## Chemistry: Molecular Approach (4th Edition)

(a) $\underline{\text{60}\text{.6}\ \text{in}}$ (b)$\underline{\text{3140}\ \text{g}}$ (c)$\underline{\text{3}\text{.7}\ \text{qt}}$ (d)$\underline{\text{4}\text{.29}\ \text{in}}$
(a) \begin{align} & 154\ cm=\left( 154\ cm \right)\left( \frac{1\ in}{\text{2}\text{.54}\ \text{cm}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \,1\,in\,=\,2.54\,\,cm \right] \\ & =60.6\ in \end{align} The given quantity$\text{154}\ \text{cm }$ in $\text{in}\text{.}$is $\underline{\text{60}\text{.6}\ \text{in}}$. (b) \begin{align} & 3.14\ kg=\left( 3.14\ kg \right)\left( \frac{{{10}^{3}}\ \text{g}}{1\ kg} \right) \\ & =3140\ g \end{align} The given quantity$\text{3}\text{.14}\ \text{kg}$ in $\text{g}$is $\underline{\text{3140}\ \text{g}}$. (c) \begin{align} & 3.5\ L=\left( 3.5\ L \right)\left( \frac{1.06\ \text{qt}}{1\ L} \right)\,\,\,\,\,\,\,\,\,\,\,\left[ \because \ 1L=\,1.06\ qt \right] \\ & =3.7\ qt \end{align} The given quantity$\text{3}\text{.5}\ \text{L}$ in $\text{qt}$is $\underline{\text{3}\text{.7}\ \text{qt}}$. (d) \begin{align} & 109\ mm=\left( 109\ mm \right)\left( \frac{0.03937\ \text{in}}{1\ \text{mm}} \right)\,\,\,\,\left[ \because \,1mm\,=\,\,0.03937\,in \right] \\ & =4.29\ in \end{align} The given quantity$\text{109}\ \text{mm }$in $\text{in}\text{.}$ is $\underline{\text{4}\text{.29}\ \text{in}}$.