Question

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1.)the number of atoms in 2.57 grams of carbon


4.)the number of atoms in 108 grams of chlorine gas (Cl2 (g))


3.)the mass of 1.00 X 10^20 atoms of sodium

Answer 1

Answer: See below

Step-by-step explanation:

1.

`\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}`

2.

`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}`

3.

`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}`