1 g equivalent of `V_2O_5` in the given reaction is equal to `V_2O_5 + Zn rarr ZnO + V` (Glven at wt. of V = A)

1 g equivalent of V_2O_5 in the given reaction is equal to V_2O_5 + Zn → ZnO + V (Given at. wt. of V = A)

1 g molecule of V_(2)O_(5) contains :

Balance the equations given below : ZnS+O_2 to ZnO + SO_2

(v) 1 2/3 x 1 3/5 is equal to

Benzene on controlled oxidation with V_2O_5 forms

If 10g of V_(2)O_(5) is dissolved in acid and is reduced to V^(2+) by zinc metal, how many mole I_(2) could be reduced by the resulting solution if it is further oxidised to VO^(2+) ions? [Assume no change in state of Zn^(2+) ions] ( V=51 , O=16 , I=127 )

For the first order reaction 2N_(2)O_(5)(g) rarr 4NO_(2)(g) + O_(2)(g)

In the reaction VO+Fe_(2)O_(3) rarr FeO+V_(2)O_(5) , the eq.wt. of V_(2)O_(5) is equal to its

E^(o) for the reaction Fe + Zn^(2+) rarr Zn + Fe^(2+) is – 0.35 V. The given cell reaction is :

If 10.0 g V_2O_5 is dissolbed in acid and reduced to V^(2+) by treatment with tin (Sn) metal how many moles of I_2 could be reduced by the resulting V^(2+) solution as it is oxidised to V^(4+) ? (Atomic weight of V is 51)