Asample of o2 gas (2.0 mmol) effused through a pinhole in 5.0 s. it will take s for the same amount of co2 to effuse under the same conditions.

If a sample of O₂ gas (2.0 mmol) effused through a pinhole in 5.0 s, it will take 5.9 s for the same amount of CO₂ to effuse under the same conditions.

A sample of O₂ gas (2.0 mmol) effused through a pinhole in 5.0 s. The rate of effusion of O₂ is:

Knowing their molar masses (M), we can calculate the rate of effusion of CO₂ using Graham's law.

The rate of effusion of CO₂ is 0.34 mmol/s. The time for 2.0 mmol to effuse is:

If a sample of O₂ gas (2.0 mmol) effused through a pinhole in 5.0 s, it will take 5.9 s for the same amount of CO₂ to effuse under the same conditions.

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The time taken for the same amount of CO₂ to effuse under the same condition is 5.9 s

Let t₁ be the time taken for O₂ to effuse

Let M₁ be thr molar mass of O₂

Let t₂ be the time taken for CO₂ to effuse

Let M₂ be the molar mass of CO₂

Time taken for O₂ to effuse (t₁) = 5 s

Molar mass of O₂ (M₁) = 2 × 16 = 32 g/mol

Molar mass of CO₂ (M₂) = 12 + (2×16) = 44 g/mol

Time taken for CO₂ to effuse (t₂) = ?

Cross multiply

Therefore, it will take 5.9 s for the same amount of CO₂ to effuse under the same condition.

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the correct answer is:

a. formation of precipitation

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