Answer
See the answer below.
Work Step by Step
a) Initial concentration: $0.50\ mol\div 9.50\ L=0.0526\ M$, reacted x:
$K=[CO][Br_2]/[COBr_2]$
$0.190=x\times x/(0.0526-x)$
$x^2+0.190x-0.01=0$
$x=0.043$
$[CO]=[Br_2]=0.043\ M,[COBr_2]=0.01\ M$
b) Initial concentration: $0.50\ mol\div 4.50\ L=0.111\ M$, reacted x:
$K=[CO][Br_2]/[COBr_2]$
$0.190=x\times x/(0.111-x)$
$x^2+0.190x-0.021=0$
$x=0.078$
$[CO]=[Br_2]=0.078\ M,[COBr_2]=0.033\ M$
c) The equilibrium shifted to the left, since $0.033\ M\times4.5\ L>0.01\ M\times9.5\ L$
