Answer:
There are approximately 2.448 x 10^23 bromine atoms present in 35.2 g of CH2Br2.
Step-by-step explanation:
The molar mass of CH2Br2 can be calculated as follows:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of 2 Br = 2 x 79.90 g/mol = 159.80 g/mol
Therefore, the molar mass of CH2Br2 = 12.01 + 1.01 + 159.80 = 172.82 g/mol
Next, we can calculate the number of moles of CH2Br2 as follows:
moles of CH2Br2 = mass of CH2Br2 / molar mass of CH2Br2
moles of CH2Br2 = 35.2 g / 172.82 g/mol
moles of CH2Br2 = 0.203 moles
Finally, we can use Avogadro's number to calculate the number of bromine atoms present:
Number of bromine atoms = moles of CH2Br2 x 2 (since there are 2 bromine atoms per molecule of CH2Br2) x Avogadro's number
Number of bromine atoms = 0.203 x 2 x 6.022 x 10^23
Number of bromine atoms = 2.448 x 10^23 bromine atoms
Therefore, there are approximately 2.448 x 10^23 bromine atoms present in 35.2 g of CH2Br2.