Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Answer:
Here the force is 10N and the mass is 5 kg. Dividing both sides by 5kg, we get a = 2 m/s^2.
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Answer:
Cu2O + H2SO4 ——> CuSO4 + Cu + H2O
Cu2O is oxidized and reduced. It lost an electron in CuSO4 for the 2+ charge, but it gained an electron in Cu to go from the 1+ charge to a neutral charge. This is dismutation (aka disproportionation)
Explanation:
Answer:
Explanation:
Hello there!
In this case, according to the given chemical reaction, it is possible for us to relate the rate of formation of C and the rate of consumption of B as shown below:
Thus, we solve for the rate of change of B as shown below:
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