Chemistry: Molecular Approach (4th Edition)

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

Chapter 2 - Exercises - Page 79: 45


\[\underline{9.4\times {{10}^{13}}}\] excess electrons and \[\underline{8.5\times {{10}^{-17}}\text{ kg}}\]

Work Step by Step

The electronic charge is \[-1.60218\times {{10}^{-19}}\text{ C}\]. Calculate the number of electrons as follows: \[\begin{align} & \text{Number of electrons}=\left( -15\,\text{ }\!\!\mu\!\!\text{ C} \right)\times \frac{{{10}^{-6}}}{1\,\text{ }\!\!\mu\!\!\text{ }}\times \left( \frac{1\text{ electron}}{-1.60218\times {{10}^{-19}}\,\text{C}} \right) \\ & =9.4\times {{10}^{13}}\,\text{electrons} \end{align}\] The electron has a mass of \[0.00091\times {{10}^{-27}}\text{ kg}\]. Calculate the total mass as follows: \[\begin{align} & \text{Mass}=\frac{0.00091\times {{10}^{-27}}\text{ kg}}{1\,\text{electron}}\times 9.4\times {{10}^{13}}\,\text{electrons} \\ & =\text{8}\text{.5}\times \text{1}{{\text{0}}^{-17}}\,\text{kg} \end{align}\] The number of excess electrons acquired is \[\underline{9.4\times {{10}^{13}}}\] and the collective mass is \[\underline{8.5\times {{10}^{-17}}\text{ kg}}\].
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