A) 0.500 V
B) 0.325 V
C) 0.650 V
D) 0.150 V
Correct Answer: B
Solution :
\[{{E}^{0}}=\frac{{{n}_{1}}E_{1}^{0}+{{n}_{2}}E_{2}^{0}}{n}=\frac{0.15\times 1+0.50\times 1}{2}=0.325\,V\]You need to login to perform this action.
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