`Fe^(2+) (aq)+NO_(3)^(-) (aq)+H_(2)SO_(4) (conc.) to` Brown ring The brown ring is due to the formation of complex, `[Fe(H_(2)O)_(5)NO]SO_(4)`. What i

The slow step for the reaction is: `NO_(2) + NO_(2) to NO + NO_(3)` (Slow) The rate law equation is: `Rate( r) = k[NO_(2)][NO_(2)]= k[NO_(2)]^(2)`.

Rate = `k[NO_(2)][F_(2)]`

Correct Answer - C Since `HNO_(3)` is also a strong oxidising agent, therefore, aniline forms tarry oxidation products along with some oxidation products.

Correct Answer - C In nitration of benzene, the reactive species in - volved is nitronium ion `(NO_(2)^(+))`.

Correct Answer - A `HNO_(3)+H_(2)SO_(4) rarr NO_(2)^(+)+HSO_(4)^(-)+H_(2)O` In this reaction nitric acid acts like a base.

Dilute `H_(2)SO_(4)` behaves as a proton acid and zinc metal displaces hydrogen from the acid. `Zn+H_(2)SO_(4)("dil")to ZnSO_(4)+H_(2)` Concentrated `H_(2)O_(4)` behaves as a strong oxidising agent and itself is reduced to...

Correct Answer - C `Au to` Noble metal `&` it is used in tonning for photography.

Correct Answer - B `2NO_(3)^(-)+4H_(2)SO_(4)+6Fe^(2+)to6Fe^(3+)+2NOuarr+4SO_(4)^(2-)+4H_(2)O` `Fe^(2+)+NOuarr +SO_(4)^(2-) to [Fe(NO)]^(2+)SO_(4)^(2-)`

Correct Answer - C Metal having more value of SRP will deposit first, but Mg does not deposition aqueous solution.

Correct Answer - B We have, `P(A)=40/100,P(B)=25/100and P(AnnB)=15/100` So, `P((B)/(A))=(P(AnnB))/(P(A))=(15//100)/(40//100)=3/8`