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dexar [7]
2 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]2 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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What pressure does 3.54 moles of chlorine gas at 376 k exert on the walls of it 51.2 l container? I'm Lazy sooooo
Lilit [14]

Answer:

The answer to your question is P = 2.13 atm

Explanation:

Data

Pressure = ?

number of moles = 3.54

Temperature = 376 °K

Volume = 51.2 L

R = 0.08205 atm L/mol°K

Formula

PV = nRT

- Solve for P

  P = nRT / V

- Substitution

  P = (3.54)(0.08205)(376) / 51.2

- Simplification

  P = 109.21 / 51.2

Result

 P = 2.13 atm  

6 0
1 year ago
A 1.50-liter sample of dry air in a cylinder exerts a pressure of 3.00 atmospheres at a temperature of 25°C. Without changing th
Aneli [31]
Hope this helps you.

3 0
2 years ago
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Two weak acids, A and B, have pKa values of 4 and 6, respectively. Which statement is true?A) Acid A dissociates to a greater ex
zalisa [80]

Answer:

A) Acid A dissociates to a greater extent in water than acid B

Explanation:

A) Acid A dissociates to a greater extent in water than acid B

We are given the pKa for both acids, and we know

pKa = - log Ka

Taking antilog to both sides of the equation we can solve for Ka

⇒ -pKa = log Ka

-antilog pKa = Ka

10 ^-pka = Ka

So ka for acid A = 10⁻⁴

and

ka for acid B = 10⁻⁶

True the equilibrium constant for acid A is greater, so it dissociates more.

B) For solutions of equal concentration, acid B will have a lower pH.

We know the stronger acid is A, and it dissociates more. Since pH is the negative log of H₃O⁺ concentration, it follows that at equal concentrations the acid A will have at equilibrium a greater [H₃O⁺] and hence a lower pH

C) B is the conjugate base of A

False:

If B were the conjugate base of A, its  Kb would have been given by:

Ka x Kb = Kw

Kb = 10⁻¹⁴/10⁻⁶ = 10⁻⁸ for the conjugate base of acid B

Kb = 10⁻¹⁴/10⁻⁴ = 10⁻¹⁰ for the conjugate base of acid A

which are not equal.

D) Acid A is more likely to be a polyprotic acid than acid B.

False

Just having the pkas for both acids one cannot know if any of the acids is polyprotic. We will need the formula for the acids.

E) The equivalence point of acid A is higher than that of acid B

False

The equivalence point depends on the the concentration of the acids  and their volumes.

The equivalence point is reached in the titration when the number of equivalents of base equals the number of equivalents of acid:

# equivalents acid = # equivalents of base  @ end point

and

# equivalents acid = Molarity of acid x Volume of acid

4 0
2 years ago
Complete combustion of a compound containing hydrogen and carbon produced 2.641 g of carbon dioxide and 1.442 grams of water as
klio [65]

  The empirical   formula  of hydrocarbon is C₃H₈

The  molecular formula of hydrocarbon   is C₆H₁₆


<u><em> Empirical  formula  calculation</em></u>

Hydrocarbon  contain  carbon and hydrogen

  Step 1:  find the  mass  carbon (C) in carbon dioxide (CO₂)  and hydrogen (H )  in water

mass of  of element = molar mass  of element/ molar mass molecule x total mass of    molecule

From periodic table the molar mass  of C =12,    for CO₂ = 12+( 16 x2) =44 g/mol,     for H = 1.00 g/mol,    for H₂O = (2 x1)+16 = 18 g/mol

mass of C = 12/44 x 2.641 =0.7203 g

since there are 2 atom  of H in H₂O the molar mass of H = 1 x2 = 2 g/mol

mass of H  is therefore =  2/18 x 1.442 =0.1602 g


Step 2:  find the moles of C and H

moles = mass÷ molar mass

moles of C = 0.7203 g÷ 12 g/mol = 0.060  moles

moles of H  =  0.1602÷ 1 g/mol = 0.1602 moles


Step 3: find the mole ratio  of C and H by dividing  each  mole by smallest mole ( 0.06)

for C = 0.06/0.06 =1

  For H = 0.1602/0.06 =2.67

multiply   by 3  to remove the decimal

For C = 1 x3 =3

For H = 2.67 x3 =8

therefore the empirical formula = C₃H₈


<u><em>The molecular formula calculation</em></u>

[C₃H₈]n  = 88.1 g/mol

[12 x 3)+( 1 x8)]n =88.1 g/mol

44 n = 88.1

divide both side by 44

n=2

therefore [C₃H₈]₂   = C₆H₁₆



7 0
1 year ago
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Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a
JulijaS [17]

Answer : The correct option is, 30.9^oC

Explanation :

Formula used :

q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

q = heat released = 24 KJ

m = mass of bomb calorimeter = 1.30 Kg

c = specific heat = 3.41J/g^oC

T_{final} = final temperature = ?

T_{initial} = initial temperature = 25.5^oC

Now put all the given values in the above formula, we get  the final temperature of the calorimeter.

q=m\times c\times (T_{final}-T_{initial})

24KJ=1.30Kg\times 3.41J/g^oC\times (T_{final}-25.5)^oC

T_{final}=30.9^oC

Therefore, the final temperature of the calorimeter is, 30.9^oC

5 0
2 years ago
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