Question

How many bromine atoms are present in 37.9 g of CH2Br2?

Answer 1

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Step-by-step explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

174 g of CH2Br2 ------------------ 160 g of Br2

37.9 g of CH2Br2 --------------- x

x = 37.9 x 160/174 = 34.85 g of Br

1 mol of Br ----------------- 160 g Br2

x ---------------- 174 g Be2

x = 174 x 1 /160 = 1.088 mol of Br2

1 mol of Br ----------------- 6.023 x 10 ²³ atoms

1.088 mol of Br ------------- x

x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms