Hbr is the answer when I work it out that’s what I got.
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:
friend me on here and imma send you the link Explanation:
Answer:
Boiling point for the solution is 100.237°C
Explanation:
We must apply colligative property of boiling point elevation
T° boiling solution - T° boiling pure solvent = Kb . m
m = molalilty (a given data)
Kb = Ebulloscopic constant (a given data)
We know that water boils at 100°C so let's replace the information in the formula.
T° boiling solution - 100°C = 0.512 °C/m . 0.464 m
T° boiliing solution = 0.512 °C/m . 0.464 m + 100°C → 100.237 °C
Answer:
480 g of oxygen.
Explanation:
C3H8 + 5O2 ---> 3CO2 + 4H2O
Using the molar masses:
3*12 + 6*16 g of CO2 were formed from 10*16 g O2
132g g CO2 from 160 g O2
1g CO2 from (160/132) g O2
396 g from (160/132) * 396
= 480 g of oxygen.