Question

"n2(g effuses at a rate that is ______ times that of br2(g under the same conditions."

Answer 1

`N_2` effuses at a rate that is 2.38 times that of
`Br_2`

Step-by-step explanation:

We have two components:
`N_2` and
`Br_2`. We need to compare the rate of effusion under the same conditions.

According to the Graham effusion law: if we have two components Component 1 (C1) and Component 2 (C2), with Molecular Weights M1 and M2 respectively. They will have a rate of effusion V1 and V2 respectively as follows:

`(V1)/(V2) =\sqrt{(M2)/(M1) }`

Now, we will replace for our components:

Component 1 - Nitrogen
`N_2`

Component 2 - Bromine
`Br_2`

Molecular weight Nitrogen M1 = M(
`N_2`) = 28 g/mol

Molecular weight Bromine M2 =M( Bromine
`Br_2`) = 159 g/mol

After that, we should replace the values:

`(V1)/(V2) =\sqrt{(159)/(28) } \\(V1)/(V2) =√(5.67)\\\\ (V1)/(V2) =2.38\\`

As V1 is the effusion rate of Nitrogen and V2 is the effusion rate of Bromine. We can reorganize the equation:

`V1 = 2.38 V2`

The conclusion is that V1 is 2.38 times V2.

It means
`N_2` effuses at a rate that is 2.38 times that of
`Br_2` .

Answer 2

Graham's law of effusion states that the rate of diffusion of a certain gas iis inversely proportional to the sqrt of the molar mass.
Molar mass of N2 is 28 g/mol and that of Br2 is 158.81 g/mol.

The sqrt of the ratio of their molar mass is equal to 5.67, the sqrt of this number is 2.38. Thus, N2 diffuses 2.38 times faster than Br2.