(Solved): The dissociation of molecular iodine into iodine atoms is represented as I2(g)2I(g) At 1000. ...

The dissociation of molecular iodine into iodine atoms is represented as $${\mathrm{I}}_2(g)?2\mathrm{I}(g)$$ At $$1000.0\mathrm{K}$$, the equilibrium constant $$K_c$$ for the reaction is $$3.80\times 10^{?5}$$. Suppose you start with $$0.0460\mathrm{mol}$$ of $${\mathrm{I}}_2$$ in a $$2.29\mathrm{L}$$ flask at $$1000.0\mathrm{K}$$. What are the concentrations of the gases at equilibrium? Part 1 of 2 What is the equilibrium concentration of $${\mathrm{I}}_2$$ ? Round your answer to 3 significant digits. $$\mathrm{M}$$ Part 2 of 2 What is the equilibrium concentration of I? Round your answer to 3 significant digits. $$\mathrm{M}$$