1.

The solubility product of AgBr is 4.9 x 10-9, Thesolubility of AgBr will be

Answer»

solubility product is s² = 4.9×10^-9=> s² = 49×10^-10=> s = √49×10^-10=> s = 7×10^-5

so, solubility is s = 7×10^-5



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