[MilkyCar]

Another equilibrium-related question I'm stuck on:

The reaction:
2NO(g) + Br2(g) 2NOBr(g)

has Kp = 109 at  25°c. If the equilibrium partial pressure of Br2 is 0.0191 atm and the equilibrium partial pressure of NOBr is 0.0605 atm, calculate the partial pressure of NO at equilibrium.

I set up the problem like this:

109 = 0.0605^2/(Partial Pressure x 0.0191)

I got 0.00175 and then rounded to 0.0018 atm as a result after I doubled 0.0605, then divided by 0.0191, and then divide by 109 on both sides. The website marked it incorrect and I'm unsure what I'm doing wrong here.

[Borek]

It is not clear to me what you did, but you never mentioned squaring or taking out the root, so there is no way your description is correct.

[MilkyCar]

I should've square-rooted both sides (both 109 and the result from dividing the two partial pressures) on the last step. I did that and I've solved it. Thanks for clarifying.