF] gas sample [=] \\pu{0.0 ℃}"},{"type":"physical unit","value":"Volume [OF] flask [=] \\pu{1.0 L}"}] | <h1 class="questionTitle" itemprop="name">If a 0.614 g sample of a gas maintains a pressure of 238 mm Hg when contained in a 1.0L flask at 0.0°C, what is the gases molecular weight?
</h1> | null | 44.0 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We first need to use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to determine the moles of the gas. Then we will divide the given mass by the calculated moles.</p>
<p>The equation for the ideal gas law is <mathjax>#PV=nRT#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#"mass"="0.614 g"#</mathjax><br/>
<mathjax>#P="238 mmHg"="238 torr"#</mathjax><br/>
<mathjax>#V="1.0 L"#</mathjax><br/>
<mathjax>#T="0.0"^"o""C"+273.15="273.15 K"#</mathjax><br/>
<mathjax>#R="62.363577 L torr K"^(-1) "mol"^(-1)"#</mathjax><br/>
<a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Gas_constant</a></p>
<p><strong>Unknown</strong><br/>
moles of gas</p>
<p><strong>Solution</strong><br/>
Rearrange the equation for the ideal gas law to isolate <mathjax>#n#</mathjax>, then solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=((238cancel"torr")xx(1.0cancel"L"))/((62.363577cancel"L" cancel"torr" cancel("K"^(-1)) "mol"^(-1))xx(273.15cancel"K"))="0.01397 mol"#</mathjax> </p>
<p>Determine molecular weight in g/mol by dividing the given mass by the calculated moles.</p>
<p><mathjax>#"molecular weight"=(0.614"g")/(0.01397"mol")="44.0 g/mol"#</mathjax></p>
<p>This could be carbon dioxide gas, as it has a molar mass of <mathjax>#"44.0 g/mol"#</mathjax>.</p></div>
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<div class="markdown"><p>The molecular weight (molar mass) in g/mol is <mathjax>#"44.0 g"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We first need to use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to determine the moles of the gas. Then we will divide the given mass by the calculated moles.</p>
<p>The equation for the ideal gas law is <mathjax>#PV=nRT#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#"mass"="0.614 g"#</mathjax><br/>
<mathjax>#P="238 mmHg"="238 torr"#</mathjax><br/>
<mathjax>#V="1.0 L"#</mathjax><br/>
<mathjax>#T="0.0"^"o""C"+273.15="273.15 K"#</mathjax><br/>
<mathjax>#R="62.363577 L torr K"^(-1) "mol"^(-1)"#</mathjax><br/>
<a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Gas_constant</a></p>
<p><strong>Unknown</strong><br/>
moles of gas</p>
<p><strong>Solution</strong><br/>
Rearrange the equation for the ideal gas law to isolate <mathjax>#n#</mathjax>, then solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=((238cancel"torr")xx(1.0cancel"L"))/((62.363577cancel"L" cancel"torr" cancel("K"^(-1)) "mol"^(-1))xx(273.15cancel"K"))="0.01397 mol"#</mathjax> </p>
<p>Determine molecular weight in g/mol by dividing the given mass by the calculated moles.</p>
<p><mathjax>#"molecular weight"=(0.614"g")/(0.01397"mol")="44.0 g/mol"#</mathjax></p>
<p>This could be carbon dioxide gas, as it has a molar mass of <mathjax>#"44.0 g/mol"#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">If a 0.614 g sample of a gas maintains a pressure of 238 mm Hg when contained in a 1.0L flask at 0.0°C, what is the gases molecular weight?
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<div class="markdown"><p>The molecular weight (molar mass) in g/mol is <mathjax>#"44.0 g"#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We first need to use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> to determine the moles of the gas. Then we will divide the given mass by the calculated moles.</p>
<p>The equation for the ideal gas law is <mathjax>#PV=nRT#</mathjax>.</p>
<p><strong>Given</strong><br/>
<mathjax>#"mass"="0.614 g"#</mathjax><br/>
<mathjax>#P="238 mmHg"="238 torr"#</mathjax><br/>
<mathjax>#V="1.0 L"#</mathjax><br/>
<mathjax>#T="0.0"^"o""C"+273.15="273.15 K"#</mathjax><br/>
<mathjax>#R="62.363577 L torr K"^(-1) "mol"^(-1)"#</mathjax><br/>
<a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Gas_constant</a></p>
<p><strong>Unknown</strong><br/>
moles of gas</p>
<p><strong>Solution</strong><br/>
Rearrange the equation for the ideal gas law to isolate <mathjax>#n#</mathjax>, then solve.</p>
<p><mathjax>#n=(PV)/(RT)#</mathjax></p>
<p><mathjax>#n=((238cancel"torr")xx(1.0cancel"L"))/((62.363577cancel"L" cancel"torr" cancel("K"^(-1)) "mol"^(-1))xx(273.15cancel"K"))="0.01397 mol"#</mathjax> </p>
<p>Determine molecular weight in g/mol by dividing the given mass by the calculated moles.</p>
<p><mathjax>#"molecular weight"=(0.614"g")/(0.01397"mol")="44.0 g/mol"#</mathjax></p>
<p>This could be carbon dioxide gas, as it has a molar mass of <mathjax>#"44.0 g/mol"#</mathjax>.</p></div>
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</article> | If a 0.614 g sample of a gas maintains a pressure of 238 mm Hg when contained in a 1.0L flask at 0.0°C, what is the gases molecular weight?
| null |
3,224 | aa1563f2-6ddd-11ea-a9ef-ccda262736ce | https://socratic.org/questions/what-is-pressure-of-a-gas-that-expands-to-a-volume-of-20-0-liters-if-the-gas-s-i | 0.63 atm | start physical_unit 15 16 pressure atm qc_end physical_unit 15 16 12 13 volume qc_end physical_unit 15 16 20 21 pressure qc_end physical_unit 15 16 26 27 volume qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] the gas [IN] atm"}] | [{"type":"physical unit","value":"0.63 atm"}] | [{"type":"physical unit","value":"Volume2 [OF] the gas [=] \\pu{20.0 liters}"},{"type":"physical unit","value":"Pressure [OF] the gas [=] \\pu{2.50 atm}"},{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{5.00 L}"}] | <h1 class="questionTitle" itemprop="name">What is pressure of a gas that expands to a volume of 20.0 liters, if the gas's initial pressure is 2.50 atm and initial volume is 5.00 L?</h1> | null | 0.63 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>At constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>, which is Boyle's law.</p>
<p><mathjax>#P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.50*atmxx5.00*L)/(20*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??atm#</mathjax></p></div>
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<div class="markdown"><p>At constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>, which is <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>.</p></div>
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<div class="markdown"><p>At constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>, which is Boyle's law.</p>
<p><mathjax>#P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.50*atmxx5.00*L)/(20*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??atm#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is pressure of a gas that expands to a volume of 20.0 liters, if the gas's initial pressure is 2.50 atm and initial volume is 5.00 L?</h1>
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<div class="markdown"><p>At constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>, which is <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>At constant temperature, <mathjax>#P_1V_1=P_2V_2#</mathjax>, which is Boyle's law.</p>
<p><mathjax>#P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(2.50*atmxx5.00*L)/(20*L)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??atm#</mathjax></p></div>
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</article> | What is pressure of a gas that expands to a volume of 20.0 liters, if the gas's initial pressure is 2.50 atm and initial volume is 5.00 L? | null |
3,225 | ac93f1e4-6ddd-11ea-953c-ccda262736ce | https://socratic.org/questions/how-do-you-balance-nh3-o2-no2-h2o-the-number-in-the-parenthasis-are-subscribts-t | 4 NH3 + 7 O2 -> 4 NO2 + 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":"4 NH3 + 7 O2 -> 4 NO2 + 6 H2O"}] | [{"type":"chemical equation","value":"NH3 + O2 -> NO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How do you balance __ NH3 + __O2--> __NO2 + __H2O?
The number in the parenthasis are subscribts, thank you all so much!!!?</h1> | null | 4 NH3 + 7 O2 -> 4 NO2 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There is no fixed method actually. You have to take a random number as the number of molecules for <mathjax>#NH_3#</mathjax>, then calculate result. </p>
<p>As for here, I have taken <mathjax>#4#</mathjax> as the number of molecules of <mathjax>#NH_3#</mathjax>. That means <mathjax>#4#</mathjax> nitrogen atoms and <mathjax>#12#</mathjax> hydrogen atoms are taking part in this reaction.</p>
<p>Now, hydrogen atoms are present only in <mathjax>#H_2O#</mathjax> as a product in this reaction. So that makes <mathjax>#6#</mathjax> molecules of <mathjax>#H_2O#</mathjax> (as there are <mathjax>#12#</mathjax> hydrogen atoms). </p>
<p>Now, when we calculated the number of molecules of <mathjax>#H_2O#</mathjax>, we also got the number of oxygen atoms present there.</p>
<p>On the other hand, as there are <mathjax>#4#</mathjax> molecules of <mathjax>#NH_3#</mathjax>, there has to be <mathjax>#4#</mathjax> molecules of <mathjax>#NO_2#</mathjax> as well to balance the number of nitrogen atoms in both side of the reaction.</p>
<p>That makes <mathjax>#8#</mathjax> another atoms of oxygen. </p>
<p>So now the total number of oxygen atoms is <mathjax>#(8+6)#</mathjax> or <mathjax>#14#</mathjax>, which means <mathjax>#14#</mathjax> atoms or <mathjax>#7#</mathjax> molecules of oxygen.</p>
<p>So the final balanced reaction is :</p>
<blockquote>
<p><mathjax>#4NH_3 + 7O_2 -> 4NO_2 + 6H_2O#</mathjax></p>
</blockquote></div>
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<div class="markdown"><p><mathjax>#4NH_3 + 7O_2 -> 4NO_2 + 6H_2O#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There is no fixed method actually. You have to take a random number as the number of molecules for <mathjax>#NH_3#</mathjax>, then calculate result. </p>
<p>As for here, I have taken <mathjax>#4#</mathjax> as the number of molecules of <mathjax>#NH_3#</mathjax>. That means <mathjax>#4#</mathjax> nitrogen atoms and <mathjax>#12#</mathjax> hydrogen atoms are taking part in this reaction.</p>
<p>Now, hydrogen atoms are present only in <mathjax>#H_2O#</mathjax> as a product in this reaction. So that makes <mathjax>#6#</mathjax> molecules of <mathjax>#H_2O#</mathjax> (as there are <mathjax>#12#</mathjax> hydrogen atoms). </p>
<p>Now, when we calculated the number of molecules of <mathjax>#H_2O#</mathjax>, we also got the number of oxygen atoms present there.</p>
<p>On the other hand, as there are <mathjax>#4#</mathjax> molecules of <mathjax>#NH_3#</mathjax>, there has to be <mathjax>#4#</mathjax> molecules of <mathjax>#NO_2#</mathjax> as well to balance the number of nitrogen atoms in both side of the reaction.</p>
<p>That makes <mathjax>#8#</mathjax> another atoms of oxygen. </p>
<p>So now the total number of oxygen atoms is <mathjax>#(8+6)#</mathjax> or <mathjax>#14#</mathjax>, which means <mathjax>#14#</mathjax> atoms or <mathjax>#7#</mathjax> molecules of oxygen.</p>
<p>So the final balanced reaction is :</p>
<blockquote>
<p><mathjax>#4NH_3 + 7O_2 -> 4NO_2 + 6H_2O#</mathjax></p>
</blockquote></div>
</div>
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<h1 class="questionTitle" itemprop="name">How do you balance __ NH3 + __O2--> __NO2 + __H2O?
The number in the parenthasis are subscribts, thank you all so much!!!?</h1>
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Jahan Psyche
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Stefan V.
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Oct 23, 2015
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<div class="markdown"><p><mathjax>#4NH_3 + 7O_2 -> 4NO_2 + 6H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>There is no fixed method actually. You have to take a random number as the number of molecules for <mathjax>#NH_3#</mathjax>, then calculate result. </p>
<p>As for here, I have taken <mathjax>#4#</mathjax> as the number of molecules of <mathjax>#NH_3#</mathjax>. That means <mathjax>#4#</mathjax> nitrogen atoms and <mathjax>#12#</mathjax> hydrogen atoms are taking part in this reaction.</p>
<p>Now, hydrogen atoms are present only in <mathjax>#H_2O#</mathjax> as a product in this reaction. So that makes <mathjax>#6#</mathjax> molecules of <mathjax>#H_2O#</mathjax> (as there are <mathjax>#12#</mathjax> hydrogen atoms). </p>
<p>Now, when we calculated the number of molecules of <mathjax>#H_2O#</mathjax>, we also got the number of oxygen atoms present there.</p>
<p>On the other hand, as there are <mathjax>#4#</mathjax> molecules of <mathjax>#NH_3#</mathjax>, there has to be <mathjax>#4#</mathjax> molecules of <mathjax>#NO_2#</mathjax> as well to balance the number of nitrogen atoms in both side of the reaction.</p>
<p>That makes <mathjax>#8#</mathjax> another atoms of oxygen. </p>
<p>So now the total number of oxygen atoms is <mathjax>#(8+6)#</mathjax> or <mathjax>#14#</mathjax>, which means <mathjax>#14#</mathjax> atoms or <mathjax>#7#</mathjax> molecules of oxygen.</p>
<p>So the final balanced reaction is :</p>
<blockquote>
<p><mathjax>#4NH_3 + 7O_2 -> 4NO_2 + 6H_2O#</mathjax></p>
</blockquote></div>
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Nikka C.
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<div class="markdown"><p>Two possible answers: (1) <mathjax>#2NH_3#</mathjax> + <mathjax>#7/2O_2#</mathjax> = <mathjax>#2NO_2#</mathjax> + <mathjax>#3H_2O#</mathjax> or (2) <mathjax>#4NH_3#</mathjax> + <mathjax>#7O_2#</mathjax> = <mathjax>#4NO_2#</mathjax> + <mathjax>#6H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First, you need to tally all the atoms.</p>
<p><mathjax>#NH_3#</mathjax> + <mathjax>#O_2#</mathjax> = <mathjax>#NO_2#</mathjax> + <mathjax>#H_2O#</mathjax></p>
<p><em>Left side:</em><br/>
N = 1<br/>
H = 3<br/>
O = 2</p>
<p><em>Right side:</em><br/>
N = 1<br/>
H = 2<br/>
O = 2 + 1 (two from the <mathjax>#NO_2#</mathjax> and one from <mathjax>#H_2O#</mathjax> ; <strong>DO NOT ADD IT UP YET</strong> )</p>
<p>You always have to find the simplest element that you can balance (in this case, the H).</p>
<p><em>Left side:</em><br/>
N = 1 x 2 = <strong>2</strong><br/>
H = 3 x 2 = <strong>6</strong><br/>
O = 2</p>
<p><em>Right side:</em><br/>
N = <strong>1</strong><br/>
H = 2 x 3 = <strong>6</strong><br/>
O = 2 + (1 x 3) from the O in <mathjax>#H_2O#</mathjax></p>
<p>Notice that with balancing the atoms, you do not forget that they are part of a substance - meaning, you have to multiply everything. </p>
<p><mathjax>#2NH_3#</mathjax> + <mathjax>#O_2#</mathjax> = <mathjax>#NO_2#</mathjax> + <mathjax>#3H_2O#</mathjax></p>
<p>Now, the N is not balanced so you need to multiply the right side N by 2.</p>
<p><em>Left side:</em><br/>
N = 1 x 2 = <strong>2</strong><br/>
H = 3 x 2 = <strong>6</strong><br/>
O = 2</p>
<p><em>Right side:</em><br/>
N = 1 x 2 = <strong>2</strong><br/>
H = 2 x 3 = <strong>6</strong><br/>
O = (2 x 2) + (1 x 3) = 7</p>
<p><mathjax>#2NH_3#</mathjax> + <mathjax>#O_2#</mathjax> = <mathjax>#2NO_2#</mathjax> + <mathjax>#3H_2O#</mathjax></p>
<p>Now, the only element left to be balanced is O. Since 7 is an odd number, you can use a fraction for the equation to be in its reduced form. </p>
<p><em>Left side:</em><br/>
N = 1 x 2 = <strong>2</strong><br/>
H = 3 x 2 = <strong>6</strong><br/>
O = 2 x 3.5 = <strong>7</strong> (the decimal 3.5 can be written as <mathjax>#7/2#</mathjax>)</p>
<p><em>Right side:</em><br/>
N = 1 x 2 = <strong>2</strong><br/>
H = 2 x 3 = <strong>6</strong><br/>
O = (2 x 2) + (1 x 3) = <strong>7</strong> </p>
<p>Hence,</p>
<p><mathjax>#2NH_3#</mathjax> + <mathjax>#7/2O_2#</mathjax> = <mathjax>#2NO_2#</mathjax> + <mathjax>#3H_2O#</mathjax></p>
<p>but if you want the equation to show whole numbers only, you can always multiply everything by the denominator in the fraction.</p>
<p><strong><mathjax>#(cancel 2)#</mathjax></strong> [<mathjax>#2NH_3#</mathjax> + <mathjax>#7/(cancel 2)O_2#</mathjax> = <mathjax>#2NO_2#</mathjax> + <mathjax>#3H_2O#</mathjax>]</p>
<p>=</p>
<p><mathjax>#4NH_3#</mathjax> + <mathjax>#7O_2#</mathjax> = <mathjax>#4NO_2#</mathjax> + <mathjax>#6H_2O#</mathjax></p></div>
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</article> | How do you balance __ NH3 + __O2--> __NO2 + __H2O?
The number in the parenthasis are subscribts, thank you all so much!!!? | null |
3,226 | ac108867-6ddd-11ea-b6ba-ccda262736ce | https://socratic.org/questions/5944d43b7c0149754a9a50ad | N2 + 3 H2 <=> 2 NH3 | start chemical_equation qc_end substance 7 7 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reduction"}] | [{"type":"chemical equation","value":"N2 + 3 H2 <=> 2 NH3"}] | [{"type":"substance name","value":"Dinitrogen"}] | <h1 class="questionTitle" itemprop="name">How do we represent the reduction of dinitrogen?</h1> | null | N2 + 3 H2 <=> 2 NH3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction is thermodynamically downhill, i.e. it is exothermic and releases energy. Of course the reaction has a very high activation energy, and requires effective catalysis to reduce this energy of activation. There are some good examples of in vitro dinitrogen activation, i.e. experiments at single metal centres, which can be conveniently studied. </p>
<p>Why should the reaction require high energy of activation? </p></div>
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<div class="markdown"><p><mathjax>#N_2(g) + 3H_2(g) rightleftharpoons2NH_3(g)#</mathjax> <mathjax>#;DeltaH^@=-92*kJ*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction is thermodynamically downhill, i.e. it is exothermic and releases energy. Of course the reaction has a very high activation energy, and requires effective catalysis to reduce this energy of activation. There are some good examples of in vitro dinitrogen activation, i.e. experiments at single metal centres, which can be conveniently studied. </p>
<p>Why should the reaction require high energy of activation? </p></div>
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<div class="markdown"><p><mathjax>#N_2(g) + 3H_2(g) rightleftharpoons2NH_3(g)#</mathjax> <mathjax>#;DeltaH^@=-92*kJ*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The reaction is thermodynamically downhill, i.e. it is exothermic and releases energy. Of course the reaction has a very high activation energy, and requires effective catalysis to reduce this energy of activation. There are some good examples of in vitro dinitrogen activation, i.e. experiments at single metal centres, which can be conveniently studied. </p>
<p>Why should the reaction require high energy of activation? </p></div>
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</article> | How do we represent the reduction of dinitrogen? | null |
3,227 | abe9c488-6ddd-11ea-9ebf-ccda262736ce | https://socratic.org/questions/what-is-the-number-of-moles-in-9-63-l-of-h-2s-gas-at-stp | 0.43 moles | start physical_unit 10 11 mole mol qc_end physical_unit 10 11 7 8 volume qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Mole [OF] H2S gas [IN] moles"}] | [{"type":"physical unit","value":"0.43 moles"}] | [{"type":"physical unit","value":"Volume [OF] H2S gas [=] \\pu{9.63 L}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the number of moles in 9.63 L of #H_2S# gas at STP? </h1> | null | 0.43 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Any gas, at STP, will have 22.4 L per mol. Taking this into account, using dimensional analysis:</p>
<p><mathjax>#9.63 L H_2S * (1 mol H_2S)/(22.4 L H_2S)#</mathjax> </p>
<p><mathjax>#(9.63 L H_2S)/(22.4 L H_2S)=0.4299 mol#</mathjax> </p>
<p>Which rounds to 0.430 mol, with three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<div class="markdown"><p>0.430 mol (rounded with significant figures)</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Any gas, at STP, will have 22.4 L per mol. Taking this into account, using dimensional analysis:</p>
<p><mathjax>#9.63 L H_2S * (1 mol H_2S)/(22.4 L H_2S)#</mathjax> </p>
<p><mathjax>#(9.63 L H_2S)/(22.4 L H_2S)=0.4299 mol#</mathjax> </p>
<p>Which rounds to 0.430 mol, with three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<h1 class="questionTitle" itemprop="name">What is the number of moles in 9.63 L of #H_2S# gas at STP? </h1>
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<div class="markdown"><p>0.430 mol (rounded with significant figures)</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Any gas, at STP, will have 22.4 L per mol. Taking this into account, using dimensional analysis:</p>
<p><mathjax>#9.63 L H_2S * (1 mol H_2S)/(22.4 L H_2S)#</mathjax> </p>
<p><mathjax>#(9.63 L H_2S)/(22.4 L H_2S)=0.4299 mol#</mathjax> </p>
<p>Which rounds to 0.430 mol, with three <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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May 19, 2016
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<div class="markdown"><p><mathjax>#P.V=n.R.T#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Use the ideal gas equation</p>
<p><mathjax>#P.V=n.R.T#</mathjax></p>
<p>Solve for <mathjax>#n#</mathjax>,</p>
<p><mathjax>#n=( P*V)/(R.T)#</mathjax></p>
<p><mathjax>#P" is the pressure of the gas (" atm")"#</mathjax><br/>
<mathjax>#V" is the volume of the gas ("L")"#</mathjax><br/>
<mathjax>#T" is the Kelvin temperature ("K")"#</mathjax><br/>
<mathjax>#R" is the universal gas constant ("0.0821L.\atm.mol^-1.K^-1")" #</mathjax></p>
<p><mathjax>#n=(1.00*atm*9.63*L) /(0.0821*L*atm*mol.^-1*K^-1*273*K#</mathjax></p>
<p><mathjax>#n=(1.00*cancel(atm)*9.63*cancel(L)) /(0.0821*cancel(L)* cancel(atm)*mol.^-1*cancel(K^-1)*273*cancel(K)#</mathjax></p>
<p><mathjax>#n=0.430* mol.#</mathjax></p></div>
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</article> | What is the number of moles in 9.63 L of #H_2S# gas at STP? | null |
3,228 | aaf82c22-6ddd-11ea-933a-ccda262736ce | https://socratic.org/questions/calculate-concentration-in-g-dm-3-of-chloride-ions-in-0-25m-molar-potassium-chlo | 8.86 g/dm^3 | start physical_unit 6 7 concentration g/dm^3 qc_end physical_unit 12 14 10 11 concentration qc_end end | [{"type":"physical unit","value":"Concentration [OF] chloride ions [IN] g/dm^3"}] | [{"type":"physical unit","value":"8.86 g/dm^3"}] | [{"type":"physical unit","value":"Concentration [OF] potassium chloride solution [=] \\pu{0.25 M}"}] | <h1 class="questionTitle" itemprop="name">Calculate the concentration in #"g dm"^-3# of chloride ions in a #"0.25-M"# potassium chloride solution?</h1> | null | 8.86 g/dm^3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to convert the concentration of chloride anions from <em>moles per cubic decimeter</em> to <em>grams per cubic decimeter</em>. </p>
<p>Now, you know that potassium chloride is <strong>soluble</strong> in water, so it will dissociate completely in aqueous solution to produce potassium cations and chloride anions.</p>
<blockquote>
<p><mathjax>#"KCl"_ ((aq)) -> "K"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, every mole of potassium chloride that you dissolve in this solution will produce <mathjax>#1#</mathjax> <strong>mole</strong> of chloride anions, so you can say that</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = ["KCl"] = "0.25 M"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the chloride anions tells the number of moles of chloride anions present <strong>for every</strong> <mathjax>#"1 L = 1dm"^3#</mathjax> of this solution. </p>
<p>In this case, you have <mathjax>#0.25#</mathjax> <strong>moles</strong> of chloride anions for every <mathjax>#"1 dm"^3#</mathjax> of the solution. To convert the number of moles to <em>grams</em>, use the <strong>molar mass</strong> of <strong>atomic chlorine</strong>, <mathjax>#"Cl"#</mathjax>--keep in mind that the mass of the chloride anion is <em>approximately equal</em> to the mass of the neutral atom! </p>
<blockquote>
<p><mathjax>#0.25 color(red)(cancel(color(black)("moles Cl"^(-)))) * "35.453 g"/(1 color(red)(cancel(color(black)("mole Cl"^(-))))) = "8.86 g"#</mathjax></p>
</blockquote>
<p>Since this represents the mass of chloride anions present in <mathjax>#"1 dm"^3#</mathjax> of this solution, you can say that the concentration of the solution is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("concentration Cl"^(-) = "8.9 g dm"^(-3))))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
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<div class="markdown"><p><mathjax>#"8.9 g dm"^(-3)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to convert the concentration of chloride anions from <em>moles per cubic decimeter</em> to <em>grams per cubic decimeter</em>. </p>
<p>Now, you know that potassium chloride is <strong>soluble</strong> in water, so it will dissociate completely in aqueous solution to produce potassium cations and chloride anions.</p>
<blockquote>
<p><mathjax>#"KCl"_ ((aq)) -> "K"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, every mole of potassium chloride that you dissolve in this solution will produce <mathjax>#1#</mathjax> <strong>mole</strong> of chloride anions, so you can say that</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = ["KCl"] = "0.25 M"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the chloride anions tells the number of moles of chloride anions present <strong>for every</strong> <mathjax>#"1 L = 1dm"^3#</mathjax> of this solution. </p>
<p>In this case, you have <mathjax>#0.25#</mathjax> <strong>moles</strong> of chloride anions for every <mathjax>#"1 dm"^3#</mathjax> of the solution. To convert the number of moles to <em>grams</em>, use the <strong>molar mass</strong> of <strong>atomic chlorine</strong>, <mathjax>#"Cl"#</mathjax>--keep in mind that the mass of the chloride anion is <em>approximately equal</em> to the mass of the neutral atom! </p>
<blockquote>
<p><mathjax>#0.25 color(red)(cancel(color(black)("moles Cl"^(-)))) * "35.453 g"/(1 color(red)(cancel(color(black)("mole Cl"^(-))))) = "8.86 g"#</mathjax></p>
</blockquote>
<p>Since this represents the mass of chloride anions present in <mathjax>#"1 dm"^3#</mathjax> of this solution, you can say that the concentration of the solution is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("concentration Cl"^(-) = "8.9 g dm"^(-3))))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
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<h1 class="questionTitle" itemprop="name">Calculate the concentration in #"g dm"^-3# of chloride ions in a #"0.25-M"# potassium chloride solution?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#"8.9 g dm"^(-3)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to convert the concentration of chloride anions from <em>moles per cubic decimeter</em> to <em>grams per cubic decimeter</em>. </p>
<p>Now, you know that potassium chloride is <strong>soluble</strong> in water, so it will dissociate completely in aqueous solution to produce potassium cations and chloride anions.</p>
<blockquote>
<p><mathjax>#"KCl"_ ((aq)) -> "K"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>Now, every mole of potassium chloride that you dissolve in this solution will produce <mathjax>#1#</mathjax> <strong>mole</strong> of chloride anions, so you can say that</p>
<blockquote>
<p><mathjax>#["Cl"^(-)] = ["KCl"] = "0.25 M"#</mathjax></p>
</blockquote>
<p>The <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the chloride anions tells the number of moles of chloride anions present <strong>for every</strong> <mathjax>#"1 L = 1dm"^3#</mathjax> of this solution. </p>
<p>In this case, you have <mathjax>#0.25#</mathjax> <strong>moles</strong> of chloride anions for every <mathjax>#"1 dm"^3#</mathjax> of the solution. To convert the number of moles to <em>grams</em>, use the <strong>molar mass</strong> of <strong>atomic chlorine</strong>, <mathjax>#"Cl"#</mathjax>--keep in mind that the mass of the chloride anion is <em>approximately equal</em> to the mass of the neutral atom! </p>
<blockquote>
<p><mathjax>#0.25 color(red)(cancel(color(black)("moles Cl"^(-)))) * "35.453 g"/(1 color(red)(cancel(color(black)("mole Cl"^(-))))) = "8.86 g"#</mathjax></p>
</blockquote>
<p>Since this represents the mass of chloride anions present in <mathjax>#"1 dm"^3#</mathjax> of this solution, you can say that the concentration of the solution is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("concentration Cl"^(-) = "8.9 g dm"^(-3))))#</mathjax></p>
</blockquote>
<p>The answer must be rounded to two <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>, the number of sig figs you have for the molarity of the solution. </p></div>
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</article> | Calculate the concentration in #"g dm"^-3# of chloride ions in a #"0.25-M"# potassium chloride solution? | null |
3,229 | ac0bc96f-6ddd-11ea-bb9d-ccda262736ce | https://socratic.org/questions/if-0-900-g-of-oxalic-acid-h-2c-2o-4-90-04-g-mol-is-completely-neutralized-with-0 | 4.00 × 10^(-3) L | start physical_unit 19 20 volume l qc_end physical_unit 6 6 1 2 mass qc_end physical_unit 6 6 7 8 molar_mass qc_end c_other OTHER qc_end physical_unit 15 15 13 14 molarity qc_end end | [{"type":"physical unit","value":"Volume [OF] lithium hydroxide [IN] L"}] | [{"type":"physical unit","value":"4.00 × 10^(-3) L"}] | [{"type":"physical unit","value":"Mass [OF] H2C2O4 [=] \\pu{0.900 g}"},{"type":"physical unit","value":"Molar mass [OF] H2C2O4 [=] \\pu{90.04 g/mol}"},{"type":"other","value":"Completely neutralize."},{"type":"physical unit","value":"Molarity [OF] LiOH solution [=] \\pu{0.500 M}"}] | <h1 class="questionTitle" itemprop="name">If 0.900 g of oxalic acid, #H_2C_2O_4# (90.04 g/mol) is completely neutralized with 0.500 M #LiOH#, what volume of lithium hydroxide is required?</h1> | null | 4.00 × 10^(-3) L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ethandioic acid is diprotic and reacts like this:</p>
<p><mathjax>#sf(H_2C_2O_4+2LiOHrarrLi_2C_2O_4+2H_2O)#</mathjax></p>
<p>This tells us that 1 mole <mathjax>#sf(H_2C_2O_4)-=#</mathjax> 2 mole <mathjax>#sf(LiOH)#</mathjax></p>
<p>The no. moles <mathjax>#sf(H_2C_2O_4)#</mathjax> is given by:</p>
<p><mathjax>#sf(n_(H_2C_2O_4)=m/M_(r)=0.900/90.04=0.001)#</mathjax></p>
<p>So the no. moles of LiOH must be 2x this amount:</p>
<p><mathjax>#sf(n_(LiOH)=2xx0.001=0.002)#</mathjax></p>
<p>Concentration = no. moles <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/volume of solution i.e:</p>
<p><mathjax>#sf(c_(LiOH)=n_(LiOH)/v_(LiOH)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(v_(LiOH)=n_(LiOH)/c_(LiOH)=0.002/0.500=0.004color(white)(x)"L")#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#sf(0.004color(white)(x)"L")#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ethandioic acid is diprotic and reacts like this:</p>
<p><mathjax>#sf(H_2C_2O_4+2LiOHrarrLi_2C_2O_4+2H_2O)#</mathjax></p>
<p>This tells us that 1 mole <mathjax>#sf(H_2C_2O_4)-=#</mathjax> 2 mole <mathjax>#sf(LiOH)#</mathjax></p>
<p>The no. moles <mathjax>#sf(H_2C_2O_4)#</mathjax> is given by:</p>
<p><mathjax>#sf(n_(H_2C_2O_4)=m/M_(r)=0.900/90.04=0.001)#</mathjax></p>
<p>So the no. moles of LiOH must be 2x this amount:</p>
<p><mathjax>#sf(n_(LiOH)=2xx0.001=0.002)#</mathjax></p>
<p>Concentration = no. moles <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/volume of solution i.e:</p>
<p><mathjax>#sf(c_(LiOH)=n_(LiOH)/v_(LiOH)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(v_(LiOH)=n_(LiOH)/c_(LiOH)=0.002/0.500=0.004color(white)(x)"L")#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">If 0.900 g of oxalic acid, #H_2C_2O_4# (90.04 g/mol) is completely neutralized with 0.500 M #LiOH#, what volume of lithium hydroxide is required?</h1>
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<div class="markdown"><p><mathjax>#sf(0.004color(white)(x)"L")#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Ethandioic acid is diprotic and reacts like this:</p>
<p><mathjax>#sf(H_2C_2O_4+2LiOHrarrLi_2C_2O_4+2H_2O)#</mathjax></p>
<p>This tells us that 1 mole <mathjax>#sf(H_2C_2O_4)-=#</mathjax> 2 mole <mathjax>#sf(LiOH)#</mathjax></p>
<p>The no. moles <mathjax>#sf(H_2C_2O_4)#</mathjax> is given by:</p>
<p><mathjax>#sf(n_(H_2C_2O_4)=m/M_(r)=0.900/90.04=0.001)#</mathjax></p>
<p>So the no. moles of LiOH must be 2x this amount:</p>
<p><mathjax>#sf(n_(LiOH)=2xx0.001=0.002)#</mathjax></p>
<p>Concentration = no. moles <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/volume of solution i.e:</p>
<p><mathjax>#sf(c_(LiOH)=n_(LiOH)/v_(LiOH)#</mathjax></p>
<p><mathjax>#:.#</mathjax><mathjax>#sf(v_(LiOH)=n_(LiOH)/c_(LiOH)=0.002/0.500=0.004color(white)(x)"L")#</mathjax></p></div>
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</article> | If 0.900 g of oxalic acid, #H_2C_2O_4# (90.04 g/mol) is completely neutralized with 0.500 M #LiOH#, what volume of lithium hydroxide is required? | null |
3,230 | ab720068-6ddd-11ea-b3b4-ccda262736ce | https://socratic.org/questions/what-is-the-volume-at-stp-of-2-66-mol-of-methane-gas | 59.62 L | start physical_unit 10 11 volume l qc_end physical_unit 10 11 7 8 mole qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume [OF] methane gas [IN] L"}] | [{"type":"physical unit","value":"59.62 L"}] | [{"type":"physical unit","value":"Mole [OF] methane gas [=] \\pu{2.66 mol}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">What is the volume at STP of 2.66 mol of methane gas?</h1> | null | 59.62 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Because we are at standard temperature and pressure and are given only one set of conditions, we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>:<br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/Aj9e7GO2Svm4efvXTi8E_slide_2.jpg"/> </p>
<p>I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.</p>
<p>List your known and unknown variables. </p>
<p><mathjax>#color(purple)("Knowns:"#</mathjax><br/>
- Number of moles<br/>
- Temperature<br/>
- Pressure</p>
<p><mathjax>#color(blue)("Unknowns:"#</mathjax><br/>
Volume of <mathjax>#CH_4(g)#</mathjax></p>
<p>At STP, the temperature is 273K and the pressure is 1 atm.</p>
<p>Let's rearrange the equation to solve for V:</p>
<p><mathjax># (nxxRxxT)/P#</mathjax></p>
<p><mathjax>#V = (2.66cancel"mol"xx0.0821Lxxcancel(atm)/cancel(molxxK)xx273cancelK)/(1cancel(atm)#</mathjax></p>
<p><mathjax>#V = 59.6 L#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#V= 59.6L#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Because we are at standard temperature and pressure and are given only one set of conditions, we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>:<br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/Aj9e7GO2Svm4efvXTi8E_slide_2.jpg"/> </p>
<p>I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.</p>
<p>List your known and unknown variables. </p>
<p><mathjax>#color(purple)("Knowns:"#</mathjax><br/>
- Number of moles<br/>
- Temperature<br/>
- Pressure</p>
<p><mathjax>#color(blue)("Unknowns:"#</mathjax><br/>
Volume of <mathjax>#CH_4(g)#</mathjax></p>
<p>At STP, the temperature is 273K and the pressure is 1 atm.</p>
<p>Let's rearrange the equation to solve for V:</p>
<p><mathjax># (nxxRxxT)/P#</mathjax></p>
