Answer: See below
Explanation:
1.
[tex]\text{2.57 grams C} \, \, \cdot \frac{1 \text{mol C}}{12.011 \text{g}} \cdot \frac{6.022 \cdot 10^{23} \text{atoms}}{1 \text{mol}} = 1.29 \cdot 10^{23} \text{ atoms of C}[/tex]
2.
[tex]108 \text{ grams Cl}_2} \,\, \cdot \frac{2 \text{ Cl atoms}}{1 \text{Cl}_2 \text{ molecule}} \cdot \frac{1 \text{ mol Cl}}{35.453 \text{g}} \cdot \frac{6.022 \cdot 10^{23} \text{atoms}}{1 \text{ mol Cl}} = 3.67 \cdot 10^{24} \text{ atoms of Cl}[/tex]
3.
[tex]1.00 \,\cdot 10^{20} \text{ atoms Na } \cdot \frac{1 \text{mol Na}}{6.022 \cdot 10^{23} \text{ atoms}} \cdot \frac{22.990 \text{g}}{1 \text{mol Na}} = 0.00382 \text{ grams Na}[/tex]
