Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 1 - Exercises - Page 42: 142


\[\underline{49 percent}\]

Work Step by Step

Since \[1\text{ kg}=1000\text{ g}\], convert the mass into grams as follows: \[\left( 4.36\text{ kg} \right)\left( \frac{1000\text{ g}}{1\text{ kg}} \right)=4360\text{ g}\] Now, assume that the volume of copper is \[x\text{ c}{{\text{m}}^{3}}\]. Thus, the volume of lead is calculated as follows: \[\begin{align} & {{V}_{\text{T}}}={{V}_{\text{Pb}}}+{{V}_{\text{Cu}}} \\ & 427\text{ c}{{\text{m}}^{3}}={{V}_{\text{Pb}}}+x\text{ c}{{\text{m}}^{3}} \\ & {{V}_{\text{Pb}}}=\left( 427-x \right)\text{ c}{{\text{m}}^{3}} \end{align}\] Now, the mass of copper is calculated as follows: \[{{m}_{\text{Cu}}}=\left( {{d}_{\text{Cu}}} \right)\left( {{V}_{\text{Cu}}} \right)\] Here, \[{{d}_{\text{Cu}}}\] is the density of copper and \[{{V}_{\text{Cu}}}\] is the volume of copper. Substitute \[8.96\text{ g/c}{{\text{m}}^{3}}\] for \[{{d}_{\text{Cu}}}\] and \[x\text{ c}{{\text{m}}^{3}}\] for \[{{V}_{\text{Cu}}}\]: \[\begin{align} & {{m}_{\text{Cu}}}=\left( 8.96\text{ g/c}{{\text{m}}^{3}} \right)\left( x\text{ c}{{\text{m}}^{3}} \right) \\ & =8.96x\text{ g} \end{align}\] Now, the density \[\left( {{d}_{\text{Pb}}} \right)\] of lead is \[\text{11}\text{.4 g/c}{{\text{m}}^{3}}\]. Calculate the mass of the lead as follows: \[{{m}_{\text{Pb}}}=\left( {{d}_{\text{Pb}}} \right)\left( {{V}_{\text{Pb}}} \right)\] Substitute \[\text{11}\text{.4 g/c}{{\text{m}}^{\text{3}}}\] for \[{{d}_{\text{Cu}}}\] and \[\left( 427-x \right)\text{ c}{{\text{m}}^{3}}\] for \[{{V}_{\text{Cu}}}\]: \[\begin{align} & {{m}_{\text{Cu}}}=\left( \text{11}\text{.4 g/c}{{\text{m}}^{3}} \right)\left( 427-x \right)\text{c}{{\text{m}}^{3}} \\ & =11.4\left( 427-x \right)\text{ g} \end{align}\] The total mass is written as follows: \[\begin{align} & {{m}_{\text{T}}}={{m}_{\text{Pb}}}+{{m}_{\text{Cu}}} \\ & 4360\text{ g}=8.96x\text{ g}+11.4\left( 427-x \right)\text{ g} \\ & x=\frac{4867.8-4360}{2.44} \\ & =208.11 \end{align}\] Thus, the volume of copper is \[208.11\text{ c}{{\text{m}}^{3}}\]. The percentage of the total volume of the sphere that is copper is as follows: \[\text{percentage}=\frac{{{V}_{\text{Cu}}}}{{{V}_{\text{T}}}}\times 100\] Substitute \[208.11\text{ c}{{\text{m}}^{3}}\] for \[{{V}_{\text{Cu}}}\] and \[\text{427 c}{{\text{m}}^{3}}\] for \[{{V}_{\text{T}}}\]: \[\begin{align} & \text{percentage}=\frac{208.11\text{ c}{{\text{m}}^{3}}}{427\text{ c}{{\text{m}}^{3}}\text{ }}\times 100 \\ & =49\,\text{percent} \end{align}\] The percentage of the total volume of the sphere is \[49\,percent\].
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