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 $N$:
$$2.39 \times 10^{21} \space molecules \space (NH_4)_2Cr_2O_7 \times \frac{2 \space atom \space N}{1 \space molecule \space (NH_4)_2Cr_2O_7} = 4.78 \times 10^{21} \space atoms \space N$$
