Chemistry: Molecular Approach (4th Edition)

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

Chapter 4 - Exercises - Page 188: 55


a. $0.150 \space M \space N{O_3}^- $ b. $0.300 \space M \space N{O_3}^-$ c. $0.450 \space M \space N{O_3}^-$

Work Step by Step

1. Determine the amount of $N{O_3}^-$ ions in each compound: a. 1 b. 2 c. 3 2. Use them as conversion factors: a. $$0.150 \space M \space KNO_3 \times \frac{1 \space mol \space N{O_3}^-}{1 \space mol \space KNO_3} = 0.150 \space M \space N{O_3}^- $$ b. $$0.150 \space M \space Ca(NO_3)_2 \times \frac{2 \space mol \space N{O_3}^-}{1 \space mol \space Ca(NO_3)_2} = 0.300 \space M \space N{O_3}^- $$ c. $$0.150 \space M \space Al(NO_3)_3 \times \frac{3 \space mol \space N{O_3}^-}{1 \space mol \space Al(NO_3)_3} = 0.450 \space M \space N{O_3}^- $$
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