write half reaction for KMnO4 and (COMM)2 and balance the equation.calculate the atomic mass of M in14.0g/dm3 of solution
The half reaction for KMnO4 is
MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O
I have no idea what (COMM)2 is
To write the half-reaction for KMnO4, we need to determine the oxidation states of each element involved.
The oxidation state of Mn in KMnO4 is +7. The oxidation state of O in a compound is generally -2 (unless it's a peroxide, in which it is -1).
Therefore, the half-reaction for KMnO4 can be written as follows:
MnO4- --> Mn(2+) (reduction half-reaction)
Next, we need to balance the equation. Start by balancing the number of Mn atoms on each side:
MnO4- --> Mn(2+)
The Mn atoms are already balanced.
Next, balance the number of O atoms:
MnO4- --> Mn(2+) + 4O2-
Lastly, balance the charge by adding electrons:
MnO4- + 8H+ + 5e- --> Mn(2+) + 4H2O
Now, to calculate the atomic mass of M in a 14.0 g/dm3 solution, we need to know the molar mass of the compound M. From the information given, it is unclear which specific compound M refers to (assuming it is not Mn).
Once you have the molar mass of compound M, you can calculate the atomic mass by using the equation:
Atomic mass = (molar mass of M) / (Avogadro's number)
Remember to convert the concentration from g/dm3 to mol/dm3 by dividing by the molar mass.