Answer:
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Answer:
ΔH°rxn = -827.5 kJ
Explanation:
Let's consider the following balanced equation.
2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)
We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]
ΔH°rxn = -827.5 kJ
24.063/8.021 = 3 half lives
(1/2)^3 =1/8
Answer:
P(N) = 38.48 mmHg
Explanation:
Given data:
Partial pressure of He = 15.22 mmHg
Partial pressure of O = 35.21 mmHg
Partial pressure of N = ?
Total pressure = 88.91 mmHg
Solution:
According to Dalton law of partial pressure,
The total pressure inside container is equal to the sum of partial pressures of individual gases present in container.
Mathematical expression:
P(total) = P₁ + P₂ + P₃+ ............+Pₙ
Now we will solve this problem by using this law.
P(total) = P(He) + P(O) + P(N)
88.91 mmHg = 15.22 mmHg + 35.21 mmHg + P(N)
88.91 mmHg = 50.43 mmHg + P(N)
P(N) = 88.91 mmHg - 50.43 mmHg
P(N) = 38.48 mmHg
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