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dexar [7]
3 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]3 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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fredd [130]
<h2>Answer : Electromagnetic energy</h2><h3>Explanation :</h3>

The solar panel usually combines multiple photo voltaic cells. Electromagnetic energy is used to power a solar panel. As the electromagnetic energy or rather radiations from the sun reaches to Earth in the form of radiation which is then collected by these photo voltaic cells. As the term solar energy simply suggests that the energy is coming from the sun. Thus, electromagnetic energy coming from the sun was produced when electric charges inside the sun change their potential energy.

6 0
2 years ago
How many kilograms of oxygen did Samantha need to utilize to react with 36.29 kg of triglycerides (C55H104O6)?
Paraphin [41]

Answer:

105.33Kg of O2.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

C55H104O6 + 78O2 —> 55CO2 + 52H2O

Step 2:

Determination of the masses of C55H104O6 and O2 that reacted from the balanced equation.

This is illustrated below:

Molar mass of C55H104O6 = (12x55) + (104x1) + (16x6) = 860g/mol

Mass of C55H104O6 from the balanced equation = 1 x 860 = 860g.

Divide the mass of C55H104O6 by 1000 to express in kg i.e

Mass of C55H104O6 = 860/1000= 0.86Kg

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 78 x 32 = 2496g.

Divide the mass of O2 by 1000 to express in kg i.e

Mass of O2 = 2496/1000 = 2.496kg.

From the balanced equation above,

0.86kg of C55H104O6 reacted with 2.496Kg of O2.

Step 3:

Determination of the mass of O2 in kg needed to react with 36.29 kg of triglycerides (C55H104O6).

This can be obtained as follow:

From the balanced equation above,

0.86kg of C55H104O6 reacted with 2.496Kg of O2.

Therefore, 36.29 kg of C55H104O6 will react with = (36.29 x 2.496)/0.86 = 105.33Kg of O2.

Therefore, 105.33Kg of O2 is needed for the reaction.

5 0
2 years ago
The following diagram shows a single cell. Is the cell eukaryotic or prokaryotic? Explain your reasoning
ANTONII [103]

Answer:

eukaryotic because it has a nucleus

Explanation:

3 0
3 years ago
No pressure but anyone who can answer these is a genius. (Please fill in the question mark)
disa [49]

use variable

1K₂MnF₆ + aSbF₅⇒ bKSbF₆ + cMnF₃ + dF₂

K, left=2,right=b⇒b=2

Mn, left=1, right=c⇒c=1

Sb, left=a, right=b⇒a=b=2

F, left=6.1+5a, right=6b+3c+2d

equation:

6+5(2) = 6(2)+3(1)+2d

16=15+2d

1=2d

d=0.5

So the reaction equation becomes:

1K₂MnF₆ + 2SbF₅⇒ 2KSbF₆ + 1MnF₃ + 0.5F₂ x2

2K₂MnF₆ + 4SbF₅⇒ 4KSbF₆ + 2MnF₃ + F₂

4 0
2 years ago
How many ions are present if 2.0 moles of hno3 are completely dissolved in water?
Arte-miy333 [17]
Nitric acid(HNO3) is a strong acid so most of its component will dissociate into an ion form. When dissolved into water, nitric acid will have this reaction
HNO3= 1 H^{+} + NO3^{-} 

For every 1 mol of HNO3, there will be 1 mol of H+ ion and NO3- ion formed, which makes a total of 2 ions(degree of dissociation = 2). 
Then for 2 moles of HNO3 it would be: 2 moles * 2= 4 moles of ion
5 0
3 years ago
Read 2 more answers
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