<p><mathjax>#V = (2.66cancel"mol"xx0.0821Lxxcancel(atm)/cancel(molxxK)xx273cancelK)/(1cancel(atm)#</mathjax></p>
<p><mathjax>#V = 59.6 L#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the volume at STP of 2.66 mol of methane gas?</h1>
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<div class="markdown"><p><mathjax>#V= 59.6L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Because we are at standard temperature and pressure and are given only one set of conditions, we have to use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law equation</a>:<br/>
<img alt="slideplayer.com" src="https://useruploads.socratic.org/Aj9e7GO2Svm4efvXTi8E_slide_2.jpg"/> </p>
<p>I should mention that the pressure does not always have units of atm, it depends on the units of pressure given in the gas constant.</p>
<p>List your known and unknown variables. </p>
<p><mathjax>#color(purple)("Knowns:"#</mathjax><br/>
- Number of moles<br/>
- Temperature<br/>
- Pressure</p>
<p><mathjax>#color(blue)("Unknowns:"#</mathjax><br/>
Volume of <mathjax>#CH_4(g)#</mathjax></p>
<p>At STP, the temperature is 273K and the pressure is 1 atm.</p>
<p>Let's rearrange the equation to solve for V:</p>
<p><mathjax># (nxxRxxT)/P#</mathjax></p>
<p><mathjax>#V = (2.66cancel"mol"xx0.0821Lxxcancel(atm)/cancel(molxxK)xx273cancelK)/(1cancel(atm)#</mathjax></p>
<p><mathjax>#V = 59.6 L#</mathjax></p></div>
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<div class="markdown"><p>Because the temperature and pressure are given as STP, all that is need is to use the molar volume.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1 mole = 22.4 liters. </p>
<p>so # 2.66 moles / 1.00 mole = </p>
<pre><code> #V/ 22.4#
</code></pre>
<p>2.66 x 22.4 = 59.58 liters. </p></div>
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</article> | What is the volume at STP of 2.66 mol of methane gas? | null |
3,231 | ac5ce0e2-6ddd-11ea-bd20-ccda262736ce | https://socratic.org/questions/the-specific-heat-of-silver-is-0-24-j-g-c-how-many-joules-of-energy-are-needed-t | 4.4 joules | start physical_unit 4 4 energy j qc_end physical_unit 4 4 6 9 specific_heat qc_end physical_unit 4 4 19 20 mass qc_end physical_unit 4 4 24 25 temperature qc_end physical_unit 4 4 27 28 temperature qc_end end | [{"type":"physical unit","value":"Needed energy [OF] silver [IN] joules"}] | [{"type":"physical unit","value":"4.4 joules"}] | [{"type":"physical unit","value":"Specific heat [OF] silver [=] \\pu{0.24 J/(g * ℃)}"},{"type":"physical unit","value":"Mass [OF] silver [=] \\pu{7.37 g}"},{"type":"physical unit","value":"Temperature1 [OF] silver [=] \\pu{25.0 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] silver [=] \\pu{27.5 ℃}"}] | <h1 class="questionTitle" itemprop="name">The specific heat of silver is 0.24 J/g°C. How many joules of energy are needed to warm 7.37 g of silver from 25.0°C to 27.5°C?</h1> | null | 4.4 joules | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation to use for this problem is written below. We need to determine which variables are known, and which is unknown. Then substitute the known values into the equation and solve.</p>
<p><img alt="http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html" src="https://useruploads.socratic.org/qcAUA2tQmaGuxDTkLxvK_shta.gif"/> <br/>
<strong>Known</strong><br/>
<mathjax>#c="0.24 J/g"^@"C")#</mathjax><br/>
<mathjax>#m="7.37 g"#</mathjax><br/>
<mathjax>#T_i=25.0^@"C"#</mathjax><br/>
<mathjax>#T_f=27.5^@"C"#</mathjax><br/>
<mathjax>#DeltaT=T_f-T_i=27.5^@"C"-25.0^@"C"=2.5^@"C"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#Q#</mathjax></p>
<p><strong>Solution</strong><br/>
<mathjax>#Q=("0.24 J"/(cancel"g"^@cancel"C"))(7.37cancel"g")(2.5^@cancel"C")="4.4 J"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>The amount of energy required is <mathjax>#"4.4 J"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation to use for this problem is written below. We need to determine which variables are known, and which is unknown. Then substitute the known values into the equation and solve.</p>
<p><img alt="http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html" src="https://useruploads.socratic.org/qcAUA2tQmaGuxDTkLxvK_shta.gif"/> <br/>
<strong>Known</strong><br/>
<mathjax>#c="0.24 J/g"^@"C")#</mathjax><br/>
<mathjax>#m="7.37 g"#</mathjax><br/>
<mathjax>#T_i=25.0^@"C"#</mathjax><br/>
<mathjax>#T_f=27.5^@"C"#</mathjax><br/>
<mathjax>#DeltaT=T_f-T_i=27.5^@"C"-25.0^@"C"=2.5^@"C"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#Q#</mathjax></p>
<p><strong>Solution</strong><br/>
<mathjax>#Q=("0.24 J"/(cancel"g"^@cancel"C"))(7.37cancel"g")(2.5^@cancel"C")="4.4 J"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<h1 class="questionTitle" itemprop="name">The specific heat of silver is 0.24 J/g°C. How many joules of energy are needed to warm 7.37 g of silver from 25.0°C to 27.5°C?</h1>
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<div class="markdown"><p>The amount of energy required is <mathjax>#"4.4 J"#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation to use for this problem is written below. We need to determine which variables are known, and which is unknown. Then substitute the known values into the equation and solve.</p>
<p><img alt="http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html" src="https://useruploads.socratic.org/qcAUA2tQmaGuxDTkLxvK_shta.gif"/> <br/>
<strong>Known</strong><br/>
<mathjax>#c="0.24 J/g"^@"C")#</mathjax><br/>
<mathjax>#m="7.37 g"#</mathjax><br/>
<mathjax>#T_i=25.0^@"C"#</mathjax><br/>
<mathjax>#T_f=27.5^@"C"#</mathjax><br/>
<mathjax>#DeltaT=T_f-T_i=27.5^@"C"-25.0^@"C"=2.5^@"C"#</mathjax></p>
<p><strong>Unknown:</strong> <mathjax>#Q#</mathjax></p>
<p><strong>Solution</strong><br/>
<mathjax>#Q=("0.24 J"/(cancel"g"^@cancel"C"))(7.37cancel"g")(2.5^@cancel"C")="4.4 J"#</mathjax> rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</article> | The specific heat of silver is 0.24 J/g°C. How many joules of energy are needed to warm 7.37 g of silver from 25.0°C to 27.5°C? | null |
3,232 | a98ad091-6ddd-11ea-b846-ccda262736ce | https://socratic.org/questions/a-gas-occupies-15-liters-at-a-pressure-of-80-0-mm-hg-what-is-the-volume-when-the | 12.00 liters | start physical_unit 1 1 volume l qc_end physical_unit 1 1 3 4 volume qc_end physical_unit 1 1 9 10 pressure qc_end physical_unit 1 1 21 22 pressure qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] liters"}] | [{"type":"physical unit","value":"12.00 liters"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{15 liters}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{80.0 mmHg}"},{"type":"physical unit","value":"Pressure2 [OF] the gas [=] \\pu{100.0 mmHg}"}] | <h1 class="questionTitle" itemprop="name">A gas occupies 15 liters at a pressure of 80.0 mm Hg. What is the volume when the pressure is increased to 100.0 mm Hg?</h1> | null | 12.00 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#V_2 =(P_1V_1)/P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#80*mm*Hg#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#15*L#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#1/(100*mm*Hg)#</mathjax></p>
<p>So the answer is <mathjax>#4/5#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#15*L#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12*L#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>We use <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>: <mathjax>#P_1V_1=P_2V_2#</mathjax>; temperature and moles are constant.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#V_2 =(P_1V_1)/P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#80*mm*Hg#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#15*L#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#1/(100*mm*Hg)#</mathjax></p>
<p>So the answer is <mathjax>#4/5#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#15*L#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12*L#</mathjax>.</p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">A gas occupies 15 liters at a pressure of 80.0 mm Hg. What is the volume when the pressure is increased to 100.0 mm Hg?</h1>
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<div class="markdown"><p>We use <a href="http://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's law</a>: <mathjax>#P_1V_1=P_2V_2#</mathjax>; temperature and moles are constant.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#V_2 =(P_1V_1)/P_2#</mathjax> <mathjax>#=#</mathjax> <mathjax>#80*mm*Hg#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#15*L#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#1/(100*mm*Hg)#</mathjax></p>
<p>So the answer is <mathjax>#4/5#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#15*L#</mathjax> <mathjax>#=#</mathjax> <mathjax>#12*L#</mathjax>.</p></div>
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</article> | A gas occupies 15 liters at a pressure of 80.0 mm Hg. What is the volume when the pressure is increased to 100.0 mm Hg? | null |
3,233 | abc43392-6ddd-11ea-801d-ccda262736ce | https://socratic.org/questions/what-is-the-pressure-in-a-19-3-l-cylinder-filled-with-16-7-g-of-oxygen-gas-at-a- | 0.70 atm | start physical_unit 14 15 pressure atm qc_end physical_unit 14 15 6 7 volume qc_end physical_unit 14 15 11 12 mass qc_end physical_unit 14 15 20 21 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] oxygen gas [IN] atm"}] | [{"type":"physical unit","value":"0.70 atm"}] | [{"type":"physical unit","value":"Volume [OF] oxygen gas [=] \\pu{19.3 L}"},{"type":"physical unit","value":"Mass [OF] oxygen gas [=] \\pu{16.7 g}"},{"type":"physical unit","value":"Temperature [OF] oxygen gas [=] \\pu{319 K}"}] | <h1 class="questionTitle" itemprop="name">What is the pressure in a 19.3-L cylinder filled with 16.7 g of oxygen gas at a temperature of 319 K? </h1> | null | 0.70 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#((16.00*g)/(32.00*g*mol^-1)xx0.0821(L*atm)/(K*mol)xx319*K)/(19.3*L)~=0.70*atm#</mathjax></p>
<p>Why did I have to divide the mass of the oxygen gas by <mathjax>#32#</mathjax>?</p></div>
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<div class="markdown"><p><mathjax>#P=(nRT)/V~=0.70*atm#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#((16.00*g)/(32.00*g*mol^-1)xx0.0821(L*atm)/(K*mol)xx319*K)/(19.3*L)~=0.70*atm#</mathjax></p>
<p>Why did I have to divide the mass of the oxygen gas by <mathjax>#32#</mathjax>?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the pressure in a 19.3-L cylinder filled with 16.7 g of oxygen gas at a temperature of 319 K? </h1>
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<div class="markdown"><p><mathjax>#P=(nRT)/V~=0.70*atm#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#((16.00*g)/(32.00*g*mol^-1)xx0.0821(L*atm)/(K*mol)xx319*K)/(19.3*L)~=0.70*atm#</mathjax></p>
<p>Why did I have to divide the mass of the oxygen gas by <mathjax>#32#</mathjax>?</p></div>
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</article> | What is the pressure in a 19.3-L cylinder filled with 16.7 g of oxygen gas at a temperature of 319 K? | null |
3,234 | abea39b4-6ddd-11ea-b615-ccda262736ce | https://socratic.org/questions/59151d3eb72cff3a6b742345 | 0.36 atm | start physical_unit 4 4 pressure atm qc_end physical_unit 4 4 9 10 pressure qc_end physical_unit 4 4 15 16 temperature qc_end physical_unit 4 4 28 29 temperature qc_end end | [{"type":"physical unit","value":"Pressure2 [OF] gas sample [IN] atm"}] | [{"type":"physical unit","value":"0.36 atm"}] | [{"type":"physical unit","value":"Pressure1 [OF] gas sample [=] \\pu{0.370 atm}"},{"type":"physical unit","value":"Temperature1 [OF] gas sample [=] \\pu{5.00 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] gas sample [=] \\pu{0 ℃}"}] | <h1 class="questionTitle" itemprop="name">A sample of a gas has a pressure of #"0.370 atm"# at a temperature of #"5.00"^@"C"#. What will the pressure be if the temperature is lowered to #"0"^@"C"#?</h1> | null | 0.36 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Gay-Lussac's gas law will be used to solve this problem. It states that the pressure of a gas is directly proportional to its Kelvin temperature, as long as mass and volume are held constant. This means that if the volume increases, so does the temperature and vice-versa. The equation for this law is:</p>
<p><mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p>The given data has temperature in degrees Celsius, so it needs to be converted to Kelvins. This is done by adding <mathjax>#"273.15"#</mathjax> to the Celsius temperature.</p>
<p><strong>Organize your data:</strong></p>
<p><strong>Given</strong></p>
<p><mathjax>#P_1="0.370 atm"#</mathjax></p>
<p><mathjax>#T_1="5.00"^@"C" + 273.15="278.15 K"#</mathjax></p>
<p><mathjax>#T_2="0.00"^@"C + 273.15 ="273.15 K"#</mathjax></p>
<p><strong>Unknown:</strong> </p>
<p><mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax>. Insert the given data and solve.</p>
<p><mathjax>#P_2=(P_1T_2)/(T_1)#</mathjax></p>
<p><mathjax>#P_2=(0.370color(white)(.)"atm"xx273.15color(red)cancel(color(black)("K")))/(278.15color(red)cancel(color(black)("K")))="0.363 atm"#</mathjax> (rounded to three significant figures)</p></div>
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<div class="answerSummary">
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<div class="markdown"><p>The pressure at <mathjax>#"273.15 K"#</mathjax> <mathjax>#(0^@"C")#</mathjax> will be <mathjax>#"0.363 atm"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Gay-Lussac's gas law will be used to solve this problem. It states that the pressure of a gas is directly proportional to its Kelvin temperature, as long as mass and volume are held constant. This means that if the volume increases, so does the temperature and vice-versa. The equation for this law is:</p>
<p><mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p>The given data has temperature in degrees Celsius, so it needs to be converted to Kelvins. This is done by adding <mathjax>#"273.15"#</mathjax> to the Celsius temperature.</p>
<p><strong>Organize your data:</strong></p>
<p><strong>Given</strong></p>
<p><mathjax>#P_1="0.370 atm"#</mathjax></p>
<p><mathjax>#T_1="5.00"^@"C" + 273.15="278.15 K"#</mathjax></p>
<p><mathjax>#T_2="0.00"^@"C + 273.15 ="273.15 K"#</mathjax></p>
<p><strong>Unknown:</strong> </p>
<p><mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax>. Insert the given data and solve.</p>
<p><mathjax>#P_2=(P_1T_2)/(T_1)#</mathjax></p>
<p><mathjax>#P_2=(0.370color(white)(.)"atm"xx273.15color(red)cancel(color(black)("K")))/(278.15color(red)cancel(color(black)("K")))="0.363 atm"#</mathjax> (rounded to three significant figures)</p></div>
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<h1 class="questionTitle" itemprop="name">A sample of a gas has a pressure of #"0.370 atm"# at a temperature of #"5.00"^@"C"#. What will the pressure be if the temperature is lowered to #"0"^@"C"#?</h1>
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<div class="markdown"><p>The pressure at <mathjax>#"273.15 K"#</mathjax> <mathjax>#(0^@"C")#</mathjax> will be <mathjax>#"0.363 atm"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Gay-Lussac's gas law will be used to solve this problem. It states that the pressure of a gas is directly proportional to its Kelvin temperature, as long as mass and volume are held constant. This means that if the volume increases, so does the temperature and vice-versa. The equation for this law is:</p>
<p><mathjax>#P_1/T_1=P_2/T_2#</mathjax></p>
<p>The given data has temperature in degrees Celsius, so it needs to be converted to Kelvins. This is done by adding <mathjax>#"273.15"#</mathjax> to the Celsius temperature.</p>
<p><strong>Organize your data:</strong></p>
<p><strong>Given</strong></p>
<p><mathjax>#P_1="0.370 atm"#</mathjax></p>
<p><mathjax>#T_1="5.00"^@"C" + 273.15="278.15 K"#</mathjax></p>
<p><mathjax>#T_2="0.00"^@"C + 273.15 ="273.15 K"#</mathjax></p>
<p><strong>Unknown:</strong> </p>
<p><mathjax>#P_2#</mathjax></p>
<p><strong>Solution</strong><br/>
Rearrange the equation to isolate <mathjax>#P_2#</mathjax>. Insert the given data and solve.</p>
<p><mathjax>#P_2=(P_1T_2)/(T_1)#</mathjax></p>
<p><mathjax>#P_2=(0.370color(white)(.)"atm"xx273.15color(red)cancel(color(black)("K")))/(278.15color(red)cancel(color(black)("K")))="0.363 atm"#</mathjax> (rounded to three significant figures)</p></div>
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</article> | A sample of a gas has a pressure of #"0.370 atm"# at a temperature of #"5.00"^@"C"#. What will the pressure be if the temperature is lowered to #"0"^@"C"#? | null |
3,235 | a92f2238-6ddd-11ea-9686-ccda262736ce | https://socratic.org/questions/a-0-500-l-solution-of-6-m-hcl-has-to-be-made-how-much-12-m-hcl-is-needed | 0.25 L | start physical_unit 3 3 volume l qc_end physical_unit 3 3 1 2 volume qc_end physical_unit 7 7 5 6 molarity qc_end physical_unit 7 7 14 15 molarity qc_end end | [{"type":"physical unit","value":"Volume2 [OF] HCl solution [IN] L"}] | [{"type":"physical unit","value":"0.25 L"}] | [{"type":"physical unit","value":"Volume1 [OF] HCl solution [=] \\pu{0.500 L}"},{"type":"physical unit","value":"Molarity1 [OF] HCl solution [=] \\pu{6 M}"},{"type":"physical unit","value":"Molarity2 [OF] HCl solution [=] \\pu{12 M}"}] | <h1 class="questionTitle" itemprop="name">A 0.500 L solution of 6 M #HCl# has to be made. How much 12 M #HCl# is needed?</h1> | null | 0.25 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>6M HCl has half the concentration of 12M HCl, <br/>
therefore a solution that is 50% 12M HCl, is a 6M solution.</p>
<p>To make 0.5L of 6M HCl, you would need 0.25 (half of 0.5) litres of 12M HCl.</p></div>
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<div class="markdown"><p>0.25 litres</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>6M HCl has half the concentration of 12M HCl, <br/>
therefore a solution that is 50% 12M HCl, is a 6M solution.</p>
<p>To make 0.5L of 6M HCl, you would need 0.25 (half of 0.5) litres of 12M HCl.</p></div>
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<h1 class="questionTitle" itemprop="name">A 0.500 L solution of 6 M #HCl# has to be made. How much 12 M #HCl# is needed?</h1>
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<div class="markdown"><p>0.25 litres</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>6M HCl has half the concentration of 12M HCl, <br/>
therefore a solution that is 50% 12M HCl, is a 6M solution.</p>
<p>To make 0.5L of 6M HCl, you would need 0.25 (half of 0.5) litres of 12M HCl.</p></div>
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</article> | A 0.500 L solution of 6 M #HCl# has to be made. How much 12 M #HCl# is needed? | null |
3,236 | ab51b911-6ddd-11ea-aae6-ccda262736ce | https://socratic.org/questions/how-many-moles-of-h-2-are-needed-to-react-with-80-mol-of-n-2 | 2.40 moles | start physical_unit 4 4 mole mol qc_end physical_unit 13 13 10 11 mole qc_end end | [{"type":"physical unit","value":"Mole [OF] H2 [IN] moles"}] | [{"type":"physical unit","value":"2.40 moles"}] | [{"type":"physical unit","value":"Mole [OF] N2 [=] \\pu{0.80 mol}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #H_2# are needed to react with .80 mol of #N_2#?</h1> | null | 2.40 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"N"_2("g")+"3H"_2("g")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NH"_3("g")#</mathjax></p>
<p>Multiply the <mathjax>#"0.80 mol N"_2"#</mathjax> times <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"H"_2"#</mathjax> and <mathjax>#"N"_2"#</mathjax> so that moles <mathjax>#"N"_2"#</mathjax> cancel, leaving moles <mathjax>#"H"_2"#</mathjax>.</p>
<p><mathjax>#0.80cancel("mol N"_2)xx(3"mol H"_2)/(1cancel("mol N"_2))="2.4 mol H"_2"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"2.4 mol H"_2"#</mathjax> are needed to react with <mathjax>#"0.80 mol N"_2"#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"N"_2("g")+"3H"_2("g")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NH"_3("g")#</mathjax></p>
<p>Multiply the <mathjax>#"0.80 mol N"_2"#</mathjax> times <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"H"_2"#</mathjax> and <mathjax>#"N"_2"#</mathjax> so that moles <mathjax>#"N"_2"#</mathjax> cancel, leaving moles <mathjax>#"H"_2"#</mathjax>.</p>
<p><mathjax>#0.80cancel("mol N"_2)xx(3"mol H"_2)/(1cancel("mol N"_2))="2.4 mol H"_2"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of #H_2# are needed to react with .80 mol of #N_2#?</h1>
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<div class="markdown"><p><mathjax>#"2.4 mol H"_2"#</mathjax> are needed to react with <mathjax>#"0.80 mol N"_2"#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"N"_2("g")+"3H"_2("g")#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2NH"_3("g")#</mathjax></p>
<p>Multiply the <mathjax>#"0.80 mol N"_2"#</mathjax> times <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <mathjax>#"H"_2"#</mathjax> and <mathjax>#"N"_2"#</mathjax> so that moles <mathjax>#"N"_2"#</mathjax> cancel, leaving moles <mathjax>#"H"_2"#</mathjax>.</p>
<p><mathjax>#0.80cancel("mol N"_2)xx(3"mol H"_2)/(1cancel("mol N"_2))="2.4 mol H"_2"#</mathjax></p></div>
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</article> | How many moles of #H_2# are needed to react with .80 mol of #N_2#? | null |
3,237 | ab2cf71a-6ddd-11ea-b782-ccda262736ce | https://socratic.org/questions/how-many-grams-of-acetic-acid-should-be-dissolved-in-100-g-of-water-to-make-a-4- | 4.17 grams | start physical_unit 4 5 mass g qc_end physical_unit 13 13 10 11 mass qc_end physical_unit 22 22 17 17 mass_percent qc_end end | [{"type":"physical unit","value":"Mass [OF] acetic acid [IN] grams"}] | [{"type":"physical unit","value":"4.17 grams"}] | [{"type":"physical unit","value":"Mass [OF] water [=] \\pu{100.0 g}"},{"type":"physical unit","value":"Percent by mass [OF] vinegar in solution [=] \\pu{4.0 percent}"}] | <h1 class="questionTitle" itemprop="name">How many grams of acetic acid should be dissolved in 100.0 g of water to make a 4.0 percent (by mass) solution of vinegar?</h1> | null | 4.17 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by mass</a> is defined as mass of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case will be acetic acid, divided by mass of solution, and multiplied by <mathjax>#100#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(blue)("% m/m" = "mass of solute"/"mass of solution" xx 100)#</mathjax></p>
</blockquote>
<p>Now, the trick here is to realize that adding the acetic acid will <strong>Increase</strong> the mass of the solution. </p>
<p>If the mass of glacial acetic acid, which is <em>pure</em> acetic acid, is equal to <mathjax>#x#</mathjax> grams, then you can say that adding this amount to your sample of water will bring the <em>mass of the solution</em> to</p>
<blockquote>
<p><mathjax>#m_"sol" = (100.0 + x)" g"#</mathjax></p>
</blockquote>
<p>You need this solution to be <mathjax>#"4.0% m/m"#</mathjax> acetic acid, so you can say that</p>
<blockquote>
<p><mathjax>#"4.0%" = (xcolor(red)(cancel(color(black)("g"))))/((100.0+x)color(red)(cancel(color(black)("g")))) xx 100#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#4.0 * (100.0+x) = 100x#</mathjax></p>
<p><mathjax>#400 = 96x implies x = 400/96 = 4.1667#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> of the target solution, the answer will be </p>
<blockquote>
<p><mathjax>#m_"acetic acid" = color(green)("4.2 g")#</mathjax></p>
</blockquote>
<p>So, if you add <mathjax>#"4.2 g"#</mathjax> of glacial acetic acid to <mathjax>#"100.0 g"#</mathjax> of water, you will get a <mathjax>#"4.0% m/m"#</mathjax> acetic acid solution, also known as <em>vinegar</em>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"4.2 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by mass</a> is defined as mass of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case will be acetic acid, divided by mass of solution, and multiplied by <mathjax>#100#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(blue)("% m/m" = "mass of solute"/"mass of solution" xx 100)#</mathjax></p>
</blockquote>
<p>Now, the trick here is to realize that adding the acetic acid will <strong>Increase</strong> the mass of the solution. </p>
<p>If the mass of glacial acetic acid, which is <em>pure</em> acetic acid, is equal to <mathjax>#x#</mathjax> grams, then you can say that adding this amount to your sample of water will bring the <em>mass of the solution</em> to</p>
<blockquote>
<p><mathjax>#m_"sol" = (100.0 + x)" g"#</mathjax></p>
</blockquote>
<p>You need this solution to be <mathjax>#"4.0% m/m"#</mathjax> acetic acid, so you can say that</p>
<blockquote>
<p><mathjax>#"4.0%" = (xcolor(red)(cancel(color(black)("g"))))/((100.0+x)color(red)(cancel(color(black)("g")))) xx 100#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#4.0 * (100.0+x) = 100x#</mathjax></p>
<p><mathjax>#400 = 96x implies x = 400/96 = 4.1667#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> of the target solution, the answer will be </p>
<blockquote>
<p><mathjax>#m_"acetic acid" = color(green)("4.2 g")#</mathjax></p>
</blockquote>
<p>So, if you add <mathjax>#"4.2 g"#</mathjax> of glacial acetic acid to <mathjax>#"100.0 g"#</mathjax> of water, you will get a <mathjax>#"4.0% m/m"#</mathjax> acetic acid solution, also known as <em>vinegar</em>. </p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How many grams of acetic acid should be dissolved in 100.0 g of water to make a 4.0 percent (by mass) solution of vinegar?</h1>
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<div class="answer" id="203626" itemprop="suggestedAnswer" itemscope="" itemtype="http://schema.org/Answer">
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Stefan V.
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<span class="dateCreated" datetime="2015-12-25T19:56:53" itemprop="dateCreated">
Dec 25, 2015
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<div class="markdown"><p><mathjax>#"4.2 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As you know, <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration by mass</a> is defined as mass of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>, which in your case will be acetic acid, divided by mass of solution, and multiplied by <mathjax>#100#</mathjax>. </p>
<blockquote>
<p><mathjax>#color(blue)("% m/m" = "mass of solute"/"mass of solution" xx 100)#</mathjax></p>
</blockquote>
<p>Now, the trick here is to realize that adding the acetic acid will <strong>Increase</strong> the mass of the solution. </p>
<p>If the mass of glacial acetic acid, which is <em>pure</em> acetic acid, is equal to <mathjax>#x#</mathjax> grams, then you can say that adding this amount to your sample of water will bring the <em>mass of the solution</em> to</p>
<blockquote>
<p><mathjax>#m_"sol" = (100.0 + x)" g"#</mathjax></p>
</blockquote>
<p>You need this solution to be <mathjax>#"4.0% m/m"#</mathjax> acetic acid, so you can say that</p>
<blockquote>
<p><mathjax>#"4.0%" = (xcolor(red)(cancel(color(black)("g"))))/((100.0+x)color(red)(cancel(color(black)("g")))) xx 100#</mathjax></p>
</blockquote>
<p>This is equivalent to </p>
<blockquote>
<p><mathjax>#4.0 * (100.0+x) = 100x#</mathjax></p>
<p><mathjax>#400 = 96x implies x = 400/96 = 4.1667#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the number of sig figs you have for the <a href="http://socratic.org/chemistry/solutions-and-their-behavior/percent-concentration">percent concentration</a> of the target solution, the answer will be </p>
<blockquote>
<p><mathjax>#m_"acetic acid" = color(green)("4.2 g")#</mathjax></p>
</blockquote>
<p>So, if you add <mathjax>#"4.2 g"#</mathjax> of glacial acetic acid to <mathjax>#"100.0 g"#</mathjax> of water, you will get a <mathjax>#"4.0% m/m"#</mathjax> acetic acid solution, also known as <em>vinegar</em>. </p></div>
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</article> | How many grams of acetic acid should be dissolved in 100.0 g of water to make a 4.0 percent (by mass) solution of vinegar? | null |
3,238 | aa604bba-6ddd-11ea-80d1-ccda262736ce | https://socratic.org/questions/a-hot-air-balloon-has-a-volume-of-500-0-l-at-what-temperature-is-the-gas-if-ther | 23.87 K | start physical_unit 14 15 temperature k qc_end physical_unit 1 3 8 9 volume qc_end physical_unit 14 15 19 20 mole qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature [OF] the gas [IN] K"}] | [{"type":"physical unit","value":"23.87 K"}] | [{"type":"physical unit","value":"Volume [OF] the hot air balloon [=] \\pu{500.0 L}"},{"type":"physical unit","value":"Mole [OF] the gas [=] \\pu{25 moles}"},{"type":"other","value":"Standard pressure."}] | <h1 class="questionTitle" itemprop="name">A hot air balloon has a volume of 500.0 L. At what temperature is the gas if there are 25 moles present at standard pressure?</h1> | null | 23.87 K | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>V = 500 ml , n = 25 moles , R = 0.0821 L atm /K mol, T = ? , P = 0.98 atm</p>
<p>PV = nRT , rearranging it , we will get T = PV/nR</p>
<p>substituting the values, </p>
<p>T = 0.98 atm x 0.5 L / 25 mol x 0.0821 L atm /K mol</p>
<p>T = 0.49 K / 2.05 </p>
<p>T = 24 K</p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p>24 K</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>V = 500 ml , n = 25 moles , R = 0.0821 L atm /K mol, T = ? , P = 0.98 atm</p>
<p>PV = nRT , rearranging it , we will get T = PV/nR</p>
<p>substituting the values, </p>
<p>T = 0.98 atm x 0.5 L / 25 mol x 0.0821 L atm /K mol</p>
<p>T = 0.49 K / 2.05 </p>
<p>T = 24 K</p></div>
</div>
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<h1 class="questionTitle" itemprop="name">A hot air balloon has a volume of 500.0 L. At what temperature is the gas if there are 25 moles present at standard pressure?</h1>
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Manish Bhardwaj
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<div class="markdown"><p>24 K</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>V = 500 ml , n = 25 moles , R = 0.0821 L atm /K mol, T = ? , P = 0.98 atm</p>
<p>PV = nRT , rearranging it , we will get T = PV/nR</p>
<p>substituting the values, </p>
<p>T = 0.98 atm x 0.5 L / 25 mol x 0.0821 L atm /K mol</p>
<p>T = 0.49 K / 2.05 </p>
<p>T = 24 K</p></div>
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</article> | A hot air balloon has a volume of 500.0 L. At what temperature is the gas if there are 25 moles present at standard pressure? | null |
3,239 | aa47b0b4-6ddd-11ea-aaf8-ccda262736ce | https://socratic.org/questions/what-is-the-percent-composition-of-water-in-na-2so-4-10h-2o | 55.92% | start physical_unit 6 8 percent_composition none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Percent composition [OF] water in Na2SO4.10H2O"}] | [{"type":"physical unit","value":"55.92%"}] | [{"type":"chemical equation","value":"Na2SO4.10H2O"}] | <h1 class="questionTitle" itemprop="name">What is the percent composition of water in #"Na"_2"SO"_4 * 10"H"_2"O"#?</h1> | null | 55.92% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to figure out how many grams of water you have in <mathjax>#"100 g"#</mathjax> of sodium sulfate decahydrate, <mathjax>#"Na"_2"SO"_4 * 10"H"_2"O"#</mathjax>.</p>
<p>To do that, start with what you know. More specifically, start with the fact that <mathjax>#1#</mathjax> <strong>mole</strong> of sodium sulfate decahydrate contains </p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of sodium sulfate</em>, <mathjax>#1 xx "Na"_2"SO"_4#</mathjax></li>
<li><em><strong>ten moles</strong> of water</em>, <mathjax>#10 xx "H"_2"O"#</mathjax></li>
</ul>
</blockquote>
<p>You also know that sodium sulfate has a <strong>molar mass</strong> of <mathjax>#"142.04 g mol"^(-1)#</mathjax> and that water has a <strong>molar mass</strong> of <mathjax>#"18.015 g mol"^(-1)#</mathjax>. This means that <mathjax>#1#</mathjax> <strong>mole</strong> of sodium sulfate decahydrate contains </p>
<blockquote>
<ul>
<li><mathjax>#1 xx "142.04 g " ->#</mathjax> <em><strong>sodium sulfate</strong></em></li>
<li><mathjax>#10 xx "18.015 g " ->#</mathjax> <em><strong>water</strong></em></li>
</ul>
</blockquote>
<p>and has a <strong>total mass</strong> of </p>
<blockquote>
<p><mathjax>#"142.04 g"color(white)(.) + 10 xx "18.105 g" = "322.19 g"#</mathjax></p>
</blockquote>
<p>This means that <mathjax>#"100 g"#</mathjax> of sodium sulfate decahydrate will contain </p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g Na"_ 2"SO"_ 4))) * (10 xx "18.015 g H" _ 2"O")/(322.19 color(red)(cancel(color(black)("g Na"_ 2"SO"_ 4)))) = "55.9 g H"_2"O"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of water in this hydrate is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% H"_2"O" = 55.9%)))#</mathjax></p>
</blockquote>
<p>Keep in mind that the <mathjax>#%#</mathjax> sign means <em>out of</em> <mathjax>#"100 g"#</mathjax> <em>of hydrate</em>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Sodium sulfate decahydrate is <mathjax>#55.9 %"H"_2"O"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to figure out how many grams of water you have in <mathjax>#"100 g"#</mathjax> of sodium sulfate decahydrate, <mathjax>#"Na"_2"SO"_4 * 10"H"_2"O"#</mathjax>.</p>
<p>To do that, start with what you know. More specifically, start with the fact that <mathjax>#1#</mathjax> <strong>mole</strong> of sodium sulfate decahydrate contains </p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of sodium sulfate</em>, <mathjax>#1 xx "Na"_2"SO"_4#</mathjax></li>
<li><em><strong>ten moles</strong> of water</em>, <mathjax>#10 xx "H"_2"O"#</mathjax></li>
</ul>
</blockquote>
<p>You also know that sodium sulfate has a <strong>molar mass</strong> of <mathjax>#"142.04 g mol"^(-1)#</mathjax> and that water has a <strong>molar mass</strong> of <mathjax>#"18.015 g mol"^(-1)#</mathjax>. This means that <mathjax>#1#</mathjax> <strong>mole</strong> of sodium sulfate decahydrate contains </p>
<blockquote>
<ul>
<li><mathjax>#1 xx "142.04 g " ->#</mathjax> <em><strong>sodium sulfate</strong></em></li>
<li><mathjax>#10 xx "18.015 g " ->#</mathjax> <em><strong>water</strong></em></li>
</ul>
</blockquote>
<p>and has a <strong>total mass</strong> of </p>
<blockquote>
<p><mathjax>#"142.04 g"color(white)(.) + 10 xx "18.105 g" = "322.19 g"#</mathjax></p>
</blockquote>
<p>This means that <mathjax>#"100 g"#</mathjax> of sodium sulfate decahydrate will contain </p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g Na"_ 2"SO"_ 4))) * (10 xx "18.015 g H" _ 2"O")/(322.19 color(red)(cancel(color(black)("g Na"_ 2"SO"_ 4)))) = "55.9 g H"_2"O"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of water in this hydrate is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% H"_2"O" = 55.9%)))#</mathjax></p>
</blockquote>
<p>Keep in mind that the <mathjax>#%#</mathjax> sign means <em>out of</em> <mathjax>#"100 g"#</mathjax> <em>of hydrate</em>. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the percent composition of water in #"Na"_2"SO"_4 * 10"H"_2"O"#?</h1>
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Stefan V.
