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(1× boMn) +(1×boO) =0
boMn+(1×(-2))=0
boMn=2
b. MnO2²-=-2
boMn+(2×(-2)=-2
boMn=4-2=2
c. MnO2=0
boMn+(-4)=0
boMn=4
d. KMnO2=0
(1×1)+boMn+(2×(-2))=0
boMn=3