Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 5 - Exercises - Page 240d: 61

Answer

a. $$ 3.54 \times 10^{22} \space atoms \space N$$ b. $$ 3.76 \times 10^{22} \space atoms \space N$$ c. $$ 4.78 \times 10^{21} \space atoms \space N$$

Work Step by Step

1. Using the answers from exercise 59: a. Each $NH_3$ has 1 $N$: $$3.54 \times 10^{22} \space molecules \space NH_3 \times \frac{1 \space atom \space N}{1 \space molecule \space NH_3} = 3.54 \times 10^{22} \space atoms \space N$$ b. Each $N_2H_4$ has 2 $N$: $$1.88 \times 10^{22} \space molecules \space N_2H_4 \times \frac{2 \space atoms \space N}{1 \space molecule \space N_2H_4} = 3.76 \times 10^{22} \space atoms \space N$$ c. Each $(NH_4)_2Cr_2O_7$ has 2 $N$: $$2.39 \times 10^{21} \space molecules \space (NH_4)_2Cr_2O_7 \times \frac{2 \space atom \space N}{1 \space molecule \space (NH_4)_2Cr_2O_7} = 4.78 \times 10^{21} \space atoms \space N$$
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