The solubility of `CuBr` is `2xx10^(-4)` at `25^(@)C`. The `K_(sp)` value for `CuBr` isA. `4xx10^(-8) mol^(2) l^(-2)`B. `4xx10^(-11) mol^(2) L^(-1)`C. `4xx10^(-4) mol^(2) l^(-2)`D. `4xx10^(-15) mol^(2) l^(-2)`

The solubility of `CuBr` is `2xx10^(-4)` at `25^(@)C`. The `K_(sp)` va

1.	The solubility of `CuBr` is `2xx10^(-4)` at `25^(@)C`. The `K_(sp)` value for `CuBr` isA. `4xx10^(-8) mol^(2) l^(-2)`B. `4xx10^(-11) mol^(2) L^(-1)`C. `4xx10^(-4) mol^(2) l^(-2)`D. `4xx10^(-15) mol^(2) l^(-2)`
Answer» Correct Answer - A `underset(K_(sp))(CuBr)hArrunderset((s))(Cu^(+))+ underset((S))(Br^(-))` `K_(sp)=S^(2)=(2xx10^(-4))^(2)=4xx10^(-8)(mol^(2))/(l^(2))`

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The solubility of `CuBr` is `2xx10^(-4)` at `25^(@)C`. The `K_(sp)` value for `CuBr` isA. `4xx10^(-8) mol^(2) l^(-2)`B. `4xx10^(-11) mol^(2) L^(-1)`C. `4xx10^(-4) mol^(2) l^(-2)`D. `4xx10^(-15) mol^(2) l^(-2)`

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