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3 years ago

1.1 Outline a method for separating the chalk from potassium chloride,

Chemistry

1 answer:

miss Akunina [59]3 years ago

5 0

Answer:

Explanation:

To separate the a mixture of chalk and potassium chloride, we must not that chalk is calcium carbonate compound, CaCO₃.

Calcium carbonate has low solubility in water. KCl is readily soluble in water and it is also an ionic compound.

To separate a mixture of compounds with various solubility, we can carryout dissolution, filtration and evaporation.

We first pour pure water into the mixture. Water will dissolve the potassium chloride readily.

Then using a filter paper we filter out the suspended chalk particles. Leave the filtrate to then dry and collect it.

The solution filtered should be evaporated to dryness. This will leave the KCl behind from the solution.

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The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C

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Answer:

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Explanation:

Given the reaction:

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Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

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T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})$

$1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})$

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

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Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

$ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})$

$ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})$

$1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})$

T₂=371.77 K= 98.62 °C

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