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<div class="markdown"><p>Sodium sulfate decahydrate is <mathjax>#55.9 %"H"_2"O"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your goal here is to figure out how many grams of water you have in <mathjax>#"100 g"#</mathjax> of sodium sulfate decahydrate, <mathjax>#"Na"_2"SO"_4 * 10"H"_2"O"#</mathjax>.</p>
<p>To do that, start with what you know. More specifically, start with the fact that <mathjax>#1#</mathjax> <strong>mole</strong> of sodium sulfate decahydrate contains </p>
<blockquote>
<ul>
<li><em><strong>one mole</strong> of sodium sulfate</em>, <mathjax>#1 xx "Na"_2"SO"_4#</mathjax></li>
<li><em><strong>ten moles</strong> of water</em>, <mathjax>#10 xx "H"_2"O"#</mathjax></li>
</ul>
</blockquote>
<p>You also know that sodium sulfate has a <strong>molar mass</strong> of <mathjax>#"142.04 g mol"^(-1)#</mathjax> and that water has a <strong>molar mass</strong> of <mathjax>#"18.015 g mol"^(-1)#</mathjax>. This means that <mathjax>#1#</mathjax> <strong>mole</strong> of sodium sulfate decahydrate contains </p>
<blockquote>
<ul>
<li><mathjax>#1 xx "142.04 g " ->#</mathjax> <em><strong>sodium sulfate</strong></em></li>
<li><mathjax>#10 xx "18.015 g " ->#</mathjax> <em><strong>water</strong></em></li>
</ul>
</blockquote>
<p>and has a <strong>total mass</strong> of </p>
<blockquote>
<p><mathjax>#"142.04 g"color(white)(.) + 10 xx "18.105 g" = "322.19 g"#</mathjax></p>
</blockquote>
<p>This means that <mathjax>#"100 g"#</mathjax> of sodium sulfate decahydrate will contain </p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g Na"_ 2"SO"_ 4))) * (10 xx "18.015 g H" _ 2"O")/(322.19 color(red)(cancel(color(black)("g Na"_ 2"SO"_ 4)))) = "55.9 g H"_2"O"#</mathjax></p>
</blockquote>
<p>Therefore, you can say that the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">percent composition</a></strong> of water in this hydrate is equal to </p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("% H"_2"O" = 55.9%)))#</mathjax></p>
</blockquote>
<p>Keep in mind that the <mathjax>#%#</mathjax> sign means <em>out of</em> <mathjax>#"100 g"#</mathjax> <em>of hydrate</em>. </p></div>
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</article> | What is the percent composition of water in #"Na"_2"SO"_4 * 10"H"_2"O"#? | null |
3,240 | abb1db80-6ddd-11ea-a8e5-ccda262736ce | https://socratic.org/questions/the-pressure-of-4-2-l-of-nitrogen-gas-in-a-flexible-container-is-decreased-to-on | 16.8 L | start physical_unit 6 7 volume l qc_end physical_unit 6 7 3 4 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume2 [OF] nitrogen gas [IN] L"}] | [{"type":"physical unit","value":"16.8 L"}] | [{"type":"physical unit","value":"Volume1 [OF] nitrogen gas [=] \\pu{4.2 L}"},{"type":"other","value":"The pressure of nitrogen gas in a flexible container is decreased to one-half its original pressure, and its absolute temperature is increased to double the original temperature."}] | <h1 class="questionTitle" itemprop="name">The pressure of 4.2 L of nitrogen gas in a flexible container is decreased to one-half its original pressure, and its absolute temperature is increased to double the original temperature. What is its volume now?</h1> | null | 16.8 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice for this problem will be the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure, volume, and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure, volume, and temperature of the gas at a final state</p>
<p>Now, even without doing any calculations, you can examine the above equation and predict what you expect the volume of the gas to be at the final state. </p>
<p>Notice that <em>pressure</em> and <em>volume</em> have an <strong>inverse relationship</strong> when <em>temperature</em> <strong>is kept constant</strong> - this is known as <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>. </p>
<p>This tells you that a <strong>decrease</strong> in pressure will cause an <strong>increase</strong> in volume. </p>
<p>On the other hand, <em>temperature</em> and <em>volume</em> have a <strong>direct relationship</strong> when <em>pressure</em> is <strong>kept constant</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a>. </p>
<p>This tells you that an <strong>increase</strong> in temperature will cause an <strong>increase</strong> in volume. </p>
<p>In your case, these two effects will actually <strong>combine</strong>. Increasing the temperature of the gas <strong>while</strong> decreasing its pressure will result in an <strong>increase</strong> in volume, regardless of the actual changes in temperature and pressure. </p>
<p>Mathematically, you can prove this by rearranging the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_2 = P_1/P_2 * T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>You know that</p>
<blockquote>
<ul>
<li><mathjax>#P_2 = P_1/2 ->#</mathjax> <em>the pressure is <strong>halved</strong></em></li>
<li><mathjax>#T_2 = 2 * T_1 ->#</mathjax> <em>the temperature is <strong>doubled</strong></em> </li>
</ul>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#V_2 = color(red)(cancel(color(black)(P_1)))/(1/2 * color(red)(cancel(color(black)(P_1)))) * (2 * color(red)(cancel(color(black)(T_1))))/color(red)(cancel(color(black)(T_1))) * V_1#</mathjax></p>
<p><mathjax>#V_2 = 2 * 2 * V_1 = 4 * V_1#</mathjax></p>
</blockquote>
<p>Indeed, both effects combine to cause an <strong>increase</strong> in volume. </p>
<p>The volume of the gas will thus be </p>
<blockquote>
<p><mathjax>#V_2 = 4 * "4.2 L" = "16.8 L"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#V_2 = color(green)("17 L")#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"17 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice for this problem will be the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure, volume, and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure, volume, and temperature of the gas at a final state</p>
<p>Now, even without doing any calculations, you can examine the above equation and predict what you expect the volume of the gas to be at the final state. </p>
<p>Notice that <em>pressure</em> and <em>volume</em> have an <strong>inverse relationship</strong> when <em>temperature</em> <strong>is kept constant</strong> - this is known as <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>. </p>
<p>This tells you that a <strong>decrease</strong> in pressure will cause an <strong>increase</strong> in volume. </p>
<p>On the other hand, <em>temperature</em> and <em>volume</em> have a <strong>direct relationship</strong> when <em>pressure</em> is <strong>kept constant</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a>. </p>
<p>This tells you that an <strong>increase</strong> in temperature will cause an <strong>increase</strong> in volume. </p>
<p>In your case, these two effects will actually <strong>combine</strong>. Increasing the temperature of the gas <strong>while</strong> decreasing its pressure will result in an <strong>increase</strong> in volume, regardless of the actual changes in temperature and pressure. </p>
<p>Mathematically, you can prove this by rearranging the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_2 = P_1/P_2 * T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>You know that</p>
<blockquote>
<ul>
<li><mathjax>#P_2 = P_1/2 ->#</mathjax> <em>the pressure is <strong>halved</strong></em></li>
<li><mathjax>#T_2 = 2 * T_1 ->#</mathjax> <em>the temperature is <strong>doubled</strong></em> </li>
</ul>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#V_2 = color(red)(cancel(color(black)(P_1)))/(1/2 * color(red)(cancel(color(black)(P_1)))) * (2 * color(red)(cancel(color(black)(T_1))))/color(red)(cancel(color(black)(T_1))) * V_1#</mathjax></p>
<p><mathjax>#V_2 = 2 * 2 * V_1 = 4 * V_1#</mathjax></p>
</blockquote>
<p>Indeed, both effects combine to cause an <strong>increase</strong> in volume. </p>
<p>The volume of the gas will thus be </p>
<blockquote>
<p><mathjax>#V_2 = 4 * "4.2 L" = "16.8 L"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
<blockquote>
<p><mathjax>#V_2 = color(green)("17 L")#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">The pressure of 4.2 L of nitrogen gas in a flexible container is decreased to one-half its original pressure, and its absolute temperature is increased to double the original temperature. What is its volume now?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2015-12-27T23:26:46" itemprop="dateCreated">
Dec 27, 2015
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<div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"17 L"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Your tool of choice for this problem will be the <a href="https://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "#</mathjax>, where</p>
</blockquote>
<p><mathjax>#P_1#</mathjax>, <mathjax>#V_1#</mathjax>, <mathjax>#T_1#</mathjax> - the pressure, volume, and temperature of the gas at an initial state<br/>
<mathjax>#P_2#</mathjax>, <mathjax>#V_2#</mathjax>, <mathjax>#T_2#</mathjax> - the pressure, volume, and temperature of the gas at a final state</p>
<p>Now, even without doing any calculations, you can examine the above equation and predict what you expect the volume of the gas to be at the final state. </p>
<p>Notice that <em>pressure</em> and <em>volume</em> have an <strong>inverse relationship</strong> when <em>temperature</em> <strong>is kept constant</strong> - this is known as <a href="https://socratic.org/chemistry/the-behavior-of-gases/boyle-s-law">Boyle's Law</a>. </p>
<p>This tells you that a <strong>decrease</strong> in pressure will cause an <strong>increase</strong> in volume. </p>
<p>On the other hand, <em>temperature</em> and <em>volume</em> have a <strong>direct relationship</strong> when <em>pressure</em> is <strong>kept constant</strong> - this is known as <a href="http://socratic.org/chemistry/the-behavior-of-gases/charles-law">Charles' Law</a>. </p>
<p>This tells you that an <strong>increase</strong> in temperature will cause an <strong>increase</strong> in volume. </p>
<p>In your case, these two effects will actually <strong>combine</strong>. Increasing the temperature of the gas <strong>while</strong> decreasing its pressure will result in an <strong>increase</strong> in volume, regardless of the actual changes in temperature and pressure. </p>
<p>Mathematically, you can prove this by rearranging the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">combined gas law</a> equation to solve for <mathjax>#V_2#</mathjax></p>
<blockquote>
<p><mathjax>#V_2 = P_1/P_2 * T_2/T_1 * V_1#</mathjax></p>
</blockquote>
<p>You know that</p>
<blockquote>
<ul>
<li><mathjax>#P_2 = P_1/2 ->#</mathjax> <em>the pressure is <strong>halved</strong></em></li>
<li><mathjax>#T_2 = 2 * T_1 ->#</mathjax> <em>the temperature is <strong>doubled</strong></em> </li>
</ul>
</blockquote>
<p>This means that you have </p>
<blockquote>
<p><mathjax>#V_2 = color(red)(cancel(color(black)(P_1)))/(1/2 * color(red)(cancel(color(black)(P_1)))) * (2 * color(red)(cancel(color(black)(T_1))))/color(red)(cancel(color(black)(T_1))) * V_1#</mathjax></p>
<p><mathjax>#V_2 = 2 * 2 * V_1 = 4 * V_1#</mathjax></p>
</blockquote>
<p>Indeed, both effects combine to cause an <strong>increase</strong> in volume. </p>
<p>The volume of the gas will thus be </p>
<blockquote>
<p><mathjax>#V_2 = 4 * "4.2 L" = "16.8 L"#</mathjax></p>
</blockquote>
<p>Rounded to two <a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a>, the answer will be </p>
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<p><mathjax>#V_2 = color(green)("17 L")#</mathjax></p>
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</article> | The pressure of 4.2 L of nitrogen gas in a flexible container is decreased to one-half its original pressure, and its absolute temperature is increased to double the original temperature. What is its volume now? | null |
3,241 | aae0f526-6ddd-11ea-81a2-ccda262736ce | https://socratic.org/questions/according-to-the-following-reaction-how-many-moles-of-sulfur-trioxide-will-be-fo | 0.97 moles | start physical_unit 9 10 mole mol qc_end physical_unit 22 23 19 20 mole qc_end c_other OTHER qc_end c_other OTHER qc_end chemical_equation 28 36 qc_end end | [{"type":"physical unit","value":"Mole [OF] sulfur trioxide [IN] moles"}] | [{"type":"physical unit","value":"0.97 moles"}] | [{"type":"physical unit","value":"Mole [OF] sulfur dioxide [=] \\pu{0.971 moles}"},{"type":"other","value":"Complete reaction."},{"type":"other","value":"Excess oxygen gas."},{"type":"chemical equation","value":"sulfer dioxide (g) + oxygen (g) -> sulfer trioxide"}] | <h1 class="questionTitle" itemprop="name">According to the following reaction, how many moles of sulfur trioxide will be formed upon the complete reaction of 0.971 moles of sulfur dioxide with excess oxygen gas? sulfer dioxide (g) + oxygen (g) #-># sulfer trioxide</h1> | null | 0.97 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need a stoichiometic equation:</p>
<p><mathjax>#SO_2(g) + 1/2O_2(g) rarr SO_3(g)#</mathjax></p>
<p>The equation unequivocally tells us that the reaction of <mathjax>#64*g#</mathjax> <mathjax>#SO_2(g)#</mathjax> with <mathjax>#16*g#</mathjax> <mathjax>#O_2(g)#</mathjax> gives <mathjax>#80*g#</mathjax> <mathjax>#SO_3(g)#</mathjax>. The given masses are the molar equivalents of each gas. Again, from the stoichometric equation, sulfur was the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> (because excess dioxygen gas was specified); there was a given molar quantity of sulfur (as its dioxide), and thus a given molar quantity of the trioxide. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Surely it will be <mathjax>#0.971*mol#</mathjax> of <mathjax>#SO_3#</mathjax>?</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need a stoichiometic equation:</p>
<p><mathjax>#SO_2(g) + 1/2O_2(g) rarr SO_3(g)#</mathjax></p>
<p>The equation unequivocally tells us that the reaction of <mathjax>#64*g#</mathjax> <mathjax>#SO_2(g)#</mathjax> with <mathjax>#16*g#</mathjax> <mathjax>#O_2(g)#</mathjax> gives <mathjax>#80*g#</mathjax> <mathjax>#SO_3(g)#</mathjax>. The given masses are the molar equivalents of each gas. Again, from the stoichometric equation, sulfur was the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> (because excess dioxygen gas was specified); there was a given molar quantity of sulfur (as its dioxide), and thus a given molar quantity of the trioxide. </p></div>
</div>
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<h1 class="questionTitle" itemprop="name">According to the following reaction, how many moles of sulfur trioxide will be formed upon the complete reaction of 0.971 moles of sulfur dioxide with excess oxygen gas? sulfer dioxide (g) + oxygen (g) #-># sulfer trioxide</h1>
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<div class="markdown"><p>Surely it will be <mathjax>#0.971*mol#</mathjax> of <mathjax>#SO_3#</mathjax>?</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need a stoichiometic equation:</p>
<p><mathjax>#SO_2(g) + 1/2O_2(g) rarr SO_3(g)#</mathjax></p>
<p>The equation unequivocally tells us that the reaction of <mathjax>#64*g#</mathjax> <mathjax>#SO_2(g)#</mathjax> with <mathjax>#16*g#</mathjax> <mathjax>#O_2(g)#</mathjax> gives <mathjax>#80*g#</mathjax> <mathjax>#SO_3(g)#</mathjax>. The given masses are the molar equivalents of each gas. Again, from the stoichometric equation, sulfur was the <a href="https://socratic.org/chemistry/stoichiometry/limiting-reagent">limiting reagent</a> (because excess dioxygen gas was specified); there was a given molar quantity of sulfur (as its dioxide), and thus a given molar quantity of the trioxide. </p></div>
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</article> | According to the following reaction, how many moles of sulfur trioxide will be formed upon the complete reaction of 0.971 moles of sulfur dioxide with excess oxygen gas? sulfer dioxide (g) + oxygen (g) #-># sulfer trioxide | null |
3,242 | ad20e2b0-6ddd-11ea-a2bc-ccda262736ce | https://socratic.org/questions/57f3b63511ef6b3558fc9bec | 1,2-C6H4(OC(=O)CH3)(CO2H) + KOH -> 1,2-C6H4(OC(=O)CH3)(CO2-K+) + H2O | start chemical_equation qc_end chemical_equation 8 8 qc_end substance 5 6 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"1,2-C6H4(OC(=O)CH3)(CO2H) + KOH -> 1,2-C6H4(OC(=O)CH3)(CO2-K+) + H2O"}] | [{"type":"chemical equation","value":"1,2-C6H4(OC(=O)CH3)(CO2H)"},{"type":"substance name","value":"Potassium hydroxide"}] | <h1 class="questionTitle" itemprop="name">What is the reaction between potassium hydroxide and #1,2−C_6H_4(OC(=O)CH_3)(CO_2H)#?</h1> | null | 1,2-C6H4(OC(=O)CH3)(CO2H) + KOH -> 1,2-C6H4(OC(=O)CH3)(CO2-K+) + H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1,2-C_6H_4(O(O=)C-CH_3)(CO_2H) + KOHrarr#</mathjax></p>
<p><mathjax>#1,2-C_6H_4(O(O=)C-CH_3)(CO_2^(-)K^+) + H_2O#</mathjax></p>
<p>Aspirin is in fact one of the most useful drugs known to modern medicine, and is currently widely used in low doses in order to prevent heart disease.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>This is a representation of aspirin, <mathjax>#1,2-C_6H_4(O(O=)C-CH_3)(CO_2H)#</mathjax>, which would undergo an acid base reaction with <mathjax>#KOH(aq)#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1,2-C_6H_4(O(O=)C-CH_3)(CO_2H) + KOHrarr#</mathjax></p>
<p><mathjax>#1,2-C_6H_4(O(O=)C-CH_3)(CO_2^(-)K^+) + H_2O#</mathjax></p>
<p>Aspirin is in fact one of the most useful drugs known to modern medicine, and is currently widely used in low doses in order to prevent heart disease.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the reaction between potassium hydroxide and #1,2−C_6H_4(OC(=O)CH_3)(CO_2H)#?</h1>
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<div class="markdown"><p>This is a representation of aspirin, <mathjax>#1,2-C_6H_4(O(O=)C-CH_3)(CO_2H)#</mathjax>, which would undergo an acid base reaction with <mathjax>#KOH(aq)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#1,2-C_6H_4(O(O=)C-CH_3)(CO_2H) + KOHrarr#</mathjax></p>
<p><mathjax>#1,2-C_6H_4(O(O=)C-CH_3)(CO_2^(-)K^+) + H_2O#</mathjax></p>
<p>Aspirin is in fact one of the most useful drugs known to modern medicine, and is currently widely used in low doses in order to prevent heart disease.</p></div>
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</article> | What is the reaction between potassium hydroxide and #1,2−C_6H_4(OC(=O)CH_3)(CO_2H)#? | null |
3,243 | a8a054e8-6ddd-11ea-a081-ccda262736ce | https://socratic.org/questions/a-second-order-reaction-is-40-completed-after-one-hour-what-is-the-rate-constant | 1.1 × 10^(-2) (M * min)^(-1) | start physical_unit 16 17 reaction_rate_constant (M_*_min)^(-1) qc_end physical_unit 0 3 5 5 rate_law qc_end physical_unit 0 3 8 9 time qc_end end | [{"type":"physical unit","value":"rate constant [OF] this reaction [IN] (M * min)^(-1)"}] | [{"type":"physical unit","value":"1.1 × 10^(-2) (M * min)^(-1)"}] | [{"type":"physical unit","value":"rate law [OF] a second order reaction [=] \\pu{40%}"},{"type":"physical unit","value":"completed time [OF] a second order reaction [=] \\pu{1 hour}"}] | <h1 class="questionTitle" itemprop="name">A second order reaction is 40% completed after one hour. What is the rate constant for this reaction?</h1> | null | 1.1 × 10^(-2) (M * min)^(-1) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/chemical-kinetics/rate-law">rate law</a> for a second-order reaction is</p>
<p><mathjax>#1/([A])=1/([A]_0)+kt#</mathjax></p>
<p>We can say that the current concentration <mathjax>#[A]#</mathjax> is <mathjax>#0.6#</mathjax> and the original concentration <mathjax>#[A]_0#</mathjax> is <mathjax>#1#</mathjax>. This is true because the reaction is <mathjax>#40%#</mathjax> complete, which means that <mathjax>#60%#</mathjax> still has to be reacted, and <mathjax>#60%#</mathjax> of an original concentration <mathjax>#1#</mathjax> is <mathjax>#0.6#</mathjax>.</p>
<p><mathjax>#t=60#</mathjax> since <mathjax>#60#</mathjax> minutes have elapsed.</p>
<p><mathjax>#1/(0.6)=1/1+k(60)#</mathjax></p>
<p><mathjax>#1.1xx10^-2=k#</mathjax></p>
<p>The rate constant is <mathjax>#1.1xx10^-2#</mathjax> <mathjax>#"M"^-1#</mathjax> <mathjax>#"min"^-1#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#1.1xx10^-2#</mathjax> <mathjax>#"M"^-1#</mathjax> <mathjax>#"min"^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/chemical-kinetics/rate-law">rate law</a> for a second-order reaction is</p>
<p><mathjax>#1/([A])=1/([A]_0)+kt#</mathjax></p>
<p>We can say that the current concentration <mathjax>#[A]#</mathjax> is <mathjax>#0.6#</mathjax> and the original concentration <mathjax>#[A]_0#</mathjax> is <mathjax>#1#</mathjax>. This is true because the reaction is <mathjax>#40%#</mathjax> complete, which means that <mathjax>#60%#</mathjax> still has to be reacted, and <mathjax>#60%#</mathjax> of an original concentration <mathjax>#1#</mathjax> is <mathjax>#0.6#</mathjax>.</p>
<p><mathjax>#t=60#</mathjax> since <mathjax>#60#</mathjax> minutes have elapsed.</p>
<p><mathjax>#1/(0.6)=1/1+k(60)#</mathjax></p>
<p><mathjax>#1.1xx10^-2=k#</mathjax></p>
<p>The rate constant is <mathjax>#1.1xx10^-2#</mathjax> <mathjax>#"M"^-1#</mathjax> <mathjax>#"min"^-1#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">A second order reaction is 40% completed after one hour. What is the rate constant for this reaction?</h1>
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mason m
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<span class="dateCreated" datetime="2015-12-19T16:51:51" itemprop="dateCreated">
Dec 19, 2015
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<div class="markdown"><p><mathjax>#1.1xx10^-2#</mathjax> <mathjax>#"M"^-1#</mathjax> <mathjax>#"min"^-1#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The <a href="http://socratic.org/chemistry/chemical-kinetics/rate-law">rate law</a> for a second-order reaction is</p>
<p><mathjax>#1/([A])=1/([A]_0)+kt#</mathjax></p>
<p>We can say that the current concentration <mathjax>#[A]#</mathjax> is <mathjax>#0.6#</mathjax> and the original concentration <mathjax>#[A]_0#</mathjax> is <mathjax>#1#</mathjax>. This is true because the reaction is <mathjax>#40%#</mathjax> complete, which means that <mathjax>#60%#</mathjax> still has to be reacted, and <mathjax>#60%#</mathjax> of an original concentration <mathjax>#1#</mathjax> is <mathjax>#0.6#</mathjax>.</p>
<p><mathjax>#t=60#</mathjax> since <mathjax>#60#</mathjax> minutes have elapsed.</p>
<p><mathjax>#1/(0.6)=1/1+k(60)#</mathjax></p>
<p><mathjax>#1.1xx10^-2=k#</mathjax></p>
<p>The rate constant is <mathjax>#1.1xx10^-2#</mathjax> <mathjax>#"M"^-1#</mathjax> <mathjax>#"min"^-1#</mathjax>.</p></div>
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</article> | A second order reaction is 40% completed after one hour. What is the rate constant for this reaction? | null |
3,244 | aaa4d14b-6ddd-11ea-9e99-ccda262736ce | https://socratic.org/questions/a-3-6-mole-sample-of-methane-gas-is-kept-in-a-1-50-liter-container-at-a-temperat | 7442.69 kPa | start physical_unit 25 26 pressure kpa qc_end physical_unit 5 6 1 2 mole qc_end physical_unit 13 13 11 12 volume qc_end physical_unit 13 13 18 19 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] the gas [IN] kPa"}] | [{"type":"physical unit","value":"7442.69 kPa"}] | [{"type":"physical unit","value":"Mole [OF] methane gas [=] \\pu{3.6 mole}"},{"type":"physical unit","value":"Volume [OF] container [=] \\pu{1.50 liter}"},{"type":"physical unit","value":"Temperature [OF] container [=] \\pu{100 ℃}"}] | <h1 class="questionTitle" itemprop="name">A 3.6 mole sample of methane gas is kept in a 1.50 liter container at a temperature of 100°C. What is the pressure of the gas?</h1> | null | 7442.69 kPa | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>, which states that,</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<blockquote>
<ul>
<li>
<p><mathjax>#P#</mathjax> is the pressure </p>
</li>
<li>
<p><mathjax>#V#</mathjax> is the volume in liters (for this case)</p>
</li>
<li>
<p><mathjax>#n#</mathjax> is the number of moles of the substance</p>
</li>
<li>
<p><mathjax>#R#</mathjax> is the ideal gas constant, which varies</p>
</li>
<li>
<p><mathjax>#T#</mathjax> is the temperature in Kelvin </p>
</li>
</ul>
</blockquote>
<p>Since we need to find pressure, we can rearrange the equation into:</p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p>Now, we need to convert the temperature into Kelvin. We know that <mathjax>#"K"=""^@"C"+273.15#</mathjax>, and so <mathjax>#100^@"C"=100+273.15=373 \ "K"#</mathjax>.</p>
<p>Since our temperature is in <mathjax>#"K"#</mathjax> and volume in liters, let's use <mathjax>#R=8.314 \ "L" \ "kPa" \ "K"^-1 \ "mol"^-1#</mathjax>. Taken from: <a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Gas_constant</a></p>
<p>And so, we find that the pressure is:</p>
<p><mathjax>#P=(3.6 \ "mol"*8.314 \ "L" \ "kPa" \ "K"^-1 \ "mol"^-1*373 \ "K")/(1.5 \ "L")#</mathjax></p>
<p><mathjax>#=(11164.0392color(red)cancelcolor(black)"mol"color(red)cancelcolor(black)"L" \ "kPa"color(red)cancelcolor(black)"K"^-1color(red)cancelcolor(black)"mol"^-1color(red)cancelcolor(black)"K")/(1.5color(red)cancelcolor(black)"L")#</mathjax></p>
<p><mathjax>#=7442.6928 \ "kPa"#</mathjax></p>
<p><mathjax>#~~7442.7 \ "kPa"#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>I get <mathjax>#7442.7 \ "kPa"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>, which states that,</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<blockquote>
<ul>
<li>
<p><mathjax>#P#</mathjax> is the pressure </p>
</li>
<li>
<p><mathjax>#V#</mathjax> is the volume in liters (for this case)</p>
</li>
<li>
<p><mathjax>#n#</mathjax> is the number of moles of the substance</p>
</li>
<li>
<p><mathjax>#R#</mathjax> is the ideal gas constant, which varies</p>
</li>
<li>
<p><mathjax>#T#</mathjax> is the temperature in Kelvin </p>
</li>
</ul>
</blockquote>
<p>Since we need to find pressure, we can rearrange the equation into:</p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p>Now, we need to convert the temperature into Kelvin. We know that <mathjax>#"K"=""^@"C"+273.15#</mathjax>, and so <mathjax>#100^@"C"=100+273.15=373 \ "K"#</mathjax>.</p>
<p>Since our temperature is in <mathjax>#"K"#</mathjax> and volume in liters, let's use <mathjax>#R=8.314 \ "L" \ "kPa" \ "K"^-1 \ "mol"^-1#</mathjax>. Taken from: <a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Gas_constant</a></p>
<p>And so, we find that the pressure is:</p>
<p><mathjax>#P=(3.6 \ "mol"*8.314 \ "L" \ "kPa" \ "K"^-1 \ "mol"^-1*373 \ "K")/(1.5 \ "L")#</mathjax></p>
<p><mathjax>#=(11164.0392color(red)cancelcolor(black)"mol"color(red)cancelcolor(black)"L" \ "kPa"color(red)cancelcolor(black)"K"^-1color(red)cancelcolor(black)"mol"^-1color(red)cancelcolor(black)"K")/(1.5color(red)cancelcolor(black)"L")#</mathjax></p>
<p><mathjax>#=7442.6928 \ "kPa"#</mathjax></p>
<p><mathjax>#~~7442.7 \ "kPa"#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A 3.6 mole sample of methane gas is kept in a 1.50 liter container at a temperature of 100°C. What is the pressure of the gas?</h1>
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<div class="markdown"><p>I get <mathjax>#7442.7 \ "kPa"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We use the <a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>, which states that,</p>
<p><mathjax>#PV=nRT#</mathjax></p>
<blockquote>
<ul>
<li>
<p><mathjax>#P#</mathjax> is the pressure </p>
</li>
<li>
<p><mathjax>#V#</mathjax> is the volume in liters (for this case)</p>
</li>
<li>
<p><mathjax>#n#</mathjax> is the number of moles of the substance</p>
</li>
<li>
<p><mathjax>#R#</mathjax> is the ideal gas constant, which varies</p>
</li>
<li>
<p><mathjax>#T#</mathjax> is the temperature in Kelvin </p>
</li>
</ul>
</blockquote>
<p>Since we need to find pressure, we can rearrange the equation into:</p>
<p><mathjax>#P=(nRT)/V#</mathjax></p>
<p>Now, we need to convert the temperature into Kelvin. We know that <mathjax>#"K"=""^@"C"+273.15#</mathjax>, and so <mathjax>#100^@"C"=100+273.15=373 \ "K"#</mathjax>.</p>
<p>Since our temperature is in <mathjax>#"K"#</mathjax> and volume in liters, let's use <mathjax>#R=8.314 \ "L" \ "kPa" \ "K"^-1 \ "mol"^-1#</mathjax>. Taken from: <a href="https://en.wikipedia.org/wiki/Gas_constant" rel="nofollow" target="_blank">https://en.wikipedia.org/wiki/Gas_constant</a></p>
<p>And so, we find that the pressure is:</p>
<p><mathjax>#P=(3.6 \ "mol"*8.314 \ "L" \ "kPa" \ "K"^-1 \ "mol"^-1*373 \ "K")/(1.5 \ "L")#</mathjax></p>
<p><mathjax>#=(11164.0392color(red)cancelcolor(black)"mol"color(red)cancelcolor(black)"L" \ "kPa"color(red)cancelcolor(black)"K"^-1color(red)cancelcolor(black)"mol"^-1color(red)cancelcolor(black)"K")/(1.5color(red)cancelcolor(black)"L")#</mathjax></p>
<p><mathjax>#=7442.6928 \ "kPa"#</mathjax></p>
<p><mathjax>#~~7442.7 \ "kPa"#</mathjax></p></div>
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</article> | A 3.6 mole sample of methane gas is kept in a 1.50 liter container at a temperature of 100°C. What is the pressure of the gas? | null |
3,245 | acb1d3d2-6ddd-11ea-bdec-ccda262736ce | https://socratic.org/questions/a-container-holds-6-4-moles-of-gas-hydrogen-gas-makes-up-25-of-the-total-moles-i | 0.31 atm | start physical_unit 7 7 partial_pressure atm qc_end physical_unit 6 6 3 4 mole qc_end physical_unit 6 6 24 25 total_pressure qc_end end | [{"type":"physical unit","value":"Partial pressure [OF] hydrogen [IN] atm"}] | [{"type":"physical unit","value":"0.31 atm"}] | [{"type":"physical unit","value":"Mole [OF] the gas [=] \\pu{6.4 moles}"},{"type":"physical unit","value":"Mole percent [OF] hydrogen gas in the container [=] \\pu{25%}"},{"type":"physical unit","value":"Total pressure [OF] the gas [=] \\pu{1.24 atm}"}] | <h1 class="questionTitle" itemprop="name">A container holds 6.4 moles of gas. Hydrogen gas makes up 25% of the total moles in the container if the total pressure is 1.24 atm. What is the partial pressure of hydrogen? </h1> | null | 0.31 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_"Total"=(n_"Total"RT)/V#</mathjax></p>
<p><mathjax>#=(n_(H_2)+n_"other gases")/V#</mathjax></p>
<p>But <mathjax>#n_(H_2)=25%xxn_"Total"#</mathjax>.</p>
<p>And thus <mathjax>#P_(H_2)=25%xxP_"Total"=25%xx1.24*atm=??#</mathjax></p>
<p>For more spray, see <a href="https://socratic.org/questions/how-would-you-define-and-explain-dalton-s-law-of-partial-pressure">here and links.</a> </p>
<p>This treatment relies on <mathjax>#"Dalton's Law of Partial Presssures"#</mathjax>, which states that in </p>
<p><mathjax>#"in a gaseous mixture, the partial pressure exerted"#</mathjax><br/>
<mathjax>#"by a gaseous component, is the same as the pressure it would"#</mathjax><br/>
<mathjax>#"exert if it ALONE occupied the container."#</mathjax></p>
<p><mathjax>#"The total Pressure is the sum of the individual partial pressures."#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#P_(H_2)=25%xxP_"Total"=0.31*atm#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_"Total"=(n_"Total"RT)/V#</mathjax></p>
<p><mathjax>#=(n_(H_2)+n_"other gases")/V#</mathjax></p>
<p>But <mathjax>#n_(H_2)=25%xxn_"Total"#</mathjax>.</p>
<p>And thus <mathjax>#P_(H_2)=25%xxP_"Total"=25%xx1.24*atm=??#</mathjax></p>
<p>For more spray, see <a href="https://socratic.org/questions/how-would-you-define-and-explain-dalton-s-law-of-partial-pressure">here and links.</a> </p>
<p>This treatment relies on <mathjax>#"Dalton's Law of Partial Presssures"#</mathjax>, which states that in </p>
<p><mathjax>#"in a gaseous mixture, the partial pressure exerted"#</mathjax><br/>
<mathjax>#"by a gaseous component, is the same as the pressure it would"#</mathjax><br/>
<mathjax>#"exert if it ALONE occupied the container."#</mathjax></p>
<p><mathjax>#"The total Pressure is the sum of the individual partial pressures."#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">A container holds 6.4 moles of gas. Hydrogen gas makes up 25% of the total moles in the container if the total pressure is 1.24 atm. What is the partial pressure of hydrogen? </h1>
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anor277
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Mar 27, 2017
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<div class="markdown"><p><mathjax>#P_(H_2)=25%xxP_"Total"=0.31*atm#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P_"Total"=(n_"Total"RT)/V#</mathjax></p>
<p><mathjax>#=(n_(H_2)+n_"other gases")/V#</mathjax></p>
<p>But <mathjax>#n_(H_2)=25%xxn_"Total"#</mathjax>.</p>
<p>And thus <mathjax>#P_(H_2)=25%xxP_"Total"=25%xx1.24*atm=??#</mathjax></p>
<p>For more spray, see <a href="https://socratic.org/questions/how-would-you-define-and-explain-dalton-s-law-of-partial-pressure">here and links.</a> </p>
<p>This treatment relies on <mathjax>#"Dalton's Law of Partial Presssures"#</mathjax>, which states that in </p>
<p><mathjax>#"in a gaseous mixture, the partial pressure exerted"#</mathjax><br/>
<mathjax>#"by a gaseous component, is the same as the pressure it would"#</mathjax><br/>
<mathjax>#"exert if it ALONE occupied the container."#</mathjax></p>
<p><mathjax>#"The total Pressure is the sum of the individual partial pressures."#</mathjax></p></div>
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</article> | A container holds 6.4 moles of gas. Hydrogen gas makes up 25% of the total moles in the container if the total pressure is 1.24 atm. What is the partial pressure of hydrogen? | null |
3,246 | a8956efa-6ddd-11ea-8580-ccda262736ce | https://socratic.org/questions/how-would-you-use-the-henderson-hasselbalch-equation-to-calculate-the-ph-of-a-bu-1 | 4.01 | start physical_unit 12 14 ph none qc_end physical_unit 22 22 17 18 molarity qc_end physical_unit 29 29 24 25 molarity qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"pH [OF] a buffer solution"}] | [{"type":"physical unit","value":"4.01"}] | [{"type":"physical unit","value":"Molarity [OF] HCO2H [=] \\pu{0.27 M}"},{"type":"physical unit","value":"Molarity [OF] HCOONa [=] \\pu{0.50 M}"},{"type":"other","value":"Henderson-Hasselbalch equation"}] | <h1 class="questionTitle" itemprop="name">How would you use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution that is 0.27 M in formic acid (#HCO_2H#) and 0.50 M in sodium formate (HCO_2Na)?</h1> | null | 4.01 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before I introduce Henderson-Hasselbalch's equation, we should identify the acid and base. <strong>Formate <mathjax>#(HCO_2^(-))#</mathjax> is the conjugate base</strong> of the acid, <strong>formic acid</strong>, <mathjax>#(HCO_2H)#</mathjax>. A conjugate base has one less proton <mathjax>#(H^+)#</mathjax>than the acid you started with.</p>
<p>Now, we can use this equation:<br/>
<img alt="www.chemteam.info" src="https://useruploads.socratic.org/voZKqTbLQTSan1ihzJEn_HH-equation3.gif"/> </p>
<p><strong>For this type of question we need a pKa, which is not given in your question. Therefore, I Googled the Ka and it is<br/>
(<mathjax>#1.8xx10^(-4))#</mathjax></strong></p>
<p>To obtain the pKa from the Ka value, all we have to do is take the <br/>
negative logarithm (-log) of the Ka:</p>
<p><mathjax>#pKa= -log(1.8xx10^(-4))#</mathjax> = 3.74</p>
<p>Thus, <strong>your pKa is 3.74</strong></p>
<p>Next, we can obtain the concentration of base and concentration of acid from the question.</p>
<p><strong>[<mathjax>#HCO_2^(-)#</mathjax>] = .50 M [<mathjax>#HCO_2H#</mathjax>] =.27 M</strong></p>
<p>We're not really concerned with the sodium cation that attached to the formate ion because it's a spectator ion and it has no effect on the buffer system. </p>
<p>Now, we have all of the information to determine the pH. Let's plug our values into the equation:</p>
<p><mathjax>#pH = 3.74 + log( [0.50]/[0.27])#</mathjax></p>
<p><mathjax>#pH = 4.01#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>That buffer system has a <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of 4.01</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before I introduce Henderson-Hasselbalch's equation, we should identify the acid and base. <strong>Formate <mathjax>#(HCO_2^(-))#</mathjax> is the conjugate base</strong> of the acid, <strong>formic acid</strong>, <mathjax>#(HCO_2H)#</mathjax>. A conjugate base has one less proton <mathjax>#(H^+)#</mathjax>than the acid you started with.</p>
<p>Now, we can use this equation:<br/>
<img alt="www.chemteam.info" src="https://useruploads.socratic.org/voZKqTbLQTSan1ihzJEn_HH-equation3.gif"/> </p>
<p><strong>For this type of question we need a pKa, which is not given in your question. Therefore, I Googled the Ka and it is<br/>
(<mathjax>#1.8xx10^(-4))#</mathjax></strong></p>
<p>To obtain the pKa from the Ka value, all we have to do is take the <br/>
negative logarithm (-log) of the Ka:</p>
<p><mathjax>#pKa= -log(1.8xx10^(-4))#</mathjax> = 3.74</p>
<p>Thus, <strong>your pKa is 3.74</strong></p>
<p>Next, we can obtain the concentration of base and concentration of acid from the question.</p>
<p><strong>[<mathjax>#HCO_2^(-)#</mathjax>] = .50 M [<mathjax>#HCO_2H#</mathjax>] =.27 M</strong></p>
<p>We're not really concerned with the sodium cation that attached to the formate ion because it's a spectator ion and it has no effect on the buffer system. </p>
<p>Now, we have all of the information to determine the pH. Let's plug our values into the equation:</p>
<p><mathjax>#pH = 3.74 + log( [0.50]/[0.27])#</mathjax></p>
<p><mathjax>#pH = 4.01#</mathjax></p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">How would you use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution that is 0.27 M in formic acid (#HCO_2H#) and 0.50 M in sodium formate (HCO_2Na)?</h1>
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<div class="markdown"><p>That buffer system has a <a href="https://socratic.org/chemistry/acids-and-bases/the-ph-concept">pH</a> of 4.01</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Before I introduce Henderson-Hasselbalch's equation, we should identify the acid and base. <strong>Formate <mathjax>#(HCO_2^(-))#</mathjax> is the conjugate base</strong> of the acid, <strong>formic acid</strong>, <mathjax>#(HCO_2H)#</mathjax>. A conjugate base has one less proton <mathjax>#(H^+)#</mathjax>than the acid you started with.</p>
<p>Now, we can use this equation:<br/>
<img alt="www.chemteam.info" src="https://useruploads.socratic.org/voZKqTbLQTSan1ihzJEn_HH-equation3.gif"/> </p>
<p><strong>For this type of question we need a pKa, which is not given in your question. Therefore, I Googled the Ka and it is<br/>
(<mathjax>#1.8xx10^(-4))#</mathjax></strong></p>
<p>To obtain the pKa from the Ka value, all we have to do is take the <br/>
negative logarithm (-log) of the Ka:</p>
<p><mathjax>#pKa= -log(1.8xx10^(-4))#</mathjax> = 3.74</p>
<p>Thus, <strong>your pKa is 3.74</strong></p>
<p>Next, we can obtain the concentration of base and concentration of acid from the question.</p>
<p><strong>[<mathjax>#HCO_2^(-)#</mathjax>] = .50 M [<mathjax>#HCO_2H#</mathjax>] =.27 M</strong></p>
<p>We're not really concerned with the sodium cation that attached to the formate ion because it's a spectator ion and it has no effect on the buffer system. </p>
<p>Now, we have all of the information to determine the pH. Let's plug our values into the equation:</p>
<p><mathjax>#pH = 3.74 + log( [0.50]/[0.27])#</mathjax></p>
<p><mathjax>#pH = 4.01#</mathjax></p></div>
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</article> | How would you use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution that is 0.27 M in formic acid (#HCO_2H#) and 0.50 M in sodium formate (HCO_2Na)? | null |
3,247 | a9dc6b5b-6ddd-11ea-aa00-ccda262736ce | https://socratic.org/questions/how-much-energy-is-released-when-a-40-0-g-sample-of-liquid-water-completely-free | 133.89 kJ | start physical_unit 9 12 heat_energy kj qc_end physical_unit 9 12 7 8 mass qc_end physical_unit 9 12 16 17 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Released energy [OF] liquid water sample [IN] kJ"}] | [{"type":"physical unit","value":"133.89 kJ"}] | [{"type":"physical unit","value":"Mass [OF] liquid water sample [=] \\pu{40.0 g}"},{"type":"physical unit","value":"Temperature [OF] liquid water sample [=] \\pu{0 ℃}"},{"type":"other","value":"Completely freeze."}] | <h1 class="questionTitle" itemprop="name">How much energy is released when a 40.0 g sample of liquid water completely freezes at 0 °C? </h1> | null | 133.89 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Calculate the energy released when 40 grams of liquid water freezes.<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/qRYs6zP3QGmCvdlo6QcG_Heat%20of%20Fusion%20Problem%2040%20grams%20Ice.jpg"/> </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>Q = 3,200 calories = 133,888 joules = 134 Kilojoules</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Calculate the energy released when 40 grams of liquid water freezes.<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/qRYs6zP3QGmCvdlo6QcG_Heat%20of%20Fusion%20Problem%2040%20grams%20Ice.jpg"/> </p></div>
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<div class="markdown"><p>Q = 3,200 calories = 133,888 joules = 134 Kilojoules</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Calculate the energy released when 40 grams of liquid water freezes.<br/>
<img alt="enter image source here" src="https://useruploads.socratic.org/qRYs6zP3QGmCvdlo6QcG_Heat%20of%20Fusion%20Problem%2040%20grams%20Ice.jpg"/> </p></div>
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</article> | How much energy is released when a 40.0 g sample of liquid water completely freezes at 0 °C? | null |
3,248 | a8934b6d-6ddd-11ea-885e-ccda262736ce | https://socratic.org/questions/when-heated-above-100-c-nitrosyl-chloride-noci-partly-decomposes-to-form-nitroge | NOCl(g) ->[\Delta] NO(g) + 1/2 Cl2 | start chemical_equation qc_end substance 12 13 qc_end substance 15 15 qc_end c_other OTHER qc_end chemical_equation 7 7 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this reaction"}] | [{"type":"chemical equation","value":"NOCl(g) ->[\\Delta] NO(g) + 1/2 Cl2"}] | [{"type":"substance name","value":"Nitrogen monoxide"},{"type":"substance name","value":"Chlorine"},{"type":"other","value":"Heat above 100 ℃."},{"type":"chemical equation","value":"NOCl"}] | <h1 class="questionTitle" itemprop="name">When heated above 100°C, nitrosyl chloride (#NOCl#) partly decomposes to form nitrogen monoxide and chlorine. What is the equation for this reaction?</h1> | null | NOCl(g) ->[\Delta] NO(g) + 1/2 Cl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is mass balanced? Is charge balanced? Formally, is this an example of a <mathjax>#"redox reaction"#</mathjax>? What is reduced; what is oxidized?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#NOCl(g) + Delta rarr NO(g) + 1/2Cl_2#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is mass balanced? Is charge balanced? Formally, is this an example of a <mathjax>#"redox reaction"#</mathjax>? What is reduced; what is oxidized?</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">When heated above 100°C, nitrosyl chloride (#NOCl#) partly decomposes to form nitrogen monoxide and chlorine. What is the equation for this reaction?</h1>
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<div class="markdown"><p><mathjax>#NOCl(g) + Delta rarr NO(g) + 1/2Cl_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Is mass balanced? Is charge balanced? Formally, is this an example of a <mathjax>#"redox reaction"#</mathjax>? What is reduced; what is oxidized?</p></div>
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</article> | When heated above 100°C, nitrosyl chloride (#NOCl#) partly decomposes to form nitrogen monoxide and chlorine. What is the equation for this reaction? | null |
3,249 | abda6a26-6ddd-11ea-96a8-ccda262736ce | https://socratic.org/questions/what-is-the-number-of-representative-particles-of-a-substance-present-in-1-mole- | 6.02 × 10^23 | start physical_unit 5 6 number none qc_end physical_unit 15 16 12 13 mole qc_end end | [{"type":"physical unit","value":"Number [OF] representative particles"}] | [{"type":"physical unit","value":"6.02 × 10^23"}] | [{"type":"physical unit","value":"Mole [OF] that substance [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">What is the number of representative particles of a substance present in 1 mole of that substance?</h1> | null | 6.02 × 10^23 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Yes, I know it's the <mathjax>#"Avogadro Number,"#</mathjax> <mathjax>#N_A#</mathjax>.</p>
<p>Why should we use such an absurdly large number? Well, it is the link between the micro world of atoms and molecules, and the macro world of grams, and litres.</p>
<p>See this <a href="https://socratic.org/questions/what-is-meant-by-the-term-one-mole-of-molecules?answerSuccess=1">older answer and links.</a> If you can assimilate this knowledge, you will have almost mastered A level and undergraduate chemistry. So see if you can spend a little time on the concept. We will be available to help. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>The <mathjax>#"Avocado number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Yes, I know it's the <mathjax>#"Avogadro Number,"#</mathjax> <mathjax>#N_A#</mathjax>.</p>
<p>Why should we use such an absurdly large number? Well, it is the link between the micro world of atoms and molecules, and the macro world of grams, and litres.</p>
