Question

How many bromine atoms are present 35.2 g of CH2Br2?

Can someone explain how to get answers with steps.

Answer 1

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.