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Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
Well the the answer is 70.8c but if you round it up it is 71c which I choice and got it correct so the answer is 71c
Answer:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL
¹/3 C3H8(g) + ⁵/3 O2(g)
Explanation:
The coefficient before every molecule is representative of the number of moles. We can represent it in ration form so as to calculate the question;
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) means;
For every 1 mole of C₃H₈(g) and 5 moles of O₂(g) produces 3 moles of CO₂(g) and 4 moles of H₂O(l).
Therefore to produce 1.00 mole of CO₂(g);
We represent it in ratio;
C₃H₈(g) : CO₂(g)
1 : 3
What about ;
? (x) : 1
We cross multiply;
3x = 1 * 1
X = 1/3
We evaluate the same for O₂;
O₂(g) : CO₂(g)
5 : 3
What about
? (x) : 1
3x = 5 * 1
x = 5/3
Learn More:
For more on evaluating moles in chemical reactions check out;
brainly.com/question/13967925
brainly.com/question/13969737
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