The answer I would choose is the third one
We need to use the following formula
Δ


n= 4 moles
F= constant= 96500C/mol
let's plug in the values.
ΔG= -(4)(96500)(0.24)=
-92640 J or -92.6 kJ
<em><u>the</u></em><em><u> </u></em><em><u>number</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u> </u></em><em><u>i</u></em><em><u>n</u></em><em><u> </u></em><em><u>aluminium</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>1</u></em><em><u>4</u></em>
Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108