Question

If a 3.30 m sample of a is heated to 500 k, what is the concentration of b at equilibrium?

Answer 1

(Missing in your question ):
we have the following  reaction:
A(aq) ↔ 2 B(aq) 
and Kc = 7.02 x 10^-6 at 500K
So at equilibrium,
Kc = [Products] / [ reactants]
     = [B]^2 / [A]
we have [A] = 3.3 m and Kc is given= 7.02 x10^-6
by substitution:
7.02x10^-6 = [B]^2 / 3.3
∴[B]^2 = 2.3 x 10^-5
∴[B] = 0.005 m