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Equilibrium process at 686 deg. C02(g)+H2CO(g)+H20(g) [CO2] is 0.086 M [H2] is 0.045 M [CO] is 0.050 M [H20] is 0.040 M Calculate Kc for the reaction...

Equilibrium process at 686 deg. C.
C02(g)+H2CO(g)+H20(g)
[CO2] is 0.086 M
[H2] is 0.045 M
[CO] is 0.050 M
[H20] is 0.040 M
Calculate Kc for the reaction at 686 deg. C.
Which I did
0.050*0.040/.086*.045 = approx. .52
If add CO2 to increase it concentration to 0.55 mol/L, what will the concentrations
off all the gases be when equilibrium is reestablished.

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