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