Chemistry: Molecular Approach (4th Edition)

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

Chapter 1 - Exercises - Page 40: 93


(a) \[\underline{\text{60}\text{.6}\ \text{in}}\] (b)\[\underline{\text{3140}\ \text{g}}\] (c)\[\underline{\text{3}\text{.7}\ \text{qt}}\] (d)\[\underline{\text{4}\text{.29}\ \text{in}}\]

Work Step by Step

(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 \[\underline{\text{4}\text{.29}\ \text{in}}\].
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