Answer: 6.Explanation:1) Aluminum

So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.
2) Manganesium

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.
3) Balance
Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:

4) Net equation
Add the two half-equations:

As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.
The right side has 2 Al, 3 Mn, and 2*3 positive charges.
So, the equation is balanced.
5) Count the number of electrons involved.
As you see 2 atoms of aluminum lost 6 electrons (3 each).
That is the answer to the question. 6 electrons will be lost.
Answer:
0.68 V
Explanation:
For anode;
3Mg(s) ---->3Mg^2+(aq) + 6e
For cathode;
2Al^3+(aq) + 6e -----> 2Al(s)
Overall balanced reaction equation;
3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)
Since
E°anode = -2.356 V
E°cathode = -1.676 V
E°cell=-1.676 -(-2.356)
E°cell= 0.68 V
Answer:
1.395J/g°C
Explanation:
The following were obtained from the question:
Q = 6527J
M = 312g
ΔT = 15°C
C =?
Q = MCΔT
C = Q/MΔT
C = 6527/(312 x 15)
C = 1.395J/g°C
The specific heat capacity of the substance is 1.395J/g°C
Info: Al(oh)3 might be an improperly capitalized: Al(OH)3
Error: Some elements or groups in the reagents are not present in the products: O
Error: equation Al4C3+H2O=Al(oh)3+CH4 is an impossible reaction
Please correct your reaction or click on one of the suggestions below:
Al4C3 + H2O = Al(OH)3 + CH4