<p>See this <a href="https://socratic.org/questions/what-is-meant-by-the-term-one-mole-of-molecules?answerSuccess=1">older answer and links.</a> If you can assimilate this knowledge, you will have almost mastered A level and undergraduate chemistry. So see if you can spend a little time on the concept. We will be available to help. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the number of representative particles of a substance present in 1 mole of that substance?</h1>
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<div class="markdown"><p>The <mathjax>#"Avocado number"#</mathjax>, <mathjax>#6.022xx10^23*mol^-1#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Yes, I know it's the <mathjax>#"Avogadro Number,"#</mathjax> <mathjax>#N_A#</mathjax>.</p>
<p>Why should we use such an absurdly large number? Well, it is the link between the micro world of atoms and molecules, and the macro world of grams, and litres.</p>
<p>See this <a href="https://socratic.org/questions/what-is-meant-by-the-term-one-mole-of-molecules?answerSuccess=1">older answer and links.</a> If you can assimilate this knowledge, you will have almost mastered A level and undergraduate chemistry. So see if you can spend a little time on the concept. We will be available to help. </p></div>
</div>
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</article> | What is the number of representative particles of a substance present in 1 mole of that substance? | null |
3,250 | a9af2a8c-6ddd-11ea-8dc3-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-oxygen-in-potassium-oxide-k-2o | -2 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 10 10 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] oxygen"}] | [{"type":"physical unit","value":"-2"}] | [{"type":"chemical equation","value":"K2O"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of oxygen in potassium oxide, #K_2O#?</h1> | null | -2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxidation number is the charge left on the atom of interest, when all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons are removed with the charge devolved to the more electronegative atom.</p>
<p>We have the ionic salt, <mathjax>#K_2O#</mathjax>. Clearly, oxygen is more electronegative than potassium, and so we have <mathjax>#2xxK^(+)+O^(2-)#</mathjax>.</p>
<p>The oxidation number of <mathjax>#O#</mathjax> in oxide anion is thus <mathjax>#-II#</mathjax>.</p>
<p>What about in <mathjax>#"potassium peroxide"#</mathjax>, <mathjax>#K_2O_2#</mathjax>. We still have <mathjax>#K^+#</mathjax>. What's the oxidation state of oxygen?</p></div>
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<div class="markdown"><p>We have <mathjax>#O^(2-)#</mathjax>, and thus <mathjax>#O(-II)#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxidation number is the charge left on the atom of interest, when all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons are removed with the charge devolved to the more electronegative atom.</p>
<p>We have the ionic salt, <mathjax>#K_2O#</mathjax>. Clearly, oxygen is more electronegative than potassium, and so we have <mathjax>#2xxK^(+)+O^(2-)#</mathjax>.</p>
<p>The oxidation number of <mathjax>#O#</mathjax> in oxide anion is thus <mathjax>#-II#</mathjax>.</p>
<p>What about in <mathjax>#"potassium peroxide"#</mathjax>, <mathjax>#K_2O_2#</mathjax>. We still have <mathjax>#K^+#</mathjax>. What's the oxidation state of oxygen?</p></div>
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<h1 class="questionTitle" itemprop="name">What is the oxidation number of oxygen in potassium oxide, #K_2O#?</h1>
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anor277
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<div class="markdown"><p>We have <mathjax>#O^(2-)#</mathjax>, and thus <mathjax>#O(-II)#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Oxidation number is the charge left on the atom of interest, when all the <a href="https://socratic.org/chemistry/bonding-basics/bonding">bonding</a> electrons are removed with the charge devolved to the more electronegative atom.</p>
<p>We have the ionic salt, <mathjax>#K_2O#</mathjax>. Clearly, oxygen is more electronegative than potassium, and so we have <mathjax>#2xxK^(+)+O^(2-)#</mathjax>.</p>
<p>The oxidation number of <mathjax>#O#</mathjax> in oxide anion is thus <mathjax>#-II#</mathjax>.</p>
<p>What about in <mathjax>#"potassium peroxide"#</mathjax>, <mathjax>#K_2O_2#</mathjax>. We still have <mathjax>#K^+#</mathjax>. What's the oxidation state of oxygen?</p></div>
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</article> | What is the oxidation number of oxygen in potassium oxide, #K_2O#? | null |
3,251 | a95bed9b-6ddd-11ea-a83b-ccda262736ce | https://socratic.org/questions/how-many-moles-of-naoh-are-present-in-19-0-ml-of-0-150-m-naoh | 2.85 × 10^(-3) moles | start physical_unit 4 4 mole mol qc_end physical_unit 4 4 8 9 volume qc_end physical_unit 4 4 11 12 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] NaOH [IN] moles"}] | [{"type":"physical unit","value":"2.85 × 10^(-3) moles"}] | [{"type":"physical unit","value":"Volume [OF] NaOH [=] \\pu{19.0 mL}"},{"type":"physical unit","value":"Molarity [OF] NaOH [=] \\pu{0.150 M}"}] | <h1 class="questionTitle" itemprop="name">How many moles of NaOH are present in 19.0 mL of 0.150 M NaOH?</h1> | null | 2.85 × 10^(-3) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as <em>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>, which in your case is sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, divided by <strong>liters</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>SImply put, a <mathjax>#"1-M"#</mathjax> solution will have <mathjax>#1#</mathjax> <strong>mole</strong> of solute dissolved in <mathjax>#1#</mathjax> <strong>liter</strong> of solution. </p>
<p>Now, you know that your solution has a <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of <mathjax>#"0.150 M"#</mathjax> and a volume of <mathjax>#"19.0 mL"#</mathjax>. </p>
<p>Since the volume is much <strong>smaller</strong> than <mathjax>#"1 L"#</mathjax>, you can expect to have <em>fewer moles</em> of sodium hydroxide in this sample than you would have had in a full liter of solution. </p>
<p>Convert the volume of the sample from mililiters to liters by using the conversion factor</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#19.0color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 19.0 * 10^(-3)"L"#</mathjax></p>
</blockquote>
<p>This means that the number of moles of solute you get in this sample will be equal to </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n = 0.150"moles"/color(red)(cancel(color(black)("L"))) * 19.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = color(green)("0.00285 moles NaOH")#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#"0.00285 moles"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as <em>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>, which in your case is sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, divided by <strong>liters</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>SImply put, a <mathjax>#"1-M"#</mathjax> solution will have <mathjax>#1#</mathjax> <strong>mole</strong> of solute dissolved in <mathjax>#1#</mathjax> <strong>liter</strong> of solution. </p>
<p>Now, you know that your solution has a <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of <mathjax>#"0.150 M"#</mathjax> and a volume of <mathjax>#"19.0 mL"#</mathjax>. </p>
<p>Since the volume is much <strong>smaller</strong> than <mathjax>#"1 L"#</mathjax>, you can expect to have <em>fewer moles</em> of sodium hydroxide in this sample than you would have had in a full liter of solution. </p>
<p>Convert the volume of the sample from mililiters to liters by using the conversion factor</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#19.0color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 19.0 * 10^(-3)"L"#</mathjax></p>
</blockquote>
<p>This means that the number of moles of solute you get in this sample will be equal to </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n = 0.150"moles"/color(red)(cancel(color(black)("L"))) * 19.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = color(green)("0.00285 moles NaOH")#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">How many moles of NaOH are present in 19.0 mL of 0.150 M NaOH?</h1>
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<div class="markdown"><p><mathjax>#"0.00285 moles"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> is defined as <em>moles of <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a></em>, which in your case is sodium hydroxide, <mathjax>#"NaOH"#</mathjax>, divided by <strong>liters</strong> of solution. </p>
<blockquote>
<p><mathjax>#color(blue)("molarity" = "moles of solute"/"liters of solution")#</mathjax></p>
</blockquote>
<p>SImply put, a <mathjax>#"1-M"#</mathjax> solution will have <mathjax>#1#</mathjax> <strong>mole</strong> of solute dissolved in <mathjax>#1#</mathjax> <strong>liter</strong> of solution. </p>
<p>Now, you know that your solution has a <a href="http://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a> of <mathjax>#"0.150 M"#</mathjax> and a volume of <mathjax>#"19.0 mL"#</mathjax>. </p>
<p>Since the volume is much <strong>smaller</strong> than <mathjax>#"1 L"#</mathjax>, you can expect to have <em>fewer moles</em> of sodium hydroxide in this sample than you would have had in a full liter of solution. </p>
<p>Convert the volume of the sample from mililiters to liters by using the conversion factor</p>
<blockquote>
<p><mathjax>#"1 L" = 10^3"mL"#</mathjax></p>
</blockquote>
<p>to get</p>
<blockquote>
<p><mathjax>#19.0color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 19.0 * 10^(-3)"L"#</mathjax></p>
</blockquote>
<p>This means that the number of moles of solute you get in this sample will be equal to </p>
<blockquote>
<p><mathjax>#color(blue)(c = n/V implies n = c * V)#</mathjax></p>
<p><mathjax>#n = 0.150"moles"/color(red)(cancel(color(black)("L"))) * 19.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = color(green)("0.00285 moles NaOH")#</mathjax></p>
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</article> | How many moles of NaOH are present in 19.0 mL of 0.150 M NaOH? | null |
3,252 | ac5cb668-6ddd-11ea-9fcc-ccda262736ce | https://socratic.org/questions/5900e95911ef6b57e2fb5698 | 2.5 g | start physical_unit 3 3 mass g qc_end physical_unit 12 12 8 9 volume qc_end physical_unit 20 21 15 16 molarity qc_end end | [{"type":"physical unit","value":"Mass [OF] solute [IN] g"}] | [{"type":"physical unit","value":"2.5 g"}] | [{"type":"physical unit","value":"Volume [OF] sodium sulfate solution [=] \\pu{675 mL}"},{"type":"physical unit","value":"Molarity [OF] sodium sulfate solution [=] \\pu{0.025 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What mass of solute is contained in a #675*mL# volume of solution that was #0.025*mol*L^-1# with respect to #"sodium sulfate"#? </h1> | null | 2.5 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax>, and thus, more symbolically, <mathjax>#C=n/V#</mathjax>. </p>
<p>As you progress in chemistry, you will use this relationship <mathjax>#"ad nauseum"#</mathjax>.</p>
<p>Given the formula:</p>
<p><mathjax>#"moles"="concentration"xx"volume".#</mathjax> And thus..........</p>
<p><mathjax>#n_(Na_2SO_4)=0.025*mol*cancel(L^-1)xx675*cancel(mL)xx10^-3cancel(L*mL^-1)#</mathjax></p>
<p><mathjax>#=0.016875*mol#</mathjax>, note that dimensionally, the units we got were <mathjax>#"moles"#</mathjax>. </p>
<p>And for the mass of this quantity, we perform the product, <mathjax>#"mass"xx"molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.016875*cancel(mol)xx142.04*g*cancel(mol^-1)#</mathjax><br/>
<mathjax>#=??g#</mathjax></p></div>
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<div class="answerSummary">
<div>
<div class="markdown"><p>We get a mass of under <mathjax>#2.5*g#</mathjax> of dissolved <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax>, and thus, more symbolically, <mathjax>#C=n/V#</mathjax>. </p>
<p>As you progress in chemistry, you will use this relationship <mathjax>#"ad nauseum"#</mathjax>.</p>
<p>Given the formula:</p>
<p><mathjax>#"moles"="concentration"xx"volume".#</mathjax> And thus..........</p>
<p><mathjax>#n_(Na_2SO_4)=0.025*mol*cancel(L^-1)xx675*cancel(mL)xx10^-3cancel(L*mL^-1)#</mathjax></p>
<p><mathjax>#=0.016875*mol#</mathjax>, note that dimensionally, the units we got were <mathjax>#"moles"#</mathjax>. </p>
<p>And for the mass of this quantity, we perform the product, <mathjax>#"mass"xx"molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.016875*cancel(mol)xx142.04*g*cancel(mol^-1)#</mathjax><br/>
<mathjax>#=??g#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What mass of solute is contained in a #675*mL# volume of solution that was #0.025*mol*L^-1# with respect to #"sodium sulfate"#? </h1>
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<div class="markdown"><p>We get a mass of under <mathjax>#2.5*g#</mathjax> of dissolved <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"="Moles of solute"/"Volume of solution"#</mathjax>, and thus, more symbolically, <mathjax>#C=n/V#</mathjax>. </p>
<p>As you progress in chemistry, you will use this relationship <mathjax>#"ad nauseum"#</mathjax>.</p>
<p>Given the formula:</p>
<p><mathjax>#"moles"="concentration"xx"volume".#</mathjax> And thus..........</p>
<p><mathjax>#n_(Na_2SO_4)=0.025*mol*cancel(L^-1)xx675*cancel(mL)xx10^-3cancel(L*mL^-1)#</mathjax></p>
<p><mathjax>#=0.016875*mol#</mathjax>, note that dimensionally, the units we got were <mathjax>#"moles"#</mathjax>. </p>
<p>And for the mass of this quantity, we perform the product, <mathjax>#"mass"xx"molar mass"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#0.016875*cancel(mol)xx142.04*g*cancel(mol^-1)#</mathjax><br/>
<mathjax>#=??g#</mathjax></p></div>
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</article> | What mass of solute is contained in a #675*mL# volume of solution that was #0.025*mol*L^-1# with respect to #"sodium sulfate"#? | null |
3,253 | a99c5d38-6ddd-11ea-9ae3-ccda262736ce | https://socratic.org/questions/what-is-the-molar-mass-of-nh4-2co3 | 96.09 g/mol | start physical_unit 6 6 molar_mass g/mol qc_end chemical_equation 6 6 qc_end end | [{"type":"physical unit","value":"Molar mass [OF] (NH4)2CO3 [IN] g/mol"}] | [{"type":"physical unit","value":"96.09 g/mol"}] | [{"type":"chemical equation","value":"(NH4)2CO3"}] | <h1 class="questionTitle" itemprop="name">What is the molar mass of (NH4)2CO3? </h1> | null | 96.09 g/mol | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You just need to first get the atomic weights of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> involved. You can easily get these from your periodic table. </p>
<p>If you are going to do this properly, please use the weight with <em>at least two decimal places</em> for accuracy (e.g. 15.99 g/mol).</p>
<p>Also, please take note that I will be using the unit g/mol for all the weights. Thus,</p>
<p><em>Step 1</em></p>
<p><mathjax>#"N = 14.01 g/mol"#</mathjax><br/>
<mathjax>#"H = 1.008 g/mol"#</mathjax><br/>
<mathjax>#"O = 16.00 g/mol"#</mathjax><br/>
<mathjax>#"C = 12.01 g/mol"#</mathjax></p>
<p>Since your compound is <mathjax>#(NH_4)_2CO_3#</mathjax>, you need to multiply the atomic weights by their subscripts. Therefore,</p>
<p><em>Step 2</em></p>
<p><mathjax>#"N = 14.01 g/mol × 2 =" bb"28.02 g/mol"#</mathjax><br/>
<mathjax>#"H = 1.008 g/mol × (4×2) =" bb"8.064 g/mol"#</mathjax> <br/>
<mathjax>#"O = 16.00 g/mol × 3 =" bb"48.00 g/mol"#</mathjax><br/>
<mathjax>#"C = 12.01 g/mol × 1 =" bb"12.00 g/mol"#</mathjax></p>
<p>To get the mass of the substance, we need to add all the weights from Step 2.</p>
<p><em>Step 3</em></p>
<p><mathjax>#"molar mass of" ("NH"_4)_2"CO"_3 = "(28.02 + 8.064 + 48.00 + 12.01) g/mol" = bb"96.09 g/mol"#</mathjax></p></div>
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<div class="markdown"><p>96.09 g/mol</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You just need to first get the atomic weights of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> involved. You can easily get these from your periodic table. </p>
<p>If you are going to do this properly, please use the weight with <em>at least two decimal places</em> for accuracy (e.g. 15.99 g/mol).</p>
<p>Also, please take note that I will be using the unit g/mol for all the weights. Thus,</p>
<p><em>Step 1</em></p>
<p><mathjax>#"N = 14.01 g/mol"#</mathjax><br/>
<mathjax>#"H = 1.008 g/mol"#</mathjax><br/>
<mathjax>#"O = 16.00 g/mol"#</mathjax><br/>
<mathjax>#"C = 12.01 g/mol"#</mathjax></p>
<p>Since your compound is <mathjax>#(NH_4)_2CO_3#</mathjax>, you need to multiply the atomic weights by their subscripts. Therefore,</p>
<p><em>Step 2</em></p>
<p><mathjax>#"N = 14.01 g/mol × 2 =" bb"28.02 g/mol"#</mathjax><br/>
<mathjax>#"H = 1.008 g/mol × (4×2) =" bb"8.064 g/mol"#</mathjax> <br/>
<mathjax>#"O = 16.00 g/mol × 3 =" bb"48.00 g/mol"#</mathjax><br/>
<mathjax>#"C = 12.01 g/mol × 1 =" bb"12.00 g/mol"#</mathjax></p>
<p>To get the mass of the substance, we need to add all the weights from Step 2.</p>
<p><em>Step 3</em></p>
<p><mathjax>#"molar mass of" ("NH"_4)_2"CO"_3 = "(28.02 + 8.064 + 48.00 + 12.01) g/mol" = bb"96.09 g/mol"#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molar mass of (NH4)2CO3? </h1>
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<div class="markdown"><p>96.09 g/mol</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You just need to first get the atomic weights of the <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a> involved. You can easily get these from your periodic table. </p>
<p>If you are going to do this properly, please use the weight with <em>at least two decimal places</em> for accuracy (e.g. 15.99 g/mol).</p>
<p>Also, please take note that I will be using the unit g/mol for all the weights. Thus,</p>
<p><em>Step 1</em></p>
<p><mathjax>#"N = 14.01 g/mol"#</mathjax><br/>
<mathjax>#"H = 1.008 g/mol"#</mathjax><br/>
<mathjax>#"O = 16.00 g/mol"#</mathjax><br/>
<mathjax>#"C = 12.01 g/mol"#</mathjax></p>
<p>Since your compound is <mathjax>#(NH_4)_2CO_3#</mathjax>, you need to multiply the atomic weights by their subscripts. Therefore,</p>
<p><em>Step 2</em></p>
<p><mathjax>#"N = 14.01 g/mol × 2 =" bb"28.02 g/mol"#</mathjax><br/>
<mathjax>#"H = 1.008 g/mol × (4×2) =" bb"8.064 g/mol"#</mathjax> <br/>
<mathjax>#"O = 16.00 g/mol × 3 =" bb"48.00 g/mol"#</mathjax><br/>
<mathjax>#"C = 12.01 g/mol × 1 =" bb"12.00 g/mol"#</mathjax></p>
<p>To get the mass of the substance, we need to add all the weights from Step 2.</p>
<p><em>Step 3</em></p>
<p><mathjax>#"molar mass of" ("NH"_4)_2"CO"_3 = "(28.02 + 8.064 + 48.00 + 12.01) g/mol" = bb"96.09 g/mol"#</mathjax></p></div>
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</article> | What is the molar mass of (NH4)2CO3? | null |
3,254 | a8de2b1c-6ddd-11ea-b912-ccda262736ce | https://socratic.org/questions/what-is-the-complete-combustion-of-propanol-c3h7oh-with-oxygen | 2 C3H7OH + 9 O2 -> 6 CO2 + 8 H2O | start chemical_equation qc_end chemical_equation 7 7 qc_end substance 9 9 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the complete combustion"}] | [{"type":"chemical equation","value":"2 C3H7OH + 9 O2 -> 6 CO2 + 8 H2O"}] | [{"type":"chemical equation","value":"C3H7OH"},{"type":"substance name","value":"Oxygen"}] | <h1 class="questionTitle" itemprop="name">What is the complete combustion of propanol (C3H7OH) with oxygen? </h1> | null | 2 C3H7OH + 9 O2 -> 6 CO2 + 8 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Alcohols react with oxygen in a similar way to hydrocarbons, where they produce carbon dioxide (<mathjax>#CO_2#</mathjax>) and water (<mathjax>#H_2O#</mathjax>).</p>
<p>Therefore, the reaction of the complete combustion of propanol is:</p>
<p><mathjax>#C_(color(blue)(3))H_(color(red)(7))OH + color(green)(9/2)O_2 -> color(blue)(3)CO_2 + color(red)(4)H_2O#</mathjax></p>
<p>Or the equation can be re-written as:</p>
<p><mathjax>#color(orange)(2)C_(color(blue)(3))H_(color(red)(7))OH + color(green)(9)O_2 -> color(blue)(6)CO_2 + color(red)(8)H_2O#</mathjax></p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#color(orange)(2)C_(color(blue)(3))H_(color(red)(7))OH + color(green)(9)O_2 -> color(blue)(6)CO_2 + color(red)(8)H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Alcohols react with oxygen in a similar way to hydrocarbons, where they produce carbon dioxide (<mathjax>#CO_2#</mathjax>) and water (<mathjax>#H_2O#</mathjax>).</p>
<p>Therefore, the reaction of the complete combustion of propanol is:</p>
<p><mathjax>#C_(color(blue)(3))H_(color(red)(7))OH + color(green)(9/2)O_2 -> color(blue)(3)CO_2 + color(red)(4)H_2O#</mathjax></p>
<p>Or the equation can be re-written as:</p>
<p><mathjax>#color(orange)(2)C_(color(blue)(3))H_(color(red)(7))OH + color(green)(9)O_2 -> color(blue)(6)CO_2 + color(red)(8)H_2O#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the complete combustion of propanol (C3H7OH) with oxygen? </h1>
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<div class="markdown"><p><mathjax>#color(orange)(2)C_(color(blue)(3))H_(color(red)(7))OH + color(green)(9)O_2 -> color(blue)(6)CO_2 + color(red)(8)H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Alcohols react with oxygen in a similar way to hydrocarbons, where they produce carbon dioxide (<mathjax>#CO_2#</mathjax>) and water (<mathjax>#H_2O#</mathjax>).</p>
<p>Therefore, the reaction of the complete combustion of propanol is:</p>
<p><mathjax>#C_(color(blue)(3))H_(color(red)(7))OH + color(green)(9/2)O_2 -> color(blue)(3)CO_2 + color(red)(4)H_2O#</mathjax></p>
<p>Or the equation can be re-written as:</p>
<p><mathjax>#color(orange)(2)C_(color(blue)(3))H_(color(red)(7))OH + color(green)(9)O_2 -> color(blue)(6)CO_2 + color(red)(8)H_2O#</mathjax></p></div>
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</article> | What is the complete combustion of propanol (C3H7OH) with oxygen? | null |
3,255 | aa5bde82-6ddd-11ea-8be8-ccda262736ce | https://socratic.org/questions/a-balloon-filled-with-oxygen-gas-occupies-a-volume-of-5-5-l-at-25-c-what-volume- | 6.9 L | start physical_unit 18 19 volume l qc_end physical_unit 4 5 10 11 volume qc_end physical_unit 4 5 13 14 temperature qc_end physical_unit 18 19 22 23 temperature qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] L"}] | [{"type":"physical unit","value":"6.9 L"}] | [{"type":"physical unit","value":"Volume1 [OF] oxygen gas [=] \\pu{5.5 L}"},{"type":"physical unit","value":"Temperature1 [OF] oxygen gas [=] \\pu{25 ℃}"},{"type":"physical unit","value":"Temperature2 [OF] the gas [=] \\pu{100.0 ℃}"}] | <h1 class="questionTitle" itemprop="name">A balloon filled with oxygen gas occupies a volume of 5.5 L at 25°C. What volume will the gas occupy at 100.0 °C?</h1> | null | 6.9 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To answer this question, we will need to assume the following:</p>
<ol>
<li>Oxygen is behaving ideally, so we can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax></li>
<li>During the process, the pressure is remaining constant.</li>
</ol>
<p>Since we are heating the ballon, the number of mole is not changing and therefore, <mathjax>#n#</mathjax> is constant too.</p>
<p>Thus <mathjax>#V=underbrace((nR)/P)_(color(blue)("constant k"))T=>V/T=k=>(V_1)/(T_1)=(V_2)/(T_2)#</mathjax></p>
<p><mathjax>#=>V_2=(V_1)/(T_1)xxT_2=(5.5L)/(298cancel(K))xx373cancel(K)=6.9L#</mathjax></p>
<p>Therefore, oxygen will occupy a volume of <mathjax>#6.9L#</mathjax> at <mathjax>#100^@C#</mathjax>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#V_2=6.9L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To answer this question, we will need to assume the following:</p>
<ol>
<li>Oxygen is behaving ideally, so we can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax></li>
<li>During the process, the pressure is remaining constant.</li>
</ol>
<p>Since we are heating the ballon, the number of mole is not changing and therefore, <mathjax>#n#</mathjax> is constant too.</p>
<p>Thus <mathjax>#V=underbrace((nR)/P)_(color(blue)("constant k"))T=>V/T=k=>(V_1)/(T_1)=(V_2)/(T_2)#</mathjax></p>
<p><mathjax>#=>V_2=(V_1)/(T_1)xxT_2=(5.5L)/(298cancel(K))xx373cancel(K)=6.9L#</mathjax></p>
<p>Therefore, oxygen will occupy a volume of <mathjax>#6.9L#</mathjax> at <mathjax>#100^@C#</mathjax>.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">A balloon filled with oxygen gas occupies a volume of 5.5 L at 25°C. What volume will the gas occupy at 100.0 °C?</h1>
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<div class="markdown"><p><mathjax>#V_2=6.9L#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To answer this question, we will need to assume the following:</p>
<ol>
<li>Oxygen is behaving ideally, so we can use the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>: <mathjax>#PV=nRT#</mathjax></li>
<li>During the process, the pressure is remaining constant.</li>
</ol>
<p>Since we are heating the ballon, the number of mole is not changing and therefore, <mathjax>#n#</mathjax> is constant too.</p>
<p>Thus <mathjax>#V=underbrace((nR)/P)_(color(blue)("constant k"))T=>V/T=k=>(V_1)/(T_1)=(V_2)/(T_2)#</mathjax></p>
<p><mathjax>#=>V_2=(V_1)/(T_1)xxT_2=(5.5L)/(298cancel(K))xx373cancel(K)=6.9L#</mathjax></p>
<p>Therefore, oxygen will occupy a volume of <mathjax>#6.9L#</mathjax> at <mathjax>#100^@C#</mathjax>.</p></div>
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</article> | A balloon filled with oxygen gas occupies a volume of 5.5 L at 25°C. What volume will the gas occupy at 100.0 °C? | null |
3,256 | aa7511d4-6ddd-11ea-86ca-ccda262736ce | https://socratic.org/questions/how-do-you-write-iron-iii-chloride-plus-aluminum-yields-aluminum-chloride-and-ir | FeCl3 + Al -> Fe + AlCl3 | start chemical_equation qc_end substance 4 6 qc_end substance 8 8 qc_end substance 10 11 qc_end substance 4 4 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the chemical equation"}] | [{"type":"chemical equation","value":"FeCl3 + Al -> Fe + AlCl3"}] | [{"type":"substance name","value":"Iron (III) chloride"},{"type":"substance name","value":"Aluminum"},{"type":"substance name","value":"Aluminum chloride"},{"type":"substance name","value":"Iron"}] | <h1 class="questionTitle" itemprop="name">How do you write "Iron (III) chloride plus aluminum yields aluminum chloride and iron" in a chemical equation?</h1> | null | FeCl3 + Al -> Fe + AlCl3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You don't need to note charges in the equation, if that's what you meant. The charge is implied because of the three Chlorides.</p>
<p>Hope this helps!</p></div>
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<div class="markdown"><p><mathjax>#FeCl_3 + Al rarr Fe + AlCl_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You don't need to note charges in the equation, if that's what you meant. The charge is implied because of the three Chlorides.</p>
<p>Hope this helps!</p></div>
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<h1 class="questionTitle" itemprop="name">How do you write "Iron (III) chloride plus aluminum yields aluminum chloride and iron" in a chemical equation?</h1>
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<div class="markdown"><p><mathjax>#FeCl_3 + Al rarr Fe + AlCl_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You don't need to note charges in the equation, if that's what you meant. The charge is implied because of the three Chlorides.</p>
<p>Hope this helps!</p></div>
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3,257 | ad210334-6ddd-11ea-8750-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-of-naoh-if-2-liters-of-the-solution-contains- | 2.00 M | start physical_unit 8 8 molarity mol/l qc_end physical_unit 6 6 10 11 volume qc_end physical_unit 8 8 16 17 mole qc_end end | [{"type":"physical unit","value":"Molarity [OF] NaOH solution [IN] M"}] | [{"type":"physical unit","value":"2.00 M"}] | [{"type":"physical unit","value":"Volume [OF] NaOH solution [=] \\pu{2 liters}"},{"type":"physical unit","value":"Mole [OF] NaOH [=] \\pu{4 moles}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution of #NaOH# if 2 liters of the solution contains 4 moles of #NaOH#? </h1> | null | 2.00 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(4*mol" Sodium hydroxide")/(2*L" of solution")#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#2*mol*L^-1#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#2#</mathjax> <mathjax>#mol*L^-1#</mathjax> with respect to <mathjax>#NaOH#</mathjax>.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(4*mol" Sodium hydroxide")/(2*L" of solution")#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#2*mol*L^-1#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution of #NaOH# if 2 liters of the solution contains 4 moles of #NaOH#? </h1>
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anor277
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<div class="markdown"><p><mathjax>#2#</mathjax> <mathjax>#mol*L^-1#</mathjax> with respect to <mathjax>#NaOH#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#(4*mol" Sodium hydroxide")/(2*L" of solution")#</mathjax></p>
<p><mathjax>#=#</mathjax> <mathjax>#2*mol*L^-1#</mathjax></p></div>
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</article> | What is the molarity of a solution of #NaOH# if 2 liters of the solution contains 4 moles of #NaOH#? | null |
3,258 | a95bcd6e-6ddd-11ea-90ef-ccda262736ce | https://socratic.org/questions/what-is-the-mass-in-grams-of-1-mol-of-li | 6.94 grams | start physical_unit 10 10 mass g qc_end physical_unit 10 10 7 8 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] Li [IN] grams"}] | [{"type":"physical unit","value":"6.94 grams"}] | [{"type":"physical unit","value":"Mole [OF] Li [=] \\pu{1 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the mass in grams of 1 mol of Li?</h1> | null | 6.94 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you look on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, there is a number at the bottom of each element, there is a number. That number is the atomic weight of an individual atom in u, or <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> unit. This number is also the weight in grams per mole (pretty convenient right).</p>
<p>Hope this helps! </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>6.941 g/mol</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you look on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, there is a number at the bottom of each element, there is a number. That number is the atomic weight of an individual atom in u, or <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> unit. This number is also the weight in grams per mole (pretty convenient right).</p>
<p>Hope this helps! </p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass in grams of 1 mol of Li?</h1>
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<div class="markdown"><p>6.941 g/mol</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>If you look on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, there is a number at the bottom of each element, there is a number. That number is the atomic weight of an individual atom in u, or <a href="http://socratic.org/chemistry/a-first-introduction-to-matter/atomic-mass-and-isotope-abundance">atomic mass</a> unit. This number is also the weight in grams per mole (pretty convenient right).</p>
<p>Hope this helps! </p></div>
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</article> | What is the mass in grams of 1 mol of Li? | null |
3,259 | aa6f6c38-6ddd-11ea-916f-ccda262736ce | https://socratic.org/questions/what-is-k-a-of-zncl-2 | 1.94 × 10^(-8) | start physical_unit 4 4 ka none qc_end chemical_equation 4 4 qc_end end | [{"type":"physical unit","value":"Ka [OF] ZnCl2"}] | [{"type":"physical unit","value":"1.94 × 10^(-8)"}] | [{"type":"chemical equation","value":"ZnCl2"}] | <h1 class="questionTitle" itemprop="name">What is #K_a# of #"ZnCl"_2# ?</h1> | <div class="questionDetailsContainer">
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<div class="markdown"><p>I know its small but what is the exact . Tell me the source too </p></div>
</h2>
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</div> | 1.94 × 10^(-8) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I was able to find a value for the acid dissociation constant of the <mathjax>#"Zn"^(2+)#</mathjax> cations here</p>
<p><a href="http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm" rel="nofollow" target="_blank">http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm</a></p>
<p>Of course, seeing how <mathjax>#"ZnCl"_2#</mathjax> is a <strong>soluble</strong> ionic compound and knowing that the chloride anion does not hydrolyze in water, the acid dissociation constant actually belongs to the <mathjax>#"Zn"^(2+)#</mathjax> cations.</p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>followed by</p>
<blockquote>
<p><mathjax>#"Zn"_ ((aq))^(2+) + 6"H"_ 2"O"_ ((l)) -> ["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+)#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+) + "H"_ 2"O"_ ((l)) rightleftharpoons ["Zn"("H"_ 2"O")_ 5("OH")]_ ((aq))^(+) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
</blockquote></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#2.5 * 10^(-10)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I was able to find a value for the acid dissociation constant of the <mathjax>#"Zn"^(2+)#</mathjax> cations here</p>
<p><a href="http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm" rel="nofollow" target="_blank">http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm</a></p>
<p>Of course, seeing how <mathjax>#"ZnCl"_2#</mathjax> is a <strong>soluble</strong> ionic compound and knowing that the chloride anion does not hydrolyze in water, the acid dissociation constant actually belongs to the <mathjax>#"Zn"^(2+)#</mathjax> cations.</p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>followed by</p>
<blockquote>
<p><mathjax>#"Zn"_ ((aq))^(2+) + 6"H"_ 2"O"_ ((l)) -> ["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+)#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+) + "H"_ 2"O"_ ((l)) rightleftharpoons ["Zn"("H"_ 2"O")_ 5("OH")]_ ((aq))^(+) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
</blockquote></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What is #K_a# of #"ZnCl"_2# ?</h1>
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Stefan V.
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Mar 12, 2017
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<div class="markdown"><p><mathjax>#2.5 * 10^(-10)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>I was able to find a value for the acid dissociation constant of the <mathjax>#"Zn"^(2+)#</mathjax> cations here</p>
<p><a href="http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm" rel="nofollow" target="_blank">http://bilbo.chm.uri.edu/CHM112/tables/KaTable.htm</a></p>
<p>Of course, seeing how <mathjax>#"ZnCl"_2#</mathjax> is a <strong>soluble</strong> ionic compound and knowing that the chloride anion does not hydrolyze in water, the acid dissociation constant actually belongs to the <mathjax>#"Zn"^(2+)#</mathjax> cations.</p>
<p>You will have</p>
<blockquote>
<p><mathjax>#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#</mathjax></p>
</blockquote>
<p>followed by</p>
<blockquote>
<p><mathjax>#"Zn"_ ((aq))^(2+) + 6"H"_ 2"O"_ ((l)) -> ["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+)#</mathjax></p>
</blockquote>
<p>and</p>
<blockquote>
<p><mathjax>#["Zn"("H"_ 2"O")_ 6]_ ((aq))^(2+) + "H"_ 2"O"_ ((l)) rightleftharpoons ["Zn"("H"_ 2"O")_ 5("OH")]_ ((aq))^(+) + "H"_ 3"O"_ ((aq))^(+)#</mathjax></p>
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Professor Sam
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<div class="markdown"><p>The Ka is nearly <mathjax>#1.936E-8#</mathjax> </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="http://www.webassign.net/labsgraceperiod/ucscgencheml1/lab_8/manual.pdf" rel="nofollow" target="_blank">http://www.webassign.net/labsgraceperiod/ucscgencheml1/lab_8/manual.pdf</a></p>
<p>In this website I checked that pH of 0.1M of ZnCl2 is 4.4<br/>
I tried to solve for Ka (<mathjax>#H^+#</mathjax> = <mathjax>#10^-4.4#</mathjax>)<br/>
<mathjax>#sqrt(0.1x) = H^+ or 10^-4.4#</mathjax><br/>
or<br/>
<mathjax>#0.1x = 0.000044^2#</mathjax></p>
<p><mathjax>#0.1x= 1.936E-9#</mathjax></p>
<p><mathjax>#x = Ka = 1.936E-8 = \frac{1.936E-9}{0.1}#</mathjax></p>
<p>You can solve this by ICE table. Answer would be the same</p>
<p>The Ka is <mathjax>#1.936E-8..#</mathjax> and If you check in this website what is the Ka of Zn(H2O)6 it would be the same . Just they have rounded it off to simpler numbers. These dont matter much</p></div>
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</article> | What is #K_a# of #"ZnCl"_2# ? |
I know its small but what is the exact . Tell me the source too
|
3,260 | a85c3b74-6ddd-11ea-8953-ccda262736ce | https://socratic.org/questions/hydrogen-reacts-with-771-g-of-carbon-to-form-90-g-of-a-compound-what-is-the-mass | 14.3% | start physical_unit 20 23 mass_percent none qc_end physical_unit 6 6 3 4 mass qc_end physical_unit 22 23 9 10 mass qc_end end | [{"type":"physical unit","value":"Mass percent [OF] hydrogen in the compound"}] | [{"type":"physical unit","value":"14.3%"}] | [{"type":"physical unit","value":"Mass [OF] carbon [=] \\pu{0.771 g}"},{"type":"physical unit","value":"Mass [OF] the compound [=] \\pu{0.90 g}"}] | <h1 class="questionTitle" itemprop="name">Hydrogen reacts with .771 g of carbon to form .90 g of a compound. What is the mass percent of hydrogen in the compound?</h1> | null | 14.3% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">mass percent</a></strong> of hydrogen in this compound, you must determine how many grams of hydrogen you'd get in <mathjax>#"100 g"#</mathjax> of compound. </p>
<p>In your case, you know that an unknown mass of hydrogen reacts with <mathjax>#"0.771 g"#</mathjax> of carbon to form <mathjax>#"0.90 g"#</mathjax> of <em>hydrocarbon</em>, which is a compound that contains only carbon and hydrogen. </p>
<p>Use the total mass of the hydrocarbon to determine how many grams of hydrogen reacted with the carbon </p>
<blockquote>
<p><mathjax>#m_"hydrocarbon" = m_(C) + m_(H)#</mathjax></p>
<p><mathjax>#m_(H) = "0.90 g" - "0.771 g" = "0.129 g"#</mathjax></p>
</blockquote>
<p>Now, if <mathjax>#"0.90 g"#</mathjax> of this compound contain <mathjax>#"0.129 g"#</mathjax> of hydrogen, it follows that <mathjax>#"100 g"#</mathjax> of this compound will contain </p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g hydrocarbon"))) * "0.129 g H"/(0.90color(red)(cancel(color(black)("g hydrocarbon")))) = "14.33 g H"#</mathjax></p>
</blockquote>
<p>So, if <mathjax>#"100 g"#</mathjax> of this compound contain <mathjax>#"14.33 g"#</mathjax> of hydrogen, it follows that the mass percent of hydrogen is </p>
<blockquote>
<p><mathjax>#"mass percent H" = color(green)(|bar(ul(color(white)(a/a)"14.3%"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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<div class="markdown"><p><mathjax>#14.3%#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">mass percent</a></strong> of hydrogen in this compound, you must determine how many grams of hydrogen you'd get in <mathjax>#"100 g"#</mathjax> of compound. </p>
<p>In your case, you know that an unknown mass of hydrogen reacts with <mathjax>#"0.771 g"#</mathjax> of carbon to form <mathjax>#"0.90 g"#</mathjax> of <em>hydrocarbon</em>, which is a compound that contains only carbon and hydrogen. </p>
<p>Use the total mass of the hydrocarbon to determine how many grams of hydrogen reacted with the carbon </p>
<blockquote>
<p><mathjax>#m_"hydrocarbon" = m_(C) + m_(H)#</mathjax></p>
<p><mathjax>#m_(H) = "0.90 g" - "0.771 g" = "0.129 g"#</mathjax></p>
</blockquote>
<p>Now, if <mathjax>#"0.90 g"#</mathjax> of this compound contain <mathjax>#"0.129 g"#</mathjax> of hydrogen, it follows that <mathjax>#"100 g"#</mathjax> of this compound will contain </p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g hydrocarbon"))) * "0.129 g H"/(0.90color(red)(cancel(color(black)("g hydrocarbon")))) = "14.33 g H"#</mathjax></p>
</blockquote>
<p>So, if <mathjax>#"100 g"#</mathjax> of this compound contain <mathjax>#"14.33 g"#</mathjax> of hydrogen, it follows that the mass percent of hydrogen is </p>
<blockquote>
<p><mathjax>#"mass percent H" = color(green)(|bar(ul(color(white)(a/a)"14.3%"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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<h1 class="questionTitle" itemprop="name">Hydrogen reacts with .771 g of carbon to form .90 g of a compound. What is the mass percent of hydrogen in the compound?</h1>
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<div class="markdown"><p><mathjax>#14.3%#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find the <strong><a href="https://socratic.org/chemistry/the-mole-concept/percent-composition">mass percent</a></strong> of hydrogen in this compound, you must determine how many grams of hydrogen you'd get in <mathjax>#"100 g"#</mathjax> of compound. </p>
<p>In your case, you know that an unknown mass of hydrogen reacts with <mathjax>#"0.771 g"#</mathjax> of carbon to form <mathjax>#"0.90 g"#</mathjax> of <em>hydrocarbon</em>, which is a compound that contains only carbon and hydrogen. </p>
<p>Use the total mass of the hydrocarbon to determine how many grams of hydrogen reacted with the carbon </p>
<blockquote>
<p><mathjax>#m_"hydrocarbon" = m_(C) + m_(H)#</mathjax></p>
<p><mathjax>#m_(H) = "0.90 g" - "0.771 g" = "0.129 g"#</mathjax></p>
</blockquote>
<p>Now, if <mathjax>#"0.90 g"#</mathjax> of this compound contain <mathjax>#"0.129 g"#</mathjax> of hydrogen, it follows that <mathjax>#"100 g"#</mathjax> of this compound will contain </p>
<blockquote>
<p><mathjax>#100 color(red)(cancel(color(black)("g hydrocarbon"))) * "0.129 g H"/(0.90color(red)(cancel(color(black)("g hydrocarbon")))) = "14.33 g H"#</mathjax></p>
</blockquote>
<p>So, if <mathjax>#"100 g"#</mathjax> of this compound contain <mathjax>#"14.33 g"#</mathjax> of hydrogen, it follows that the mass percent of hydrogen is </p>
<blockquote>
<p><mathjax>#"mass percent H" = color(green)(|bar(ul(color(white)(a/a)"14.3%"color(white)(a/a)|)))#</mathjax></p>
</blockquote>
<p>I'll leave the answer rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>. </p></div>
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</article> | Hydrogen reacts with .771 g of carbon to form .90 g of a compound. What is the mass percent of hydrogen in the compound? | null |
3,261 | a9ebdc1c-6ddd-11ea-a3b6-ccda262736ce | https://socratic.org/questions/580f8cd211ef6b568c8f5b18 | 0.5 molar | start physical_unit 4 5 mole mol/l qc_end physical_unit 16 16 10 11 volume qc_end physical_unit 16 16 14 15 molarity qc_end end | [{"type":"physical unit","value":"Mole [OF] magnesium ion [IN] molar"}] | [{"type":"physical unit","value":"0.5 molar"}] | [{"type":"physical unit","value":"Volume [OF] MgSO4(aq) [=] \\pu{1000 mL}"},{"type":"physical unit","value":"Molarity [OF] MgSO4(aq) [=] \\pu{0.50 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What molar quantity of magnesium ion is present in a #1000*mL# volume of #0.50*mol*L^-1# #MgSO_4(aq)#?</h1> | null | 0.5 molar | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You specify a <mathjax>#0.5*mol*L^-1#</mathjax> solution of <mathjax>#MgSO_4(aq)#</mathjax>.</p>
<p>Since <mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>, </p>
<p><mathjax>#"Moles of solute"="Concentration"xx"Volume of solution"#</mathjax>.</p>
<p>If we have a <mathjax>#1000*mL#</mathjax> volume, then.......</p>
<p><mathjax>#"Moles of solute"#</mathjax> </p>
<p><mathjax>#0.5*mol*cancel(L^-1)xx1000*cancel(mL)xx10^-3cancel(L)*cancel(mL^-1)=0.5*mol " with respect to " Mg^(2+)(aq)#</mathjax>.</p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>You have <mathjax>#0.5#</mathjax> <mathjax>#"mol"#</mathjax> of <mathjax>#Mg^(2+)(aq)#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You specify a <mathjax>#0.5*mol*L^-1#</mathjax> solution of <mathjax>#MgSO_4(aq)#</mathjax>.</p>
<p>Since <mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>, </p>
<p><mathjax>#"Moles of solute"="Concentration"xx"Volume of solution"#</mathjax>.</p>
<p>If we have a <mathjax>#1000*mL#</mathjax> volume, then.......</p>
<p><mathjax>#"Moles of solute"#</mathjax> </p>
<p><mathjax>#0.5*mol*cancel(L^-1)xx1000*cancel(mL)xx10^-3cancel(L)*cancel(mL^-1)=0.5*mol " with respect to " Mg^(2+)(aq)#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What molar quantity of magnesium ion is present in a #1000*mL# volume of #0.50*mol*L^-1# #MgSO_4(aq)#?</h1>
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anor277
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Stefan V.
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<div class="markdown"><p>You have <mathjax>#0.5#</mathjax> <mathjax>#"mol"#</mathjax> of <mathjax>#Mg^(2+)(aq)#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You specify a <mathjax>#0.5*mol*L^-1#</mathjax> solution of <mathjax>#MgSO_4(aq)#</mathjax>.</p>
<p>Since <mathjax>#"Concentration"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#"Moles of solute"/"Volume of solution"#</mathjax>, </p>
<p><mathjax>#"Moles of solute"="Concentration"xx"Volume of solution"#</mathjax>.</p>
<p>If we have a <mathjax>#1000*mL#</mathjax> volume, then.......</p>
<p><mathjax>#"Moles of solute"#</mathjax> </p>
<p><mathjax>#0.5*mol*cancel(L^-1)xx1000*cancel(mL)xx10^-3cancel(L)*cancel(mL^-1)=0.5*mol " with respect to " Mg^(2+)(aq)#</mathjax>.</p></div>
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Stefan V.
