The equilibrium constant for the reaction H2(g) + I2(g) <=> 2HI(g) at a certain temperature is 6.0. What would be the equilibrium concentration of HI at that temperature, when the equilibrium concentration of H2 and I2 are 0.2 mol dm-3 and 0.3 mol dm-3 respectively?

The reaction is $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$.
The equilibrium constant expression, $K_c$, is:

$$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$$

We are given:
$K_c = 6.0$
$[\text{H}_2] = 0.2 \text{ mol dm}^{-3}$
$[\text{I}_2] = 0.3 \text{ mol dm}^{-3}$
Substituting these values:

$$6.0 = \frac{[\text{HI}]^2}{(0.2 \text{ mol dm}^{-3})(0.3 \text{ mol dm}^{-3})}$$

$$(0.2)(0.3) = 0.06 \text{ (mol dm}^{-3})^2$$

So the equation becomes:

$$6.0 = \frac{[\text{HI}]^2}{0.06}$$

$$[\text{HI}]^2 = 6.0 \times 0.06$$ $$[\text{HI}]^2 = 0.36 \text{ (mol dm}^{-3})^2$$

$$[\text{HI}] = \sqrt{0.36 \text{ (mol dm}^{-3})^2}$$ $$[\text{HI}] = 0.6 \text{ mol dm}^{-3}$$

Comparing this result with the given options, option D is $0.60 \text{ mol dm}^{-3}$.

The final answer is $\text{D. } 0.60 \text{ mol dm}^{-3}$.