## Chemistry: Molecular Approach (4th Edition)

Convert the heights as follows: \begin{align} & \text{4 ft, 9 in = }\left\{ \text{4 ft}\left( \frac{1\text{ m}}{\text{3}\text{.2808 ft}} \right) \right\}+\left\{ \text{9 in}\left( \frac{1\text{ m}}{\text{39}\text{.3701 in}} \right) \right\} \\ & =1.4478\text{ m} \end{align} \begin{align} & \text{4 ft, 8 in = }\left\{ \text{4 ft}\left( \frac{1\text{ m}}{\text{3}\text{.2808 ft}} \right) \right\}+\left\{ \text{8 in}\left( \frac{1\text{ m}}{\text{39}\text{.3701 in}} \right) \right\} \\ & =\text{1}\text{.4197 m} \end{align} \begin{align} & \text{5 ft, 1 in = }\left\{ \text{5 ft}\left( \frac{1\text{ m}}{\text{3}\text{.2808 ft}} \right) \right\}+\left\{ \text{1 in}\left( \frac{1\text{ m}}{\text{39}\text{.3701 in}} \right) \right\} \\ & =\text{1}\text{.5494 m} \end{align} \begin{align} & \text{5 ft, 5 in = }\left\{ \text{5 ft}\left( \frac{1\text{ m}}{\text{3}\text{.2808 ft}} \right) \right\}+\left\{ \text{5 in}\left( \frac{1\text{ m}}{\text{39}\text{.3701 in}} \right) \right\} \\ & =\text{1}\text{.6510 m} \end{align} \begin{align} & \text{5 ft, 8 in = }\left\{ \text{5 ft}\left( \frac{1\text{ m}}{\text{3}\text{.2808 ft}} \right) \right\}+\left\{ \text{8 in}\left( \frac{1\text{ m}}{\text{39}\text{.3701 in}} \right) \right\} \\ & =\text{1}\text{.7272 m} \end{align} Calculate the sum of heights as follows: \begin{align} & \text{Sum of heights = 1}\text{.4478 m + 1}\text{.4197 m}+\text{1}\text{.5494 m}+\text{1}\text{.6510 m}+\text{1}\text{.7272 m} \\ & =\text{7}\text{.7951 m} \end{align} Heights in $\left( \text{m} \right)$ are $\underline{\text{1}\text{.4478 m,} \text{1}\text{.4197 m,} \text{1}\text{.5494 m},\text{1}\text{.6510 m}, and \text{1}\text{.7272 m}}$ The sum of heights is $\underline{\text{7}\text{.7951 m}}$.