Chemistry: Molecular Approach (4th Edition)

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

Chapter 1 - Exercises - Page 40: 121


\[\underline{3.11\times {{10}^{5}}\text{ lb}}\]

Work Step by Step

Volume of water in the pool is 185 cubic yards. \[1\text{ y}{{\text{d}}^{3}}=764,555\text{ c}{{\text{m}}^{3}}\] Thus, volume of 185 cubic yards in cubic centimeter will be written as follows: \[\begin{align} & V=\left( 764,555\times 185 \right)\text{ c}{{\text{m}}^{3}} \\ & =141,442,675\text{ c}{{\text{m}}^{3}} \end{align}\] The relation between density \[\left( d \right)\] and mass \[\left( m \right)\] is as follows: \[m=\left( V \right)\left( d \right)\] Here, V is volume. Density of water is \[1\ g\text{/c}{{\text{m}}^{3}}\]. Thus, mass of water is calculated as follows: \[\begin{align} & m=\left( V \right)\left( d \right) \\ & =\left( 141,442,675\text{ c}{{\text{m}}^{3}} \right)\left( 1\text{ g/c}{{\text{m}}^{3}} \right) \\ & =141,442,675\text{ g} \end{align}\] Now, \[\text{1 g}=0.0022\text{ lb}\] Thus, mass of water in pounds is written as follows: \[\begin{align} & m=\left( 141,442,675\times 0.0022 \right)\text{ lb} \\ & =311,173.89\text{ lb} \\ & =\text{3}\text{.11}\times \text{1}{{\text{0}}^{5}}\text{ lb} \end{align}\] The mass of water in pounds is \[\underline{3.11\times {{10}^{5}}\text{ lb}}\].
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