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Dec 6, 2016
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<div class="markdown"><p>Here's what I got. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>It's a good idea to keep in mind that <strong>equivalents</strong> have various definitions depending on the type of reaction you're working with. </p>
<p>In this context, the number of equivalents of a given ion is given by the <strong>number of moles</strong> of opposite unitary charge ions that can combine with <mathjax>#1#</mathjax> <strong>mole</strong> of the given ion. </p>
<p>For example, <mathjax>#"Mg"^(2+)#</mathjax> has a charge of <mathjax>#2+#</mathjax>, which means that it can combine with <mathjax>#2#</mathjax> ions that carry a <mathjax>#1-#</mathjax> charge. </p>
<p>So if <mathjax>#1#</mathjax> <strong>mole</strong> of <mathjax>#"Mg"^(2+)#</mathjax> can combine with <mathjax>#2#</mathjax> <strong>moles</strong> of <mathjax>#1-#</mathjax> ions, it follows that you get <mathjax>#2#</mathjax> <strong>equivalents</strong> of <mathjax>#"Mg"^(2+)#</mathjax> for every <mathjax>#1#</mathjax> <strong>mole</strong> of <mathjax>#"Mg"^(2+)#</mathjax> added to the solution. </p>
<p>In this particular case, your solution will contain</p>
<blockquote>
<p><mathjax>#1000 color(red)(cancel(color(black)("mL solution"))) * (0.5color(red)(cancel(color(black)("moles MgSO"_4))))/(1000color(red)(cancel(color(black)("mL solution")))) * (1 color(red)(cancel(color(black)("mole Mg"^(2+)))))/(1color(red)(cancel(color(black)("mole MgSO"_4)))) * "2 eq Mg"^(2+)/(1color(red)(cancel(color(black)("mole Mg"^(2+))#</mathjax></p>
<blockquote>
<blockquote>
<p><mathjax># = color(darkgreen)(ul(color(black)("1 eq Mg"^(2+))))#</mathjax></p>
</blockquote>
</blockquote>
</blockquote>
<p>Remember, this depends on what exactly you're about to do with your solution. If you're going to mix this solution with a solution that contains <mathjax>#1-#</mathjax> ions, then you can talk about it having <mathjax>#"2 eq"#</mathjax> of <mathjax>#"Mg"^(2+)#</mathjax>.</p>
<p>In and of itself, the solution simply contains <mathjax>#0.5#</mathjax> <strong>moles</strong> of <mathjax>#"Mg"^(2+)#</mathjax> and <mathjax>#0.5#</mathjax> <strong>moles</strong> of <mathjax>#"SO"_4^(2-)#</mathjax>.</p>
<p>The equivalents come into play when dealing with reactions. Equivalents are defined as an ion's <em>capacity</em> to combine with opposite unitary charge ions, so you need an actual reaction in order for them to make sense at all. </p></div>
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</article> | What molar quantity of magnesium ion is present in a #1000*mL# volume of #0.50*mol*L^-1# #MgSO_4(aq)#? | null |
3,262 | ab0bc88a-6ddd-11ea-afaf-ccda262736ce | https://socratic.org/questions/58d4162fb72cff3d921dcbd6 | 2.37 | start physical_unit 5 6 ph none qc_end physical_unit 5 6 11 12 [nh4+] qc_end end | [{"type":"physical unit","value":"pH [OF] aqueous solution"}] | [{"type":"physical unit","value":"2.37"}] | [{"type":"physical unit","value":"[NH4+] [OF] aqueous solution [=] \\pu{1.0 mol/L}"}] | <h1 class="questionTitle" itemprop="name">What is #pH# of an aqueous solution for which #[NH_4^+]=1.0*mol*L^-1#?</h1> | null | 2.37 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax>.</p>
<p>And we know that ammonium salts are weakly acidic, and undergo the acid-base reaction:</p>
<p><mathjax>#NH_4^(+) + H_2O(l) rightleftharpoonsNH_3(aq) + H_3O^+#</mathjax></p>
<p>For ammonium ion, <mathjax>#K_a=1.80xx10^-5#</mathjax></p>
<p>And if there were a <mathjax>#1.0*mol*L^-1#</mathjax> solution of ammonium ion, we would solve the equilibrium expression in the usual way.</p>
<p>i.e. <mathjax>#K_a=1.80xx10^-5=([H_3O^+][NH_3(aq)])/([NH_4^(+)])#</mathjax></p>
<p>And if <mathjax>#x*mol*L^-1#</mathjax> ammonium ion ionizes, then.........</p>
<p><mathjax>#K_a=1.80xx10^-5=x^2/((1.0-x)*mol*L^-1)#</mathjax>, a quadratic in <mathjax>#x#</mathjax>, which may be simplified given the reasonable assumption that <mathjax>#1.0">>"x#</mathjax>, and thus <mathjax>#1.0-x~=1.0#</mathjax>.</p>
<p>And thus <mathjax>#x~=sqrt(1.80^-5)#</mathjax>,</p>
<p>so <mathjax>#x_1=4.2xx10^-3#</mathjax>, and recycling this approximation of <mathjax>#x#</mathjax> back into the expression:</p>
<p><mathjax>#x_2=4.2xx10^-3#</mathjax>, and since <mathjax>#x#</mathjax> has converged, we may accept this value. </p>
<p>Thus <mathjax>#[H_3O^+]=4.2xx10^-3*mol*L^-1#</mathjax>, and <mathjax>#pH=-log_10[H_3O^+]=-log_10(4.2xx10^-3)=2.37#</mathjax>.</p></div>
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<div class="markdown"><p>It should be under <mathjax>#7#</mathjax>, should it not?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax>.</p>
<p>And we know that ammonium salts are weakly acidic, and undergo the acid-base reaction:</p>
<p><mathjax>#NH_4^(+) + H_2O(l) rightleftharpoonsNH_3(aq) + H_3O^+#</mathjax></p>
<p>For ammonium ion, <mathjax>#K_a=1.80xx10^-5#</mathjax></p>
<p>And if there were a <mathjax>#1.0*mol*L^-1#</mathjax> solution of ammonium ion, we would solve the equilibrium expression in the usual way.</p>
<p>i.e. <mathjax>#K_a=1.80xx10^-5=([H_3O^+][NH_3(aq)])/([NH_4^(+)])#</mathjax></p>
<p>And if <mathjax>#x*mol*L^-1#</mathjax> ammonium ion ionizes, then.........</p>
<p><mathjax>#K_a=1.80xx10^-5=x^2/((1.0-x)*mol*L^-1)#</mathjax>, a quadratic in <mathjax>#x#</mathjax>, which may be simplified given the reasonable assumption that <mathjax>#1.0">>"x#</mathjax>, and thus <mathjax>#1.0-x~=1.0#</mathjax>.</p>
<p>And thus <mathjax>#x~=sqrt(1.80^-5)#</mathjax>,</p>
<p>so <mathjax>#x_1=4.2xx10^-3#</mathjax>, and recycling this approximation of <mathjax>#x#</mathjax> back into the expression:</p>
<p><mathjax>#x_2=4.2xx10^-3#</mathjax>, and since <mathjax>#x#</mathjax> has converged, we may accept this value. </p>
<p>Thus <mathjax>#[H_3O^+]=4.2xx10^-3*mol*L^-1#</mathjax>, and <mathjax>#pH=-log_10[H_3O^+]=-log_10(4.2xx10^-3)=2.37#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is #pH# of an aqueous solution for which #[NH_4^+]=1.0*mol*L^-1#?</h1>
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<div class="markdown"><p>It should be under <mathjax>#7#</mathjax>, should it not?</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH=-log_10[H_3O^+]#</mathjax>.</p>
<p>And we know that ammonium salts are weakly acidic, and undergo the acid-base reaction:</p>
<p><mathjax>#NH_4^(+) + H_2O(l) rightleftharpoonsNH_3(aq) + H_3O^+#</mathjax></p>
<p>For ammonium ion, <mathjax>#K_a=1.80xx10^-5#</mathjax></p>
<p>And if there were a <mathjax>#1.0*mol*L^-1#</mathjax> solution of ammonium ion, we would solve the equilibrium expression in the usual way.</p>
<p>i.e. <mathjax>#K_a=1.80xx10^-5=([H_3O^+][NH_3(aq)])/([NH_4^(+)])#</mathjax></p>
<p>And if <mathjax>#x*mol*L^-1#</mathjax> ammonium ion ionizes, then.........</p>
<p><mathjax>#K_a=1.80xx10^-5=x^2/((1.0-x)*mol*L^-1)#</mathjax>, a quadratic in <mathjax>#x#</mathjax>, which may be simplified given the reasonable assumption that <mathjax>#1.0">>"x#</mathjax>, and thus <mathjax>#1.0-x~=1.0#</mathjax>.</p>
<p>And thus <mathjax>#x~=sqrt(1.80^-5)#</mathjax>,</p>
<p>so <mathjax>#x_1=4.2xx10^-3#</mathjax>, and recycling this approximation of <mathjax>#x#</mathjax> back into the expression:</p>
<p><mathjax>#x_2=4.2xx10^-3#</mathjax>, and since <mathjax>#x#</mathjax> has converged, we may accept this value. </p>
<p>Thus <mathjax>#[H_3O^+]=4.2xx10^-3*mol*L^-1#</mathjax>, and <mathjax>#pH=-log_10[H_3O^+]=-log_10(4.2xx10^-3)=2.37#</mathjax>.</p></div>
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</article> | What is #pH# of an aqueous solution for which #[NH_4^+]=1.0*mol*L^-1#? | null |
3,263 | ab553c46-6ddd-11ea-9faf-ccda262736ce | https://socratic.org/questions/a-compound-contains-14-5-g-of-magnesium-and-3-5-g-of-oxygen-what-is-the-mass-per | 19.44% | start physical_unit 18 21 mass_percent none qc_end physical_unit 6 6 3 4 mass qc_end physical_unit 11 11 8 9 mass qc_end end | [{"type":"physical unit","value":"Mass percent [OF] oxygen in the compound"}] | [{"type":"physical unit","value":"19.44%"}] | [{"type":"physical unit","value":"Mass [OF] magnesium [=] \\pu{14.5 g}"},{"type":"physical unit","value":"Mass [OF] oxygen [=] \\pu{3.5 g}"}] | <h1 class="questionTitle" itemprop="name">A compound contains 14.5 g of magnesium and 3.5 g of oxygen. What is the mass percent of oxygen in the compound?</h1> | null | 19.44% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Mass ercentage states how much of a compound's mass is made up of something. In this case <mathjax>#m_p=("mass of element")/("mass of whole compound")*100=3.5/(14.5+3.5)*100=3.5/18*100=7/36*100=19.dot4%#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#19.dot4#</mathjax><mathjax>#%#</mathjax></p></div>
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<div class="markdown"><p>Mass ercentage states how much of a compound's mass is made up of something. In this case <mathjax>#m_p=("mass of element")/("mass of whole compound")*100=3.5/(14.5+3.5)*100=3.5/18*100=7/36*100=19.dot4%#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">A compound contains 14.5 g of magnesium and 3.5 g of oxygen. What is the mass percent of oxygen in the compound?</h1>
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<div class="markdown"><p><mathjax>#19.dot4#</mathjax><mathjax>#%#</mathjax></p></div>
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<div class="markdown"><p>Mass ercentage states how much of a compound's mass is made up of something. In this case <mathjax>#m_p=("mass of element")/("mass of whole compound")*100=3.5/(14.5+3.5)*100=3.5/18*100=7/36*100=19.dot4%#</mathjax>.</p></div>
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</article> | A compound contains 14.5 g of magnesium and 3.5 g of oxygen. What is the mass percent of oxygen in the compound? | null |
3,264 | ac1dade4-6ddd-11ea-91ee-ccda262736ce | https://socratic.org/questions/how-would-you-balance-this-equation-c-2h-6-o-2-co-2-h-2o | 2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O | start chemical_equation qc_end chemical_equation 6 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] this equation"}] | [{"type":"chemical equation","value":"2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O"}] | [{"type":"chemical equation","value":"C2H6 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">How would you balance this equation: #C_2H_6+O_2 -> CO_2+H_2O#?</h1> | null | 2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"C"_2"H"_6" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + H"_2"O"#</mathjax></p>
<p><strong>Balance H first.</strong></p>
<p>There are 6 H atoms on the left and 2 on the right. Place a coefficient of 3 in front of the <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#"C"_2"H"_6" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 3H"_2"O"#</mathjax></p>
<p><strong>Balance C next.</strong></p>
<p>There are 2 C atoms on the left and 1 C atom on the right. Add a coefficient of 2 in front of the <mathjax>#"CO"_2"#</mathjax>.</p>
<p><mathjax>#"C"_2"H"_6" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CO"_2" + 3H"_2"O"#</mathjax></p>
<p><strong>Balance O last.</strong></p>
<p>There are 2 O atoms on the left and 7 O atoms on the right.</p>
<p>Place a coefficient of 7/2 in front of <mathjax>#"O"_2"#</mathjax> on the left.</p>
<p><mathjax>#"C"_2"H"_6" + "7/2"O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CO"_2" + 3H"_2"O"#</mathjax></p>
<p>All of the coefficients in a chemical equation must be whole numbers. To change <mathjax>#7/2#</mathjax> to a whole number, multiply all coefficients times 2.</p>
<p><mathjax>#"2C"_2"H"_6" + 7O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4CO"_2" + 6H"_2"O"#</mathjax></p>
<p>There are now 4 C atoms on both sides, 12 H atoms on both sides, and 14 O atoms on both sides. The equation is balanced.</p></div>
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<div class="markdown"><p><mathjax>#"2C"_2"H"_6" + 7O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4CO"_2" + 6H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"C"_2"H"_6" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + H"_2"O"#</mathjax></p>
<p><strong>Balance H first.</strong></p>
<p>There are 6 H atoms on the left and 2 on the right. Place a coefficient of 3 in front of the <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#"C"_2"H"_6" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 3H"_2"O"#</mathjax></p>
<p><strong>Balance C next.</strong></p>
<p>There are 2 C atoms on the left and 1 C atom on the right. Add a coefficient of 2 in front of the <mathjax>#"CO"_2"#</mathjax>.</p>
<p><mathjax>#"C"_2"H"_6" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CO"_2" + 3H"_2"O"#</mathjax></p>
<p><strong>Balance O last.</strong></p>
<p>There are 2 O atoms on the left and 7 O atoms on the right.</p>
<p>Place a coefficient of 7/2 in front of <mathjax>#"O"_2"#</mathjax> on the left.</p>
<p><mathjax>#"C"_2"H"_6" + "7/2"O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CO"_2" + 3H"_2"O"#</mathjax></p>
<p>All of the coefficients in a chemical equation must be whole numbers. To change <mathjax>#7/2#</mathjax> to a whole number, multiply all coefficients times 2.</p>
<p><mathjax>#"2C"_2"H"_6" + 7O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4CO"_2" + 6H"_2"O"#</mathjax></p>
<p>There are now 4 C atoms on both sides, 12 H atoms on both sides, and 14 O atoms on both sides. The equation is balanced.</p></div>
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<h1 class="questionTitle" itemprop="name">How would you balance this equation: #C_2H_6+O_2 -> CO_2+H_2O#?</h1>
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<div class="markdown"><p><mathjax>#"2C"_2"H"_6" + 7O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4CO"_2" + 6H"_2"O"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"C"_2"H"_6" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + H"_2"O"#</mathjax></p>
<p><strong>Balance H first.</strong></p>
<p>There are 6 H atoms on the left and 2 on the right. Place a coefficient of 3 in front of the <mathjax>#"H"_2"O"#</mathjax>.</p>
<p><mathjax>#"C"_2"H"_6" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"CO"_2" + 3H"_2"O"#</mathjax></p>
<p><strong>Balance C next.</strong></p>
<p>There are 2 C atoms on the left and 1 C atom on the right. Add a coefficient of 2 in front of the <mathjax>#"CO"_2"#</mathjax>.</p>
<p><mathjax>#"C"_2"H"_6" + O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CO"_2" + 3H"_2"O"#</mathjax></p>
<p><strong>Balance O last.</strong></p>
<p>There are 2 O atoms on the left and 7 O atoms on the right.</p>
<p>Place a coefficient of 7/2 in front of <mathjax>#"O"_2"#</mathjax> on the left.</p>
<p><mathjax>#"C"_2"H"_6" + "7/2"O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"2CO"_2" + 3H"_2"O"#</mathjax></p>
<p>All of the coefficients in a chemical equation must be whole numbers. To change <mathjax>#7/2#</mathjax> to a whole number, multiply all coefficients times 2.</p>
<p><mathjax>#"2C"_2"H"_6" + 7O"_2#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"4CO"_2" + 6H"_2"O"#</mathjax></p>
<p>There are now 4 C atoms on both sides, 12 H atoms on both sides, and 14 O atoms on both sides. The equation is balanced.</p></div>
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</article> | How would you balance this equation: #C_2H_6+O_2 -> CO_2+H_2O#? | null |
3,265 | a8d5ad9d-6ddd-11ea-a10a-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-complete-ionic-equation-for-nh-4cl-aq-naoh-aq-h-2o-nh-3-g-n | NH4+ + Cl- + Na+ + OH- -> H2O + NH3 + Na+ + Cl- | start chemical_equation qc_end chemical_equation 8 16 qc_end end | [{"type":"other","value":"Chemical Equation [OF] ionic equation"}] | [{"type":"chemical equation","value":"NH4+ + Cl- + Na+ + OH- -> H2O + NH3 + Na+ + Cl-"}] | [{"type":"chemical equation","value":"NH4Cl(aq) + NaOH(aq) -> H2O + NH3(g) + NaCl(aq)"}] | <h1 class="questionTitle" itemprop="name">What is the balanced complete ionic equation for #NH_4Cl(aq) + NaOH(aq) -> H_2O + NH_3(g) + NaCl(aq)#?</h1> | null | NH4+ + Cl- + Na+ + OH- -> H2O + NH3 + Na+ + Cl- | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation is already balanced, all we need to do is break down the aqueous <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> in their ions.</p></div>
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<div class="markdown"><p><mathjax>#NH_4^+ + Cl^(-)+ Na^+ + OH^(-) rarr H_2O + NH_3 + Na^+ + Cl^-#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equation is already balanced, all we need to do is break down the aqueous <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> in their ions.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the balanced complete ionic equation for #NH_4Cl(aq) + NaOH(aq) -> H_2O + NH_3(g) + NaCl(aq)#?</h1>
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<div class="markdown"><p><mathjax>#NH_4^+ + Cl^(-)+ Na^+ + OH^(-) rarr H_2O + NH_3 + Na^+ + Cl^-#</mathjax></p></div>
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<div class="markdown"><p>The equation is already balanced, all we need to do is break down the aqueous <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/compounds">compounds</a> in their ions.</p></div>
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</article> | What is the balanced complete ionic equation for #NH_4Cl(aq) + NaOH(aq) -> H_2O + NH_3(g) + NaCl(aq)#? | null |
3,266 | a9909d6a-6ddd-11ea-92b6-ccda262736ce | https://socratic.org/questions/how-many-grams-are-found-in-5-20-mol-of-cobalt | 306 grams | start physical_unit 9 9 mass g qc_end physical_unit 9 9 6 7 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] cobalt [IN] grams"}] | [{"type":"physical unit","value":"306 grams"}] | [{"type":"physical unit","value":"Mole [OF] cobalt [=] \\pu{5.20 mol}"}] | <h1 class="questionTitle" itemprop="name">How many grams are found in 5.20 mol of cobalt? </h1> | null | 306 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of <mathjax>#Co#</mathjax> is <mathjax>#58.933#</mathjax></p>
<p><mathjax>#5.20 cancel("mol""Co")#</mathjax> x <mathjax>#(58.933g " Co")/(1cancel("mol""Co))#</mathjax> <mathjax>#=306 g " Co" #</mathjax></p></div>
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<div class="markdown"><p><mathjax>#306g" Co"#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The molar mass of <mathjax>#Co#</mathjax> is <mathjax>#58.933#</mathjax></p>
<p><mathjax>#5.20 cancel("mol""Co")#</mathjax> x <mathjax>#(58.933g " Co")/(1cancel("mol""Co))#</mathjax> <mathjax>#=306 g " Co" #</mathjax></p></div>
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May 2, 2017
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<div class="markdown"><p><mathjax>#306g" Co"#</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>#Co#</mathjax> is <mathjax>#58.933#</mathjax></p>
<p><mathjax>#5.20 cancel("mol""Co")#</mathjax> x <mathjax>#(58.933g " Co")/(1cancel("mol""Co))#</mathjax> <mathjax>#=306 g " Co" #</mathjax></p></div>
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</article> | How many grams are found in 5.20 mol of cobalt? | null |
3,267 | aae62f36-6ddd-11ea-a78d-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-a-compound-composed-of-43-64-p-and-56-36-o-by-m | P2O5 | start chemical_formula qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] empirical"}] | [{"type":"chemical equation","value":"P2O5"}] | [{"type":"physical unit","value":"Percent by mass [OF] P in the compound [=] \\pu{43.64%}"},{"type":"physical unit","value":"Percent by mass [OF] O in the compound [=] \\pu{56.36%}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a compound composed of 43.64% P and 56.36% O by mass?</h1> | null | P2O5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume that there are <mathjax>#100#</mathjax> <mathjax>#g#</mathjax> unknown, and work out the proportion of phosphorus and oxygen atoms.</p>
<p>In <mathjax>#100*g#</mathjax> of stuff there are <mathjax>#43.64*g#</mathjax> phosphorus and <mathjax>#56.36*g#</mathjax> oxygen.</p>
<p>We divide thru by the ATOMIC masses of each constituent:</p>
<p><mathjax>#O:#</mathjax> <mathjax>#(56.36*g)/(16.00*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.52*mol" oxygen"#</mathjax></p>
<p><mathjax>#P:#</mathjax> <mathjax>#(43.64*g)/(31.00*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.41*mol" phosphorus"#</mathjax>.</p>
<p>We divide thru by the SMALLEST molar quantity, we want to get whole numbers:</p>
<p><mathjax>#(3.52*mol)/(1.41*mol)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.5" oxygen"#</mathjax></p>
<p><mathjax>#(1.41*mol)/(1.41*mol)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1" phosphorus"#</mathjax></p>
<p>A formula of <mathjax>#PO_(2.5)#</mathjax> is no good to us at all. Why, because as we have specified, the empirical formula is the simplest <mathjax>#"WHOLE NUMBER RATIO"#</mathjax> defining constituent atoms in a species. So, we simply double the ration to give <mathjax>#P_2O_5#</mathjax> as required. If the molecular mass of this material is <mathjax>#283.89#</mathjax> <mathjax>#g*mol^-1#</mathjax>, what is the molecular formula of this material??? </p></div>
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<div class="markdown"><p><mathjax>#P_2O_5#</mathjax> is the empirical formula, the simplest WHOLE number ratio that defines constituent atoms in a species. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume that there are <mathjax>#100#</mathjax> <mathjax>#g#</mathjax> unknown, and work out the proportion of phosphorus and oxygen atoms.</p>
<p>In <mathjax>#100*g#</mathjax> of stuff there are <mathjax>#43.64*g#</mathjax> phosphorus and <mathjax>#56.36*g#</mathjax> oxygen.</p>
<p>We divide thru by the ATOMIC masses of each constituent:</p>
<p><mathjax>#O:#</mathjax> <mathjax>#(56.36*g)/(16.00*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.52*mol" oxygen"#</mathjax></p>
<p><mathjax>#P:#</mathjax> <mathjax>#(43.64*g)/(31.00*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.41*mol" phosphorus"#</mathjax>.</p>
<p>We divide thru by the SMALLEST molar quantity, we want to get whole numbers:</p>
<p><mathjax>#(3.52*mol)/(1.41*mol)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.5" oxygen"#</mathjax></p>
<p><mathjax>#(1.41*mol)/(1.41*mol)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1" phosphorus"#</mathjax></p>
<p>A formula of <mathjax>#PO_(2.5)#</mathjax> is no good to us at all. Why, because as we have specified, the empirical formula is the simplest <mathjax>#"WHOLE NUMBER RATIO"#</mathjax> defining constituent atoms in a species. So, we simply double the ration to give <mathjax>#P_2O_5#</mathjax> as required. If the molecular mass of this material is <mathjax>#283.89#</mathjax> <mathjax>#g*mol^-1#</mathjax>, what is the molecular formula of this material??? </p></div>
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<div class="markdown"><p><mathjax>#P_2O_5#</mathjax> is the empirical formula, the simplest WHOLE number ratio that defines constituent atoms in a species. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We assume that there are <mathjax>#100#</mathjax> <mathjax>#g#</mathjax> unknown, and work out the proportion of phosphorus and oxygen atoms.</p>
<p>In <mathjax>#100*g#</mathjax> of stuff there are <mathjax>#43.64*g#</mathjax> phosphorus and <mathjax>#56.36*g#</mathjax> oxygen.</p>
<p>We divide thru by the ATOMIC masses of each constituent:</p>
<p><mathjax>#O:#</mathjax> <mathjax>#(56.36*g)/(16.00*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#3.52*mol" oxygen"#</mathjax></p>
<p><mathjax>#P:#</mathjax> <mathjax>#(43.64*g)/(31.00*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.41*mol" phosphorus"#</mathjax>.</p>
<p>We divide thru by the SMALLEST molar quantity, we want to get whole numbers:</p>
<p><mathjax>#(3.52*mol)/(1.41*mol)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#2.5" oxygen"#</mathjax></p>
<p><mathjax>#(1.41*mol)/(1.41*mol)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1" phosphorus"#</mathjax></p>
<p>A formula of <mathjax>#PO_(2.5)#</mathjax> is no good to us at all. Why, because as we have specified, the empirical formula is the simplest <mathjax>#"WHOLE NUMBER RATIO"#</mathjax> defining constituent atoms in a species. So, we simply double the ration to give <mathjax>#P_2O_5#</mathjax> as required. If the molecular mass of this material is <mathjax>#283.89#</mathjax> <mathjax>#g*mol^-1#</mathjax>, what is the molecular formula of this material??? </p></div>
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</article> | What is the empirical formula of a compound composed of 43.64% P and 56.36% O by mass? | null |
3,268 | abb250cd-6ddd-11ea-990a-ccda262736ce | https://socratic.org/questions/how-many-oxygen-atoms-does-1-mol-of-o-2-gas-have | 1.20 × 10^24 | start physical_unit 2 3 number none qc_end physical_unit 8 9 5 6 mole qc_end end | [{"type":"physical unit","value":"Number [OF] oxygen atoms"}] | [{"type":"physical unit","value":"1.20 × 10^24"}] | [{"type":"physical unit","value":"Mole [OF] O2 gas [=] \\pu{1 mol}"}] | <h1 class="questionTitle" itemprop="name">How many oxygen atoms does 1 mol of #O_2# gas have? </h1> | null | 1.20 × 10^24 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You specify a mole of oxygen molecules. This is a mole, Avogadro's number, of <mathjax>#O_2#</mathjax> species.</p>
<p>Clearly there are <mathjax>#2xxN_A#</mathjax> oxygen atoms, where <mathjax>#N_A="Avogadro's number", #</mathjax> <mathjax>#6.022xx10^23#</mathjax>.</p></div>
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<div class="markdown"><p>If I buy a pair of socks, how many socks do I have? Well I have the one pair, but I do have TWO socks!</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You specify a mole of oxygen molecules. This is a mole, Avogadro's number, of <mathjax>#O_2#</mathjax> species.</p>
<p>Clearly there are <mathjax>#2xxN_A#</mathjax> oxygen atoms, where <mathjax>#N_A="Avogadro's number", #</mathjax> <mathjax>#6.022xx10^23#</mathjax>.</p></div>
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<div class="markdown"><p>If I buy a pair of socks, how many socks do I have? Well I have the one pair, but I do have TWO socks!</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>You specify a mole of oxygen molecules. This is a mole, Avogadro's number, of <mathjax>#O_2#</mathjax> species.</p>
<p>Clearly there are <mathjax>#2xxN_A#</mathjax> oxygen atoms, where <mathjax>#N_A="Avogadro's number", #</mathjax> <mathjax>#6.022xx10^23#</mathjax>.</p></div>
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</article> | How many oxygen atoms does 1 mol of #O_2# gas have? | null |
3,269 | ab46958c-6ddd-11ea-a2e4-ccda262736ce | https://socratic.org/questions/a-rigid-cylinder-with-a-movable-piston-contains-500-liters-of-a-gas-at-30-0-c-wi | 456 liters | start physical_unit 27 28 volume l qc_end physical_unit 27 28 8 9 volume qc_end physical_unit 27 28 14 15 temperature qc_end physical_unit 27 28 20 21 pressure qc_end c_other STP qc_end end | [{"type":"physical unit","value":"Volume2 [OF] the gas [IN] liters"}] | [{"type":"physical unit","value":"456 liters"}] | [{"type":"physical unit","value":"Volume1 [OF] the gas [=] \\pu{500 liters}"},{"type":"physical unit","value":"Temperature1 [OF] the gas [=] \\pu{30.0 ℃}"},{"type":"physical unit","value":"Pressure1 [OF] the gas [=] \\pu{1.00 atmosphere}"},{"type":"other","value":"STP"}] | <h1 class="questionTitle" itemprop="name">A rigid cylinder with a movable piston contains 500 liters of a gas at 30.0 C with a pressure of 1.00 atmosphere. What is the volume of the gas in the cylinder at STP?</h1> | null | 456 liters | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This is an example of a <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Laws</a> problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)(p_1V_1)/T_1 = (p_2V_2)/T_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_2 = V_1 × p_1/p_2 × T_2/T_1#</mathjax></p>
</blockquote>
</blockquote>
<p>STP is defined as 1 bar and 0 °C.</p>
<p>We should therefore convert your original pressure of 1.00 atm to bars.</p>
<p><mathjax>#p_1 = 1.00 color(red)(cancel(color(black)("atm"))) × "1.013 bar"/(1 color(red)(cancel(color(black)("atm")))) = "1.013 bar"#</mathjax></p>
<p>Your data are then:</p>
<p><mathjax>#p_1 = "1.013 bar"; V_1 = "500 L"; T_1 = "(30.0 + 273.15) K" = "303.15 K"#</mathjax></p>
<p><mathjax>#p_2 = "1 bar"; color(white)(mm)V_2 = "?";color(white)(mml) T_2 = "(0 +273.15) K" = "273.15 K"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#V_2 = V_1 × p_1/p_2 × T_2/T_1 = "500 L" × (1.013 color(red)(cancel(color(black)("bar"))))/(1 color(red)(cancel(color(black)("bar")))) × (273.15 color(red)(cancel(color(black)("K"))))/(303.15 color(red)(cancel(color(black)("K")))) = "456 L"#</mathjax></p>
<p>The volume is 456 L.</p></div>
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<div class="markdown"><p>The volume of gas in the cylinder is 456 L.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This is an example of a <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Laws</a> problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)(p_1V_1)/T_1 = (p_2V_2)/T_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_2 = V_1 × p_1/p_2 × T_2/T_1#</mathjax></p>
</blockquote>
</blockquote>
<p>STP is defined as 1 bar and 0 °C.</p>
<p>We should therefore convert your original pressure of 1.00 atm to bars.</p>
<p><mathjax>#p_1 = 1.00 color(red)(cancel(color(black)("atm"))) × "1.013 bar"/(1 color(red)(cancel(color(black)("atm")))) = "1.013 bar"#</mathjax></p>
<p>Your data are then:</p>
<p><mathjax>#p_1 = "1.013 bar"; V_1 = "500 L"; T_1 = "(30.0 + 273.15) K" = "303.15 K"#</mathjax></p>
<p><mathjax>#p_2 = "1 bar"; color(white)(mm)V_2 = "?";color(white)(mml) T_2 = "(0 +273.15) K" = "273.15 K"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#V_2 = V_1 × p_1/p_2 × T_2/T_1 = "500 L" × (1.013 color(red)(cancel(color(black)("bar"))))/(1 color(red)(cancel(color(black)("bar")))) × (273.15 color(red)(cancel(color(black)("K"))))/(303.15 color(red)(cancel(color(black)("K")))) = "456 L"#</mathjax></p>
<p>The volume is 456 L.</p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">A rigid cylinder with a movable piston contains 500 liters of a gas at 30.0 C with a pressure of 1.00 atmosphere. What is the volume of the gas in the cylinder at STP?</h1>
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<div class="markdown"><p>The volume of gas in the cylinder is 456 L.</p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><blockquote></blockquote>
<p>This is an example of a <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Laws</a> problem.</p>
<blockquote>
<blockquote>
<p><mathjax>#color(blue)(bar(ul(|color(white)(a/a)(p_1V_1)/T_1 = (p_2V_2)/T_2color(white)(a/a)|)))" "#</mathjax></p>
</blockquote>
</blockquote>
<p>We can rearrange this formula to get</p>
<blockquote>
<blockquote>
<p><mathjax>#V_2 = V_1 × p_1/p_2 × T_2/T_1#</mathjax></p>
</blockquote>
</blockquote>
<p>STP is defined as 1 bar and 0 °C.</p>
<p>We should therefore convert your original pressure of 1.00 atm to bars.</p>
<p><mathjax>#p_1 = 1.00 color(red)(cancel(color(black)("atm"))) × "1.013 bar"/(1 color(red)(cancel(color(black)("atm")))) = "1.013 bar"#</mathjax></p>
<p>Your data are then:</p>
<p><mathjax>#p_1 = "1.013 bar"; V_1 = "500 L"; T_1 = "(30.0 + 273.15) K" = "303.15 K"#</mathjax></p>
<p><mathjax>#p_2 = "1 bar"; color(white)(mm)V_2 = "?";color(white)(mml) T_2 = "(0 +273.15) K" = "273.15 K"#</mathjax></p>
<blockquote></blockquote>
<p>∴ <mathjax>#V_2 = V_1 × p_1/p_2 × T_2/T_1 = "500 L" × (1.013 color(red)(cancel(color(black)("bar"))))/(1 color(red)(cancel(color(black)("bar")))) × (273.15 color(red)(cancel(color(black)("K"))))/(303.15 color(red)(cancel(color(black)("K")))) = "456 L"#</mathjax></p>
<p>The volume is 456 L.</p></div>
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</article> | A rigid cylinder with a movable piston contains 500 liters of a gas at 30.0 C with a pressure of 1.00 atmosphere. What is the volume of the gas in the cylinder at STP? | null |
3,270 | aad22d2e-6ddd-11ea-b4f5-ccda262736ce | https://socratic.org/questions/a-solution-of-potassium-chlorate-kcio-3-has-20-grams-of-the-salt-dissolved-in-10 | 10 grams | start physical_unit 10 11 mass g qc_end physical_unit 10 11 7 8 mass qc_end c_other OTHER qc_end physical_unit 17 17 14 15 mass qc_end physical_unit 17 17 19 20 temperature qc_end chemical_equation 5 5 qc_end end | [{"type":"physical unit","value":"Added mass [OF] the salt [IN] grams"}] | [{"type":"physical unit","value":"10 grams"}] | [{"type":"physical unit","value":"Mass1 [OF] the salt [=] \\pu{20 grams}"},{"type":"other","value":"Reaching the saturation point."},{"type":"physical unit","value":"Mass [OF] water [=] \\pu{100 grams}"},{"type":"physical unit","value":"Temperature [OF] water [=] \\pu{70 ℃}"},{"type":"chemical equation","value":"KCIO3"}] | <h1 class="questionTitle" itemprop="name">A solution of potassium chlorate, #KCIO_3#, has 20 grams of the salt dissolved in 100 grams of water at 70 C. Approximately how many more grams of the salt can be added to the solution before reaching the saturation point?</h1> | null | 10 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this, let's look at a <strong>solubility curve graph</strong>:</p>
<blockquote>
<p><img alt="http://www.dynamicscience.com.au" src="https://useruploads.socratic.org/yMpRvIaHSOmdPgwlQSUA_solcrv.gif"/> </p>
</blockquote>
<p>The curve for <mathjax>#"KClO"_3#</mathjax> is the brown line, and according to the <mathjax>#70#</mathjax> <mathjax>#""^"o""C"#</mathjax> vertical line, a saturated solution of <mathjax>#"KClO"_3#</mathjax> contains <mathjax>#color(red)(30color(white)(l)"g"#</mathjax> in <mathjax>#100#</mathjax> <mathjax>#"g H"_2"O"#</mathjax>.</p>
<blockquote></blockquote>
<p>The question says there are <mathjax>#20#</mathjax> <mathjax>#"g KClO"_3#</mathjax> dissolved, which means </p>
<blockquote>
<p><mathjax>#color(red)(30color(white)(l)"g") - 20color(white)(l)"g" = color(blue)(ulbar(|stackrel(" ")(" "10color(white)(l)"g KClO"_3" ")|)#</mathjax></p>
</blockquote>
<p>can further dissolve before the solution is <em>saturated</em>.</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#10#</mathjax> more grams of <mathjax>#"KClO"_3#</mathjax> can dissolve.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this, let's look at a <strong>solubility curve graph</strong>:</p>
<blockquote>
<p><img alt="http://www.dynamicscience.com.au" src="https://useruploads.socratic.org/yMpRvIaHSOmdPgwlQSUA_solcrv.gif"/> </p>
</blockquote>
<p>The curve for <mathjax>#"KClO"_3#</mathjax> is the brown line, and according to the <mathjax>#70#</mathjax> <mathjax>#""^"o""C"#</mathjax> vertical line, a saturated solution of <mathjax>#"KClO"_3#</mathjax> contains <mathjax>#color(red)(30color(white)(l)"g"#</mathjax> in <mathjax>#100#</mathjax> <mathjax>#"g H"_2"O"#</mathjax>.</p>
<blockquote></blockquote>
<p>The question says there are <mathjax>#20#</mathjax> <mathjax>#"g KClO"_3#</mathjax> dissolved, which means </p>
<blockquote>
<p><mathjax>#color(red)(30color(white)(l)"g") - 20color(white)(l)"g" = color(blue)(ulbar(|stackrel(" ")(" "10color(white)(l)"g KClO"_3" ")|)#</mathjax></p>
</blockquote>
<p>can further dissolve before the solution is <em>saturated</em>.</p></div>
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<h1 class="questionTitle" itemprop="name">A solution of potassium chlorate, #KCIO_3#, has 20 grams of the salt dissolved in 100 grams of water at 70 C. Approximately how many more grams of the salt can be added to the solution before reaching the saturation point?</h1>
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Nathan L.
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<div class="markdown"><p><mathjax>#10#</mathjax> more grams of <mathjax>#"KClO"_3#</mathjax> can dissolve.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>To solve this, let's look at a <strong>solubility curve graph</strong>:</p>
<blockquote>
<p><img alt="http://www.dynamicscience.com.au" src="https://useruploads.socratic.org/yMpRvIaHSOmdPgwlQSUA_solcrv.gif"/> </p>
</blockquote>
<p>The curve for <mathjax>#"KClO"_3#</mathjax> is the brown line, and according to the <mathjax>#70#</mathjax> <mathjax>#""^"o""C"#</mathjax> vertical line, a saturated solution of <mathjax>#"KClO"_3#</mathjax> contains <mathjax>#color(red)(30color(white)(l)"g"#</mathjax> in <mathjax>#100#</mathjax> <mathjax>#"g H"_2"O"#</mathjax>.</p>
<blockquote></blockquote>
<p>The question says there are <mathjax>#20#</mathjax> <mathjax>#"g KClO"_3#</mathjax> dissolved, which means </p>
<blockquote>
<p><mathjax>#color(red)(30color(white)(l)"g") - 20color(white)(l)"g" = color(blue)(ulbar(|stackrel(" ")(" "10color(white)(l)"g KClO"_3" ")|)#</mathjax></p>
</blockquote>
<p>can further dissolve before the solution is <em>saturated</em>.</p></div>
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</article> | A solution of potassium chlorate, #KCIO_3#, has 20 grams of the salt dissolved in 100 grams of water at 70 C. Approximately how many more grams of the salt can be added to the solution before reaching the saturation point? | null |
3,271 | ac2f076d-6ddd-11ea-ab70-ccda262736ce | https://socratic.org/questions/a-student-collects-450-ml-of-hcl-g-at-a-pressure-of-100-kpa-and-a-temperature-of | 0.42 L | start physical_unit 6 6 volume l qc_end physical_unit 6 6 3 4 volume qc_end physical_unit 6 6 11 12 pressure qc_end physical_unit 6 6 17 18 temperature qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Volume2 [OF] HCl [IN] L"}] | [{"type":"physical unit","value":"0.42 L"}] | [{"type":"physical unit","value":"Volume1 [OF] HCl(g) [=] \\pu{450 mL}"},{"type":"physical unit","value":"Pressure1 [OF] HCl(g) [=] \\pu{100 kPa}"},{"type":"physical unit","value":"Temperature1 [OF] HCl(g) [=] \\pu{17 ℃}"},{"type":"other","value":"At standard temperature and pressure."}] | <h1 class="questionTitle" itemprop="name">A student collects #450# #mL# of #HCl (g)# at a pressure of #100# #kPa# and a temperature of #17°C#. What is the new volume of the #HCl# at standard temperature and pressure?</h1> | null | 0.42 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First convert the temperature to kelvin: <mathjax>#17+273=290#</mathjax> <mathjax>#K#</mathjax>. Standard temperature is <mathjax>#273#</mathjax> <mathjax>#K#</mathjax>.</p>
<p>Now <mathjax>#100#</mathjax> <mathjax>#kPa#</mathjax> is already standard pressure, so the pressure is constant. </p>
<p>We should work in SI units, so <mathjax>#450#</mathjax> <mathjax>#mL#</mathjax> = <mathjax>#0.450#</mathjax> <mathjax>#L#</mathjax>.</p>
<p>From the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a>, but with pressure omitted because it is constant, take <mathjax>#V_1/T_1=V_2/T_2#</mathjax> and rearrange to <mathjax>#V_2=(V_1T_2)/T_1=(0.450xx273)/290=0.424#</mathjax> <mathjax>#L#</mathjax> = <mathjax>#424#</mathjax> <mathjax>#mL#</mathjax>. </p></div>
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<div class="markdown"><p>Take <mathjax>#V_1/T_1=V_2/T_2#</mathjax> and rearrange to:</p>
<p><mathjax>#V_2=(V_1T_2)/T_1=(0.450xx273)/290=0.424#</mathjax> <mathjax>#L#</mathjax> = <mathjax>#424#</mathjax> <mathjax>#mL#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First convert the temperature to kelvin: <mathjax>#17+273=290#</mathjax> <mathjax>#K#</mathjax>. Standard temperature is <mathjax>#273#</mathjax> <mathjax>#K#</mathjax>.</p>
<p>Now <mathjax>#100#</mathjax> <mathjax>#kPa#</mathjax> is already standard pressure, so the pressure is constant. </p>
<p>We should work in SI units, so <mathjax>#450#</mathjax> <mathjax>#mL#</mathjax> = <mathjax>#0.450#</mathjax> <mathjax>#L#</mathjax>.</p>
<p>From the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a>, but with pressure omitted because it is constant, take <mathjax>#V_1/T_1=V_2/T_2#</mathjax> and rearrange to <mathjax>#V_2=(V_1T_2)/T_1=(0.450xx273)/290=0.424#</mathjax> <mathjax>#L#</mathjax> = <mathjax>#424#</mathjax> <mathjax>#mL#</mathjax>. </p></div>
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<h1 class="questionTitle" itemprop="name">A student collects #450# #mL# of #HCl (g)# at a pressure of #100# #kPa# and a temperature of #17°C#. What is the new volume of the #HCl# at standard temperature and pressure?</h1>
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David G.
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<span class="dateCreated" datetime="2016-03-28T03:46:24" itemprop="dateCreated">
Mar 28, 2016
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<div class="markdown"><p>Take <mathjax>#V_1/T_1=V_2/T_2#</mathjax> and rearrange to:</p>
<p><mathjax>#V_2=(V_1T_2)/T_1=(0.450xx273)/290=0.424#</mathjax> <mathjax>#L#</mathjax> = <mathjax>#424#</mathjax> <mathjax>#mL#</mathjax>. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>First convert the temperature to kelvin: <mathjax>#17+273=290#</mathjax> <mathjax>#K#</mathjax>. Standard temperature is <mathjax>#273#</mathjax> <mathjax>#K#</mathjax>.</p>
<p>Now <mathjax>#100#</mathjax> <mathjax>#kPa#</mathjax> is already standard pressure, so the pressure is constant. </p>
<p>We should work in SI units, so <mathjax>#450#</mathjax> <mathjax>#mL#</mathjax> = <mathjax>#0.450#</mathjax> <mathjax>#L#</mathjax>.</p>
<p>From the <a href="http://socratic.org/chemistry/the-behavior-of-gases/combined-gas-law">Combined Gas Law</a>, but with pressure omitted because it is constant, take <mathjax>#V_1/T_1=V_2/T_2#</mathjax> and rearrange to <mathjax>#V_2=(V_1T_2)/T_1=(0.450xx273)/290=0.424#</mathjax> <mathjax>#L#</mathjax> = <mathjax>#424#</mathjax> <mathjax>#mL#</mathjax>. </p></div>
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</article> | A student collects #450# #mL# of #HCl (g)# at a pressure of #100# #kPa# and a temperature of #17°C#. What is the new volume of the #HCl# at standard temperature and pressure? | null |
3,272 | ab44909a-6ddd-11ea-99a2-ccda262736ce | https://socratic.org/questions/what-is-the-balanced-equation-of-c6h6-o2-co2-h2o | 2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O | start chemical_equation qc_end chemical_equation 6 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O"}] | [{"type":"chemical equation","value":"C6H6 + O2 -> CO2 + H2O"}] | <h1 class="questionTitle" itemprop="name">What is the balanced equation of C6H6+O2 = CO2+H2O?</h1> | null | 2 C6H6 + 15 O2 -> 12 CO2 + 6 H2O | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let the variables <mathjax>#color(red)(p, q, r, s)#</mathjax> represent integers such that:<br/>
<mathjax>#color(white)("XX")color(red)p * C_6H_6 +color(red)q * O_2 = color(red)r * CO_2 +color(red)s * H_2O#</mathjax></p>
<p>Considering the element <mathjax>#C#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 =color(red)r * 1#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)r = 6color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#H#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 = color(red)s * 2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)s = 3color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#O#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)q=(15/2)color(red)p#</mathjax></p>
<p>The smallest value of <mathjax>#color(red)p > 0#</mathjax> for which <mathjax>#color(red)q#</mathjax> is an integer is:<br/>
<mathjax>#color(white)("XX")color(red)p=color(red)2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr {color(red)q=color(red)15; color(red)r=color(red)12; color(red)s=color(red)6}#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#color(red)2C_6H_6+color(red)15O_2 =color(red)12CO_2+color(red)6H_2O#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Let the variables <mathjax>#color(red)(p, q, r, s)#</mathjax> represent integers such that:<br/>
<mathjax>#color(white)("XX")color(red)p * C_6H_6 +color(red)q * O_2 = color(red)r * CO_2 +color(red)s * H_2O#</mathjax></p>
<p>Considering the element <mathjax>#C#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 =color(red)r * 1#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)r = 6color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#H#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 = color(red)s * 2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)s = 3color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#O#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)q=(15/2)color(red)p#</mathjax></p>
<p>The smallest value of <mathjax>#color(red)p > 0#</mathjax> for which <mathjax>#color(red)q#</mathjax> is an integer is:<br/>
<mathjax>#color(white)("XX")color(red)p=color(red)2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr {color(red)q=color(red)15; color(red)r=color(red)12; color(red)s=color(red)6}#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">What is the balanced equation of C6H6+O2 = CO2+H2O?</h1>
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Alan P.
