A) \[AgCl\]
B) \[PbC{{l}_{2}}\]
C) \[H{{g}_{2}}C{{l}_{2}}\]
D) \[HgC{{l}_{2}}\]
Correct Answer: C
Solution :
\[H{{g}_{2}}C{{l}_{2}}+2N{{H}_{4}}OH\to Hg+Hg(N{{H}_{2}})Cl\] \[K=\frac{E}{2}\]
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