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svet-max [94.6K]
4 years ago
14

The following set of solutions are prepared by dissolving the requisite amount of solute in water to obtain the desired concentr

ations. The pressure of the system is then manipulated such that movement is from the solution listed first to the solution listed second.Classify each set of solutions based on the movement of solvent from the first solution listed to the second solution listed as osmosis or reverse osmosis. 1. 4.2 M HNO3 to 2.4 M HNO3 2. 0.4 M NaCl to 0.8 M NaCl 3. 1 M KOH to 2 M KOH 4. 0.6 M KCl to 0.3 M KCl 5. 1 M CH3COOH to 0.4 M CH3COOH 6. 0.01 M Sucrose to 1 M Sucrose
Chemistry
1 answer:
Oduvanchick [21]4 years ago
3 0

Answer:

1. Reverse osmosis

2. Osmosis

3. Osmosis

4. Reverse osmosis

5. Reverse osmosis

6. Osmosis

Explanation:

The osmosis is the process when the solvent goes through a membrane from a side with a low concentration of solute to a higher concentration, to achieve an equilibrium. When the movement is doing on the opposite side (from higher concentration to low), the process is called reverse osmosis.

1. It is going from a higher to a low concentration, so it's reverse osmosis.

2. It's going from a low to a higher concentration, so it's osmosis.

3. It's going from a low to a higher concentration, so it's osmosis.

4. It is going from a higher to a low concentration, so it's reverse osmosis.

5. It is going from a higher to a low concentration, so it's reverse osmosis.

6. It's going from a low to a higher concentration, so it's osmosis.

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The new volume of a gas at STP that occupies 16L of space is 27.43L.

<h3>HOW TO CALCULATE VOLUME?</h3>

The volume of a given gas can be calculated using the following expression:

P1V1/T1 = P2V2/T2

Where;

  • P1 = initial pressure
  • P2 = final pressure
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1 × 16/273 = 2.6 × V2/468

16/273 = 2.6V2/468

0.0586 × 468 = 2.6V2

V2 = 27.43

Therefore, the new volume of a gas at STP that occupies 16L of space is 27.43L.

Learn more about volume at: brainly.com/question/1578538

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2 years ago
A certain substance X has a normal freezing point of −3.1°C and a molal freezing point depression constant =Kf·6.23°C·kgmol−1. C
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We replace data in the formula:

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Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1 + 3.1°C

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Given :

Number of atoms of an element, n = 2.072 × 10⁴³ atoms.

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So, number of moles in given number of atoms are :

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