Number of neutrons = Atomic mass - Atomic number (number of protons)
Hope that helps!
Answer:
[C] = 0.4248M
Explanation:
A + B ⇄ 2C
C(i) 1.68M 1.68M 0.00
ΔC -x -x +2x
C(eq) 1.68-x 1.68-x 2x
Keq = [C]²/[A][B] = (2x)²/(1.68 - x)²= 8.98 x 10⁻²
Take SqrRt of both sides => √(2x)²/(1.68 - x)² = √8.98 x 10⁻²
=> 2x/1.68 - x = 0.2895
=> 2x = 0.2895(1.68 - x)
=> 2x = 0.4863 - 0.2895x
=> 2x + 0.2895x = 0.4863
=> 2.2895x = 0.4863
=> x = 0.4863/2.2895 = 0.2124
[C] = 2x = 2(0.2124)M = 0.4248M in 'C'
Answer:
-1.9 KJ/mol
Explanation:
In order to solve the problem, we have to rearrange the equations in a way in which all molecules of O₂ and CO₂ are eliminated:
2C(diamond) + 2O₂(g) → 2CO₂(g) ΔH₁= 2 x (-395.4 KJ) ------> we multiply by 2 both reactants and products
2 CO₂(g) → 2CO(g) + O₂(g) ΔH₂= 566.0 KJ
CO₂(g) → C(graphite) + O₂(g) ΔH₃= -1 x (-393.5 KJ) ------> we use reverse rxn
2CO(g) → C(graphite) + CO₂(g) ΔH₄= -172.5 KJ
When we cancel the molecules that appear both in reactants and products, the total reaction is the following:
2C(diamond) → 2C(graphite)
ΔHt= ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ = 2 x (-395.4 KJ) + 566.0 KJ + (-1 x (-393.5 KJ)) - 172.5 KJ
ΔHt= 347.2 KJ
This is for 2 mol of C(diamond) which are converted in 2 mol of C(graphite). To obtain ΔH for the reaction of 1 mol C(diamond) to 1 mol (graphite) we have to divide into 2:
ΔH= -3.8 KJ/2mol= -1.9 KJ/mol
2,2,1,1/2 -- NOT (l too high)
3,2,1,1/2 -- Valid
2,1,-1,1/2 -- Valid
3,3,-2,-1/2 -- NOT (l too high)
1,-1,-1,-1/2 -- NOT (l too low)
4,3,3,-1/2 -- Valid
1,1,0,1/2 -- NOT (l too high)
0,2,1,1/2 -- NOT (no such thing as n = 0, l too high)
2,0,0,-1/2 -- Valid
1,0,0,1/2 -- Valid
3,2,1,0 -- NOT (cannot have ms = 0)
4,3,4,-1/2 -- NOT (ml too high)
