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The reaction O2 ⇄ 2 O has a K=1.2x10-10. If a 1.7 L container holds 3.8 moles
of O atoms (and nothing else), what will the concentration of O be when
equilibrium is reached?

So far, I have 1.2x10^-10 = 2.2^2 / 0

but I don't know how to do the rest. When it says "when equilibrium is reached" does it mean that the left side and the right side has to match? Help!!

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