Question

How many atoms of fluorine are present in 1.9 × 10
-6 g of fluorine

Answer 1

Answer:

6.02 * 10^16 atoms

Explanation:

1 mole F = 19 g

6.02 * 10^23 = 19 g

6.02 * 10^23 * 1.9/ 10^6 *19 =  1.9 × 10 ^-6 g

6.02 * 10^16 atoms = 1.9 × 10 ^-6 g