A) \[(6.3\times {{10}^{-3}})\times (6.3\times {{10}^{-3}})\]
B) \[(6.3\times {{10}^{-3}})\times (12.6\times {{10}^{-3}})\]
C) \[(6.3\times {{10}^{-3}})\times {{(12.6\times {{10}^{-3}})}^{2}}\]
D) \[(12.6\times {{10}^{-3}})\times (12.6\times {{10}^{-3}})\]
Correct Answer: C
Solution :
\[PbC{{l}_{2}}\to \underset{S}{\mathop{P{{b}^{++}}}}\,+\underset{2S}{\mathop{2C{{l}^{-}}}}\,\] \[{{K}_{sp}}=S\times {{(2S)}^{2}}\]\[=[6.3\times {{10}^{-3}}]\times {{[12.6\times {{10}^{-3}}]}^{2}}\].
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