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gregori [183]
3 years ago
5

In order for a solute to dissolve in a solvent, what must be true?

Chemistry
2 answers:
saw5 [17]3 years ago
8 0

<em>Answer:</em>

  • In order for a solute to dissolve in a solvent, the attractive forces must be broken.

<em>Explanation:</em>

The rule of solubility says

                                  <em>   "Like dissolve Like"</em>

There should be following process in order to dissolve solute into solvent

(1). The particles of solute should be apart.

(2). The solvent particles should move apart to extent so that solute particles should entrapped between solvent particles

(3).  The particles of solute and solvent should attract to each others so that they can mix.

<em>Summary: </em>

  • If new forces arises between the solute and solvent dominate over the forces between the particles of solute, then solute become dissolve into solvent and vice versa.
Minchanka [31]3 years ago
4 0
(1) The attractive forces in a solute need to be broken. (2) The attractive forces in a solute must be decreased. (3) The attractive forces in the solute must be greater than the attractive forces in the solvent. (4) The attractive forces in a solvent must be increased.
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<span>They lower the activation energy.</span>
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How much heat energy is required to raise the temperature of 0.368 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co
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What is the difference between a hydrogen ion and a hydroxide ion?
Sophie [7]

Answer:

there is some difference

Explanation:

hydrogen ion --

symbol-H

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5 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
1) Which term describes the conversion of substances into different substances?(1 point)
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Learn more: brainly.com/question/1527403

8 0
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