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podryga [215]
3 years ago
5

The solution in the two arms of the U-tube are separated by a membrane that is permeable to water and glucose but not to sucrose

. Side A is half filled with a solution of 2 M sucrose and 1 M glucose. Side B is half filled with 1 M sucrose and 2 M glucose. Initially, the liquid levels on both sides are equal. After the system reaches equilibrium. what changes are observed?
Chemistry
1 answer:
Gnom [1K]3 years ago
8 0

Answer:

The liquid level will rise in Side A and drop in Side B.

Step-by-step explanation:

The membrane is impermeable to sucrose, but permeable to sucrose and water.

<u>      Side A   </u>    <u>    Side B    </u>  

2 M sucrose   1 M sucrose

1 M glucose   2 M glucose

(a) Ignoring osmotic effects

The glucose will diffuse spontaneously from the side with higher concentration to that of lower concentration until equilibrium is established. There is no change in volume on either side.

At this point, we have

<u>     Side A       </u>   <u>        Side B   </u>      

2    M sucrose    1 M sucrose

1.5 M glucose    1.5 M glucose

=====

(b) With osmotic effects

The solute concentration on Side A is greater than on Side B.

Water will diffuse into Side A.

The liquid level will rise in Side A and drop in Side B.

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likoan [24]

Answer:

1.40*10⁻² M

Explanation:

We have the solubility formula

Solubility,

S = KH*P  

where

KH = measure of hardness of water / carbonate hardness = 3.50*10⁻² mol/L.atm

P = atmospheric pressure = 0.400 atm

Hence, we have

S = KH*P

= (3.50*10⁻² mol/L.atm)*(0.400 atm)

= 1.40*10⁻² mol/L

But 1 mol/L = 1 M,

Hence, the answer (1.40*10⁻² mol/L ) is equivalent to

= 1.40*10⁻² M

5 0
3 years ago
What is the formula for CO3 2- and Al3+ ?
Goryan [66]

Answer is: formula is Al₂(CO₃)₃.

Aluminium carbonate (Al₂(CO₃)₃) has neutral charge. Because aluminium cation has positive charge 3+ and carbonate anion has negative charge 2-, for right chemical formula, we need two aluminium cations and three carbonate anion:

charge of the molecule = 2 · (3+) + 3 · (-2).

charge of the molecule = 0.

7 0
3 years ago
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In a simple distillation setup, what is the sequence of equipment from the bench top to the round bottom flask
Julli [10]

In a simple distillation setup, the sequence of equipment from the bench top to the round bottom flask is:

  • Thermometer
  • Distillation flask
  • Liebig condenser
  • Round bottom flask
  • Bunsen burner

<h3>What is Distillation?</h3>

This is the process in which a mixture is separated through selective boiling and condensation.

The distillation flask and liebig condenser are usually located above the round bottom flask in the set up.

Read more about Distillation here brainly.com/question/24553469

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3 0
2 years ago
Using the formula M1V1 = M2V2 , how many milliliters of a 2. 50 M hydrochloric acid solution is required to make 100. 0 mL of a
gladu [14]

Dilution of the solution can be calculated by the formula of the molarity and volume. The initial volume of 2.50 M solution was 30 mL.

<h3>What is the relationship between molar concentration and dilution?</h3>

Molar concentration or the dilution factor is in an inverse relationship and with an increase in the dilution, the molarity of the solution decreases.

Given,

Initial molarity = 2.50 M

initial volume = ?

Final molarity = 0.750 M

Final volume = 100.0 ml

Substituting values in the formula:

\begin{aligned}\rm V_{1} &= \rm \dfrac{M_{2}V_{2}}{M_{1}}\\\\&= \dfrac{0.750 \times 100}{2.50}\\\\&= 30 \;\rm mL\end{aligned}

Therefore, 30 mL was the initial volume of the solution before it was diluted.

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3 0
2 years ago
Prandtl number in heat transfer analogues to which dimension less number in mass transfer. a) Schmidt number b) Sherwood number
Cloud [144]

Answer:

a) Schmidt number

Explanation:

Prandtl number in heat transfer is analogues to Schmidt number in mass transfer.

Prandtl number in heat transfer is the ration of momentum diffusivity to the heat diffusivity.

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Whereas, Schmidt number in mass transfer is the ratio of momentum diffusivity to the mass diffusivity.

S_c= \frac{\nu}{\nu_{AB}}

5 0
3 years ago
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