Chemistry: Molecular Approach (4th Edition)

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

Chapter 1 - Exercises - Page 40: 92


a) \[\underline{28.9\times {{10}^{3}}\ \mu \text{m}}\] b) \[\underline{1.432\ \text{L}}\] c) \[\underline{\text{1}\text{.211}\times \text{1}{{\text{0}}^{6}}\ \text{Gm}}\]

Work Step by Step

(a) \[\underline{28.9\times {{10}^{3}}\ \mu \text{m}}\] \[\begin{align} & 28.9\ nm=\left( 28.9\ nm \right)\left( \frac{{{10}^{-3}}\ \mu m}{1\ \text{nm}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & =28.9\times {{10}^{-3}}\ \mu m \end{align}\]\[\,\left[ \begin{align} & \because 1nm\,\,=\,\,{{10}^{-9}}m \\ & \,\,\,\,\,1m\,\,\,\,=\,\,{{10}^{6}}\mu m\Rightarrow 1nm\,\,\,\,=\,\,{{10}^{-3}}\mu m\, \\ \end{align} \right]\] The given quantity \[\text{28}\text{.9}\ \text{nm}\]in micrometres is \[\underline{28.9\times {{10}^{3}}\ \text{ }\!\!\mu\!\!\text{ m}}\]. (b) \[\begin{align} & 1432\ c{{m}^{3}}=\left( 1432\ c{{m}^{3}} \right)\left( \frac{0.001\ \text{L}}{1\ c{{m}^{3}}} \right) \\ & =1.432\ \text{L} \end{align}\] The given quantity\[\text{1432}\ \text{c}{{\text{m}}^{3}}\text{ }\]in litres is \[\underline{1.432\ \text{L}}\]. (c) \[\begin{align} & 1211\ Tm=\left( 1211\ Tm \right)\left( \frac{{{10}^{3}}\ \text{Gm}}{1\ Tm} \right) \\ & =1211\times {{10}^{3}}\ \text{Gm} \\ & =1.211\times {{10}^{6}}\ \text{Gm} \end{align}\] The given quantity\[\text{1211}\ \text{Tm}\] in Giga meters is \[\underline{\text{1}\text{.211}\times \text{1}{{\text{0}}^{6}}\ \text{Gm}}\].
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