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The metallic salt `(XY)` is soluble in water. (a) When the aqueous soluble of `(XY)` is treated with `NaOH` solution, a white precipitate `(A)` is formed. In excess of `NaOH` solution, a white precipitate `(A)` is formed. In excess of `NaOH` solution, white precipitate `(A)` dissolves to form a compound `(B)`. When this solution is boiled with soild `NH_(4) Cl`, a precipitate of compound `(C)` is formed. (b) An aqueous solution on treatment with `BaCl_(2)` solution gives a white precipitate `(D)` white is insoluble in conc `HCl`. ( c) The metallic salt `(XY)` forms a double salt `(E)` with potassium sulphate. Identify `(XY),(A),(B),(C),(D)` and `(E)`.A. M=MgB. M=BeC. `P+Ca(OH)_(2)`D. `G=O_(3)` |
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Answer» Correct Answer - B |
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