205k views
1 vote
How many bromine atoms are present in 39.4 g of CH2Br2

User Seaders
by
8.2k points

1 Answer

1 vote
37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2

4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms
User Nonshatter
by
7.9k points

Related questions

1 answer
4 votes
210k views
asked Jun 20, 2024 139k views
Chasidishe asked Jun 20, 2024
by Chasidishe
7.6k points
1 answer
1 vote
139k views
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.