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pashok25 [27]
3 years ago
14

Crystals are made using supersaturated solutions of solids in water. These solutions hold more of the solute than they would at

room temperature. How are these solutions made?
1. The water would need to be chilled to a lower temperature, which would give molecules and ions more kinetic energy, increasing solubility.

2. The water would need to be chilled to a lower temperature, which would take kinetic energy away from molecules and ions, increasing solubility.

3. The water would need to be heated to a higher temperature, which would take kinetic energy away from molecules and ions, increasing solubility.

4. The water would need to be heated to a higher temperature, which would give molecules and ions more kinetic energy, increasing solubility.
Chemistry
2 answers:
Rudik [331]3 years ago
3 0

Solubility controls the formation of crystals by the mechanism of oversaturation, therefore the correct answer is:

4. The water polo would need to be heated to a higher temperature, which would give molecules and ions more kinetic energy, increasing solubility.

Crystals are solid structures where there is a fixed spatial arrangement of the molecules, there are many preparation methods, a very practical method and obtaining good crystals is from supersaturated solutions.

Solubility is the mechanism by which a supersaturated solution is formed, for this it is used that the water when heated has more energy and manages to dissolve a greater amount of solute, therefore by decreasing the temperature and consequently reducing the solubility the solution is over saturated.

When introducing some means of disturbance in the solution, the excess of solute is deposited in the form of crystals of excellent quality, this growth method is called hydrothermal synthesis

To achieve controlled supersaturation and crystals the temperature must be controlled precisely, to allow the formation of the crystals.

In conclusion, solubility controls the formation of crystals by the mechanism of over saturation, therefore the correct answer is:

4. The water polo would need to be heated to a higher temperature, which would give molecules and ions more kinetic energy, increasing solubility.

Learn more about hydrothermal growth here:  brainly.com/question/7821375

jonny [76]3 years ago
3 0

Answer:

When a solution that has had more solute dissolved at a higher temperature is now cooled, it becomes supersaturated at the lower temperature. Because that supersaturated solution holds more solute than is stable at the lower temperature, crystals start to form.

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Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
Fiesta28 [93]

This is an incomplete question, here is a complete question.

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

M_3O_4(s)\rightleftharpoons 3M(s)+2O_2(g)

Substance      ΔG°f (kJ/mol)

M₃O₄                 -9.50

M(s)                       0

O₂(g)                     0

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? delta G°rxn = kJ / mol.

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

What is the equilibrium pressure of O₂(g) over M(s) at 298 K?

Answer :

The Gibbs energy of reaction is, 9.50 kJ/mol

The equilibrium constant of this reaction is, 0.0216

The equilibrium pressure of O₂(g) is, 0.147 atm

Explanation :

The given chemical reaction is:

PCl_3(l)\rightarrow PCl_3(g)

First we have to calculate the Gibbs energy of reaction (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}]

where,

\Delta G^o = Gibbs energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)]

\Delta G^o=9.50kJ/mol

The Gibbs energy of reaction is, 9.50 kJ/mol

Now we have to calculate the equilibrium constant of this reaction.

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs free energy  = 9.50kJ/mol = 9500 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K  = equilibrium constant = ?

9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)

K=0.0216

The equilibrium constant of this reaction is, 0.0216

Now we have to calculate the equilibrium pressure of O₂(g).

The expression of equilibrium constant is:

K=(P_{O_2})^2

0.0216=(P_{O_2})^2

P_{O_2}=0.147atm

The equilibrium pressure of O₂(g) is, 0.147 atm

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Answer:

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Explanation:

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