Question

the rate of effusion of oxygen to an unknown gas is 0.935. what is the other gas?

Answer 1

Answer:

N2

Explanation:

Rate of effusion is defined by Graham's Law:  

(Rate 1/Rate 2) = (sqrt (M2)/ sqrt (M1))

(Where M is the molar mass of each substance. )

Molar Mass of oxygen, O2, is 32 (M1).

Rate of effusion of O2 to an unknown gas is .935(Rate 1).  

Rate 2 is unknown so put 1.

Solve for x (M2).

.935/1 = sqrt x/ sqrt32

.935 x sqrt 32 = sqrt x

5.29 = sq rt x

5.29^2 = 27.975 = 28  

N2 has a molar mass of 28 so it is the correct gas.

Answer 2

The other gas would be nitrogen.

According to Graham's law of diffusion: the rate of diffusion of gases is inversely proportional to the square root of the molar masses.

Graham's law can be mathematically expressed as:

R1/R2 = √M2/√M1

In this case, R1/R2 = 0.935 = √M2/√M1 and M1 = 32 mol/g (molar mass of oxygen)

√M2 = 0.935 x √32

M2 = 0.935^2 x 32

                 = 27.96

Thus, the molar mass of the other gas is 27.96 which is approximately 28. The only gas with a molar mass of 28 is nitrogen (N2).

More on Graham's law can be found here: https://brainly.com/question/7244733