question_answer 47)

  The oxide that gives \[{{H}_{2}}{{O}_{2}}\] on treatment with dilute \[{{H}_{2}}S{{O}_{4}}\]is....... (a)\[Pb{{O}_{2}}\]                           (b)\[Ba{{O}_{2}}\cdot 8{{H}_{2}}O\]        (c)\[Mn{{O}_{2}}\]                          (d) \[Ti{{O}_{2}}\]

Answer:

  (b)\[Ba{{O}_{2}}\cdot 8{{H}_{2}}O+{{H}_{2}}S{{O}_{4}}\to BaS{{O}_{4}}+8{{H}_{2}}O+{{H}_{2}}{{O}_{2}}\]