Answer:
ΔSv = 0.1075 KJ/mol.K
Explanation:
Binary solution:
∴ a: solvent
∴ b: solute
in equilibrium:
- μ*(g) = μ(l) = μ* +RTLnXa....chemical potential (μ)
⇒ Ln (1 - Xb) = ΔG/RT
∴ ΔG = ΔHv - TΔSv
⇒ Ln(1 -Xb) = ΔHv/RT - ΔSv/R
∴ Xb → 0:
⇒ Ln(1) = ΔHv/RT - ΔSv/R
∴ T = T*b....normal boiling point
⇒ 0 = ΔHv/RT*b - ΔSv/R
⇒ ΔSv = (R)(ΔHv/RT*b)
⇒ ΔSv = ΔHv/T*b
∴ T*b = 80°C ≅ 353 K
⇒ ΔSv = (38 KJ/mol)/(353 K)
⇒ ΔSv = 0.1075 KJ/mol.K
Density = mass / volume
= 69g / 23 ml
= 3 g / ml.
Thus, the density of the sample is 3 grams per ml or 3g/ ml
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HOPE THIS HELPS!
I believe it is 6ml because you do the doseage times the ml and mutiply it by 1
The reaction involved here would be written as:
2N2 + 3H2 = 2NH3
The equilibrium constant of a reaction is the ratio of the concentrations of the products and the reactants when in equilibrium. The expression for the equilibrium constant of this reaction would be as follows:
Kc = [NH3]^2 / [N2]^2[H2]^3
Kc = 0.40^2 / (0.20)^2 (0.10)^3
Kc = 4000
