A) \[8\times {{10}^{-3}}\]
B) \[8\times {{10}^{-6}}\]
C) \[6.4\times {{10}^{-5}}\]
D) \[6.4\times {{10}^{-3}}\]
Correct Answer: A
Solution :
\[HgS{{O}_{4}}\]of \[{{K}_{sp}}={{S}^{2}}\] \[S=\sqrt{{{K}_{sp}}}\] ; \[S=\sqrt{6.4\times {{10}^{-5}}}\]; \[S=8\times {{10}^{-3}}\]m/l.You need to login to perform this action.
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