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Lelu [443]
2 years ago
13

What happens when a solid is dissolved into a liquid?

Chemistry
2 answers:
prohojiy [21]2 years ago
7 0

Answer:

c

Explanation:

chem ed 2022

daser333 [38]2 years ago
6 0

When a solid dissolves in a liquid to form a solution, solid particles become evenly dispersed among liquid particles.

<h3>What are solutions?</h3>

Solutions are formed when a solute particle dissolves in another substance known as a solvent.

Usually, solutions are formed when solids dissolve in liquids.

The solid is the solute while the liquid is the solvent.

When a solid dissolves in a liquid, solid particles become evenly dispersed among liquid particles.

Therefore, solutions are formed when solute particles dissolve in solvents.

Learn more about solutes and solutions at: brainly.com/question/25326161

#SPJ1

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Which of the following are true statements about equilibrium systems? For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(
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Answer:

The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

⇒ Le Chatellier says As the CaCO3 concentration is increased, the system will attempt to undo that concentration change by shifting the balance to the right. <u>This statement is true.</u>

<u />

2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

⇒ Le chatellier says that if we increase the pressure, the equilibrium will shift to the side with the least number of particles.

Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. <u>This statement is false. </u>

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

⇒ Le Chatellier says if we increase the pressure, the equilibrium will shift to the side with the most particles.

In this case we have 2 moles of H2 and 1 mole of O2 on the left side and 2 mole of H2O on the right side. This means on the left side are more particles. So the equilibrium will shift to the left, so <u>this statement is true.</u>

4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

⇒ Le chatellier says if H2 will be removed (this means the left side will get less particles) so the equilibrium will shift to the left, to increase the amount of F2.

⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

If we increase the temperature of an endotherm reaction, the equilibrium will shift to the side of the products (the right side). <u>This statement is true.</u>

4 0
3 years ago
Marking brilliant for answers
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Answer:

The first option.

Explanation:

Warm solvent and stirring promote quick dissolving . Also with small crystals more of the surface is in contact with the solvent than with  large crystals.

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