Answer:
The solution to the question is as follows
(a) The rate of ammonia formation = 0.061 M/s
(b) the rate of N₂ consumption = 0.0303 M/s
Explanation:
(a) To solve the question we note that the reaction consists of one mole of N₂ combining with three moles of H₂ to form 2 moles of NH₃
N₂(g) + 3H₂(g) → 2NH₃(g)
The rate of reaction of molecular hydrogen = 0.091 M/s, hence we have
3 moles of H₂ reacts to form 2 moles of NH₃, therefore
0.091 M of H₂ will react to form 2/3 × 0.091 M or 0.061 M of NH₃
Hence the rate of ammonia formation is 0.061 M/s
(b) From the reaction equation we have 3 moles of H₂ and one mole of N₂ being consumed at the same time hence
0.091 M of H₂ is consumed simultaneously with 1/3 × 0.091 M or 0.0303 M of N₂
Therefore the rate of consumption of N₂ = 0.0303 M/s
Answer:
16G = 1 mole ; then 4G = how many moles? 4/16 = 0.25 mole; That means 4 grams of oxygen is 0.25 moles.
Answer:
Yes
Explanation:
There are in double dash :) ( = )
When the solubility is the amount of solute that dissolved in a unit volume of the solvent.
and when the balanced reaction equation is:
so, by using the ICE table:
Cd(OH)2(s) ↔ Cd2+ + 2OH-
initial 0 0
change +X +2X
Equ X 2X
so the Ksp expression = [Cd2+] [OH-]^2
and we have Ksp = 7.2 x 10^-15
and when we assume the solubility = X
by substitution:
7.2 x 10^-15 = (X) (2X)^2
7.2 x 10^-15 = X*4X^2
7.2 x 10^-15 = 4X^3
∴X = ∛(7.2x10^-15)/4
= 1.22 x 10^-5 M
∴ the solubility = X = 1.22 x 10^-5 M
