Chemistry: Molecular Approach (4th Edition)

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

Chapter 1 - Exercises - Page 40: 118


\[\underline{\text{0}\text{.28 lb/i}{{n}^{3}}}\]

Work Step by Step

The relationship between pound and g is as follows: \[\begin{align} & 1\text{ lb}=\text{453}\text{.59237 g} \\ & \text{1 i}{{\text{n}}^{3}}=16.387\text{ c}{{\text{m}}^{3}} \\ & 1\text{ lb/i}{{n}^{3}}=27.68\text{ g/c}{{\text{m}}^{3}} \end{align}\] So, density in pounds per cubic inch will be as follows: \[\begin{align} & \text{1 g/c}{{\text{m}}^{3}}=\left( \text{1 g/c}{{\text{m}}^{\text{3}}} \right)\left( \frac{\text{1 lb/i}{{\text{n}}^{\text{3}}}}{\text{27}\text{.68 g/c}{{\text{m}}^{\text{3}}}} \right) \\ & =\text{0}\text{.036 lb/i}{{\text{n}}^{\text{3}}} \\ & \text{7}\text{.68 g/c}{{\text{m}}^{\text{3}}}=\left( \text{7}\text{.86 g/c}{{\text{m}}^{\text{3}}} \right)\left( \frac{\text{0}\text{.036 lb/i}{{\text{n}}^{\text{3}}}}{\text{1 g/c}{{\text{m}}^{\text{3}}}} \right) \\ & =\text{ 0}\text{.28 lb/i}{{\text{n}}^{\text{3}}} \end{align}\] Density of iron in pounds per cubic inch is \[\underline{\text{0}\text{.28 lb/i}{{n}^{3}}}\].
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