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

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) What is the % of free SO_(3) in an oleum that is labelled as '104.5% H_(2)SO_(4)' ?

Na_(2)O_(2) + H_(2)SO_(4) to

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

The action of H_(2)SO_(4) on KI gives I_(2) and H_(2)S Calculate the volume of 0.2 M H_(2)SO_(4) to produce 3.4g H_(2)S

The equivalent weight of H_(2)SO_(4) in the following reaction is Na_(2)Cr_(2)O_(7)+3SO_(2)+H_(2)SO_(4)rarr3Na_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+H_(2)O