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egoroff_w [7]
3 years ago
15

The lab assistant accidentally poured water in the spirit (alcohol). How would you help him/her to get back the spirit. Explain.

Chemistry
1 answer:
Reptile [31]3 years ago
6 0

The water and the spirit can be easily separated by a method called fractional distillation.

<u>Explanation:</u>

  • There are many ways of separating alcohol from water. The most common method is by heating. Water has a higher boiling point when compared to alcohol and the water can be easily evaporated into the air.
  • This method is called Fractional distillation. As it requires a rounded flash, a beaker, and a container. If the content is put under the heat for a certain degree celsius the water in the content is evaporated.
  • It is to be collected and connected to a container. At last, the alcohol and water are separated.
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Explanation:

The initial concentrations for a mixture :

Acetic acid at equilibrium = 0.15 M

Ethanol at equilibrium = 0.15 M

Ethyl acetate at equilibrium = 0.40 M

Water at equilibrium = 0.40 M

CH3COOH + C_2H_5OH\rightleftharpoons CH_3CO_2C_2H_5+H_2O

Initially:

0.15 M            0.15 M            0.40 M   0.40 M

At equilibrium

(0.15-x)M       (0.15-x) M     (0.40+x) M   (0.40+x) M

The equilibrium constant is given by expression

K_c=\frac{[CH_3CO_2C_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}

4.0=\frac{(0.40-x)\times (0.40-x)}{(0.15+x)\times (0.15+x)}

Solving for x:

x = 0.0333

The equilibrium concentrations for a mixture :

Acetic acid at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M

Ethanol at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M

Ethyl acetate at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M

Water at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M

4 0
3 years ago
Select the correct answer. Which substance is made of polymers? A. marble B. protein C. salt D. steel
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Answer:

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Explanation:

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1. Calculate the heat change associated with cooling a 350.0 g aluminum bar from
Amanda [17]

Answer:

14175 j heat released.

Explanation:

Given data:

Mass of aluminium = 350.0 g

Initial temperature = 70.0°C

Final temperature = 25.0°C

Specific heat capacity of Aluminium = 0.9 j/g.°C

Heat changed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Heat change:

ΔT = Final temperature - initial temperature

ΔT = 25.0°C - 70°C

ΔT = -45°C

Q = m.c. ΔT

Q = 350 g × 0.9 j/g.°C  × -45°C

Q = -14175 j

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2 years ago
Nitric acid (HNO3) is a strong acid, and it is titrated with a standard solution of sodium hydroxide (NaOH), a strong base. Whic
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