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Lyrx [107]
3 years ago
12

Use your understanding of the terms "soluble" and "insoluble" to explain why you must shake a bottle of salad dressing made with

oil and vinegar before you use it.
Chemistry
1 answer:
tekilochka [14]3 years ago
6 0

Answer:

See explanation

Explanation:

In chemistry, the idea of "like dissolves like" is of utmost importance. A substance is only soluble in another with which it can effectively interact.

We must note that to be "soluble" means that the solute actually interacts effectively (dissolves) in the solvent.

However, vinegar is a polar substance while oil is a non polar substance hence the two can not effectively interact. That is, the vinegar can not dissolve in oil.

The two will separate into two phases upon standing. Therefore, the bottle of salad dressing made with oil and vinegar must be shaken in order to mix the two thoroughly before it is used.

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CH4 + 2O2 → CO2 + 2H2O

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How many Oxygen atoms in: 5H2SO3
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Consider the following system at equilibrium:A(aq)+B(aq) <---> 2C(aq)Classify each of the following actions by whether it
velikii [3]

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

  • On addition of reactant at equilibrium shifts the equilibrium in forward direction.
  • On addition of product at equilibrium shifts the equilibrium in backward direction.
  • On removal of reactant at equilibrium shifts the equilibrium in backward direction.
  • On removal of product at equilibrium shifts the equilibrium in forward direction.

A(aq)+B(aq)\rightleftharpoons 2C(aq)

Reactants = A , B

Product = C

1. Increase A

On increasing the amount of A at equilibrium will shift the equilibrium in forward or rightward direction.

2. Increase B

On increasing the amount of B at equilibrium will shift the equilibrium in forward or rightward direction.

3. Increase C

On increasing the amount of C at equilibrium will shift the equilibrium in backward or leftward direction.

4. Decease A

On decreasing the amount of A at equilibrium will shift the equilibrium in backward or leftward direction.

5. Decease B

On decreasing the amount of B at equilibrium will shift the equilibrium in backward or leftward direction.

6. Decease C

On decreasing the amount of C at equilibrium will shift the equilibrium in forward or rightward direction.

7. Double A and Halve B

Equilibrium constant of the reaction = K

K=\frac{[C]^2}{[A][B]}

On doubling A and halving B, equilibrium constant of the reaction = K'

K'=\frac{[C]^2}{[2A][\frac{B}{2}]}=\frac{[C]^2}{[A][B]}

The value of equilibrium constant K' is equal to K, which means that equilibrium will not shift in any direction.

8. Double both B and C

Equilibrium constant of the reaction = K

K=\frac{[C]^2}{[A][B]}

On doubling B and C, equilibrium constant of the reaction = K'

K'=\frac{[2C]^2}{[A][2B]}=\frac{4[C]^2}{[A][2B]}=\frac{2[C]^2}{[A][B]}

K' = 2 K

The value of equilibrium constant K' is double the K, which means that product is increasing which means that equilibrium will shift in backward or leftward direction.

5 0
3 years ago
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