Question

A sample 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.

Answer 1

Answer is: 5,9.
Effusion is leakage of gas through a small hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. Relative rates of effusion is related to the molecular mass.
M(O₂) = 32g/mol.
M(CO₂) = 44g/mol.
t₁ = 5s.
t₂ = ?
t₂/t₁ = √(M(CO₂)/M(O₂)).
t₂ = 1,17·5s = 5,9s.

Answer 2

Answer:

Your answer is

M(O₂) = 32g/mol.

M(CO₂) = 44g/mol.

t₁ = 5s.

t₂ = ?

t₂/t₁ = √(M(CO₂)/M(O₂)).

t₂ = 1,17·5s = 5,9s.