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 is: 4.45 grams of methane gas <span>need to be combusted</span>.
Balanced chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.
Ideal gas law: p·V =
n·R·T.<span>
p = 1.1 atm.
T = 301 K.
V(H</span>₂O) <span>= 12.5 L.
R = 0,08206 L·atm/mol·K.
</span>n(H₂O) = <span>1.1 atm ·
12.5 L ÷ 0,08206 L·atm/mol·K · 301 K.
</span>n(H₂O) = 0.556 mol.
From chemical reaction: n(H₂O) : n(CH₄) = 2 : 1.
n(CH₄) = 0.556 mol ÷ 2 = 0.278 mol.
m(CH₄) = 0.278 mol · 16 g/mol.
m(CH₄) = 4.448 g.
C.
centi- is essentially 10^2 of one meter.
If you had 100m, multiplying 100 by 10^2 (or 100) would give you 10000 cm.
