x g of Ag was dissolved in `HNO_(3)` and the solution was precipitated. The value of x is

Xg of Ag was dissolved in HNO_3 and the solution was treated with excess of NaCl when 2.87g of AgCl was precipitated. The value of X is

i) A black mineral (A) on treatment with dilute sodium cyanide of (B) and (C ). ii) The solution of (B) on reaction with zinc gives precipitate of a metal (D). iii) (D) is dissolved in dil. HNO_(3) and the resulting solution gives a white precipitate (E) with dil. HCl. iv) (E) on fusion with sodium carbonate gives (D). v) (E) dissolves in aqueous solution ammonia giveing a colourless solution (F). The colourless solution of (F) is

i) A black mineral (A) on treatment with dilute sodium cyanide of (B) and (C ). ii) The solution of (B) on reaction with zinc gives precipitate of a metal (D). iii) (D) is dissolved in dil. HNO_(3) and the resulting solution gives a white precipitate (E) with dil. HCl. iv) (E) on fusion with sodium carbonate gives (D). v) (E) dissolves in aqueous solution ammonia giveing a colourless solution (F). The solution (B) is of

21.6 g of silver coin is dissolved in HNO_(3) . When NaCl is added to this solution, all silver is precipitated as AgCl. The weight of AgCl is found to be 14.35g then % silver in coin is: Ag + HNO_(3) overset(NaCl)to AgCl

1.60g of a metal was dissolved in HNO_(3) to prepare its nitrate.The nitrate on strong heating gives 2g oxide.The equivalent weight of metal is 4 x,then x is

VA group precipitate was dissolved in HNO_(3) and treated with excess of NCl_(5) . It gives a white ppt. because of

A white crystalline substance dissolves in hot water. On passing H_(2)S in this solution, black precipitate is obtained. The black precipitate dissolves completely in hot HNO_(3) . On adding a few drops of conc. H_(2)SO_(4) , a white precipitate is obtained. this substance is :

Tin dissolve in dilute HNO_(3) forming