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.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.

sure im new too!

explanation: