Answer
$[A] = 0.90 \space M$
$[B] =0.095 \space M$
$[C] = 0.095 \space M$
Work Step by Step
1. Write the $K_c$ expression for this reaction:
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[B][C]}{[A]}$$
2. Draw the equilibrium table:
\begin{matrix}
& [A] & [B] & [C]\\
Initial & 1.0 \space M & 0 & 0 \\
Change & -x & + x & + x\\
Equilibrium & 1.0 \space M - x & x & x
\end{matrix}
3. Substitute the concentrations into the expression and solve for x:
$$K_c = 0.010 = \frac{(x)(x)}{1.0 -x}$$ $$0.010(1.0 -x)= x^2$$ $$0.010 - 0.010x -x^2 = 0$$
$$x_1 = \frac{-(-0.010) + \sqrt {(-0.010)^2 - 4(0.010)(-1)}}{2(-1)} \approx -0.105$$ $$x_2= \frac{-(-0.010) - \sqrt {(-0.010)^2 - 4(0.010)(-1)}}{2(-1)} \approx 0.095$$
Since the concentration can't be negative, $x_1$ is invalid. So $x = 0.095$
Therefore:
$[A] = 1.0 \space M - x = (1.0 -0.095) \space M = 0.90 \space M$
$[B] = [C] =x = 0.095 \space M$
