Chemistry: Molecular Approach (4th Edition)

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

Chapter 4 - Exercises - Page 188: 58


a. 2.3 L b. 6.10 L c. 0.060 L

Work Step by Step

1. Use the molarity as a conversion factor: a. $$0.45 \space mol \times \frac{1 \space L}{0.200 \space mol} = 2.3 \space L$$ b. $$1.22 \space mol \times \frac{1 \space L}{0.200 \space mol} = 6.10 \space L$$ c. $$1.2 \times 10^{-2} \space mol \times \frac{1 \space L}{0.200 \space mol} = 0.060 \space L$$
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.