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<span class="dateCreated" datetime="2017-02-08T15:21:52" itemprop="dateCreated">
Feb 8, 2017
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<div class="markdown"><p><mathjax>#color(red)2C_6H_6+color(red)15O_2 =color(red)12CO_2+color(red)6H_2O#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Let the variables <mathjax>#color(red)(p, q, r, s)#</mathjax> represent integers such that:<br/>
<mathjax>#color(white)("XX")color(red)p * C_6H_6 +color(red)q * O_2 = color(red)r * CO_2 +color(red)s * H_2O#</mathjax></p>
<p>Considering the element <mathjax>#C#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 =color(red)r * 1#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)r = 6color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#H#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)p * 6 = color(red)s * 2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)s = 3color(red)p#</mathjax></p>
<p>Considering the element <mathjax>#O#</mathjax>, we have:<br/>
<mathjax>#color(white)("XX")color(red)q * 2 = color(red)r * 2 + color(red)s * 1 = 12color(red)p +3color(red)p = 15color(red)p#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr color(red)q=(15/2)color(red)p#</mathjax></p>
<p>The smallest value of <mathjax>#color(red)p > 0#</mathjax> for which <mathjax>#color(red)q#</mathjax> is an integer is:<br/>
<mathjax>#color(white)("XX")color(red)p=color(red)2#</mathjax><br/>
<mathjax>#color(white)("XXXX")rarr {color(red)q=color(red)15; color(red)r=color(red)12; color(red)s=color(red)6}#</mathjax></p></div>
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Nikka C.
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<span class="dateCreated" datetime="2017-02-09T02:34:22" itemprop="dateCreated">
Feb 9, 2017
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<div class="markdown"><p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#15/2#</mathjax> <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> 6<mathjax>#CO_2#</mathjax>+ 3<mathjax>#H_2#</mathjax>O</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>1.) Create a tally sheet of atoms involved in the reaction.</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = 6<br/>
<mathjax>#H#</mathjax> = 6<br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1<br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1 ( <strong>DO NOT ADD IT UP YET</strong> )</p>
<p>2.) Find the atoms that are easiest to balance. In this case, the <mathjax>#C#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1</p>
<p>3.) Remember that in the equation, the <mathjax>#C#</mathjax> atom is a part of a substance. Therefore, you have to multiply the attached <mathjax>#O#</mathjax> atom with the factor as well.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>4.) Find the next atom to balance. In this case, the <mathjax>#H#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p>Again, the <mathjax>#H#</mathjax> atom is chemically bonded to another <mathjax>#O#</mathjax> atom. Thus,</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>5.) Balance out the remaining <mathjax>#O#</mathjax> atoms. Since the <mathjax>#O#</mathjax> atoms on the right side is an odd number, you can use your knowledge of fractions to get the complete balanced equation.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2 x <mathjax>#color(blue)(15/2)#</mathjax> = <strong>15</strong></p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15/2)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>The equation is now balanced.</p></div>
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Nikka C.
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Feb 9, 2017
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<div class="markdown"><p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#15/2#</mathjax> <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> 6<mathjax>#CO_2#</mathjax>+ 3<mathjax>#H_2#</mathjax>O</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>1.) Create a tally sheet of atoms involved in the reaction.</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = 6<br/>
<mathjax>#H#</mathjax> = 6<br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1<br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1 ( <strong>DO NOT ADD IT UP YET</strong> )</p>
<p>2.) Find the atoms that are easiest to balance. In this case, the <mathjax>#C#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = 2 + 1</p>
<p>3.) Remember that in the equation, the <mathjax>#C#</mathjax> atom is a part of a substance. Therefore, you have to multiply the attached <mathjax>#O#</mathjax> atom with the factor as well.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2<br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#H_2#</mathjax>O</p>
<p>4.) Find the next atom to balance. In this case, the <mathjax>#H#</mathjax> atom.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + 1</p>
<p>Again, the <mathjax>#H#</mathjax> atom is chemically bonded to another <mathjax>#O#</mathjax> atom. Thus,</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2</p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>5.) Balance out the remaining <mathjax>#O#</mathjax> atoms. Since the <mathjax>#O#</mathjax> atoms on the right side is an odd number, you can use your knowledge of fractions to get the complete balanced equation.</p>
<p>Left side:<br/>
<mathjax>#C#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = 2 x <mathjax>#color(blue)(15/2)#</mathjax> = <strong>15</strong></p>
<p>Right side:<br/>
<mathjax>#C#</mathjax> = 1 x <mathjax>#color(red)(6)#</mathjax> = <strong>6</strong><br/>
<mathjax>#H#</mathjax> = 2 x <mathjax>#color(green)(3)#</mathjax> = <strong>6</strong><br/>
<mathjax>#O#</mathjax> = (2 x <mathjax>#color(red)(6)#</mathjax>) + (1 x <mathjax>#color(green)(3)#</mathjax>) = <strong>15</strong></p>
<p><mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15/2)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O</p>
<p>The equation is now balanced.</p>
<p>If you don't like fractions, you can always multiply the whole chemical equation by the denominator.</p>
<p>[<mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15/2)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(6)#</mathjax><mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(3)#</mathjax><mathjax>#H_2#</mathjax>O] x 2</p>
<p><mathjax>#2#</mathjax> <mathjax>#C_6#</mathjax><mathjax>#H_6#</mathjax> + <mathjax>#color(blue)(15)#</mathjax><mathjax>#O_2#</mathjax> <mathjax>#rarr#</mathjax> <mathjax>#color(red)(12)#</mathjax> <mathjax>#CO_2#</mathjax>+ <mathjax>#color(green)(6)#</mathjax> <mathjax>#H_2#</mathjax>O (also acceptable)</p></div>
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</article> | What is the balanced equation of C6H6+O2 = CO2+H2O? | null |
3,273 | ad07ef08-6ddd-11ea-9f9b-ccda262736ce | https://socratic.org/questions/the-k-a-of-a-monoprotic-weak-acid-is-4-67-x-10-3-what-is-the-percent-ionization- | 15.20% | start physical_unit 4 6 percent_ionization none qc_end physical_unit 4 6 8 10 ka qc_end physical_unit 4 6 18 19 molarity qc_end end | [{"type":"physical unit","value":"Percent ionization [OF] monoprotic weak acid in solution"}] | [{"type":"physical unit","value":"15.20%"}] | [{"type":"physical unit","value":"Ka [OF] monoprotic weak acid [=] \\pu{4.67 × 10^(−3)}"},{"type":"physical unit","value":"Molarity [OF] monoprotic weak acid solution [=] \\pu{0.171 M}"}] | <h1 class="questionTitle" itemprop="name">The #K_a# of a monoprotic weak acid is #4.67 x 10^-3#. What is the percent ionization of a 0.171 M solution of this acid?</h1> | null | 15.20% | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We address the equilibrium, </p>
<p><mathjax>#HA(aq) + H_2O(l) rightleftharpoons H_3O^(+) + A^(-)#</mathjax></p>
<p>And <mathjax>#K_a=([H_3O^+][A^-])/([HA])=4.67xx10^-3#</mathjax></p>
<p>Now if <mathjax>#x*mol*L^-1#</mathjax> <mathjax>#HA#</mathjax> dissociates then..........</p>
<p><mathjax>#4.67xx10^-3=x^2/(0.171-x)#</mathjax></p>
<p>And this is quadratic in <mathjax>#x#</mathjax>, but instead of using the tedious quadratic equation, we assume that <mathjax>#0.171">>"x#</mathjax>, and so.......</p>
<p><mathjax>#4.67xx10^-3~=x^2/(0.171)#</mathjax></p>
<p>And thus <mathjax>#x_1=sqrt(4.67xx10^-3xx0.171)=0.0283*mol*L^-1#</mathjax>, and now we have an approx. for <mathjax>#x#</mathjax>, we can recycle this back into the expression:</p>
<p><mathjax>#x_2=0.0258*mol*L^-1#</mathjax></p>
<p><mathjax>#x_3=0.0260*mol*L^-1#</mathjax></p>
<p><mathjax>#x_4=0.0260*mol*L^-1#</mathjax></p>
<p>Since the values have converged, I am willing to accept this answer. (This same answer would have resulted from the quadratic equation, had we bothered to do it.)</p>
<p>And thus percentage dissociation................</p>
<p><mathjax>#="Concentration of hydronium ion"/"Initial concentration of acid"xx100%=(0.0260*mol*L^-1)/(0.171*mol*L^-1)xx100%=15%#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"% dissociation = 15%"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We address the equilibrium, </p>
<p><mathjax>#HA(aq) + H_2O(l) rightleftharpoons H_3O^(+) + A^(-)#</mathjax></p>
<p>And <mathjax>#K_a=([H_3O^+][A^-])/([HA])=4.67xx10^-3#</mathjax></p>
<p>Now if <mathjax>#x*mol*L^-1#</mathjax> <mathjax>#HA#</mathjax> dissociates then..........</p>
<p><mathjax>#4.67xx10^-3=x^2/(0.171-x)#</mathjax></p>
<p>And this is quadratic in <mathjax>#x#</mathjax>, but instead of using the tedious quadratic equation, we assume that <mathjax>#0.171">>"x#</mathjax>, and so.......</p>
<p><mathjax>#4.67xx10^-3~=x^2/(0.171)#</mathjax></p>
<p>And thus <mathjax>#x_1=sqrt(4.67xx10^-3xx0.171)=0.0283*mol*L^-1#</mathjax>, and now we have an approx. for <mathjax>#x#</mathjax>, we can recycle this back into the expression:</p>
<p><mathjax>#x_2=0.0258*mol*L^-1#</mathjax></p>
<p><mathjax>#x_3=0.0260*mol*L^-1#</mathjax></p>
<p><mathjax>#x_4=0.0260*mol*L^-1#</mathjax></p>
<p>Since the values have converged, I am willing to accept this answer. (This same answer would have resulted from the quadratic equation, had we bothered to do it.)</p>
<p>And thus percentage dissociation................</p>
<p><mathjax>#="Concentration of hydronium ion"/"Initial concentration of acid"xx100%=(0.0260*mol*L^-1)/(0.171*mol*L^-1)xx100%=15%#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The #K_a# of a monoprotic weak acid is #4.67 x 10^-3#. What is the percent ionization of a 0.171 M solution of this acid?</h1>
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<div class="markdown"><p><mathjax>#"% dissociation = 15%"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We address the equilibrium, </p>
<p><mathjax>#HA(aq) + H_2O(l) rightleftharpoons H_3O^(+) + A^(-)#</mathjax></p>
<p>And <mathjax>#K_a=([H_3O^+][A^-])/([HA])=4.67xx10^-3#</mathjax></p>
<p>Now if <mathjax>#x*mol*L^-1#</mathjax> <mathjax>#HA#</mathjax> dissociates then..........</p>
<p><mathjax>#4.67xx10^-3=x^2/(0.171-x)#</mathjax></p>
<p>And this is quadratic in <mathjax>#x#</mathjax>, but instead of using the tedious quadratic equation, we assume that <mathjax>#0.171">>"x#</mathjax>, and so.......</p>
<p><mathjax>#4.67xx10^-3~=x^2/(0.171)#</mathjax></p>
<p>And thus <mathjax>#x_1=sqrt(4.67xx10^-3xx0.171)=0.0283*mol*L^-1#</mathjax>, and now we have an approx. for <mathjax>#x#</mathjax>, we can recycle this back into the expression:</p>
<p><mathjax>#x_2=0.0258*mol*L^-1#</mathjax></p>
<p><mathjax>#x_3=0.0260*mol*L^-1#</mathjax></p>
<p><mathjax>#x_4=0.0260*mol*L^-1#</mathjax></p>
<p>Since the values have converged, I am willing to accept this answer. (This same answer would have resulted from the quadratic equation, had we bothered to do it.)</p>
<p>And thus percentage dissociation................</p>
<p><mathjax>#="Concentration of hydronium ion"/"Initial concentration of acid"xx100%=(0.0260*mol*L^-1)/(0.171*mol*L^-1)xx100%=15%#</mathjax></p></div>
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</article> | The #K_a# of a monoprotic weak acid is #4.67 x 10^-3#. What is the percent ionization of a 0.171 M solution of this acid? | null |
3,274 | aa484e36-6ddd-11ea-9c0a-ccda262736ce | https://socratic.org/questions/563fa75211ef6b4bdfde955d | 0.63 moles | start physical_unit 4 5 mole mol qc_end physical_unit 16 16 12 13 mass qc_end chemical_equation 18 18 qc_end end | [{"type":"physical unit","value":"Mole [OF] carbon dioxide [IN] moles"}] | [{"type":"physical unit","value":"0.63 moles"}] | [{"type":"physical unit","value":"Mass [OF] Fe2O3 [=] \\pu{1000 g}"},{"type":"chemical equation","value":"CO"}] | <h1 class="questionTitle" itemprop="name">How many moles of carbon dioxide are recovered after reduction of a #1000*g# mass of #Fe_2O_3# by #CO#?</h1> | null | 0.63 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of haematite: <mathjax>#(1000*g)/(159.69*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#"moles"#</mathjax></p>
<p>Given the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation, it is a fact that each mole of <mathjax>#Fe_2O_3#</mathjax> requires <mathjax>#3#</mathjax> <mathjax>#"mol"#</mathjax> carbon monoxide gas for reduction of the oxide. And of course the oxidation product is carbon dioxide gas, <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> for every <mathjax>#2#</mathjax> mole of iron that is reduced. So all I have to do is calculate the number of moles of oxide, and multiply this figure by 3, to give the number of moles of carbon dioxide product. </p></div>
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</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#Fe_2O_3(s) + 3CO(g) rarr 2Fe(s) + 3CO_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Moles of haematite: <mathjax>#(1000*g)/(159.69*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#"moles"#</mathjax></p>
<p>Given the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation, it is a fact that each mole of <mathjax>#Fe_2O_3#</mathjax> requires <mathjax>#3#</mathjax> <mathjax>#"mol"#</mathjax> carbon monoxide gas for reduction of the oxide. And of course the oxidation product is carbon dioxide gas, <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> for every <mathjax>#2#</mathjax> mole of iron that is reduced. So all I have to do is calculate the number of moles of oxide, and multiply this figure by 3, to give the number of moles of carbon dioxide product. </p></div>
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<h1 class="questionTitle" itemprop="name">How many moles of carbon dioxide are recovered after reduction of a #1000*g# mass of #Fe_2O_3# by #CO#?</h1>
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<div class="markdown"><p><mathjax>#Fe_2O_3(s) + 3CO(g) rarr 2Fe(s) + 3CO_2(g)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Moles of haematite: <mathjax>#(1000*g)/(159.69*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#"moles"#</mathjax></p>
<p>Given the <a href="http://socratic.org/chemistry/stoichiometry/stoichiometry">stoichiometry</a> of the equation, it is a fact that each mole of <mathjax>#Fe_2O_3#</mathjax> requires <mathjax>#3#</mathjax> <mathjax>#"mol"#</mathjax> carbon monoxide gas for reduction of the oxide. And of course the oxidation product is carbon dioxide gas, <mathjax>#3#</mathjax> <mathjax>#mol#</mathjax> for every <mathjax>#2#</mathjax> mole of iron that is reduced. So all I have to do is calculate the number of moles of oxide, and multiply this figure by 3, to give the number of moles of carbon dioxide product. </p></div>
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</article> | How many moles of carbon dioxide are recovered after reduction of a #1000*g# mass of #Fe_2O_3# by #CO#? | null |
3,275 | a9d14312-6ddd-11ea-9882-ccda262736ce | https://socratic.org/questions/58d4772d7c01494e441323e1 | 0.40 atm | start physical_unit 23 23 pressure atm qc_end physical_unit 4 4 15 16 partial_pressure qc_end physical_unit 0 2 8 9 pressure qc_end end | [{"type":"physical unit","value":"P [OF] Ne [IN] atm"}] | [{"type":"physical unit","value":"0.40 atm"}] | [{"type":"physical unit","value":"Partial pressure [OF] helium [=] \\pu{0.60 atm}"},{"type":"physical unit","value":"Pressure [OF] a gaseous mixture [=] \\pu{1 atm}"}] | <h1 class="questionTitle" itemprop="name">A gaseous mixture of helium and neon at #1*atm# contains a partial pressure of #0.60*atm# with respect to helium. What is #P_"Ne"#?</h1> | null | 0.40 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Dalton's law of partial pressures holds that in the gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a component is the same as the pressure as it would exert if it ALONE occupied the container. The total pressure is the sum of the individual partial pressures.</p>
<p>And thus <mathjax>#P_"Total"=P_"He"+P_"Ne"#</mathjax>. We are given <mathjax>#P_"Total"#</mathjax> and <mathjax>#P_"He"#</mathjax>; it is thus fairly straightforward to get <mathjax>#P_"Ne"#</mathjax> by difference.</p>
<p>For a more general treatment, see <a href="https://socratic.org/questions/how-would-you-define-and-explain-dalton-s-law-of-partial-pressure">here.</a> </p>
<p>Capisce?</p></div>
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<div class="markdown"><p><mathjax>#P_"Ne"=P_"Total"-P_"He"=0.40*atm........#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Dalton's law of partial pressures holds that in the gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a component is the same as the pressure as it would exert if it ALONE occupied the container. The total pressure is the sum of the individual partial pressures.</p>
<p>And thus <mathjax>#P_"Total"=P_"He"+P_"Ne"#</mathjax>. We are given <mathjax>#P_"Total"#</mathjax> and <mathjax>#P_"He"#</mathjax>; it is thus fairly straightforward to get <mathjax>#P_"Ne"#</mathjax> by difference.</p>
<p>For a more general treatment, see <a href="https://socratic.org/questions/how-would-you-define-and-explain-dalton-s-law-of-partial-pressure">here.</a> </p>
<p>Capisce?</p></div>
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<h1 class="questionTitle" itemprop="name">A gaseous mixture of helium and neon at #1*atm# contains a partial pressure of #0.60*atm# with respect to helium. What is #P_"Ne"#?</h1>
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<div class="markdown"><p><mathjax>#P_"Ne"=P_"Total"-P_"He"=0.40*atm........#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Dalton's law of partial pressures holds that in the gaseous mixture, the <a href="https://socratic.org/chemistry/the-behavior-of-gases/partial-pressure">partial pressure</a> exerted by a component is the same as the pressure as it would exert if it ALONE occupied the container. The total pressure is the sum of the individual partial pressures.</p>
<p>And thus <mathjax>#P_"Total"=P_"He"+P_"Ne"#</mathjax>. We are given <mathjax>#P_"Total"#</mathjax> and <mathjax>#P_"He"#</mathjax>; it is thus fairly straightforward to get <mathjax>#P_"Ne"#</mathjax> by difference.</p>
<p>For a more general treatment, see <a href="https://socratic.org/questions/how-would-you-define-and-explain-dalton-s-law-of-partial-pressure">here.</a> </p>
<p>Capisce?</p></div>
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</article> | A gaseous mixture of helium and neon at #1*atm# contains a partial pressure of #0.60*atm# with respect to helium. What is #P_"Ne"#? | null |
3,276 | ac21cc42-6ddd-11ea-9098-ccda262736ce | https://socratic.org/questions/5610905211ef6b16f1fe7c3b | 2 PbS + 3 O2 -> 2 PbO + 2 SO2 | start chemical_equation qc_end substance 2 3 qc_end substance 6 7 qc_end chemical_equation 10 10 qc_end chemical_equation 12 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the equation"}] | [{"type":"chemical equation","value":"2 PbS + 3 O2 -> 2 PbO + 2 SO2"}] | [{"type":"substance name","value":"Lead sulfide"},{"type":"substance name","value":"Oxygen gas"},{"type":"chemical equation","value":"PbO"},{"type":"chemical equation","value":"SO2"}] | <h1 class="questionTitle" itemprop="name">How does lead sulfide react with oxygen gas to form #PbO#, and #SO_2#?</h1> | null | 2 PbS + 3 O2 -> 2 PbO + 2 SO2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Elemental oxygen is reduced (<mathjax>#0 rarr -II#</mathjax>) and sulfur is oxidized (<mathjax>#-II rarr +IV#</mathjax>). The oxidation state of the metal is unchanged. </p></div>
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<div class="markdown"><p><mathjax>#PbS + 3/2O_2 rarr PbO + SO_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Elemental oxygen is reduced (<mathjax>#0 rarr -II#</mathjax>) and sulfur is oxidized (<mathjax>#-II rarr +IV#</mathjax>). The oxidation state of the metal is unchanged. </p></div>
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<div class="markdown"><p><mathjax>#PbS + 3/2O_2 rarr PbO + SO_2#</mathjax></p></div>
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<div class="markdown"><p>Elemental oxygen is reduced (<mathjax>#0 rarr -II#</mathjax>) and sulfur is oxidized (<mathjax>#-II rarr +IV#</mathjax>). The oxidation state of the metal is unchanged. </p></div>
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</article> | How does lead sulfide react with oxygen gas to form #PbO#, and #SO_2#? | null |
3,277 | aa5f60d8-6ddd-11ea-a81b-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-1-39-mol-of-mgo | 56.02 grams | start physical_unit 8 8 mass g qc_end physical_unit 8 8 5 6 mole qc_end end | [{"type":"physical unit","value":"Mass [OF] MgO [IN] grams"}] | [{"type":"physical unit","value":"56.02 grams"}] | [{"type":"physical unit","value":"Mole [OF] MgO [=] \\pu{1.39 mol}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of 1.39 mol of #MgO#?</h1> | null | 56.02 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So the required mass is <mathjax>#1.39#</mathjax> <mathjax>#cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#40.30#</mathjax> <mathjax>#g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of magnesium oxide has a mass of <mathjax>#40.30#</mathjax> <mathjax>#g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So the required mass is <mathjax>#1.39#</mathjax> <mathjax>#cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#40.30#</mathjax> <mathjax>#g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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<h1 class="questionTitle" itemprop="name">What is the mass of 1.39 mol of #MgO#?</h1>
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<div class="markdown"><p><mathjax>#1#</mathjax> <mathjax>#mol#</mathjax> of magnesium oxide has a mass of <mathjax>#40.30#</mathjax> <mathjax>#g*mol^-1#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So the required mass is <mathjax>#1.39#</mathjax> <mathjax>#cancel(mol)#</mathjax> <mathjax>#xx#</mathjax> <mathjax>#40.30#</mathjax> <mathjax>#g*cancel(mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??#</mathjax> <mathjax>#g#</mathjax>.</p></div>
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</article> | What is the mass of 1.39 mol of #MgO#? | null |
3,278 | abca5327-6ddd-11ea-9778-ccda262736ce | https://socratic.org/questions/596306f811ef6b3da7cc9795 | 6 L | start physical_unit 10 11 volume l qc_end physical_unit 10 11 6 7 mass qc_end physical_unit 10 11 16 17 temperature qc_end physical_unit 10 11 22 23 pressure qc_end end | [{"type":"physical unit","value":"Volume [OF] dinitrogen gas [IN] L"}] | [{"type":"physical unit","value":"6 L"}] | [{"type":"physical unit","value":"Mass [OF] dinitrogen gas [=] \\pu{7 g}"},{"type":"physical unit","value":"Temperature [OF] dinitrogen gas [=] \\pu{300 K}"},{"type":"physical unit","value":"Pressure [OF] dinitrogen gas [=] \\pu{750 mmHg}"}] | <h1 class="questionTitle" itemprop="name">What volume is occupied by a #7*g# mass of dinitrogen gas at a temperature of #300*K#, and a pressure of #750*mm*Hg#?</h1> | null | 6 L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given <mathjax>#V=(nRT)/P#</mathjax>, we know that nitrogen (LIKE most of the elemental gases) is a DIATOM, i.e. <mathjax>#N_2#</mathjax>. We also know that <mathjax>#760*mm*Hg-=1*atm.#</mathjax> This is a key point, as we can use the height of a mercury column as an accurate representation of pressure. </p>
<p>So <mathjax>#V=((7.0*g)/(28.0*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx300*K)/((750*mm*Hg)/(760*mm*Hg*atm^-1))#</mathjax></p>
<p>All units EXCEPT <mathjax>#"litres"#</mathjax> cancel, so...........</p>
<p><mathjax>#~=6*L.......#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#V~=6*L#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given <mathjax>#V=(nRT)/P#</mathjax>, we know that nitrogen (LIKE most of the elemental gases) is a DIATOM, i.e. <mathjax>#N_2#</mathjax>. We also know that <mathjax>#760*mm*Hg-=1*atm.#</mathjax> This is a key point, as we can use the height of a mercury column as an accurate representation of pressure. </p>
<p>So <mathjax>#V=((7.0*g)/(28.0*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx300*K)/((750*mm*Hg)/(760*mm*Hg*atm^-1))#</mathjax></p>
<p>All units EXCEPT <mathjax>#"litres"#</mathjax> cancel, so...........</p>
<p><mathjax>#~=6*L.......#</mathjax></p></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What volume is occupied by a #7*g# mass of dinitrogen gas at a temperature of #300*K#, and a pressure of #750*mm*Hg#?</h1>
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Jul 10, 2017
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<div class="markdown"><p><mathjax>#V~=6*L#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Given <mathjax>#V=(nRT)/P#</mathjax>, we know that nitrogen (LIKE most of the elemental gases) is a DIATOM, i.e. <mathjax>#N_2#</mathjax>. We also know that <mathjax>#760*mm*Hg-=1*atm.#</mathjax> This is a key point, as we can use the height of a mercury column as an accurate representation of pressure. </p>
<p>So <mathjax>#V=((7.0*g)/(28.0*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx300*K)/((750*mm*Hg)/(760*mm*Hg*atm^-1))#</mathjax></p>
<p>All units EXCEPT <mathjax>#"litres"#</mathjax> cancel, so...........</p>
<p><mathjax>#~=6*L.......#</mathjax></p></div>
</div>
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<a href="https://socratic.org/answers/449850" itemprop="url">Answer link</a>
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</article> | What volume is occupied by a #7*g# mass of dinitrogen gas at a temperature of #300*K#, and a pressure of #750*mm*Hg#? | null |
3,279 | acde2e2a-6ddd-11ea-b377-ccda262736ce | https://socratic.org/questions/57eb6a6eb72cff7d17dd9f8f | Ba(OH)2(s) + 2 HCl(aq) -> BaCl2(aq) + 2 H2O(l) | start chemical_equation qc_end substance 8 9 qc_end substance 11 12 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the stoichiometric reaction"}] | [{"type":"chemical equation","value":"Ba(OH)2(s) + 2 HCl(aq) -> BaCl2(aq) + 2 H2O(l)"}] | [{"type":"substance name","value":"Barium hydroxide"},{"type":"substance name","value":"Hydrochloric acid"}] | <h1 class="questionTitle" itemprop="name">How do we represent the stoichiometric reaction between barium hydroxide and hydrochloric acid?</h1> | null | Ba(OH)2(s) + 2 HCl(aq) -> BaCl2(aq) + 2 H2O(l) | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Barium hydroxide is reasonably insoluble. Added hydrochloric acid will neutralize the hydroxide function, and bring the metal up into solution, as the <mathjax>#Ba^(2+)(aq)#</mathjax> species. </p></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
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<div class="markdown"><p><mathjax>#Ba(OH)_2(s) + 2HCl(aq) rarr BaCl_2(aq) + 2H_2O(l)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Barium hydroxide is reasonably insoluble. Added hydrochloric acid will neutralize the hydroxide function, and bring the metal up into solution, as the <mathjax>#Ba^(2+)(aq)#</mathjax> species. </p></div>
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<div class="markdown"><p><mathjax>#Ba(OH)_2(s) + 2HCl(aq) rarr BaCl_2(aq) + 2H_2O(l)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Barium hydroxide is reasonably insoluble. Added hydrochloric acid will neutralize the hydroxide function, and bring the metal up into solution, as the <mathjax>#Ba^(2+)(aq)#</mathjax> species. </p></div>
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</article> | How do we represent the stoichiometric reaction between barium hydroxide and hydrochloric acid? | null |
3,280 | ad057e34-6ddd-11ea-ba84-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-a-solution-that-contains-3-25-moles-of-kcl-in-2-25-l | 1.44 M | start physical_unit 12 12 molarity mol/l qc_end physical_unit 12 12 9 10 mole qc_end physical_unit 6 6 13 14 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] KCl solution [IN] M"}] | [{"type":"physical unit","value":"1.44 M"}] | [{"type":"physical unit","value":"Mole [OF] KCl [=] \\pu{3.25 moles}"},{"type":"physical unit","value":"Volume [OF] KCl solution [=] \\pu{2.25 L}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of a solution that contains 3.25 moles of #"KCl"# in 2.25 L of the solution?</h1> | null | 1.44 M | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that in order to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution, you need to figure out how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present for every <mathjax>#"1.00 L"#</mathjax> of the solution. </p>
<p>In your case, you know that the solution contains <mathjax>#3.25#</mathjax> <strong>moles</strong> of potassium chloride, the solute, in a total volume of <mathjax>#"2.25 L"#</mathjax>. </p>
<p>You can calculate the number of moles of potassium chloride present in <mathjax>#"1.00 L"#</mathjax> of the solution by using the solution's known composition as a <strong>conversion factor</strong>. </p>
<blockquote>
<p><mathjax>#1.00 color(red)(cancel(color(black)("L solution"))) * overbrace("3.25 moles KCl"/(2.25 color(red)(cancel(color(black)("L solution")))))^(color(blue)("the solution's composition")) = "1.44 moles KCl"#</mathjax></p>
</blockquote>
<p>So if <mathjax>#"1.00 L"#</mathjax> of this solution contains <mathjax>#1.44#</mathjax> <strong>moles</strong> of solute, you can say that the <strong>molarity</strong> of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 1.44 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<h2></h2>
<p>Notice that you can get the molarity of the solution by diving the number of moles of solute by the total volume of the solution. </p>
<p>This will get you</p>
<blockquote>
<p><mathjax>#["KCl"] = "3.25 moles"/"2.25 L" = 3.25/2.25 quad "mol"/"L" = "1.44 mol L"^(-1)#</mathjax></p>
</blockquote></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"1.44 mol L"^(-1)#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The idea here is that in order to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution, you need to figure out how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present for every <mathjax>#"1.00 L"#</mathjax> of the solution. </p>
<p>In your case, you know that the solution contains <mathjax>#3.25#</mathjax> <strong>moles</strong> of potassium chloride, the solute, in a total volume of <mathjax>#"2.25 L"#</mathjax>. </p>
<p>You can calculate the number of moles of potassium chloride present in <mathjax>#"1.00 L"#</mathjax> of the solution by using the solution's known composition as a <strong>conversion factor</strong>. </p>
<blockquote>
<p><mathjax>#1.00 color(red)(cancel(color(black)("L solution"))) * overbrace("3.25 moles KCl"/(2.25 color(red)(cancel(color(black)("L solution")))))^(color(blue)("the solution's composition")) = "1.44 moles KCl"#</mathjax></p>
</blockquote>
<p>So if <mathjax>#"1.00 L"#</mathjax> of this solution contains <mathjax>#1.44#</mathjax> <strong>moles</strong> of solute, you can say that the <strong>molarity</strong> of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 1.44 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<h2></h2>
<p>Notice that you can get the molarity of the solution by diving the number of moles of solute by the total volume of the solution. </p>
<p>This will get you</p>
<blockquote>
<p><mathjax>#["KCl"] = "3.25 moles"/"2.25 L" = 3.25/2.25 quad "mol"/"L" = "1.44 mol L"^(-1)#</mathjax></p>
</blockquote></div>
</div>
</div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of a solution that contains 3.25 moles of #"KCl"# in 2.25 L of the solution?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2018-03-31T00:21:19" itemprop="dateCreated">
Mar 31, 2018
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<div class="markdown"><p><mathjax>#"1.44 mol L"^(-1)#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The idea here is that in order to find the <strong><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">molarity</a></strong> of the solution, you need to figure out how many moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a> are present for every <mathjax>#"1.00 L"#</mathjax> of the solution. </p>
<p>In your case, you know that the solution contains <mathjax>#3.25#</mathjax> <strong>moles</strong> of potassium chloride, the solute, in a total volume of <mathjax>#"2.25 L"#</mathjax>. </p>
<p>You can calculate the number of moles of potassium chloride present in <mathjax>#"1.00 L"#</mathjax> of the solution by using the solution's known composition as a <strong>conversion factor</strong>. </p>
<blockquote>
<p><mathjax>#1.00 color(red)(cancel(color(black)("L solution"))) * overbrace("3.25 moles KCl"/(2.25 color(red)(cancel(color(black)("L solution")))))^(color(blue)("the solution's composition")) = "1.44 moles KCl"#</mathjax></p>
</blockquote>
<p>So if <mathjax>#"1.00 L"#</mathjax> of this solution contains <mathjax>#1.44#</mathjax> <strong>moles</strong> of solute, you can say that the <strong>molarity</strong> of the solution is equal to</p>
<blockquote>
<p><mathjax>#color(darkgreen)(ul(color(black)("molarity = 1.44 mol L"^(-1))))#</mathjax></p>
</blockquote>
<p>The answer is rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong>.</p>
<h2></h2>
<p>Notice that you can get the molarity of the solution by diving the number of moles of solute by the total volume of the solution. </p>
<p>This will get you</p>
<blockquote>
<p><mathjax>#["KCl"] = "3.25 moles"/"2.25 L" = 3.25/2.25 quad "mol"/"L" = "1.44 mol L"^(-1)#</mathjax></p>
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</article> | What is the molarity of a solution that contains 3.25 moles of #"KCl"# in 2.25 L of the solution? | null |
3,281 | a93d0ae9-6ddd-11ea-b03f-ccda262736ce | https://socratic.org/questions/how-do-you-balance-cr-fe-no-3-2-fe-cr-no-3-3 | 2 Cr + 3 Fe(NO3)2 -> 3 Fe + 2 Cr(NO3)3 | start chemical_equation qc_end chemical_equation 4 10 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the reaction"}] | [{"type":"chemical equation","value":"2 Cr + 3 Fe(NO3)2 -> 3 Fe + 2 Cr(NO3)3"}] | [{"type":"chemical equation","value":"Cr + Fe(NO3)2 -> Fe + Cr(NO3)3"}] | <h1 class="questionTitle" itemprop="name">How do you balance #Cr + Fe(NO_3)_2 -> Fe + Cr(NO_3)_3#?</h1> | null | 2 Cr + 3 Fe(NO3)2 -> 3 Fe + 2 Cr(NO3)3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We write the oxidation rxn:</p>
<p><mathjax>#CrrarrCr^(3+) + 3e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And then the reduction rxn:</p>
<p><mathjax>#Fe^(2+) +2e^(-) rarr Fe#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And then cross mulitply, <mathjax>#2xx(i) +3xx(ii)#</mathjax></p>
<p><mathjax>#2Cr+3Fe(NO_3)_2 rarr 3Fe + 2Cr(NO_3)_3#</mathjax></p>
<p>Is mass balanced? Is charge balanced?</p></div>
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<div class="markdown"><p><mathjax>#2Cr+3Fe(NO_3)_2 rarr 3Fe + 2Cr(NO_3)_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We write the oxidation rxn:</p>
<p><mathjax>#CrrarrCr^(3+) + 3e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And then the reduction rxn:</p>
<p><mathjax>#Fe^(2+) +2e^(-) rarr Fe#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And then cross mulitply, <mathjax>#2xx(i) +3xx(ii)#</mathjax></p>
<p><mathjax>#2Cr+3Fe(NO_3)_2 rarr 3Fe + 2Cr(NO_3)_3#</mathjax></p>
<p>Is mass balanced? Is charge balanced?</p></div>
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<h1 class="questionTitle" itemprop="name">How do you balance #Cr + Fe(NO_3)_2 -> Fe + Cr(NO_3)_3#?</h1>
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Dec 28, 2016
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<div class="markdown"><p><mathjax>#2Cr+3Fe(NO_3)_2 rarr 3Fe + 2Cr(NO_3)_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>We write the oxidation rxn:</p>
<p><mathjax>#CrrarrCr^(3+) + 3e^-#</mathjax> <mathjax>#(i)#</mathjax></p>
<p>And then the reduction rxn:</p>
<p><mathjax>#Fe^(2+) +2e^(-) rarr Fe#</mathjax> <mathjax>#(ii)#</mathjax></p>
<p>And then cross mulitply, <mathjax>#2xx(i) +3xx(ii)#</mathjax></p>
<p><mathjax>#2Cr+3Fe(NO_3)_2 rarr 3Fe + 2Cr(NO_3)_3#</mathjax></p>
<p>Is mass balanced? Is charge balanced?</p></div>
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</article> | How do you balance #Cr + Fe(NO_3)_2 -> Fe + Cr(NO_3)_3#? | null |
3,282 | a96d8822-6ddd-11ea-a510-ccda262736ce | https://socratic.org/questions/how-many-grams-of-carbon-dioxide-are-produced-when-2-50-g-of-sodium-hydrogen-car | 1.31 grams | start physical_unit 4 5 mass g qc_end physical_unit 12 14 9 10 mass qc_end c_other OTHER qc_end chemical_equation 24 35 qc_end end | [{"type":"physical unit","value":"Mass [OF] carbon dioxide [IN] grams"}] | [{"type":"physical unit","value":"1.31 grams"}] | [{"type":"physical unit","value":"Mass [OF] sodium hydrogen carbonate [=] \\pu{2.50 g}"},{"type":"other","value":"Excess citric acid."},{"type":"chemical equation","value":"3 NaHCO3 + H3C6H5O7 -> Na3C6H5O7 + 3 CO2 + 3 H2O"}] | <h1 class="questionTitle" itemprop="name">How many grams of carbon dioxide are produced when 2.50 g of sodium hydrogen carbonate react with excess citric acid according to the equation #3NaHCO_3+H_3C_6H_5O_7->Na_3C_6H_5O_7 + 3CO_2 + 3H_2O#?</h1> | null | 1.31 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"3NaHCO"_3+"H"_3"C"_6"H"_5"O"_7"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Na"_3"C"_6"H"_5"O"_7+"3CO"_2+"3H"_2"O"#</mathjax></p>
<p>We will need the molar masses of <mathjax>#"NaHCO"_3"#</mathjax> and <mathjax>#"CO"_2"#</mathjax>, and their mole ratio as well.</p>
<p><strong>Molar Masses</strong></p>
<p><mathjax>#"NaHCO"_3":#</mathjax><mathjax>#"84.006609 g/mol"#</mathjax></p>
<p><a href="http://pubchem.ncbi.nlm.nih.gov/compound/516892#section=Top" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/516892#section=Top</a></p>
<p><mathjax>#"CO"_2":#</mathjax><mathjax>#"44.0095 g/mol"#</mathjax></p>
<p><a href="http://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top</a></p>
<p><strong>Mole Ratio</strong></p>
<p>From the balanced equation, <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <br/>
<mathjax>#"NaHCO"_3"#</mathjax> and <mathjax>#"CO"_2"#</mathjax> is <mathjax>#"3 mol NaHCO"_3:#</mathjax><mathjax>#"3 mol CO"_2"#</mathjax>.</p>
<p><strong>Stoichiometric Equation</strong><br/>
Determine the moles of <mathjax>#"NaHCO"_3"#</mathjax> by dividing the given mass of <mathjax>#"NaHCO"_3"#</mathjax> by its molar mass, then determine moles of <mathjax>#"CO"_2"#</mathjax> by multiplying times the mole ratio with <mathjax>#"CO"_2"#</mathjax> in the numerator, then multiply times the molar mass of <mathjax>#"CO"_2"#</mathjax>.</p>
