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.
<u />
<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
<u />
<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>
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
<span>global wind patterns, rotation of the earth, shape of ocean basins.</span>
Answer:
26.8224 Meters per Second.
