What do the symbols, `H_(2)`, S and `O_(4)` mean in the formula `H_(2)SO_(4)` ?

What is the mole coefficient of H_(2)SO_(4) in the balanced equation. K_(2)Cr_(2)O_(7) + FeSO_(4) + H_(2)SO_(4) rarr K_(2)SO_(4) + Fe_(2)(SO_(4))_(3) + Cr_(2)(SO_(4))_(3) + H_(2)O

Equivalent weight of H_(2)SO_(4) in the following reaction (Pb(s) + PbO_(2)(s) + H_(2)SO_(4) rightarrow PbSO_(4) + H_(2)O is

Explain why in the laboratory preparation of H_(2)O_(2) : dilute H_(2)SO_(4) is used instead of concentrated H_(2)SO_(4) ,

What do you mean by the 'antichlor' property of H_(2)O_(2) ?

If S + O_(2) rarr SO_(2), Delta H = -298.2 kJ SO_(2) + (1)/(2) O_(2) rarr SO_(3), Delta H = -98.2 kJ SO_(3) + H_(2)O rarr H_(2)SO_(4), Delta H = -130.2 kJ H_(2) + (1)/(2) O_(2) rarr H_(2)O, Delta H = -287.3 kJ the enthalpy of formation of H_(2)SO_(4) at 298 K will be

Calculate DeltaH^(@) for the reaction : Na_(2) O + SO_(3) to Na_(2) SO_(4) given the following : A) Na_((s)) + underset((1))(H_(2)O) to NaOH_((s)) + (1)/(2) H_(2 (g)) , Delta H^(@) = -146kJ B) Na_(2) SO_(4 (s)) + underset((1))(H_(2)O) to 2 NaOH_((s)) + SO_(3 (g)) , Delta H^(@) = +418 kJ C) 2 Na_(2) O_((s)) + 2 H_(2 (g)) to 4 Na_((s)) + 2 H_(2) O_((l)) , DeltaH^(@) = +259 kJ

Passage - VI : Industrially sulphuric acid is produced by the following steps: Stage I : S+O_(2) overset(Delta)to SO_(2) Stage II : 2SO_(2) +O_(2) overset(V_(2)O_(5))to 2SO_(3) Stage III : SO_(3)+H_(2)O to H_(2)SO_(4) Since the reaction between SO_(3) and H_(2)O is violent, SO_(3) is passed into 98% H_(2)SO_(4) to produce oleum (H_(2)S_(2)O_(7)) Pure H_(2)SO_(4) does not react with metal because -

A compound with the formula C_(4)H_(10)O yields another compound C_(4)H_(8)O , on heating with K_(2)Cr_(2)O_(7) and H_(2)SO_(4) . The compound C_(4)H_(10)O is expected to be