Question

What is the rate of effusion for a gas that has a molar mass twice that of a gas that effuses at a rate of 3.6 mol/min.

Answer 1

Answer:

We set the molar masses:

gas A = 1  

gas B = 2

Then assign gas B to r1 (and MM1) gas A to r2 (and MM2) since we want to know the effusion rate for the gas that is twice the molar mass of the other.

Then using Graham's Law:r1 / r2 = (MM2 / MM1) ^-1/2r1/r2=(MM2/MM1)  

−

1/2

r1 / 3.6= (1 / 2) ^-1/2

r1 / 3.6 = 0.70710678

r1 = 2.55 mol/min

Explanation:

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