Eq.mass of K2MnO4 = Molar mass of K2MnO4/number of e- gain or loss in a redox reaction
so, Eq.mass of K2MnO4= 197/5=39.4
Potassium permanganate
+5
-6 is the oxidation number
Yes, Potassium manganate(VII), K2MnO4 or Potassium permanganate is an oxidizing agent
Manganese can be oxidized all the way up to +7, so conceivably 6 K2MnO4 +O3 -> 3 K20 + 6 KMnO4 is possible. However the most stable oxidation state of Manganese is +2 so you would have to 'hit it pretty hard' with a lot of energy to get it all the way up to +7.
Equivalent weight of KMnO4 is equal with molar weight of KMnO4. The some is and for K2MnO4, K2MnO4 - e +OH- --------- KMnO4 + KOH In general, Equivalent weight = Molar weight / Number of electrons that take or give one molecule Equivalent weight of KMnO4 = Molar weight of KMnO4 / 1
Potassium permanganate
+5
Empirical: K2MnO4
At low pH value this decomposition takes place: 3 K2MnO4 + 2 H2O → 2 KMnO4 + MnO2 + 4 KOH
-6 is the oxidation number
KMnO4 is the strongest oxidizer of these chemicals.
binding energy is the energy equivalent to the missing mass in the nucleus
Yes, Potassium manganate(VII), K2MnO4 or Potassium permanganate is an oxidizing agent
The neutron has a mass very close to the mass of proton (but not identical).
Manganese can be oxidized all the way up to +7, so conceivably 6 K2MnO4 +O3 -> 3 K20 + 6 KMnO4 is possible. However the most stable oxidation state of Manganese is +2 so you would have to 'hit it pretty hard' with a lot of energy to get it all the way up to +7.
36.5