exactly 1mol of n2o4 is placed in an empty 1 l container. if at equilibrium n2o4 is dissociated 20%, what is the value of equilibruim constant

K = 0.2

Explanation:

Based on the chemical dissociation of N₂O₄:

N₂O₄ ⇄ 2NO₂

The equilibrium constant, K, of the reaction is:

K = [NO₂]² / [N₂O₄]

Now, if 20% of N₂O₄ is dissociated, 80% remains as N₂O₄ = 0.8mol/L = 0.8M

as 20% is dissociated, 0.2moles of N₂O₄ were dissociated and 0.2*2 = 0.4mol/L of NO₂ are produced.

Replacing in K:

K = [0.4M]² / [0.8M]

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.