Chemistry: Molecular Approach (4th Edition)

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

Chapter 2 - Exercises - Page 79: 46


\[\underline{6.25\times {{10}^{18}}}\] excess electrons and \[\underline{5.69\times {{10}^{-9}}\text{ g}}\]

Work Step by Step

The electronic charge is \[-1.6\times {{10}^{-19}}\text{ C}\]. Calculate the number of electrons as follows: \[\begin{align} & \text{Number of electrons}=\frac{\left( -1.0\text{ C} \right)}{-1.6\times {{10}^{-19}}\text{ C}} \\ & =6.25\times {{10}^{18}} \end{align}\] The electron has a mass of \[9.10\times {{10}^{-28}}\text{ g}\]. Calculate the total mass as follows: \[\begin{align} & \text{Mass}=\left( 6.25\times {{10}^{18}} \right)\left( 9.10\times {{10}^{-28}}\text{ g} \right) \\ & =5.\text{69}\times \text{1}{{\text{0}}^{-9}}\text{ g} \end{align}\] The number of electrons necessary is \[\underline{6.25\times {{10}^{18}}}\] and the collective mass is \[\underline{5.69\times {{10}^{-9}}\text{ g}}\].
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