If Xe has an effusion rate of 1,087 moles/sec. What would the effusion rate for Ar be in moles/sec? round to the ones place.

To answer this question, we need to use Graham's law of effusion. This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The formula for Graham's law of effusion is: (Rate1/Rate2) = sqrt(M2/M1)

Given in the problem, we know:

• The rate of effusion for Xe (Rate1) is 1,087 moles/sec
• The molar mass of Xe (M1) is 131.29 g/mol
• The molar mass of Ar (M2) is 39.95 g/mol

We want to find the rate of effusion for Ar (Rate2).

Plugging the known values into the formula, we get:

1,087/Rate2 = sqrt(39.95/131.29)

Solving for Rate2, we get:

Rate2 = 1,087 / sqrt(39.95/131.29)

Calculating the above expression, we get:

Rate2 = 1,870.6 moles/sec

Rounding to the ones place, the effusion rate for Ar would be approximately 1,871 moles/sec.