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 3.62 mol/min​

Answer 1

Set rates:

**rate1 = 3.62

rate2 = x

2) The gas that has twice the molar mass is the one whose rate we are trying to determine.

MM2 = 2

MM1 = 1

These molar masses are arbitrary values, we just need MM2 to be twice the value for MM1.

3) Graham's Law:

r1/ r2 = MM2 / MM1

3.62 / x = 2 / 1

**x = 2.56 mol/min

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