Answer:
41.3 s
Explanation:
Let t₁ represent the time taken for SO₂ to effuse.
Let t₂ represent the time taken for Ar to effuse.
Let M₁ represent the molar mass of SO₂
Let M₂ represent the molar mass of Ar
From the question given above,
Time taken (t₁) for SO₂ = 52.3 s
Time taken (t₂) for Ar =?
Molar mass (M₁) of SO₂ = 32 + (16×2) = 32 + 32 = 64 g/mol
Molar mass (M₂) of Ar = 40 g/mol
Finally, we shall determine the time taken for Ar to effuse by using the Graham's law equation as shown below:
t₂ / t₁ = √(M₂ / M₁)
t₂ / 52.3 = √(40 / 64)
t₂ / 52.3 = √0.625
t₂ / 52.3 = 0.79
Cross multiply
t₂ = 52.3 × 0.79
t₂ = 41.3 s
Thus, the time taken for the amount of Ar to effuse is 41.3 s
Answer:
3
Explanation:
pH=-log(H+)
- Hope that helps! Please let me know if you need further explanation.
Answer is B. gas formation
From the reaction above, the rate is given by the following formula:
r = -(1/2) dA / dt = - dB / dt = (1/3) dC/ dt
Note that A and B charge is negative due to they decrease with time
given dA / dt = -0.110 M/s
hence dB / dt = -0.110 / 2 = -0.055 M/s
dC / dt = (-3/2) (-0.110) = 0.165 M/s
