A) \[2.0\times {{10}^{11}}\]
B) \[4.0 \times 1{{0}^{12}}\]
C) \[1.0\times 1{{0}^{2}}\]
D) \[1.0 \times 1{{0}^{10}}\]
Correct Answer: D
Solution :
\[E{}^\circ =\frac{0.0591}{n}\,\log \,K\] Here, \[\operatorname{n}=2, E{}^\circ = 0.295\] \[\therefore \,\,\,\log \,\,K=\frac{2\times 0.295}{0.591}=9.8\approx 10\,\,or\,\,K={{10}^{10}}\]You need to login to perform this action.
You will be redirected in
3 sec