compare the rate of diffusion of sf6 and SO2
Answers
Answered by
5
Answer:
Graham's law of effusion holds that the relative rate of effusion/diffusion of 2 gases is inversely proportional to the square root of the molecular masses of the two diffusing gases.....
rate 1
rate 2
=
√
Molecular mass of Gas 1
Molecular mass of Gas 2
rate
(
H
e
)
rate
(
S
O
2
)
=
√
64
⋅
g
⋅
m
o
l
−
1
4
⋅
g
⋅
m
o
l
−
1
rate
(
H
e
)
rate
(
S
O
2
)
=
{
8
2
}
=
4
And so helium should diffuse/effuse FOUR TIMES as FAST as sulfur dioxide. What is special about the molecular mass of each gas;
Similar questions