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Andrew Pfeiffer 2E
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Joined: Sat Sep 28, 2019 12:16 am


Postby Andrew Pfeiffer 2E » Sun Jan 19, 2020 4:15 pm

Does anyone know how to do 5I.29? Thanks!

Christine Honda 2I
Posts: 116
Joined: Sat Sep 14, 2019 12:17 am

Re: 5I.29

Postby Christine Honda 2I » Sun Jan 19, 2020 6:37 pm

Using the reaction given, set up an ICE table. You should get a Kp equation that looks like x^2/(0.22-2x)^2 and set this equal to the Kp given. Because the equilibrium constant is small, we can assume that X is negligible compared to 0.22, therefore we can ignore it in the denominator. We end up having 3.2x10^-34=x^2/(0.22)^2. Solving for X, I get 3.9x10^-18!

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