AIF3
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Answer:
16.4 °C
Explanation:
Boiling point elevation is the phenomenon in which the boiling point of a solvent will increase when another compound is added to it; meaning that athe resultant solution has a higher boiling point than its pure solvent.
Using the ebullioscopic constant,
ΔT = m * i * Kb
Where,
Δ T is the temperature difference between the boiling point of the solution, Temp.f and boiling point of the pure solvent, Temp.i
Kb is the ebulliscope factor of water = 0.510 °C.kg/mol
i is the van hoffs number = 1
m is the molality in mol/kg.
Calculating the molality of the solution,
Temp.i = 100°C
Temp.f = 104.5 °C
= 4.5/(1*0.510)
= 8.8235 mol/kg
Freezing point depression is defined as the decrease in the freezing point of a solvent on the addition of a solute.
Using the same equation, but kf = 1.86 °C.kg/mol
ΔT = m * i * Kf
Temp.i = freezing point of water = 0°C
Temp.f = (8.8235*1.86) - 0
= 16.412 °C
Freezing point of the solution = 16.4 °C
Answer:
[NaCl] = 1.72 M
Explanation:
First of all, we balance the equation:
BaCl₂ + Na₂SO₄ → 2NaCl + BaSO₄
We convert mass of barium chloride to moles
45 g . 1mol / 208.23g = 0.216 moles
Ratio is 1:2. This means, that our moles of reactant may produce the double of moles, of product.
0.216 . 2 = 0.432 moles of NaCl are been produced.
Molarity is mol/L. We convert volume of water from mL to L
250 mL . 1L /1000 mL = 0.250 L
[NaCl] = 0.432 mol/0.250L = 1.72 M
It is to make food. I hope this helps tell me if i'm wrong.
Use stoich
3.01 x 10^23 Au* 1m / 6.022 x 10^23
= .499m Au