• question_answer The oxide that gives hydrogen peroxide $({{H}_{2}}{{O}_{2}})$ on the treatment with a dilute acid $({{H}_{2}}S{{O}_{4}})$ is [Pb. PMT 1999] A) $Mn{{O}_{2}}$ B) $Pb{{O}_{2}}$ C) $N{{a}_{2}}{{O}_{2}}$ D) $Ti{{O}_{2}}$

$N{{a}_{2}}{{O}_{2}}+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}{{O}_{2}}$