HEY everyone!
Here's a link to my practice exam (my exam is this tuesday.) http://www.smccd.edu/accounts/mcomberj/c…
It is #43.) & 47.) ...please help because I don't know how to do it! thank you
Here's a link to my practice exam (my exam is this tuesday.) http://www.smccd.edu/accounts/mcomberj/c…
It is #43.) & 47.) ...please help because I don't know how to do it! thank you
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2BrCl ---> Br2 + Cl2
I 0.02M 0M 0.03M
C -2x +x +x
E 0.02 - 2x x 0.03 + x
It says that K = 0.141. So set products/reactants equal to it (to the power of their coefficients of course)
0.141=(x)(0.03+x)/(0.02-2x)^2
x is equal to 0.00135, the answer is A
CO2(g) --> 2CO(g) (I excluded the C(s) since it is a solid)
(partial pressure of CO2) + (partial pressure of 2CO) = 1
(partial pressure of CO)^2/(partial pressure of CO2)=10.7
(partial pressure of 2CO)=1-(partial pressure of CO2)
Using the system of equations, you find that the partial pressure of 2CO is equal to 0.921, the answer is E
I 0.02M 0M 0.03M
C -2x +x +x
E 0.02 - 2x x 0.03 + x
It says that K = 0.141. So set products/reactants equal to it (to the power of their coefficients of course)
0.141=(x)(0.03+x)/(0.02-2x)^2
x is equal to 0.00135, the answer is A
CO2(g) --> 2CO(g) (I excluded the C(s) since it is a solid)
(partial pressure of CO2) + (partial pressure of 2CO) = 1
(partial pressure of CO)^2/(partial pressure of CO2)=10.7
(partial pressure of 2CO)=1-(partial pressure of CO2)
Using the system of equations, you find that the partial pressure of 2CO is equal to 0.921, the answer is E