A) \[N{{a}_{2}}S{{O}_{3}}\]
B) \[PbS\]
C) \[KI\]
D) \[{{O}_{3}}\]
Correct Answer: D
Solution :
\[\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{3}}}\]is oxidized by \[{{H}_{2}}{{O}_{2}}\]to \[N{{a}_{2}}S{{O}_{4}}\] PbS is oxidized by \[{{H}_{2}}{{O}_{2}}\]to\[PbS{{O}_{4}}\] KI is oxidized by \[{{H}_{2}}{{O}_{2}}\]to \[{{I}_{2}}\] \[{{\text{O}}_{\text{3}}}\]cannot be oxidized by \[{{\text{H}}_{2}}{{O}_{2}}\]but [t is reduced to \[{{O}_{2}}\]by \[{{H}_{2}}{{O}_{2}}\] \[{{H}_{2}}{{O}_{2}}+{{O}_{3}}\xrightarrow{{}}{{H}_{2}}O+2{{O}_{2}}\]
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