<p><mathjax>#2.50cancel"g NaHCO"_3xx(1cancel"mol NaHCO"_3)/(84.006609cancel"g NaHCO"_3)xx(cancel3^1cancel"mol CO"_2)/(cancel3^1cancel"mol NaHCO"_3)xx(44.0095"g CO"_2)/(1cancel"mol CO"_2)="1.31 g CO"_2"#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<div class="markdown"><p><mathjax>#"1.31 g CO"_2"#</mathjax> will be produced.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"3NaHCO"_3+"H"_3"C"_6"H"_5"O"_7"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Na"_3"C"_6"H"_5"O"_7+"3CO"_2+"3H"_2"O"#</mathjax></p>
<p>We will need the molar masses of <mathjax>#"NaHCO"_3"#</mathjax> and <mathjax>#"CO"_2"#</mathjax>, and their mole ratio as well.</p>
<p><strong>Molar Masses</strong></p>
<p><mathjax>#"NaHCO"_3":#</mathjax><mathjax>#"84.006609 g/mol"#</mathjax></p>
<p><a href="http://pubchem.ncbi.nlm.nih.gov/compound/516892#section=Top" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/516892#section=Top</a></p>
<p><mathjax>#"CO"_2":#</mathjax><mathjax>#"44.0095 g/mol"#</mathjax></p>
<p><a href="http://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top</a></p>
<p><strong>Mole Ratio</strong></p>
<p>From the balanced equation, <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <br/>
<mathjax>#"NaHCO"_3"#</mathjax> and <mathjax>#"CO"_2"#</mathjax> is <mathjax>#"3 mol NaHCO"_3:#</mathjax><mathjax>#"3 mol CO"_2"#</mathjax>.</p>
<p><strong>Stoichiometric Equation</strong><br/>
Determine the moles of <mathjax>#"NaHCO"_3"#</mathjax> by dividing the given mass of <mathjax>#"NaHCO"_3"#</mathjax> by its molar mass, then determine moles of <mathjax>#"CO"_2"#</mathjax> by multiplying times the mole ratio with <mathjax>#"CO"_2"#</mathjax> in the numerator, then multiply times the molar mass of <mathjax>#"CO"_2"#</mathjax>.</p>
<p><mathjax>#2.50cancel"g NaHCO"_3xx(1cancel"mol NaHCO"_3)/(84.006609cancel"g NaHCO"_3)xx(cancel3^1cancel"mol CO"_2)/(cancel3^1cancel"mol NaHCO"_3)xx(44.0095"g CO"_2)/(1cancel"mol CO"_2)="1.31 g CO"_2"#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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<h1 class="questionTitle" itemprop="name">How many grams of carbon dioxide are produced when 2.50 g of sodium hydrogen carbonate react with excess citric acid according to the equation #3NaHCO_3+H_3C_6H_5O_7->Na_3C_6H_5O_7 + 3CO_2 + 3H_2O#?</h1>
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Meave60
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<span class="dateCreated" datetime="2016-01-29T12:30:38" itemprop="dateCreated">
Jan 29, 2016
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<div class="markdown"><p><mathjax>#"1.31 g CO"_2"#</mathjax> will be produced.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><strong>Balanced Equation</strong></p>
<p><mathjax>#"3NaHCO"_3+"H"_3"C"_6"H"_5"O"_7"#</mathjax><mathjax>#rarr#</mathjax><mathjax>#"Na"_3"C"_6"H"_5"O"_7+"3CO"_2+"3H"_2"O"#</mathjax></p>
<p>We will need the molar masses of <mathjax>#"NaHCO"_3"#</mathjax> and <mathjax>#"CO"_2"#</mathjax>, and their mole ratio as well.</p>
<p><strong>Molar Masses</strong></p>
<p><mathjax>#"NaHCO"_3":#</mathjax><mathjax>#"84.006609 g/mol"#</mathjax></p>
<p><a href="http://pubchem.ncbi.nlm.nih.gov/compound/516892#section=Top" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/516892#section=Top</a></p>
<p><mathjax>#"CO"_2":#</mathjax><mathjax>#"44.0095 g/mol"#</mathjax></p>
<p><a href="http://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top" rel="nofollow" target="_blank">http://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top</a></p>
<p><strong>Mole Ratio</strong></p>
<p>From the balanced equation, <a href="http://socratic.org/chemistry/the-mole-concept/the-mole">the mole</a> ratio between <br/>
<mathjax>#"NaHCO"_3"#</mathjax> and <mathjax>#"CO"_2"#</mathjax> is <mathjax>#"3 mol NaHCO"_3:#</mathjax><mathjax>#"3 mol CO"_2"#</mathjax>.</p>
<p><strong>Stoichiometric Equation</strong><br/>
Determine the moles of <mathjax>#"NaHCO"_3"#</mathjax> by dividing the given mass of <mathjax>#"NaHCO"_3"#</mathjax> by its molar mass, then determine moles of <mathjax>#"CO"_2"#</mathjax> by multiplying times the mole ratio with <mathjax>#"CO"_2"#</mathjax> in the numerator, then multiply times the molar mass of <mathjax>#"CO"_2"#</mathjax>.</p>
<p><mathjax>#2.50cancel"g NaHCO"_3xx(1cancel"mol NaHCO"_3)/(84.006609cancel"g NaHCO"_3)xx(cancel3^1cancel"mol CO"_2)/(cancel3^1cancel"mol NaHCO"_3)xx(44.0095"g CO"_2)/(1cancel"mol CO"_2)="1.31 g CO"_2"#</mathjax> rounded to three <a href="http://socratic.org/chemistry/measurement-in-chemistry/significant-figures">significant figures</a></p></div>
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</article> | How many grams of carbon dioxide are produced when 2.50 g of sodium hydrogen carbonate react with excess citric acid according to the equation #3NaHCO_3+H_3C_6H_5O_7->Na_3C_6H_5O_7 + 3CO_2 + 3H_2O#? | null |
3,283 | aa7e3fdb-6ddd-11ea-a0ab-ccda262736ce | https://socratic.org/questions/the-volume-of-a-sample-of-gas-initially-at-35-deg-c-and-58-ml-increases-to-450-m | 2117.67 degrees C | start physical_unit 4 6 temperature °c qc_end physical_unit 4 6 9 11 temperature qc_end physical_unit 4 6 13 14 volume qc_end physical_unit 4 6 17 18 volume qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Temperature2 [OF] the gas sample [IN] degrees C"}] | [{"type":"physical unit","value":"2117.67 degrees C"}] | [{"type":"physical unit","value":"Temperature1 [OF] the gas sample [=] \\pu{35 degrees C}"},{"type":"physical unit","value":"Volume1 [OF] the gas sample [=] \\pu{58 mL}"},{"type":"physical unit","value":"Volume2 [OF] the gas sample [=] \\pu{450 mL}"},{"type":"other","value":"The pressure in the container is kept constant."}] | <h1 class="questionTitle" itemprop="name">The volume of a sample of gas, initially at 35 deg C and 58 mL increases to 450 mL. What is the final temperature of the sample of gas if the pressure in the container is kept constant?</h1> | null | 2117.67 degrees C | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>...this assumes (i) a given molar quantity, and (ii) units of <mathjax>#"temperatura assoluta"#</mathjax></p>
<p>We solve for <mathjax>#T_2=V_2/V_1xxT_1#</mathjax></p>
<p><mathjax>#(450*mL)/(58*mL)xx308.15*K=??*K#</mathjax></p></div>
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<div class="markdown"><p>Well, we use old Charles' law...<mathjax>#V_1/T_1=V_2/T_2#</mathjax> for a constant molar quantity, and a constant pressure....</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>...this assumes (i) a given molar quantity, and (ii) units of <mathjax>#"temperatura assoluta"#</mathjax></p>
<p>We solve for <mathjax>#T_2=V_2/V_1xxT_1#</mathjax></p>
<p><mathjax>#(450*mL)/(58*mL)xx308.15*K=??*K#</mathjax></p></div>
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<h1 class="questionTitle" itemprop="name">The volume of a sample of gas, initially at 35 deg C and 58 mL increases to 450 mL. What is the final temperature of the sample of gas if the pressure in the container is kept constant?</h1>
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<div class="markdown"><p>Well, we use old Charles' law...<mathjax>#V_1/T_1=V_2/T_2#</mathjax> for a constant molar quantity, and a constant pressure....</p></div>
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<div class="markdown"><p>...this assumes (i) a given molar quantity, and (ii) units of <mathjax>#"temperatura assoluta"#</mathjax></p>
<p>We solve for <mathjax>#T_2=V_2/V_1xxT_1#</mathjax></p>
<p><mathjax>#(450*mL)/(58*mL)xx308.15*K=??*K#</mathjax></p></div>
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</article> | The volume of a sample of gas, initially at 35 deg C and 58 mL increases to 450 mL. What is the final temperature of the sample of gas if the pressure in the container is kept constant? | null |
3,284 | ab43a620-6ddd-11ea-bd70-ccda262736ce | https://socratic.org/questions/what-is-a-chemical-reaction-that-shows-what-happens-when-hcn-dissolves-in-water | H-C#N + H2O <=> H3O+ + C#N- | start chemical_equation qc_end chemical_equation 10 10 qc_end substance 13 13 qc_end end | [{"type":"other","value":"Chemical Equation [OF] the chemical reaction"}] | [{"type":"chemical equation","value":"H-C#N + H2O <=> H3O+ + C#N-"}] | [{"type":"chemical equation","value":"HCN"},{"type":"substance name","value":"Water"}] | <h1 class="questionTitle" itemprop="name">What is a chemical reaction that shows what happens when HCN dissolves in water?</h1> | null | H-C#N + H2O <=> H3O+ + C#N- | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Hydrogen cyanide behaves as a weak Bronsted acid: <mathjax>#pK_a=9.21#</mathjax>.</p>
<p>In water, cyanide salts such as <mathjax>#Na^(+)""^(-)C-=N#</mathjax> give rise to a slightly basic solution. Why so?</p></div>
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<div class="markdown"><p><mathjax>#H-C-=N + H_2OrightleftharpoonsH_3O^+ +""^(-)C-=N#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Hydrogen cyanide behaves as a weak Bronsted acid: <mathjax>#pK_a=9.21#</mathjax>.</p>
<p>In water, cyanide salts such as <mathjax>#Na^(+)""^(-)C-=N#</mathjax> give rise to a slightly basic solution. Why so?</p></div>
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<h1 class="questionTitle" itemprop="name">What is a chemical reaction that shows what happens when HCN dissolves in water?</h1>
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<div class="markdown"><p><mathjax>#H-C-=N + H_2OrightleftharpoonsH_3O^+ +""^(-)C-=N#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>Hydrogen cyanide behaves as a weak Bronsted acid: <mathjax>#pK_a=9.21#</mathjax>.</p>
<p>In water, cyanide salts such as <mathjax>#Na^(+)""^(-)C-=N#</mathjax> give rise to a slightly basic solution. Why so?</p></div>
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</article> | What is a chemical reaction that shows what happens when HCN dissolves in water? | null |
3,285 | a98f6c1b-6ddd-11ea-9f24-ccda262736ce | https://socratic.org/questions/what-is-the-mass-of-one-mole-of-helium-gas | 4.00 grams | 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] helium gas [IN] grams"}] | [{"type":"physical unit","value":"4.00 grams"}] | [{"type":"physical unit","value":"Mole [OF] helium gas [=] \\pu{1 mole}"}] | <h1 class="questionTitle" itemprop="name">What is the mass of one mole of helium gas?</h1> | null | 4.00 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since helium is a monatomic gas, its molar mass is its atomic weight (relative atomic mass) on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol.</p></div>
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<div class="markdown"><p>The mass of one mole (molar mass) of helium gas is <mathjax>#"4.002602 g/mol"#</mathjax>.</p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since helium is a monatomic gas, its molar mass is its atomic weight (relative atomic mass) on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol.</p></div>
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<div class="markdown"><p>The mass of one mole (molar mass) of helium gas is <mathjax>#"4.002602 g/mol"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>Since helium is a monatomic gas, its molar mass is its atomic weight (relative atomic mass) on <a href="http://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a> in g/mol.</p></div>
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</article> | What is the mass of one mole of helium gas? | null |
3,286 | ac27c062-6ddd-11ea-bcd0-ccda262736ce | https://socratic.org/questions/591e6090b72cff6a8619d48e | 149.4 pounds per square inch | start physical_unit 9 10 pressure psi qc_end physical_unit 9 10 7 8 mole qc_end physical_unit 9 10 15 16 volume qc_end physical_unit 9 10 21 22 temperature qc_end end | [{"type":"physical unit","value":"Pressure [OF] oxygen gas [IN] pounds per square inch"}] | [{"type":"physical unit","value":"149.4 pounds per square inch"}] | [{"type":"physical unit","value":"Mole [OF] oxygen gas [=] \\pu{9750 mol}"},{"type":"physical unit","value":"Volume [OF] oxygen gas [=] \\pu{9750 L}"},{"type":"physical unit","value":"Temperature [OF] oxygen gas [=] \\pu{34.9 ℃}"}] | <h1 class="questionTitle" itemprop="name">A tank used for compressed gas contains #9750*mol# oxygen gas, in a volume of #9750*L#, at a temperature of #34.9# #""^@C#...?</h1> | <div class="questionDetailsContainer">
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<h2 class="questionDetails" itemprop="text">
<div class="markdown"><p>What is the pressure inside the tank in <mathjax>#"pounds per square inch"#</mathjax>?</p></div>
</h2>
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</div> | 149.4 pounds per square inch | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P=(nRT)/V=(9750*molxx0.0821*(L*atm)/(K*mol)xx308*K)/(1650*L)#</mathjax></p>
<p><mathjax>#=149.4*atm#</mathjax></p>
<p>And then we convert this to <mathjax>#"psi"#</mathjax>; <mathjax>#1*atm-=14.6*"psi"#</mathjax>.</p>
<p>And so <mathjax>#P=149.4*atmxx14.6*"psi"*atm^-1#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#"Approx........2000 psi"#</mathjax>. Note that this is on the VERY high end for oxygen tanks for medical use. Why would you NOT such a tank for scuba diving?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p>Well, first we use standard units and get, <mathjax>#P=149.4*atm#</mathjax>, and then convert this into <mathjax>#"psi"#</mathjax>.</p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#P=(nRT)/V=(9750*molxx0.0821*(L*atm)/(K*mol)xx308*K)/(1650*L)#</mathjax></p>
<p><mathjax>#=149.4*atm#</mathjax></p>
<p>And then we convert this to <mathjax>#"psi"#</mathjax>; <mathjax>#1*atm-=14.6*"psi"#</mathjax>.</p>
<p>And so <mathjax>#P=149.4*atmxx14.6*"psi"*atm^-1#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#"Approx........2000 psi"#</mathjax>. Note that this is on the VERY high end for oxygen tanks for medical use. Why would you NOT such a tank for scuba diving?</p></div>
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<h1 class="questionTitle" itemprop="name">A tank used for compressed gas contains #9750*mol# oxygen gas, in a volume of #9750*L#, at a temperature of #34.9# #""^@C#...?</h1>
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<div class="markdown"><p>Well, first we use standard units and get, <mathjax>#P=149.4*atm#</mathjax>, and then convert this into <mathjax>#"psi"#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#P=(nRT)/V=(9750*molxx0.0821*(L*atm)/(K*mol)xx308*K)/(1650*L)#</mathjax></p>
<p><mathjax>#=149.4*atm#</mathjax></p>
<p>And then we convert this to <mathjax>#"psi"#</mathjax>; <mathjax>#1*atm-=14.6*"psi"#</mathjax>.</p>
<p>And so <mathjax>#P=149.4*atmxx14.6*"psi"*atm^-1#</mathjax> <mathjax>#=#</mathjax> </p>
<p><mathjax>#"Approx........2000 psi"#</mathjax>. Note that this is on the VERY high end for oxygen tanks for medical use. Why would you NOT such a tank for scuba diving?</p></div>
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</article> | A tank used for compressed gas contains #9750*mol# oxygen gas, in a volume of #9750*L#, at a temperature of #34.9# #""^@C#...? |
What is the pressure inside the tank in #"pounds per square inch"#?
|
3,287 | ac9a3554-6ddd-11ea-8c04-ccda262736ce | https://socratic.org/questions/how-many-moles-of-feo-are-produced-if-8-moles-of-so-2-are-produced-in-2cufes-2-5 | 4.00 moles | start physical_unit 4 4 mole mol qc_end physical_unit 11 11 8 9 mole qc_end chemical_equation 15 28 qc_end end | [{"type":"physical unit","value":"Mole [OF] FeO [IN] moles"}] | [{"type":"physical unit","value":"4.00 moles"}] | [{"type":"physical unit","value":"Mole [OF] SO2 [=] \\pu{8 moles}"},{"type":"chemical equation","value":"2 CuFeS2 + 5 O2 -> 2 Cu + 2 FeO + 4 SO2"}] | <h1 class="questionTitle" itemprop="name">How many moles of #FeO# are produced if 8 moles of #SO_2# are produced in #2CuFeS_2 + 5O_2 -> 2Cu + 2FeO + 4SO_2#?</h1> | null | 4.00 moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So: <mathjax>#8 mol SO_2harr4mol FeO#</mathjax></p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio <mathjax>#FeOdivSO_2=2div4=1div2#</mathjax>, as you can see from the equation.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>So: <mathjax>#8 mol SO_2harr4mol FeO#</mathjax></p></div>
</div>
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</div> | <article>
<h1 class="questionTitle" itemprop="name">How many moles of #FeO# are produced if 8 moles of #SO_2# are produced in #2CuFeS_2 + 5O_2 -> 2Cu + 2FeO + 4SO_2#?</h1>
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<div class="markdown"><p><a href="https://socratic.org/chemistry/the-mole-concept/the-mole">The mole</a> ratio <mathjax>#FeOdivSO_2=2div4=1div2#</mathjax>, as you can see from the equation.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>So: <mathjax>#8 mol SO_2harr4mol FeO#</mathjax></p></div>
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<div class="markdown"><p><mathjax>#"4 moles of FeO"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"In every reaction we can determine amount of reactants and"#</mathjax> <mathjax>#"products by using ratio formulas."#</mathjax></p>
<p>#"In this reaction the ratio formula = </p>
<p><mathjax>#2CuFeS_2 : 5O_2 = 2Cu : 2FeO : 4SO_2#</mathjax></p>
<p>By solving the ratio we see </p>
<p><mathjax>#(2 * 2CuFeS_2) : (5/4 * 5O_2) = (2* 2 Cu) : (2 * 2FeO) : (4 * 2 SO_2)#</mathjax></p>
<p><mathjax>#(4CuFeS_2) : ( 6.2O_2) = (4Cu) : (4FeO) : (8SO_2)#</mathjax></p>
<p>Therefore FeO produced is 4 moles</p></div>
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</article> | How many moles of #FeO# are produced if 8 moles of #SO_2# are produced in #2CuFeS_2 + 5O_2 -> 2Cu + 2FeO + 4SO_2#? | null |
3,288 | abfba8f4-6ddd-11ea-9f9a-ccda262736ce | https://socratic.org/questions/what-is-the-oxidation-number-of-iodine-in-i-2 | 0 | start physical_unit 6 6 oxidation_number none qc_end chemical_equation 8 8 qc_end end | [{"type":"physical unit","value":"Oxidation number [OF] iodine"}] | [{"type":"physical unit","value":"0"}] | [{"type":"chemical equation","value":"I2"}] | <h1 class="questionTitle" itemprop="name">What is the oxidation number of iodine in #I_2#?</h1> | null | 0 | <div class="answerDescription">
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<div class="markdown"><p>By definition, oxidation number is the charge left on the central atom when all its bonds are (conceptually) cleaved with the charge going to the most electronegative atom. So let's cleave the <mathjax>#I-I#</mathjax> bond:</p>
<p><mathjax>#I-I rarr 2I*#</mathjax> . We get two iodine atoms, two <em>neutral</em> iodine radicals, <mathjax>#2xxI*#</mathjax>. Since there is no charge, the oxidation number is a big fat <mathjax>#0#</mathjax>. On the other hand, when we do the same operation with the interhalogen <mathjax>#I-Cl#</mathjax>, the charge goes to the more electronegative atom, <mathjax>#Cl#</mathjax>:</p>
<p><mathjax>#I-Cl rarr I^+ + Cl^-#</mathjax>. To give formal oxidation states of <mathjax>#I^+#</mathjax> of iodine in the molecule, and <mathjax>#I^-#</mathjax> for chlorine (note these are Roman numerals). The same oxidation states would pertain for <mathjax>#Br-Cl#</mathjax>, as chlorine is more electronegative than bromine. What are the <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of fluorine and oxygen in <mathjax>#OF_2#</mathjax>? Please report back the answers here.</p></div>
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<div class="markdown"><p>The oxidation number of <mathjax>#I#</mathjax> in the iodine molecule is <mathjax>#0#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>By definition, oxidation number is the charge left on the central atom when all its bonds are (conceptually) cleaved with the charge going to the most electronegative atom. So let's cleave the <mathjax>#I-I#</mathjax> bond:</p>
<p><mathjax>#I-I rarr 2I*#</mathjax> . We get two iodine atoms, two <em>neutral</em> iodine radicals, <mathjax>#2xxI*#</mathjax>. Since there is no charge, the oxidation number is a big fat <mathjax>#0#</mathjax>. On the other hand, when we do the same operation with the interhalogen <mathjax>#I-Cl#</mathjax>, the charge goes to the more electronegative atom, <mathjax>#Cl#</mathjax>:</p>
<p><mathjax>#I-Cl rarr I^+ + Cl^-#</mathjax>. To give formal oxidation states of <mathjax>#I^+#</mathjax> of iodine in the molecule, and <mathjax>#I^-#</mathjax> for chlorine (note these are Roman numerals). The same oxidation states would pertain for <mathjax>#Br-Cl#</mathjax>, as chlorine is more electronegative than bromine. What are the <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of fluorine and oxygen in <mathjax>#OF_2#</mathjax>? Please report back the answers here.</p></div>
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<div class="markdown"><p>The oxidation number of <mathjax>#I#</mathjax> in the iodine molecule is <mathjax>#0#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>By definition, oxidation number is the charge left on the central atom when all its bonds are (conceptually) cleaved with the charge going to the most electronegative atom. So let's cleave the <mathjax>#I-I#</mathjax> bond:</p>
<p><mathjax>#I-I rarr 2I*#</mathjax> . We get two iodine atoms, two <em>neutral</em> iodine radicals, <mathjax>#2xxI*#</mathjax>. Since there is no charge, the oxidation number is a big fat <mathjax>#0#</mathjax>. On the other hand, when we do the same operation with the interhalogen <mathjax>#I-Cl#</mathjax>, the charge goes to the more electronegative atom, <mathjax>#Cl#</mathjax>:</p>
<p><mathjax>#I-Cl rarr I^+ + Cl^-#</mathjax>. To give formal oxidation states of <mathjax>#I^+#</mathjax> of iodine in the molecule, and <mathjax>#I^-#</mathjax> for chlorine (note these are Roman numerals). The same oxidation states would pertain for <mathjax>#Br-Cl#</mathjax>, as chlorine is more electronegative than bromine. What are the <a href="http://socratic.org/chemistry/electrochemistry/oxidation-numbers">oxidation numbers</a> of fluorine and oxygen in <mathjax>#OF_2#</mathjax>? Please report back the answers here.</p></div>
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</article> | What is the oxidation number of iodine in #I_2#? | null |
3,289 | a8d7f855-6ddd-11ea-b59d-ccda262736ce | https://socratic.org/questions/how-many-neutrons-are-found-in-one-atom-of-36cl | 392.60 kJ | start physical_unit 14 14 heat_energy kj qc_end physical_unit 14 14 8 11 mass qc_end physical_unit 14 14 26 27 enthalpy_of_sublimation qc_end c_other OTHER qc_end end | [{"type":"physical unit","value":"Absorbed energy [OF] CO2 [IN] kJ"}] | [{"type":"physical unit","value":"392.60 kJ"}] | [{"type":"physical unit","value":"Mass [OF] CO2 [=] \\pu{2.0 × 10^3 g }"},{"type":"physical unit","value":"Enthalpy of sublimation [OF] CO2 [=] \\pu{196.3 J/g}"},{"type":"other","value":"Ice (CO2) sublimate at the normal dry sublimation point."}] | <h1 class="questionTitle" itemprop="name">How many neutrons are found in one atom of #""^36Cl#?</h1> | null | 392.60 kJ | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The symbol <mathjax>#""^36"Cl"#</mathjax> is the nuclear notation for chlorine-36, which is the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mass-number">mass number</a>, <mathjax>#color(red)("A")#</mathjax>, for this particular chlorine isotope. <strong>The mass number is the sum of the protons and neutrons in the atomic nuclei of an isotope.</strong> </p>
<p><mathjax>#color(red)("Z")+color(purple)("N")=color(blue)("A")#</mathjax></p>
<p>You can determine the number of protons in a chlorine atom by looking it up on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, where you will see that its <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a>, <mathjax>#color(red)("Z")#</mathjax>, is <mathjax>#17#</mathjax>. </p>
<p><strong>The number of neutrons, <mathjax>#color(purple)("N")#</mathjax>, in an isotope of an element is the difference between the mass number and atomic number.</strong></p>
<p><mathjax>#color(blue)("A")-color(red)("Z")=color(purple)("N")#</mathjax></p>
<p><mathjax>#color(blue)(36)-color(red)(17)=color(purple)(19)"#</mathjax></p>
<p>The number of neutrons in an atom of <mathjax>#""^36"Cl"#</mathjax> isotope is <mathjax>#19#</mathjax>.</p>
<p>You can write the symbol for the isotope chlorine-36 in nuclear notation (also called isotopic notation) as follows.</p>
<p><mathjax>#""_17^36"Cl"#</mathjax></p>
<p>You can subtract the bottom number from the top number to determine the number of neutrons</p>
<p><img alt="http://chemistry-batz.wikispaces.com/04+Atomic+Structure" src="https://useruploads.socratic.org/POFGYY5VRyWT3gaIbwNm_nuclear%20notation.jpg%22%3B+size%3D%2210716"/> </p></div>
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<div class="markdown"><p>The number of neutrons in an atom of <mathjax>#""^36"Cl"#</mathjax> is <mathjax>#19#</mathjax>.</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The symbol <mathjax>#""^36"Cl"#</mathjax> is the nuclear notation for chlorine-36, which is the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mass-number">mass number</a>, <mathjax>#color(red)("A")#</mathjax>, for this particular chlorine isotope. <strong>The mass number is the sum of the protons and neutrons in the atomic nuclei of an isotope.</strong> </p>
<p><mathjax>#color(red)("Z")+color(purple)("N")=color(blue)("A")#</mathjax></p>
<p>You can determine the number of protons in a chlorine atom by looking it up on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, where you will see that its <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a>, <mathjax>#color(red)("Z")#</mathjax>, is <mathjax>#17#</mathjax>. </p>
<p><strong>The number of neutrons, <mathjax>#color(purple)("N")#</mathjax>, in an isotope of an element is the difference between the mass number and atomic number.</strong></p>
<p><mathjax>#color(blue)("A")-color(red)("Z")=color(purple)("N")#</mathjax></p>
<p><mathjax>#color(blue)(36)-color(red)(17)=color(purple)(19)"#</mathjax></p>
<p>The number of neutrons in an atom of <mathjax>#""^36"Cl"#</mathjax> isotope is <mathjax>#19#</mathjax>.</p>
<p>You can write the symbol for the isotope chlorine-36 in nuclear notation (also called isotopic notation) as follows.</p>
<p><mathjax>#""_17^36"Cl"#</mathjax></p>
<p>You can subtract the bottom number from the top number to determine the number of neutrons</p>
<p><img alt="http://chemistry-batz.wikispaces.com/04+Atomic+Structure" src="https://useruploads.socratic.org/POFGYY5VRyWT3gaIbwNm_nuclear%20notation.jpg%22%3B+size%3D%2210716"/> </p></div>
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<div class="markdown"><p>The number of neutrons in an atom of <mathjax>#""^36"Cl"#</mathjax> is <mathjax>#19#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p>The symbol <mathjax>#""^36"Cl"#</mathjax> is the nuclear notation for chlorine-36, which is the <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/mass-number">mass number</a>, <mathjax>#color(red)("A")#</mathjax>, for this particular chlorine isotope. <strong>The mass number is the sum of the protons and neutrons in the atomic nuclei of an isotope.</strong> </p>
<p><mathjax>#color(red)("Z")+color(purple)("N")=color(blue)("A")#</mathjax></p>
<p>You can determine the number of protons in a chlorine atom by looking it up on <a href="https://socratic.org/chemistry/the-periodic-table/the-periodic-table">the periodic table</a>, where you will see that its <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/atomic-number">atomic number</a>, <mathjax>#color(red)("Z")#</mathjax>, is <mathjax>#17#</mathjax>. </p>
<p><strong>The number of neutrons, <mathjax>#color(purple)("N")#</mathjax>, in an isotope of an element is the difference between the mass number and atomic number.</strong></p>
<p><mathjax>#color(blue)("A")-color(red)("Z")=color(purple)("N")#</mathjax></p>
<p><mathjax>#color(blue)(36)-color(red)(17)=color(purple)(19)"#</mathjax></p>
<p>The number of neutrons in an atom of <mathjax>#""^36"Cl"#</mathjax> isotope is <mathjax>#19#</mathjax>.</p>
<p>You can write the symbol for the isotope chlorine-36 in nuclear notation (also called isotopic notation) as follows.</p>
<p><mathjax>#""_17^36"Cl"#</mathjax></p>
<p>You can subtract the bottom number from the top number to determine the number of neutrons</p>
<p><img alt="http://chemistry-batz.wikispaces.com/04+Atomic+Structure" src="https://useruploads.socratic.org/POFGYY5VRyWT3gaIbwNm_nuclear%20notation.jpg%22%3B+size%3D%2210716"/> </p></div>
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</article> | How many neutrons are found in one atom of #""^36Cl#? | null |
3,290 | a96dd12e-6ddd-11ea-8b09-ccda262736ce | https://socratic.org/questions/what-is-the-molarity-of-416-48-g-bacl-2-dissolved-in-2-l-of-water | 1.00 mol/L | start physical_unit 7 7 molarity mol/l qc_end physical_unit 7 7 5 6 mass qc_end physical_unit 13 13 10 11 volume qc_end end | [{"type":"physical unit","value":"Molarity [OF] BaCl2 solution [IN] mol/L"}] | [{"type":"physical unit","value":"1.00 mol/L"}] | [{"type":"physical unit","value":"Mass [OF] BaCl2 [=] \\pu{416.48 g}"},{"type":"physical unit","value":"Volume [OF] water [=] \\pu{2 L}"}] | <h1 class="questionTitle" itemprop="name">What is the molarity of 416.48 g #BaCl_2# dissolved in #2# #L# of water?</h1> | null | 1.00 mol/L | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> = # of moles of solute/ Volume of solution in ( liter)</p>
<p>Number of moles of Ba<mathjax>#Cl_2#</mathjax> =<br/>
mass of Ba<mathjax>#Cl_2#</mathjax> dissolved/ molar mass of Ba<mathjax>#Cl_2#</mathjax> </p>
<p>mass of Ba<mathjax>#Cl_2#</mathjax> dissolved = 416.48 g </p>
<p>molar mass of Ba<mathjax>#Cl_2#</mathjax> = 208.23 g/mol</p>
<p>Number of moles of Ba<mathjax>#Cl_2#</mathjax> = 416.48 g / 208.23 g /mol = 2 mol</p>
<p>Molarity = # of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/ Volume of solution in ( liter)</p>
<pre><code> = 2 moles / 2 L
= 1 mol /L
</code></pre></div>
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</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>1 mol / L</p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> = # of moles of solute/ Volume of solution in ( liter)</p>
<p>Number of moles of Ba<mathjax>#Cl_2#</mathjax> =<br/>
mass of Ba<mathjax>#Cl_2#</mathjax> dissolved/ molar mass of Ba<mathjax>#Cl_2#</mathjax> </p>
<p>mass of Ba<mathjax>#Cl_2#</mathjax> dissolved = 416.48 g </p>
<p>molar mass of Ba<mathjax>#Cl_2#</mathjax> = 208.23 g/mol</p>
<p>Number of moles of Ba<mathjax>#Cl_2#</mathjax> = 416.48 g / 208.23 g /mol = 2 mol</p>
<p>Molarity = # of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/ Volume of solution in ( liter)</p>
<pre><code> = 2 moles / 2 L
= 1 mol /L
</code></pre></div>
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<h1 class="questionTitle" itemprop="name">What is the molarity of 416.48 g #BaCl_2# dissolved in #2# #L# of water?</h1>
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<div class="markdown"><p>1 mol / L</p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><a href="https://socratic.org/chemistry/solutions-and-their-behavior/molarity">Molarity</a> = # of moles of solute/ Volume of solution in ( liter)</p>
<p>Number of moles of Ba<mathjax>#Cl_2#</mathjax> =<br/>
mass of Ba<mathjax>#Cl_2#</mathjax> dissolved/ molar mass of Ba<mathjax>#Cl_2#</mathjax> </p>
<p>mass of Ba<mathjax>#Cl_2#</mathjax> dissolved = 416.48 g </p>
<p>molar mass of Ba<mathjax>#Cl_2#</mathjax> = 208.23 g/mol</p>
<p>Number of moles of Ba<mathjax>#Cl_2#</mathjax> = 416.48 g / 208.23 g /mol = 2 mol</p>
<p>Molarity = # of moles of <a href="https://socratic.org/chemistry/solutions-and-their-behavior/solute">solute</a>/ Volume of solution in ( liter)</p>
<pre><code> = 2 moles / 2 L
= 1 mol /L
</code></pre></div>
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</article> | What is the molarity of 416.48 g #BaCl_2# dissolved in #2# #L# of water? | null |
3,291 | ac486595-6ddd-11ea-a780-ccda262736ce | https://socratic.org/questions/what-is-the-ph-of-a-solution-if-the-h-3-45-10-9-m | 8.46 | start physical_unit 6 6 ph none qc_end physical_unit 6 6 11 14 [h+] qc_end end | [{"type":"physical unit","value":"pH [OF] the solution"}] | [{"type":"physical unit","value":"8.46"}] | [{"type":"physical unit","value":"[H+] [OF] the solution [=] \\pu{3.45 × 10^(-9) M}"}] | <h1 class="questionTitle" itemprop="name">What is the pH of a solution if the #[H^+] = 3.45 * 10^-9 M?#</h1> | null | 8.46 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10){3.45xx10^-9}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#8.5#</mathjax>.</p>
<p>This value is reasonable in that since <mathjax>#[H^+]<10^(-7)#</mathjax>, the solution is slightly basic, and the <mathjax>#pH#</mathjax> of course reflects this. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
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<div class="markdown"><p><mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H^+]#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10){3.45xx10^-9}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#8.5#</mathjax>.</p>
<p>This value is reasonable in that since <mathjax>#[H^+]<10^(-7)#</mathjax>, the solution is slightly basic, and the <mathjax>#pH#</mathjax> of course reflects this. </p></div>
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<h1 class="questionTitle" itemprop="name">What is the pH of a solution if the #[H^+] = 3.45 * 10^-9 M?#</h1>
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anor277
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<div class="markdown"><p><mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10)[H^+]#</mathjax></p></div>
</div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#pH#</mathjax> <mathjax>#=#</mathjax> <mathjax>#-log_(10){3.45xx10^-9}#</mathjax> <mathjax>#=#</mathjax> <mathjax>#8.5#</mathjax>.</p>
<p>This value is reasonable in that since <mathjax>#[H^+]<10^(-7)#</mathjax>, the solution is slightly basic, and the <mathjax>#pH#</mathjax> of course reflects this. </p></div>
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</article> | What is the pH of a solution if the #[H^+] = 3.45 * 10^-9 M?# | null |
3,292 | aad44eb9-6ddd-11ea-b50c-ccda262736ce | https://socratic.org/questions/what-is-the-empirical-formula-of-a-substance-with-1-587-g-arsenic-and-3-755-g-of | AsCl5 | start chemical_formula qc_end physical_unit 11 11 9 10 mass qc_end physical_unit 16 16 13 14 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the substance [IN] empirical"}] | [{"type":"chemical equation","value":"AsCl5"}] | [{"type":"physical unit","value":"Mass [OF] arsenic [=] \\pu{1.587 g}"},{"type":"physical unit","value":"Mass [OF] chlorine [=] \\pu{3.755 g}"}] | <h1 class="questionTitle" itemprop="name">What is the empirical formula of a substance with 1.587 g arsenic and 3.755 g of chlorine?</h1> | null | AsCl5 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find a compound's <strong>empirical formula</strong>, you must find the smallest whole number ratio that exists between its constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. </p>
<p>In your case, the compound is said to contain arsenic, <mathjax>#"As"#</mathjax>, and chlorine, <mathjax>#"Cl"#</mathjax>. The problem provides you with <em>grams</em> of arsenic and <em>grams</em> of chlorine, so your first goal will be to convert these to <strong>moles</strong>.</p>
<p>To do that, use the <strong>molar masses</strong> of the two elements</p>
<blockquote>
<p><mathjax>#M_("M As") = "74.922 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M Cl") = "35.453 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>Your sample will thus contain</p>
<blockquote>
<p><mathjax>#"For As: " 1.587 color(red)(cancel(color(black)("g"))) * "1 mole As"/(74.922 color(red)(cancel(color(black)("g")))) = "0.02118 moles As"#</mathjax></p>
<p><mathjax>#"For Cl: " 3.755 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453color(red)(cancel(color(black)("g")))) = "0.1059 moles Cl"#</mathjax></p>
</blockquote>
<p>To find the <strong>mole ratio</strong> that exists between the two elements, divide both values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For As: " (0.02118 color(red)(cancel(color(black)("moles"))))/(0.02118color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For Cl: " (0.1059color(red)(cancel(color(black)("moles"))))/(0.02118color(red)(cancel(color(black)("moles")))) = 5#</mathjax></p>
</blockquote>
<p>Since a <mathjax>#1:5#</mathjax> already represents the <strong>smallest whole number ratio</strong> that can exist between arsenic and chlorine in the compound, you can say that the empirical formula will be </p>
<blockquote>
<p><mathjax>#"As"_1"Cl"_5 implies color(green)(|bar(ul(color(white)(a/a)color(black)("AsCl"_5)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>#"AsCl"_5#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find a compound's <strong>empirical formula</strong>, you must find the smallest whole number ratio that exists between its constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. </p>
<p>In your case, the compound is said to contain arsenic, <mathjax>#"As"#</mathjax>, and chlorine, <mathjax>#"Cl"#</mathjax>. The problem provides you with <em>grams</em> of arsenic and <em>grams</em> of chlorine, so your first goal will be to convert these to <strong>moles</strong>.</p>
<p>To do that, use the <strong>molar masses</strong> of the two elements</p>
<blockquote>
<p><mathjax>#M_("M As") = "74.922 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M Cl") = "35.453 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>Your sample will thus contain</p>
<blockquote>
<p><mathjax>#"For As: " 1.587 color(red)(cancel(color(black)("g"))) * "1 mole As"/(74.922 color(red)(cancel(color(black)("g")))) = "0.02118 moles As"#</mathjax></p>
<p><mathjax>#"For Cl: " 3.755 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453color(red)(cancel(color(black)("g")))) = "0.1059 moles Cl"#</mathjax></p>
</blockquote>
<p>To find the <strong>mole ratio</strong> that exists between the two elements, divide both values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For As: " (0.02118 color(red)(cancel(color(black)("moles"))))/(0.02118color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For Cl: " (0.1059color(red)(cancel(color(black)("moles"))))/(0.02118color(red)(cancel(color(black)("moles")))) = 5#</mathjax></p>
</blockquote>
<p>Since a <mathjax>#1:5#</mathjax> already represents the <strong>smallest whole number ratio</strong> that can exist between arsenic and chlorine in the compound, you can say that the empirical formula will be </p>
<blockquote>
<p><mathjax>#"As"_1"Cl"_5 implies color(green)(|bar(ul(color(white)(a/a)color(black)("AsCl"_5)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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<h1 class="questionTitle" itemprop="name">What is the empirical formula of a substance with 1.587 g arsenic and 3.755 g of chlorine?</h1>
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Stefan V.
