Chemistry: Molecular Approach (4th Edition)

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

Chapter 20 - Exercises - Page 975: 92

Answer

$3.4\times10^{9}\,dis/h$

Work Step by Step

$k= \frac{0.693}{t_{1/2}}=\frac{0.693}{1.4\times10^{10}\times365\times24\,h}$ $=5.650685\times10^{-15}\,h^{-1}$ $N=6.022\times10^{23}$ Rate= $kN=5.650685\times10^{-15}\,h^{-1}\times6.022\times10^{23}$ $= 3.4\times10^{9}\,dis/h$
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