Problem 5I #1
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Problem 5I #1
Hello! I'm having difficulty getting the correct answer for problem 5I #1 and am not sure why. I'm creating an ICE table and using the quadratic equation. Maybe I am missing something when calculating? If someone could explain the steps of their work that would be great!
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Re: Problem 5I #1
Hi Kate, I plugged the 2 initial conditions that it gave us into the Kc equation, as it also gave us the value of Kc. I got (.145)^2 / (.495) (x) = .031 and multiplied both sides by x and divided both sides by .031 to get x= 1.37, which they rounded up so Br2= 1.4 mol*L-1
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Re: Problem 5I #1
kate cassiano 1F wrote:Hello! I'm having difficulty getting the correct answer for problem 5I #1 and am not sure why. I'm creating an ICE table and using the quadratic equation. Maybe I am missing something when calculating? If someone could explain the steps of their work that would be great!
hi, for my first setup I did:
Kc= (BrCL)^2/(Cl2)(Br2) = (0.145)^2/(0.495)(x) = 0.031
And to solve for x, my setup was:
(0.031)(0.495x)=(0.145)^2
x=1.4 mol.L^-1
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