`H_(2)SO_(4)` is added to `20%` cold aqueous solution of `"BaO"_(2)`. The product formed is

H_(2)O_(4) is added to 20% cold aqueous solution of BaO_(2) . The product formed is

State, the equations, what happens when : A freshly prepared solution of FeSO_(4) is added to the aqueous solution of NaNO_(3) and then conc. H_(2)SO_(4) is added to the mixture dropwise along the walls of the test tube.

when an aqueous solution of H_2SO_4 is electrolysed the product at anodes is :

when an aqueous solution of H_2SO_4 is electrolysed the product at anodes is :

when an aqueous solution of H_2SO_4 is electrolysed the product at anodes is :

(i) An aqueous solution of a white coloured compound (A) on reaction with HCl gives a white precipitate of compound (B) . (ii) (B) becomes soluble in chlorine water with the formation of (C) (iii) (C) reacts with KI to give a precipitate which becomes solube in excess of it forming a compount (D) . the compound (D) is used for detecting ammonium salts. (iv) (B) and (C) both, on treatement with SnCl_(2) give a grey precipitate of (E) . (v) When conc. H_(2)SO_(4) is added slowly into a mixture of cold solutions of (A) and FeSO_(4) is added slowly into a mixture of cold solutions of (A) and FeSO_(4) , a brown ring of compound (F) is formed. (A) on dissociation in H_(2)SO_(4) gives:

Suppose 60% w // w aqueous solution of glucose ( C_(6)H_(12) O_(6) ) and 20% w // w aqueous solution of urea ( NH_(2)CONH_(2) ) have equal molarity, then which solution has higher density :

Suppose 60% w // w aqueous solution of glucose ( C_(6)H_(12) O_(6) ) and 20% w // w aqueous solution of urea ( NH_(2)CONH_(2) ) have equal molarity, then which solution has higher density :