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5.33 mg of salt [Cr(H_(2)O)_(5)Cl].Cl_(2).H_(2)O is treated with excess of AgNO_(3)(aq) then mass of AgCl precipitate obtained will be : Given : [Cr = 52, Cl = 35.5 ,Ag = 108 ] |
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Answer» `[CR(H_(2)O)_(5)Cl].Cl_(2).H_(2)Ounderset(0.02"MOL")+2AgNO_(3)(a)to 2Ag Cldarr+[Cr(H_(2)O)_(5)Cl]UNDERSET(0.04"mol")(NO_(3))_(2)` |
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