Answer:
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Explanation:
The lowering of the freezing point of a solvent is a colligative property ruled by the formula:
Where:
- ΔTf is the lowering of the freezing point
- Kf is the molal freezing constant of the solvent: 1.86 °C/m
- m is the molality of the solution
- i is the van't Hoff factor: the number of particles (ions) per unit of ionic compound.
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<u>a) molality, m</u>
- m = number of moles of solute/ kg of solvent
- number of moles of CaI₂ = mass in grams/ molar mass
- number of moles of CaI₂ = 25.00g / 293.887 g/mol = 0.0850667mol
- m = 0.0850667mol/1.25 kg = 0.068053m
<u>b) i</u>
- Each unit of CaI₂, ideally, dissociates into 1 Ca⁺ ion and 2 I⁻ ions. Thus, i = 1 + 2 = 3
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<u>c) Freezing point lowering</u>
- ΔTf = 1.86 °C/m × 0.068053m × 3 = 0.3797ºC ≈ 0.380ºC
<h2>I have problems to upload the full answer in here, so I attach a pdf file with the whole answer.</h2>
Answer:
Ke = 34570.707
Explanation:
- H2(g) + Br2(g) → 2 HBr(g)
equilibrium constant (Ke):
⇒ Ke = [HBr]² / [Br2] [H2]
∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L
∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L
∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L
⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)
⇒ Ke = 34570.707
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