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dexar [7]
4 years ago
6

In addition to the separation techniques used in this lab (magnetism, evaporation, and filtering), there are other commonly used

separation techniques. Some of these techniques are:
Distillation – this process is used to separate components that have significantly different boiling points. The solution is heated and the lower boiling point substance is vaporized first. The vapor can be collected and condensed and the component recovered as a pure liquid. If the temperature of the mixture is then raised, the next higher boiling component will come off and be collected. Eventually only non-volatile components will be left in the original solution.
Centrifugation – a centrifuge will separate mixtures based on their mass. The mixture is placed in a centrifuge tube which is then spun at a high speed. Heavier components will settle at the bottom of the tube while lighter components will be at the top. This is the technique used to separate red blood cells from blood plasma.
Sieving – this is similar to filtration, but the sample is passed through a screen which allows smaller particles to go through and retains the larger particles.
Paper chromatography – this is a technique that separates of mixture based on the individual substance’s tendency to travel across a paper surface. This technique is used frequently to separate different dyes.
Separatory Funnel – this is a glass container with a stopcock on the bottom. Two immiscible solutions are put in it. Since the compounds do not mix, they will separate into two layers, the heavier one will be on the bottom. The stopcock can then be opened and the heavier liquid will flow out and can be collected.
Consider the following separation problems. Decide which of the above techniques, including the ones you used in your lab, would be the best to use for separating the solutions. You may need to use more than one technique for a given sample. Then give a brief explanation as to how you would do the separation.
A. Water and Sugar
B. Mixture of Hexane (Boiling Point = 68.7oC) and Octone (Boiling Point = 125oC)
C. Solid I2 (non-polar solid) and NaCl
D. "Sharpie" permanent marking pen
E. Nickel shavings and copper pellets
Chemistry
1 answer:
DENIUS [597]4 years ago
4 0

Answer:

A. Water and Sugar  can be separated by evaporation and then crystallization

B. Mixture of Hexane and Octane can be separated by distillation

C. Solid Iodine, I₂ and NaCl  can be separated by filtration and then evaporation

D. "Sharpie" permanent marking pen  can be separated by  chromatography

E. Nickel shavings and copper pellets can be separated by magnetic separation

Explanation:

A. A mixture of water and sugar can be separated by employing two separation techniques, evaporation and crystallization. First the sugar solution  is heated to evaporate most of the water. When the solution becomes very saturated, it is allowed to cool and then the sugar molecules are obtained through crystallization induced by seeding or scratching the walls of the container.

B. A mixture of hexane (boiling point = 68 °C) and Octane (boiling point = 125 °C) can be separated by distillation due to their significant difference in boiling points.

The mixture is heated in a flask connected to a Liebig condenser. Hexane with the lower boiling point will distill over first and is collected. Afterwards, octane next distills over and is collected as well.

C. A mixture of solid iodine and NaCl can be seperated by first dissolving in water. Iodine being non- polar does not dissolve and is collected as a residue from filtration using a filter paper, while the NaCl solution is collected as the filtrate. The NaCl is recovered from solution by evaporating to dryness in an evaporating dish.

D. "Sharpie" permanent marking pen contains a mixture of dyes which can be separated by paper chromatography.

A drop of the marker ink is placed on a spot above the solvent level on the paper strip used for the separation. The paper strip is  held vertically inside a jar containing a solvent which serves as the mobile phase. The jar is covered and the different dyes move along the paper which serves as the stationary phase, and is thus separated. The paper strip is removed from the jar when the ascending front of the solvent is approaching the top of the paper. The paper is dried and the various dyes can be identified by comparing the distance each has traveled with those of standards.

E. A mixture of nickel shavings and copper pellets can be separated by magnetic separation.

A magnet is brought near the mixture and the nickel shavings being magnetic is attracted to the magnet leaving copper pellets behind since copper is not magnetic.

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Explanation:

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Mole Fraction = number of moles / total number of moles

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Mole Fraction = 4.46 / 7.35 = 0.607

Partial Pressure = 0.607 * 2 = 1.214 atm

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Mole Fraction = 2.15 / 7.35 = 0.293

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Answer: K_{eq} at  the temperature of the experiment is 0.56.

Explanation:

Moles of  CO = 0.35 mole

Moles of  H_2O = 0.40 mole

Volume of solution = 1.00 L

Initial concentration of CO = \frac{0.35mol}{1.00L}=0.35M

Initial concentration of H_2O = \frac{0.40mol}{1.00L}=0.40M

Equilibrium concentration of CO = \frac{0.19mol}{1.00L}=0.19M

The given balanced equilibrium reaction is,

                      CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial conc.          0.35 M         0.40 M                  0 M        0M

    At eqm. conc.     (0.35-x) M   (0.40-x) M   (x) M      (x) M

Given:  (0.35-x) = 0.19

x= 0.16 M

The expression for equilibrium constant for this reaction will be,

K_{eq}=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}  

Now put all the given values in this expression, we get :

K_{eq}=\frac{0.16\times 0.16}{(0.35-0.16)\times (0.40-0.16)}

K_{eq}=\frac{0.16\times 0.16}{(0.19)\times (0.24)}=0.56

Thus K_{eq} at  the temperature of the experiment is 0.56.

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