`H_(2)SO_(4)+HIto` ............. ..

Anhydride of H_(2)SO_(4) is _______ .

In the reaction Cu+ underset((Conc.))(H_(2)SO_(4))to CuSO_(4) +SO_(2) +H_(2)O H_(2)SO_(4) behaves w.r.t. Cu as

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

H_(2)SO_(4) is manufactured by

Properties of H_(2)SO_(4) are listed below . Choose the property A,B,C or D which is responsible for the reactions (i) to v] . A : Acid B : Dehydrating agent C: Non - volatile acid D : Oxidizing agent (i) C_(12)H_(22)O_(11)+nH_(2)SO_(4)to12C+11H_(2)O+nH_(2)SO_(4) , (ii) S+2H_(2)SO_(4)to3SO_(2)+2H_(2)O , (iii) NaCl+H_(2)SO_(4)toNaHSO_(4)+HCl , (iv) CuO+H_(2)SO_(4)toCuSO_(4)+H_2O , (v) Na_(2)CO_(3)+H_(2)SO_(4)toNa_(2)SO_(4)+H_(2)O+CO_(2) [some properties may be repeated]

How many grams of H_(2)SO_(4) are present in 500ml of 0.2M H_(2)SO_(4) solution ? .

Assertion : HCOOH formic acid react with H_(2)SO_(4) to form CO . Reason : H_(2)SO_(4) is mild (moderate) oxidizing agent.

Oleum is mixture of H_(2)SO_(4) and SO_(3) i.e. H_(2)S_(2)O_(7) which is obtained by passing SO_(3) is solution of H_(2)SO_(4) . In order to dissolve SO_(3) in oleum, dilution of oleum is done by water in which oleum is converted into pure H_(2)SO as shown below: H_(2)SO_(4)+SO_(3)+H_(2)Oto2H_(2)SO_(4) (pure) When 100 gm oleum is diluted with water then total mass of diluted oleum is known as percentage labelling in oleum. For example: 109% H_(2)SO_(4) labelling of oleum sample means that 109 gm pure H_(2)SO_(4) is obtained on diluting 100 gm oleum with 9 gm H_(2)O which dissolves al free SO_(3) in oleum. If the number of moles of free SO_(3), H_(2)SO_(4) , and H_(2)O be x, y and z respectively in 118% H_(2)SO_(4) labelled oleum, the value of (x+y+z) is

When conc. H_(2)SO_(4) was teated with K_(4)[Fe(CN)_(6)] , CO gas was evolved. By mistake, somebody used dilute H_(2)SO_(4) instead of conc. H_(2)SO_(4) then the gas evolved was

Chromic anhydride in H_(2)SO_(4) is not blue by: