A) \[Mn{{O}_{2}}\]
B) \[Pb{{O}_{2}}\]
C) \[N{{a}_{2}}{{O}_{2}}\]
D) \[Ti{{O}_{2}}\]
Correct Answer: C
Solution :
\[N{{a}_{2}}{{O}_{2}}+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}{{O}_{2}}\]
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