Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 18 - Nuclear Chemistry - Questions for Review and Thought - Topical Questions - Page 817b: 31b


2.4 mg

Work Step by Step

The amount of radionuclide at the beginning is $A_{0}=9.6\,mg$ Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{14.3\,day}=0.04846/day$ Time $t=28.6\,days$ $\ln(\frac{A_{0}}{A})=kt$ where $A$ is the amount of radionuclide remaining. $\implies \ln(\frac{9.6\,mg}{A})=0.04846\times28.6=1.386$ Taking the inverse $\ln$ of both sides, we have $\frac{9.6\,mg}{A}=e^{1.386}=3.9988$ Or $A= \frac{9.6\,mg}{3.9988}=2.4\,mg $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.