MnO4–(aq) + Cl–(aq) Mn2+ +

  1. Balance the following equations and show the steps:1. MnO4- + C2O42- = Mn2+ + CO2 2. NO2- + MnO4- = NO3- + Mn2+ (in acid
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  2. Balance the following equations:1. MnO4- + C2O42- = Mn2+ + CO2 2. NO2- + MnO4- = NO3- + Mn2+ (in acid Solution) 3. I- + MnO4- =
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  3. How to make into compounds to get a formula?Li+: Br-, O2-, CN-, SO3^2-, PO3^3-,MnO4- Cu:Br-, O2-, CN-, SO3^2-, PO3^3-,MnO4-
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    2. Jessica asked by Jessica
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  4. How to make into compounds to get a formula?Li+: Br-, O2-, CN-, SO3^2-, PO3^3-,MnO4- Cu:Br-, O2-, CN-, SO3^2-, PO3^3-,MnO4-
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    2. Jessica asked by Jessica
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  5. Using the change in oxidation number method, balance the redox reaction shown below.(NO2^2)+ (MnO4^-)---> (NO3^-)+ Mn^2+ Divide
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  6. MnO4^-(aq) + H20(l) ==> MnO2 + OH^-net charg is -1 +7 (-8) ==> 4(-4) Manganese is reduced MnO4^- +3e- ==> MnO2 H2) is the
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  7. What is the reduction half-reaction for the reaction between iron(II) sulfate and potassium permanganate in a sulfuric acid
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  8. What is the reduction half-reaction for the reaction between iron(II) sulfate and potassium permanganate in a sulfuric acid
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  9. Potassium permanganate, KMnO4(permanganate ion, MnO4-)• MnO4-(aq) + 8H+(aq)+ 5e → Mn2+ (aq)+ 4H2O (l) Eo = +1.51V • purple
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  10. Potassium permanganate, KMnO4(permanganate ion, MnO4-)• MnO4-(aq) + 8H+(aq)+ 5e → Mn2+ (aq)+ 4H2O (l) Eo = +1.51V • purple
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