A) \[4.9\times {{10}^{-5}}\text{J}\]
B) \[\text{F}=\text{1}0+0.\text{5x}\]
C) \[(\text{take},\text{g}=0\text{m}/{{\text{s}}^{\text{2}}}\]
D) \[R=6.4\times {{10}^{3}}m\]
Correct Answer: B
Solution :
Given \[\underset{\begin{smallmatrix} (Greater \\ Volume) \end{smallmatrix}}{\mathop{Ice}}\,\underset{\begin{smallmatrix} (Lesser \\ Volume) \end{smallmatrix}}{\mathop{Water}}\,-XKcal\] On integrating within the limit \[Ni(s)+2A{{g}^{+}}(0.002M)\to N{{i}^{2+}}(0.160M)+2Ag(s)\] \[(Give\,that\,{{E}^{0}}_{cell}=1.05V)\] \[\text{12}00\text{ kJ}\,\text{mo}{{l}^{\text{-1}}}\] \[\text{145}0\text{ kJ mo}{{\text{l}}^{\text{-1}}}\] \[\text{86}%\text{M}{{\text{g}}^{+}}+\text{14}%\text{M}{{\text{g}}^{\text{2}}}^{+}\] \[69%M{{g}^{+}}+31%M{{g}^{2+}}\] \[\text{14}%\text{M}{{\text{g}}^{+}}+\text{86}%\text{M}{{\text{g}}^{\text{2+}}}\]
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