White vitriol `overset(100^(@)C)underset(Delta)rarrXoverset(250^(@)C)rarrY` . If loss in weight in the first step is 'A' gm and that of in second step is 'B' gm , then `(B)/(A)=`

"White" "vitrol "overset(100^(@)C)underset(Delta)rarrXoverset(250^(@)C)(rightarrow)Y . If loss in weight in the first step is A gm and that of in second step is B gm,then (B)/(A)=

4P_(4)O_(6) overset(Delta)underset(250^(@)C)rarr 3X +P_(4) then X =______

In the reaction 2 CH_(3)underset(O)underset(||)(C)CH_(3)overset("Base")(rarr)A overset(Delta)(rarr)B the product B is :

The mechanism of the reaction A+2B rarr C+D is: Step I: A+B underset(k_(2))overset(k_(1))hArr I Step II: B+I overset(k_(3))rarr C+D In the first step is a fast equilibruim , then the incorrect statement is :

CH_(4)overset(1000^(@)C)(rarr) A+B_((g)) , CH_(3)CH_(2)COONaoverset(Sodalime)underset(Delta)(rarr) C+Na_(2)CO_(3) C overset(Delta)underset(450^(@)C)(rarr) B+E,E+B overset(Ni)rarrX If E is a hydrocarbon, then X is identical with

Ph-NH_(2)underset(0^(@)C)overset(HNO_(2))rarr A underset(BF_(3))overset(HF)rarr B overset(Delta)rarr C , C is :

The mechanism of SN1 reaction is given below : CH_(3)-underset(CH_(3)) underset(|) overset(CH_(3)) overset(|)C-Broverset("step-1")toCH_(3)-underset(CH_(3)) underset(|)overset(o+)C-CH_(3)+ overset(o+)(Br) overset(Nu)underset("step-II")to CH_(3)- underset(CH_(3)) underset(|)overset(CH_(3))overset(|)C-Nu Given answer of the following questions on the basis of the machanism given below. Which statement is NOT correct ?