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sulfur dioxide was allowed to effuse through a small opening under constant pressure conditions. It required 179 s for 1.00 L of this gas to effuse. Calculate the effusion time of arsine (AsH3) under identical experimental conditions. (Remember that the faster the rate of effusion, the shorter the time required for effusion; that is, rate and time are inversely proportional.)

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Lelia Lubowitz
Lelia LubowitzLv2
22 Sep 2019

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Related questions

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 105 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 31 s for 1.0 L of O2 gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; in other words, rate is the amount that diffuses over the time it takes to diffuse.)

A gas of unknown molar mass was allowed to effuse through a smallopening under constant pressure conditions. It required 184 s for1.00 L of this gas to effuse. Under identical experimentalconditions it required 506 s for 1.00 L of chlorotrifluoromethane(CClF3) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter thetime required for effusion; that is, rate and time are inverselyproportional.)

Options:
a) The molar mass is 287.3 g/mol.
b) The molar mass is 790.0 g/mol.
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d) The molar mass is 173.2 g/mol.
e) The molar mass is 13.8 g/mol.

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Calculate the molar mass of the unknown gas. (Remember that thefaster the rate of effusion, the shorter the time required foreffusion of 1.0 L; that is, rate and time are inverselyproportional.)

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