100 cm^3 of oxygen diffuses through a Porous in 3second how long will it take 150 cm^3 of sulphur (iv) oxide diffuse through the same pot? ( oxygen= 16 sulphur = 32)

3.18 seconds

Explanation:

Given the following :

Volume of oxygen (V1) = 100cm^3

Time taken (t1) = 3 seconds

Volume of Sulphur (iv) oxide (v2) = 150cm^3

From Graham's Law of diffusion:

(r1/r2) = √(m1/m2)

Where r = rate of diffusion

m = molar mass

Note rate (r) = (volume / time)

[(V1/t1) ÷ (v2/t2)] = √(m1/m2)

(v1/t1) * (t2/v2) = √(m1/m2)

m1 = 02 = (16 * 2) = 32

m2 = SO2 = (32 + (16 * 2)) = 64

(100/3) * (t2/150) = √(32/64)

100t2 / 450 = √(32/64)

100t2 / 450 = √32 / 8

100t2 / 450 = √32 / 8

100t2 * 8 = 450 * √32

800t2 = 2545.5844

t2 = 2545.5844 / 800

t2 = 3.1819805

t2 = 3.18 seconds

It will take 3.18s for 150cm^3 of Sulphur (iv) oxide to diffuse through the same pot.

periodic trends is an atom that gets larger as the number of electronic shells increase, the radius of atoms increases as you go down a certain group in the periodic table of elements. in general, the size of an atom will decrease as you move from left to right of a certain period.

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