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<span class="dateCreated" datetime="2016-07-30T02:02:00" itemprop="dateCreated">
Jul 30, 2016
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<div class="markdown"><p><mathjax>#"AsCl"_5#</mathjax></p></div>
</div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to find a compound's <strong>empirical formula</strong>, you must find the smallest whole number ratio that exists between its constituent <a href="https://socratic.org/chemistry/a-first-introduction-to-matter/elements">elements</a>. </p>
<p>In your case, the compound is said to contain arsenic, <mathjax>#"As"#</mathjax>, and chlorine, <mathjax>#"Cl"#</mathjax>. The problem provides you with <em>grams</em> of arsenic and <em>grams</em> of chlorine, so your first goal will be to convert these to <strong>moles</strong>.</p>
<p>To do that, use the <strong>molar masses</strong> of the two elements</p>
<blockquote>
<p><mathjax>#M_("M As") = "74.922 g mol"^(-1)#</mathjax></p>
<p><mathjax>#M_("M Cl") = "35.453 g mol"^(-1)#</mathjax></p>
</blockquote>
<p>Your sample will thus contain</p>
<blockquote>
<p><mathjax>#"For As: " 1.587 color(red)(cancel(color(black)("g"))) * "1 mole As"/(74.922 color(red)(cancel(color(black)("g")))) = "0.02118 moles As"#</mathjax></p>
<p><mathjax>#"For Cl: " 3.755 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453color(red)(cancel(color(black)("g")))) = "0.1059 moles Cl"#</mathjax></p>
</blockquote>
<p>To find the <strong>mole ratio</strong> that exists between the two elements, divide both values by the <em>smallest one</em></p>
<blockquote>
<p><mathjax>#"For As: " (0.02118 color(red)(cancel(color(black)("moles"))))/(0.02118color(red)(cancel(color(black)("moles")))) = 1#</mathjax></p>
<p><mathjax>#"For Cl: " (0.1059color(red)(cancel(color(black)("moles"))))/(0.02118color(red)(cancel(color(black)("moles")))) = 5#</mathjax></p>
</blockquote>
<p>Since a <mathjax>#1:5#</mathjax> already represents the <strong>smallest whole number ratio</strong> that can exist between arsenic and chlorine in the compound, you can say that the empirical formula will be </p>
<blockquote>
<p><mathjax>#"As"_1"Cl"_5 implies color(green)(|bar(ul(color(white)(a/a)color(black)("AsCl"_5)color(white)(a/a)|)))#</mathjax></p>
</blockquote></div>
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</article> | What is the empirical formula of a substance with 1.587 g arsenic and 3.755 g of chlorine? | null |
3,293 | ab593b3a-6ddd-11ea-bed9-ccda262736ce | https://socratic.org/questions/57e863e111ef6b7f26033348 | 3.24 g | start physical_unit 3 4 mass g qc_end physical_unit 15 15 11 12 mass qc_end chemical_equation 18 18 qc_end end | [{"type":"physical unit","value":"Mass [OF] chlorine gas [IN] g"}] | [{"type":"physical unit","value":"3.24 g"}] | [{"type":"physical unit","value":"Mass [OF] phosphorus [=] \\pu{0.567 g}"},{"type":"chemical equation","value":"PCl5"}] | <h1 class="questionTitle" itemprop="name">What mass of chlorine gas is required to react with a #0.567*g# mass of phosphorus to make #PCl_5#?</h1> | null | 3.24 g | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a chemical reaction:</p>
<p><mathjax>#P(s) + 5/2Cl_2(g) rarr PCl_5(s)#</mathjax></p>
<p>Charge and mass are balanced as required. And we need (ii) equivalents weights of the reactants:</p>
<p><mathjax>#"Moles of phosphorus"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.567*g)/(31.00*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.83xx10^-2*mol#</mathjax>.</p>
<p>And thus we need <mathjax>#5/2#</mathjax> equiv of chlorine gas; i.e. <mathjax>#5/2xx1.83xx10^-2*molxx70.90*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p>
<p>Note that I treated phosphorus as <mathjax>#P#</mathjax>, when white phosphorus is actually <mathjax>#P_4#</mathjax>; I also knew that the elemental halogens are bimolecular. Capisce?</p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p>A bit over <mathjax>#3*g#</mathjax> of <mathjax>#Cl_2#</mathjax> gas. </p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a chemical reaction:</p>
<p><mathjax>#P(s) + 5/2Cl_2(g) rarr PCl_5(s)#</mathjax></p>
<p>Charge and mass are balanced as required. And we need (ii) equivalents weights of the reactants:</p>
<p><mathjax>#"Moles of phosphorus"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.567*g)/(31.00*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.83xx10^-2*mol#</mathjax>.</p>
<p>And thus we need <mathjax>#5/2#</mathjax> equiv of chlorine gas; i.e. <mathjax>#5/2xx1.83xx10^-2*molxx70.90*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p>
<p>Note that I treated phosphorus as <mathjax>#P#</mathjax>, when white phosphorus is actually <mathjax>#P_4#</mathjax>; I also knew that the elemental halogens are bimolecular. Capisce?</p></div>
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<h1 class="questionTitle" itemprop="name">What mass of chlorine gas is required to react with a #0.567*g# mass of phosphorus to make #PCl_5#?</h1>
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<div class="markdown"><p>A bit over <mathjax>#3*g#</mathjax> of <mathjax>#Cl_2#</mathjax> gas. </p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>We need (i) a chemical reaction:</p>
<p><mathjax>#P(s) + 5/2Cl_2(g) rarr PCl_5(s)#</mathjax></p>
<p>Charge and mass are balanced as required. And we need (ii) equivalents weights of the reactants:</p>
<p><mathjax>#"Moles of phosphorus"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.567*g)/(31.00*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1.83xx10^-2*mol#</mathjax>.</p>
<p>And thus we need <mathjax>#5/2#</mathjax> equiv of chlorine gas; i.e. <mathjax>#5/2xx1.83xx10^-2*molxx70.90*g*mol^-1#</mathjax> <mathjax>#=#</mathjax> <mathjax>#??g#</mathjax></p>
<p>Note that I treated phosphorus as <mathjax>#P#</mathjax>, when white phosphorus is actually <mathjax>#P_4#</mathjax>; I also knew that the elemental halogens are bimolecular. Capisce?</p></div>
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</article> | What mass of chlorine gas is required to react with a #0.567*g# mass of phosphorus to make #PCl_5#? | null |
3,294 | ac9a1ee9-6ddd-11ea-88f1-ccda262736ce | https://socratic.org/questions/how-many-grams-of-salt-can-be-dissolved-in-100-g-of-h2o-at-80-degrees-celsius | 40 grams | start physical_unit 4 4 mass g qc_end physical_unit 12 12 9 10 mass qc_end physical_unit 12 12 14 16 temperature qc_end end | [{"type":"physical unit","value":"Mass [OF] salt [IN] grams"}] | [{"type":"physical unit","value":"40 grams"}] | [{"type":"physical unit","value":"Mass [OF] H2O [=] \\pu{100 g}"},{"type":"physical unit","value":"Temperature [OF] H2O [=] \\pu{80 degrees Celsius}"}] | <h1 class="questionTitle" itemprop="name">How many grams of salt can be dissolved in 100 g of H2O at 80 degrees Celsius?</h1> | null | 40 grams | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can determine how many grams of <em>sodium chloride</em>, commonly known as table salt, can be dissolved in <mathjax>#"100 g"#</mathjax> of water at <mathjax>#80^@"C"#</mathjax> by taking a look at its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solubility-graphs">solubility graph</a>.</p>
<p>A substance's <em>solubility graph</em> tells you how its solubility changes, let's say starting from room temperature, when temperature is either decreased or increased. </p>
<p><img alt="http://www.sciencegeek.net/Chemistry/taters/solubility.htm" src="https://useruploads.socratic.org/XpeZ5i1lTjq4vNDJVsoQ_solubility.gif"/> </p>
<p>As you can see, at room temperature sodium chloride has a solubility of approximately <mathjax>#"35 g/100 g H"_2"O"#</mathjax>. </p>
<p>As temperature <strong>increases</strong>, its solubility <strong>increases</strong> as well. Notice, however, that it does not increase significantly. In fact, you can expect to be able to dissolve no more than <mathjax>#"40 g"#</mathjax> of sodium chloride <em>per</em> <mathjax>#"100 g"#</mathjax> of water at <mathjax>#80^@"C"#</mathjax>. </p>
<p>So, at <mathjax>#80^@"C"#</mathjax>, you will ahve an <strong>unsaturated</strong> solution if you dissolve less than <mathjax>#"40 g"#</mathjax> of sodium chloride, and a <strong>saturated</strong> solution if you dissolve about <mathjax>#"40 g"#</mathjax> of sodium chloride. </p>
<p>
<iframe src="https://www.youtube.com/embed/D2NAw-A0V1s?origin=https://socratic.org&wmode=transparent" type="text/html"></iframe>
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<div class="markdown"><p>About <mathjax>#"40 g"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can determine how many grams of <em>sodium chloride</em>, commonly known as table salt, can be dissolved in <mathjax>#"100 g"#</mathjax> of water at <mathjax>#80^@"C"#</mathjax> by taking a look at its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solubility-graphs">solubility graph</a>.</p>
<p>A substance's <em>solubility graph</em> tells you how its solubility changes, let's say starting from room temperature, when temperature is either decreased or increased. </p>
<p><img alt="http://www.sciencegeek.net/Chemistry/taters/solubility.htm" src="https://useruploads.socratic.org/XpeZ5i1lTjq4vNDJVsoQ_solubility.gif"/> </p>
<p>As you can see, at room temperature sodium chloride has a solubility of approximately <mathjax>#"35 g/100 g H"_2"O"#</mathjax>. </p>
<p>As temperature <strong>increases</strong>, its solubility <strong>increases</strong> as well. Notice, however, that it does not increase significantly. In fact, you can expect to be able to dissolve no more than <mathjax>#"40 g"#</mathjax> of sodium chloride <em>per</em> <mathjax>#"100 g"#</mathjax> of water at <mathjax>#80^@"C"#</mathjax>. </p>
<p>So, at <mathjax>#80^@"C"#</mathjax>, you will ahve an <strong>unsaturated</strong> solution if you dissolve less than <mathjax>#"40 g"#</mathjax> of sodium chloride, and a <strong>saturated</strong> solution if you dissolve about <mathjax>#"40 g"#</mathjax> of sodium chloride. </p>
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<h1 class="questionTitle" itemprop="name">How many grams of salt can be dissolved in 100 g of H2O at 80 degrees Celsius?</h1>
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Stefan V.
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<div class="markdown"><p>About <mathjax>#"40 g"#</mathjax></p></div>
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<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>You can determine how many grams of <em>sodium chloride</em>, commonly known as table salt, can be dissolved in <mathjax>#"100 g"#</mathjax> of water at <mathjax>#80^@"C"#</mathjax> by taking a look at its <a href="http://socratic.org/chemistry/solutions-and-their-behavior/solubility-graphs">solubility graph</a>.</p>
<p>A substance's <em>solubility graph</em> tells you how its solubility changes, let's say starting from room temperature, when temperature is either decreased or increased. </p>
<p><img alt="http://www.sciencegeek.net/Chemistry/taters/solubility.htm" src="https://useruploads.socratic.org/XpeZ5i1lTjq4vNDJVsoQ_solubility.gif"/> </p>
<p>As you can see, at room temperature sodium chloride has a solubility of approximately <mathjax>#"35 g/100 g H"_2"O"#</mathjax>. </p>
<p>As temperature <strong>increases</strong>, its solubility <strong>increases</strong> as well. Notice, however, that it does not increase significantly. In fact, you can expect to be able to dissolve no more than <mathjax>#"40 g"#</mathjax> of sodium chloride <em>per</em> <mathjax>#"100 g"#</mathjax> of water at <mathjax>#80^@"C"#</mathjax>. </p>
<p>So, at <mathjax>#80^@"C"#</mathjax>, you will ahve an <strong>unsaturated</strong> solution if you dissolve less than <mathjax>#"40 g"#</mathjax> of sodium chloride, and a <strong>saturated</strong> solution if you dissolve about <mathjax>#"40 g"#</mathjax> of sodium chloride. </p>
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</article> | How many grams of salt can be dissolved in 100 g of H2O at 80 degrees Celsius? | null |
3,295 | aca8fd41-6ddd-11ea-a2ef-ccda262736ce | https://socratic.org/questions/consider-the-following-reaction-xe-g-2f-2-g-xef-4-g-a-reaction-mixture-initially | 25.3 | start physical_unit 40 41 kp none qc_end chemical_equation 4 9 qc_end physical_unit 4 4 15 16 pressure qc_end physical_unit 7 7 19 20 pressure qc_end physical_unit 4 4 29 30 pressure qc_end end | [{"type":"physical unit","value":"Kp [OF] the reaction"}] | [{"type":"physical unit","value":"25.3"}] | [{"type":"chemical equation","value":"Xe(g) + 2 F2(g) -> XeF4(g)"},{"type":"physical unit","value":"Pressure [OF] Xe [=] \\pu{2.24 atm}"},{"type":"physical unit","value":"Pressure [OF] F2 [=] \\pu{4.27 atm}"},{"type":"physical unit","value":"Equilibrium pressure [OF] Xe [=] \\pu{0.34 atm}"}] | <h1 class="questionTitle" itemprop="name">Consider the following reaction: #Xe(g) + 2F_2(g) -> XeF_4(g)#. A reaction mixture initially contains 2.24 atm #Xe# and 4.27 atm #F_2#. lf the equilibrium pressure of #Xe# is 0.34 atm, how do you find the equilibrium constant (#K_p#) for the reaction?</h1> | null | 25.3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equilibrium reaction given to you looks like this </p>
<blockquote>
<p><mathjax>#"Xe"_ ((g)) + color(red)(2)"F"_ (2(g)) rightleftharpoons "XeF"_ (4(g))#</mathjax></p>
</blockquote>
<p>You know that the mixture initially contains <mathjax>#"2.24 atm"#</mathjax> of xenon, <mathjax>#"Xe"#</mathjax>, and <mathjax>#"4.27 atm"#</mathjax> of fluorine gas, <mathjax>#"F"_2#</mathjax>. </p>
<p>It's worth mentioning that the problem provides you with <em>pressures</em> instead of <em>moles</em> because when the volume of the container <strong>remains unchanged</strong>, and the temperature of the reaction is <strong>kept constant</strong>, the pressure of a gas is <strong>directly proportional</strong> to the number of moles of gas present. </p>
<p>Now, notice that the pressure of xenon is <strong>decreasing</strong> to an equilibrium value of <mathjax>#"0.34 atm"#</mathjax>. This is a <em>significant decrease</em>, which can only mean that the equilibrium constant, <mathjax>#K_p#</mathjax>, is <strong>greater than</strong> <mathjax>#1#</mathjax>. </p>
<p>In other words, the <em>forward reaction</em> is <strong>favored</strong> at the temperature at which the reaction takes place. </p>
<p>Consequently, you can expect the <strong>equilibrium pressure</strong> of fluorine gas to be <em>significantly lower</em> than its initial value. </p>
<p>So, use an <strong>ICE table</strong> to find the equilibrium pressures of fluorine gas and xenon tetrafluoride</p>
<blockquote>
<blockquote>
<p><mathjax>#" ""Xe"_ ((g)) " "+" " color(red)(2)"F"_ (2(g)) " "rightleftharpoons" " "XeF"_ (4(g))#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")color(white)(aaaaacolor(black)(2.24)aaaaaaaacolor(black)(4.27)aaaaaaaaaaacolor(black)(0)#</mathjax><br/>
<mathjax>#color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaacolor(black)((-color(red)(2)x))aaaaaaaaacolor(black)((+x))#</mathjax><br/>
<mathjax>#color(purple)("E")color(white)(aaacolor(black)(2.24-x)aaaacolor(black)(4.27-color(red)(2)x)aaaaaaaaacolor(black)(x)#</mathjax></p>
</blockquote>
<p>Now, you know that the equilibrium concentration of xenon is <mathjax>#"0.34 atm"#</mathjax>. This means that you have</p>
<blockquote>
<p><mathjax>#2.24 - x = 0.34 implies x = 1.90#</mathjax></p>
</blockquote>
<p>The equilibrium pressures of fluorine gas and xenon tetrafluoride will thus be </p>
<blockquote>
<p><mathjax>#("F"_2) = 4.27 - color(red)(2) * 1.90 = "0.47 atm"#</mathjax></p>
<p><mathjax>#("XeF"_4) = "1.90 atm"#</mathjax></p>
</blockquote>
<p>By definition, <mathjax>#K_p#</mathjax> will be equal to </p>
<blockquote>
<p><mathjax>#K_ p = (("XeF"_4))/(("Xe") * ("F"_2)^color(red)(2))#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#K_p = (1.90 color(red)(cancel(color(black)("atm"))))/(0.34color(red)(cancel(color(black)("atm"))) * ("0.47 atm")^color(red)(2)) = "25.3 atm"^(-2)#</mathjax></p>
</blockquote>
<p>Equilibrium constant are usually given <em>without added units</em>, which means that your answer will be </p>
<blockquote>
<p><mathjax>#K_p = color(green)(|bar(ul(color(white)(a/a)color(black)(25.3)color(white)(a/a)|))) ->#</mathjax> <em>rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote>
<p>As predicted, the equilibrium constant turned out to be <em>grater than</em> <mathjax>#1#</mathjax>. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#K_ p = 25.3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equilibrium reaction given to you looks like this </p>
<blockquote>
<p><mathjax>#"Xe"_ ((g)) + color(red)(2)"F"_ (2(g)) rightleftharpoons "XeF"_ (4(g))#</mathjax></p>
</blockquote>
<p>You know that the mixture initially contains <mathjax>#"2.24 atm"#</mathjax> of xenon, <mathjax>#"Xe"#</mathjax>, and <mathjax>#"4.27 atm"#</mathjax> of fluorine gas, <mathjax>#"F"_2#</mathjax>. </p>
<p>It's worth mentioning that the problem provides you with <em>pressures</em> instead of <em>moles</em> because when the volume of the container <strong>remains unchanged</strong>, and the temperature of the reaction is <strong>kept constant</strong>, the pressure of a gas is <strong>directly proportional</strong> to the number of moles of gas present. </p>
<p>Now, notice that the pressure of xenon is <strong>decreasing</strong> to an equilibrium value of <mathjax>#"0.34 atm"#</mathjax>. This is a <em>significant decrease</em>, which can only mean that the equilibrium constant, <mathjax>#K_p#</mathjax>, is <strong>greater than</strong> <mathjax>#1#</mathjax>. </p>
<p>In other words, the <em>forward reaction</em> is <strong>favored</strong> at the temperature at which the reaction takes place. </p>
<p>Consequently, you can expect the <strong>equilibrium pressure</strong> of fluorine gas to be <em>significantly lower</em> than its initial value. </p>
<p>So, use an <strong>ICE table</strong> to find the equilibrium pressures of fluorine gas and xenon tetrafluoride</p>
<blockquote>
<blockquote>
<p><mathjax>#" ""Xe"_ ((g)) " "+" " color(red)(2)"F"_ (2(g)) " "rightleftharpoons" " "XeF"_ (4(g))#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")color(white)(aaaaacolor(black)(2.24)aaaaaaaacolor(black)(4.27)aaaaaaaaaaacolor(black)(0)#</mathjax><br/>
<mathjax>#color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaacolor(black)((-color(red)(2)x))aaaaaaaaacolor(black)((+x))#</mathjax><br/>
<mathjax>#color(purple)("E")color(white)(aaacolor(black)(2.24-x)aaaacolor(black)(4.27-color(red)(2)x)aaaaaaaaacolor(black)(x)#</mathjax></p>
</blockquote>
<p>Now, you know that the equilibrium concentration of xenon is <mathjax>#"0.34 atm"#</mathjax>. This means that you have</p>
<blockquote>
<p><mathjax>#2.24 - x = 0.34 implies x = 1.90#</mathjax></p>
</blockquote>
<p>The equilibrium pressures of fluorine gas and xenon tetrafluoride will thus be </p>
<blockquote>
<p><mathjax>#("F"_2) = 4.27 - color(red)(2) * 1.90 = "0.47 atm"#</mathjax></p>
<p><mathjax>#("XeF"_4) = "1.90 atm"#</mathjax></p>
</blockquote>
<p>By definition, <mathjax>#K_p#</mathjax> will be equal to </p>
<blockquote>
<p><mathjax>#K_ p = (("XeF"_4))/(("Xe") * ("F"_2)^color(red)(2))#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#K_p = (1.90 color(red)(cancel(color(black)("atm"))))/(0.34color(red)(cancel(color(black)("atm"))) * ("0.47 atm")^color(red)(2)) = "25.3 atm"^(-2)#</mathjax></p>
</blockquote>
<p>Equilibrium constant are usually given <em>without added units</em>, which means that your answer will be </p>
<blockquote>
<p><mathjax>#K_p = color(green)(|bar(ul(color(white)(a/a)color(black)(25.3)color(white)(a/a)|))) ->#</mathjax> <em>rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote>
<p>As predicted, the equilibrium constant turned out to be <em>grater than</em> <mathjax>#1#</mathjax>. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">Consider the following reaction: #Xe(g) + 2F_2(g) -> XeF_4(g)#. A reaction mixture initially contains 2.24 atm #Xe# and 4.27 atm #F_2#. lf the equilibrium pressure of #Xe# is 0.34 atm, how do you find the equilibrium constant (#K_p#) for the reaction?</h1>
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Stefan V.
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<div class="markdown"><p><mathjax>#K_ p = 25.3#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>The equilibrium reaction given to you looks like this </p>
<blockquote>
<p><mathjax>#"Xe"_ ((g)) + color(red)(2)"F"_ (2(g)) rightleftharpoons "XeF"_ (4(g))#</mathjax></p>
</blockquote>
<p>You know that the mixture initially contains <mathjax>#"2.24 atm"#</mathjax> of xenon, <mathjax>#"Xe"#</mathjax>, and <mathjax>#"4.27 atm"#</mathjax> of fluorine gas, <mathjax>#"F"_2#</mathjax>. </p>
<p>It's worth mentioning that the problem provides you with <em>pressures</em> instead of <em>moles</em> because when the volume of the container <strong>remains unchanged</strong>, and the temperature of the reaction is <strong>kept constant</strong>, the pressure of a gas is <strong>directly proportional</strong> to the number of moles of gas present. </p>
<p>Now, notice that the pressure of xenon is <strong>decreasing</strong> to an equilibrium value of <mathjax>#"0.34 atm"#</mathjax>. This is a <em>significant decrease</em>, which can only mean that the equilibrium constant, <mathjax>#K_p#</mathjax>, is <strong>greater than</strong> <mathjax>#1#</mathjax>. </p>
<p>In other words, the <em>forward reaction</em> is <strong>favored</strong> at the temperature at which the reaction takes place. </p>
<p>Consequently, you can expect the <strong>equilibrium pressure</strong> of fluorine gas to be <em>significantly lower</em> than its initial value. </p>
<p>So, use an <strong>ICE table</strong> to find the equilibrium pressures of fluorine gas and xenon tetrafluoride</p>
<blockquote>
<blockquote>
<p><mathjax>#" ""Xe"_ ((g)) " "+" " color(red)(2)"F"_ (2(g)) " "rightleftharpoons" " "XeF"_ (4(g))#</mathjax></p>
</blockquote>
<p><mathjax>#color(purple)("I")color(white)(aaaaacolor(black)(2.24)aaaaaaaacolor(black)(4.27)aaaaaaaaaaacolor(black)(0)#</mathjax><br/>
<mathjax>#color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaacolor(black)((-color(red)(2)x))aaaaaaaaacolor(black)((+x))#</mathjax><br/>
<mathjax>#color(purple)("E")color(white)(aaacolor(black)(2.24-x)aaaacolor(black)(4.27-color(red)(2)x)aaaaaaaaacolor(black)(x)#</mathjax></p>
</blockquote>
<p>Now, you know that the equilibrium concentration of xenon is <mathjax>#"0.34 atm"#</mathjax>. This means that you have</p>
<blockquote>
<p><mathjax>#2.24 - x = 0.34 implies x = 1.90#</mathjax></p>
</blockquote>
<p>The equilibrium pressures of fluorine gas and xenon tetrafluoride will thus be </p>
<blockquote>
<p><mathjax>#("F"_2) = 4.27 - color(red)(2) * 1.90 = "0.47 atm"#</mathjax></p>
<p><mathjax>#("XeF"_4) = "1.90 atm"#</mathjax></p>
</blockquote>
<p>By definition, <mathjax>#K_p#</mathjax> will be equal to </p>
<blockquote>
<p><mathjax>#K_ p = (("XeF"_4))/(("Xe") * ("F"_2)^color(red)(2))#</mathjax></p>
</blockquote>
<p>Plug in your values to find</p>
<blockquote>
<p><mathjax>#K_p = (1.90 color(red)(cancel(color(black)("atm"))))/(0.34color(red)(cancel(color(black)("atm"))) * ("0.47 atm")^color(red)(2)) = "25.3 atm"^(-2)#</mathjax></p>
</blockquote>
<p>Equilibrium constant are usually given <em>without added units</em>, which means that your answer will be </p>
<blockquote>
<p><mathjax>#K_p = color(green)(|bar(ul(color(white)(a/a)color(black)(25.3)color(white)(a/a)|))) ->#</mathjax> <em>rounded to three <strong><a href="https://socratic.org/chemistry/measurement-in-chemistry/significant-figures">sig figs</a></strong></em></p>
</blockquote>
<p>As predicted, the equilibrium constant turned out to be <em>grater than</em> <mathjax>#1#</mathjax>. </p></div>
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</article> | Consider the following reaction: #Xe(g) + 2F_2(g) -> XeF_4(g)#. A reaction mixture initially contains 2.24 atm #Xe# and 4.27 atm #F_2#. lf the equilibrium pressure of #Xe# is 0.34 atm, how do you find the equilibrium constant (#K_p#) for the reaction? | null |
3,296 | abfb50a5-6ddd-11ea-9bd2-ccda262736ce | https://socratic.org/questions/what-pressure-will-be-exerted-by-2-01-mol-hydrogen-gas-in-a-6-5-l-cylinder-at-20 | 7.4 atm | start physical_unit 8 9 pressure atm qc_end physical_unit 8 9 6 7 mole qc_end physical_unit 8 9 16 17 temperature qc_end physical_unit 8 9 12 13 volume qc_end end | [{"type":"physical unit","value":"Pressure [OF] hydrogen gas [IN] atm"}] | [{"type":"physical unit","value":"7.4 atm"}] | [{"type":"physical unit","value":"Mole [OF] hydrogen gas [=] \\pu{2.01 mol}"},{"type":"physical unit","value":"Temperature [OF] hydrogen gas [=] \\pu{20 ℃}"},{"type":"physical unit","value":"Volume [OF] hydrogen gas [=] \\pu{6.5 L}"}] | <h1 class="questionTitle" itemprop="name">What pressure will be exerted by 2.01 mol hydrogen gas in a 6.5 L cylinder at 20°C?</h1> | null | 7.4 atm | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to solve this problem, you must be familiar with the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>In your case, you must find the pressure exerted by the gas, so rearrange the ideal gas law equation to isolate <mathjax>#P#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies P = (nRT)/V#</mathjax></p>
</blockquote>
<p>Now, before plugging in your values, make sure that the units given to you by the problem <strong>match</strong> those used in the expression of the universal gas constant. </p>
<p>In this case, you have</p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#ul(color(white)(aaaacolor(black)("What you have")aaaaaaaaaacolor(black)("What you need")aaaaa))#</mathjax></p>
<p><mathjax>#color(white)(aaaaaacolor(black)("liters " ["L"])aaaaaaaaaaaaaaacolor(black)("liters " ["L"])aaaa)color(darkgreen)(sqrt())#</mathjax></p>
<p><mathjax>#color(white)(aaaaacolor(black)("moles " ["mol"])aaaaaaaaaaaaacolor(black)("moles " ["mol"])aaa)color(darkgreen)(sqrt())#</mathjax></p>
<p><mathjax>#color(white)(acolor(black)("degrees Celsius " [""^@"C"])aaaaaaaaaacolor(black)("Kelvin " ["K"])aaaa)color(red)(xx)#</mathjax></p>
</blockquote>
<p>To convert the temperature from degrees Celsius to Kelvin, use the conversion factor </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + "273.15")))#</mathjax></p>
</blockquote>
<p>Plug in your values into the above equation and solve for <mathjax>#P#</mathjax></p>
<blockquote>
<p><mathjax>#P = (2.01 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (20 + 273.15) color(red)(cancel(color(black)("K"))))/(6.5 color(red)(cancel(color(black)("L"))))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(P = "7.4 atm")))#</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 temperature of the gas. </p></div>
</div>
</div> | <div class="answerText" itemprop="text">
<div class="answerSummary">
<div>
<div class="markdown"><p><mathjax>#"7.4 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to solve this problem, you must be familiar with the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>In your case, you must find the pressure exerted by the gas, so rearrange the ideal gas law equation to isolate <mathjax>#P#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies P = (nRT)/V#</mathjax></p>
</blockquote>
<p>Now, before plugging in your values, make sure that the units given to you by the problem <strong>match</strong> those used in the expression of the universal gas constant. </p>
<p>In this case, you have</p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#ul(color(white)(aaaacolor(black)("What you have")aaaaaaaaaacolor(black)("What you need")aaaaa))#</mathjax></p>
<p><mathjax>#color(white)(aaaaaacolor(black)("liters " ["L"])aaaaaaaaaaaaaaacolor(black)("liters " ["L"])aaaa)color(darkgreen)(sqrt())#</mathjax></p>
<p><mathjax>#color(white)(aaaaacolor(black)("moles " ["mol"])aaaaaaaaaaaaacolor(black)("moles " ["mol"])aaa)color(darkgreen)(sqrt())#</mathjax></p>
<p><mathjax>#color(white)(acolor(black)("degrees Celsius " [""^@"C"])aaaaaaaaaacolor(black)("Kelvin " ["K"])aaaa)color(red)(xx)#</mathjax></p>
</blockquote>
<p>To convert the temperature from degrees Celsius to Kelvin, use the conversion factor </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + "273.15")))#</mathjax></p>
</blockquote>
<p>Plug in your values into the above equation and solve for <mathjax>#P#</mathjax></p>
<blockquote>
<p><mathjax>#P = (2.01 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (20 + 273.15) color(red)(cancel(color(black)("K"))))/(6.5 color(red)(cancel(color(black)("L"))))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(P = "7.4 atm")))#</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 temperature of the gas. </p></div>
</div>
</div>
</div> | <article>
<h1 class="questionTitle" itemprop="name">What pressure will be exerted by 2.01 mol hydrogen gas in a 6.5 L cylinder at 20°C?</h1>
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Stefan V.
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Feb 17, 2017
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<div class="markdown"><p><mathjax>#"7.4 atm"#</mathjax></p></div>
</div>
</div>
<div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>In order to be able to solve this problem, you must be familiar with the <strong><a href="https://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a> equation</strong>, which looks like this</p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(PV = nRT)))#</mathjax></p>
</blockquote>
<p>Here</p>
<blockquote>
<ul>
<li><mathjax>#P#</mathjax> is the pressure of the gas</li>
<li><mathjax>#V#</mathjax> is the volume it occupies</li>
<li><mathjax>#n#</mathjax> is the number of moles of gas present in the sample</li>
<li><mathjax>#R#</mathjax> is the <em>universal gas constant</em>, equal to <mathjax>#0.0821("atm L")/("mol K")#</mathjax></li>
<li><mathjax>#T#</mathjax> is the <strong>absolute temperature</strong> of the gas</li>
</ul>
</blockquote>
<p>In your case, you must find the pressure exerted by the gas, so rearrange the ideal gas law equation to isolate <mathjax>#P#</mathjax></p>
<blockquote>
<p><mathjax>#PV = nRT implies P = (nRT)/V#</mathjax></p>
</blockquote>
<p>Now, before plugging in your values, make sure that the units given to you by the problem <strong>match</strong> those used in the expression of the universal gas constant. </p>
<p>In this case, you have</p>
<p>In this case, you have</p>
<blockquote>
<p><mathjax>#ul(color(white)(aaaacolor(black)("What you have")aaaaaaaaaacolor(black)("What you need")aaaaa))#</mathjax></p>
<p><mathjax>#color(white)(aaaaaacolor(black)("liters " ["L"])aaaaaaaaaaaaaaacolor(black)("liters " ["L"])aaaa)color(darkgreen)(sqrt())#</mathjax></p>
<p><mathjax>#color(white)(aaaaacolor(black)("moles " ["mol"])aaaaaaaaaaaaacolor(black)("moles " ["mol"])aaa)color(darkgreen)(sqrt())#</mathjax></p>
<p><mathjax>#color(white)(acolor(black)("degrees Celsius " [""^@"C"])aaaaaaaaaacolor(black)("Kelvin " ["K"])aaaa)color(red)(xx)#</mathjax></p>
</blockquote>
<p>To convert the temperature from degrees Celsius to Kelvin, use the conversion factor </p>
<blockquote>
<p><mathjax>#color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + "273.15")))#</mathjax></p>
</blockquote>
<p>Plug in your values into the above equation and solve for <mathjax>#P#</mathjax></p>
<blockquote>
<p><mathjax>#P = (2.01 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (20 + 273.15) color(red)(cancel(color(black)("K"))))/(6.5 color(red)(cancel(color(black)("L"))))#</mathjax></p>
<p><mathjax>#color(darkgreen)(ul(color(black)(P = "7.4 atm")))#</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 temperature of the gas. </p></div>
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</article> | What pressure will be exerted by 2.01 mol hydrogen gas in a 6.5 L cylinder at 20°C? | null |
3,297 | abc8efba-6ddd-11ea-b3ed-ccda262736ce | https://socratic.org/questions/a-student-found-that-1-19-g-of-chromium-cr-formed-1-74-g-f-chromium-oxide-the-mo | Cr2O3 | start chemical_formula qc_end physical_unit 8 8 4 5 mass qc_end physical_unit 13 14 10 11 mass qc_end physical_unit 8 8 21 22 molar_mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] chromium oxide [IN] empirical"}] | [{"type":"chemical equation","value":"Cr2O3"}] | [{"type":"physical unit","value":"Mass [OF] Cr [=] \\pu{1.19 g}"},{"type":"physical unit","value":"Mass [OF] chromium oxide [=] \\pu{1.74 g}"},{"type":"physical unit","value":"Molar mass [OF] Cr [=] \\pu{52.00 g/mol}"}] | <h1 class="questionTitle" itemprop="name">A student found that 1.19 g of chromium (#Cr#) formed 1.74 g f chromium oxide. The molar mass of #Cr# is 52.00 g/mol. What is the empirical formula of chromium oxide?</h1> | null | Cr2O3 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all these problems, we calculate the molar quantities of each substituent, and normalize them according to the masses of the atomic constituents:</p>
<p><mathjax>#"Moles of metal"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.19*g)/(52.00*g*mol^-1)=0.0229*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.74*g-1.19*g)/(16.00*g*mol^-1)=0.0343*mol#</mathjax></p>
<p>And we divide thru by the SMALLEST molar quantity, that of chromium, to give: </p>
<p><mathjax>#Cr, (0.0229*mol)/(0.0229*mol)=1; O, (0.0343*mol)/(0.0229*mol)=1.50#</mathjax>.</p>
<p>But by definition, the empirical formula is the smallest WHOLE number ratio that defines constituent atoms in a species. To get a whole number ratio, clearly we mulitply the empirical ratio by <mathjax>#2#</mathjax> to <mathjax>#Cr_2O_3#</mathjax> as required. Capisce?</p>
<p>How did I know there were <mathjax>#0.55*g#</mathjax> of oxygen present?</p></div>
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<div class="markdown"><p><mathjax>#Cr_2O_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all these problems, we calculate the molar quantities of each substituent, and normalize them according to the masses of the atomic constituents:</p>
<p><mathjax>#"Moles of metal"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.19*g)/(52.00*g*mol^-1)=0.0229*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.74*g-1.19*g)/(16.00*g*mol^-1)=0.0343*mol#</mathjax></p>
<p>And we divide thru by the SMALLEST molar quantity, that of chromium, to give: </p>
<p><mathjax>#Cr, (0.0229*mol)/(0.0229*mol)=1; O, (0.0343*mol)/(0.0229*mol)=1.50#</mathjax>.</p>
<p>But by definition, the empirical formula is the smallest WHOLE number ratio that defines constituent atoms in a species. To get a whole number ratio, clearly we mulitply the empirical ratio by <mathjax>#2#</mathjax> to <mathjax>#Cr_2O_3#</mathjax> as required. Capisce?</p>
<p>How did I know there were <mathjax>#0.55*g#</mathjax> of oxygen present?</p></div>
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<h1 class="questionTitle" itemprop="name">A student found that 1.19 g of chromium (#Cr#) formed 1.74 g f chromium oxide. The molar mass of #Cr# is 52.00 g/mol. What is the empirical formula of chromium oxide?</h1>
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<div class="markdown"><p><mathjax>#Cr_2O_3#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>As with all these problems, we calculate the molar quantities of each substituent, and normalize them according to the masses of the atomic constituents:</p>
<p><mathjax>#"Moles of metal"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.19*g)/(52.00*g*mol^-1)=0.0229*mol#</mathjax></p>
<p><mathjax>#"Moles of oxygen"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(1.74*g-1.19*g)/(16.00*g*mol^-1)=0.0343*mol#</mathjax></p>
<p>And we divide thru by the SMALLEST molar quantity, that of chromium, to give: </p>
<p><mathjax>#Cr, (0.0229*mol)/(0.0229*mol)=1; O, (0.0343*mol)/(0.0229*mol)=1.50#</mathjax>.</p>
<p>But by definition, the empirical formula is the smallest WHOLE number ratio that defines constituent atoms in a species. To get a whole number ratio, clearly we mulitply the empirical ratio by <mathjax>#2#</mathjax> to <mathjax>#Cr_2O_3#</mathjax> as required. Capisce?</p>
<p>How did I know there were <mathjax>#0.55*g#</mathjax> of oxygen present?</p></div>
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</article> | A student found that 1.19 g of chromium (#Cr#) formed 1.74 g f chromium oxide. The molar mass of #Cr# is 52.00 g/mol. What is the empirical formula of chromium oxide? | null |
3,298 | ad11daa4-6ddd-11ea-9537-ccda262736ce | https://socratic.org/questions/a-compound-has-the-empirical-formula-chcl-a-256-ml-flask-at-373-k-and-750-torr-c | C2H2Cl2 | start chemical_formula qc_end c_other OTHER qc_end physical_unit 21 22 8 9 volume qc_end physical_unit 21 22 12 13 temperature qc_end physical_unit 21 22 15 16 pressure qc_end physical_unit 21 22 18 19 mass qc_end end | [{"type":"other","value":"Chemical Formula [OF] the compound [IN] molecular"}] | [{"type":"chemical equation","value":"C2H2Cl2"}] | [{"type":"other","value":"A compound has the empirical formula CHCl."},{"type":"physical unit","value":"Volume [OF] the compound [=] \\pu{256 mL}"},{"type":"physical unit","value":"Temperature [OF] the compound [=] \\pu{373 K}"},{"type":"physical unit","value":"Pressure [OF] the compound [=] \\pu{750 torr}"},{"type":"physical unit","value":"Mass [OF] the compound [=] \\pu{0.800 g}"}] | <h1 class="questionTitle" itemprop="name">A compound has the empirical formula CHCl. A 256 mL flask, at 373 K and 750 torr, contains 0.800 g of the compound, what is the molecular formula?</h1> | null | C2H2Cl2 | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>, <mathjax>#MM = (mRT)/(PV)#</mathjax>; where <mathjax>#MM#</mathjax> <mathjax>#=#</mathjax> molecular mass; <mathjax>#m#</mathjax> <mathjax>#=#</mathjax> mass; <mathjax>#P#</mathjax> <mathjax>#=#</mathjax> pressure in atmospheres; <mathjax>#V =#</mathjax> volume in litres; <mathjax>#R#</mathjax> <mathjax>#=#</mathjax> gas constant with appropriate units.</p>
<p>So, <mathjax>#(0.800*gxx0.0821*cancelL*cancel(atm)*cancelK^(-1)*mol^(-1)xx373*cancelK)/(0.256*cancelLxx0.987*cancel(atm))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#97.0#</mathjax> <mathjax>#g*mol^(-1)#</mathjax>.</p>
<p><mathjax>#nxx(12.01+1.01+2xx35.45)*g*mol^(-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#97.0*g*mol^(-1)#</mathjax>.</p>
<p>Clearly, <mathjax>#n#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax>. And molecular formula <mathjax>#=#</mathjax> <mathjax>#C_2H_2Cl_2#</mathjax>.</p>
<p>I seem to recall (but can't be bothered to look up) that vinylidene chloride, <mathjax>#H_2C=C(Cl)_2#</mathjax> is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required. </p></div>
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<div class="markdown"><p>Assuming ideality, the molecular formula is <mathjax>#C_2H_2Cl_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>, <mathjax>#MM = (mRT)/(PV)#</mathjax>; where <mathjax>#MM#</mathjax> <mathjax>#=#</mathjax> molecular mass; <mathjax>#m#</mathjax> <mathjax>#=#</mathjax> mass; <mathjax>#P#</mathjax> <mathjax>#=#</mathjax> pressure in atmospheres; <mathjax>#V =#</mathjax> volume in litres; <mathjax>#R#</mathjax> <mathjax>#=#</mathjax> gas constant with appropriate units.</p>
<p>So, <mathjax>#(0.800*gxx0.0821*cancelL*cancel(atm)*cancelK^(-1)*mol^(-1)xx373*cancelK)/(0.256*cancelLxx0.987*cancel(atm))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#97.0#</mathjax> <mathjax>#g*mol^(-1)#</mathjax>.</p>
<p><mathjax>#nxx(12.01+1.01+2xx35.45)*g*mol^(-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#97.0*g*mol^(-1)#</mathjax>.</p>
<p>Clearly, <mathjax>#n#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax>. And molecular formula <mathjax>#=#</mathjax> <mathjax>#C_2H_2Cl_2#</mathjax>.</p>
<p>I seem to recall (but can't be bothered to look up) that vinylidene chloride, <mathjax>#H_2C=C(Cl)_2#</mathjax> is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required. </p></div>
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<h1 class="questionTitle" itemprop="name">A compound has the empirical formula CHCl. A 256 mL flask, at 373 K and 750 torr, contains 0.800 g of the compound, what is the molecular formula?</h1>
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<div class="markdown"><p>Assuming ideality, the molecular formula is <mathjax>#C_2H_2Cl_2#</mathjax></p></div>
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<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p>From the <a href="http://socratic.org/chemistry/the-behavior-of-gases/ideal-gas-law">ideal gas law</a>, <mathjax>#MM = (mRT)/(PV)#</mathjax>; where <mathjax>#MM#</mathjax> <mathjax>#=#</mathjax> molecular mass; <mathjax>#m#</mathjax> <mathjax>#=#</mathjax> mass; <mathjax>#P#</mathjax> <mathjax>#=#</mathjax> pressure in atmospheres; <mathjax>#V =#</mathjax> volume in litres; <mathjax>#R#</mathjax> <mathjax>#=#</mathjax> gas constant with appropriate units.</p>
<p>So, <mathjax>#(0.800*gxx0.0821*cancelL*cancel(atm)*cancelK^(-1)*mol^(-1)xx373*cancelK)/(0.256*cancelLxx0.987*cancel(atm))#</mathjax> <mathjax>#=#</mathjax> <mathjax>#97.0#</mathjax> <mathjax>#g*mol^(-1)#</mathjax>.</p>
<p><mathjax>#nxx(12.01+1.01+2xx35.45)*g*mol^(-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#97.0*g*mol^(-1)#</mathjax>.</p>
<p>Clearly, <mathjax>#n#</mathjax> <mathjax>#=#</mathjax> <mathjax>#1#</mathjax>. And molecular formula <mathjax>#=#</mathjax> <mathjax>#C_2H_2Cl_2#</mathjax>.</p>
<p>I seem to recall (but can't be bothered to look up) that vinylidene chloride, <mathjax>#H_2C=C(Cl)_2#</mathjax> is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required. </p></div>
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</article> | A compound has the empirical formula CHCl. A 256 mL flask, at 373 K and 750 torr, contains 0.800 g of the compound, what is the molecular formula? | null |
3,299 | aaf08b0a-6ddd-11ea-8d6a-ccda262736ce | https://socratic.org/questions/how-many-moles-of-n-are-in-0-217-g-of-n-2o | 1.00 × 10^(-2) moles | start physical_unit 4 4 mole mol qc_end physical_unit 10 10 7 8 mass qc_end end | [{"type":"physical unit","value":"Mole [OF] N [IN] moles"}] | [{"type":"physical unit","value":"1.00 × 10^(-2) moles"}] | [{"type":"physical unit","value":"Mass [OF] N2O [=] \\pu{0.217 g}"}] | <h1 class="questionTitle" itemprop="name">How many moles of #N# are in 0.217 g of #N_2O#?</h1> | null | 1.00 × 10^(-2) moles | <div class="answerDescription">
<h4 class="answerHeader">Explanation:</h4>
<div>
<div class="markdown"><p><mathjax>#"Moles of nitrous oxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.217*g)/(44.013*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5xx10^-3*mol#</mathjax>.</p>
<p>Given the composition of <mathjax>#N_2O#</mathjax>, there are this <mathjax>#2xx5xx10^-3*mol#</mathjax> of nitrogen atoms. </p></div>
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<div class="markdown"><p>There are twice the molar quantity of nitrogen atoms in nitrous oxide, i.e. <mathjax>#1xx10^-2*mol#</mathjax>.</p></div>
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<h4 class="answerHeader">Explanation:</h4>
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<div class="markdown"><p><mathjax>#"Moles of nitrous oxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.217*g)/(44.013*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5xx10^-3*mol#</mathjax>.</p>
<p>Given the composition of <mathjax>#N_2O#</mathjax>, there are this <mathjax>#2xx5xx10^-3*mol#</mathjax> of nitrogen atoms. </p></div>
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anor277
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Oct 14, 2016
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<div class="markdown"><p>There are twice the molar quantity of nitrogen atoms in nitrous oxide, i.e. <mathjax>#1xx10^-2*mol#</mathjax>.</p></div>
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<div class="markdown"><p><mathjax>#"Moles of nitrous oxide"#</mathjax> <mathjax>#=#</mathjax> <mathjax>#(0.217*g)/(44.013*g*mol^-1)#</mathjax> <mathjax>#=#</mathjax> <mathjax>#5xx10^-3*mol#</mathjax>.</p>
<p>Given the composition of <mathjax>#N_2O#</mathjax>, there are this <mathjax>#2xx5xx10^-3*mol#</mathjax> of nitrogen atoms. </p></div>
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</article> | How many moles of #N# are in 0.217 g of #N_2O#? | null |
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