Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 2 - Sections 2.1-2.9 - Exercises - Problems by Topic: 45

Answer

$9.4 \times 10^{13}$ excess electrons have been acquired. The collective mass is $8.5 \times 10^{-17}$ kg

Work Step by Step

Charge on one electron = $-1.6 \times 10^{-19}$ C Total accumulated charge = $-15 \times 10^{-6}$ C Hence, number of excess electrons = $\frac{(-15 \times 10^{-6})}{(-1.6 \times 10^{-19})}$ = $9.4 \times 10^{13}$ Mass of one electron = $9.10 \times 10^{-31}$ kg Hence, collective mass = $9.4 \times 10^{13}\times9.10 \times 10^{-31}$ kg = $8.5 \times 10^{-17}$ kg
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