Answer:
Mn^2+(aq) + 4H2O(l) + 5[VO2]^+(aq) + 10H^+(aq) ---------->MnO4^-(aq) + 8H^+(aq) + 5[VO]^2+(aq) + 5H2O(l)
Explanation:
Oxidation half equation:
Mn^2+(aq) + 4H2O(l) ------------> MnO4^-(aq) + 8H^+(aq) + 5e
Reduction half equation:
5[VO2]^+(aq) + 10H^+(aq) + 5e --------> 5[VO]^2+(aq) + 5H2O(l)
Overall redox reaction equation:
Mn^2+(aq) + 4H2O(l) + 5[VO2]^+(aq) + 10H^+(aq) ---------->MnO4^-(aq) + 8H^+(aq) + 5[VO]^2+(aq) + 5H2O(l)
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Answer:
The number of atoms contained by one molecule of Iron (II) Sulfate are 6.
Explanation:
Iron (II) Sulfate is mage up of two parts. One is the Positive part which constitutes of Fe⁺² and a negative part which constitutes of a polyatomic anion i.e. SO₄²⁻. As there are four Oxygen and one sulfur atom in sulfate Ion so sulfate ion contains 5 atoms in total. Therefore, five atoms from sulfate iona dn one atom of Iron ion makes a total of 6 atoms in one molecule of Iron (II) Sulfate.
