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oee [108]
3 years ago
12

During an endothermic phase change, what happens to the potential energy and the kinetic energy?

Chemistry
2 answers:
tester [92]3 years ago
5 0

Answer:

d. Potential energy increases, and kinetic energy stays the same.

Explanation:

This is correct on edge.

The answers to the Enthalpy and Phase Changes quiz are:

1. D 25.0 kJ

2. A Delta Hfus

3. C heat of fusion

4. D Potential energy increases, and kinetic energy stays the same.

5. B This is an exothermic reaction that involves freezing.

6. B a thermometer

7. C 549 g

8. B Energy is absorbed, and potential energy increases.

9. D a solid to a liquid

10. C the periodic table

antiseptic1488 [7]3 years ago
4 0

During endothermic phase change, the potential energy of the system always increases while the kinetic energy of the system remains constant. The potential energy of the reaction increases because energy is been added to the system from the external environment.

<u>Explanation</u>:

  • Those are three distinct methods for demonstrating a specific energy condition of an object. They don't affect one another.
  • "Potential Energy" is a relative term showing a release of possible energy to the environment. If we accept its pattern as the overall energy state of a compound, at that point, an endothermic phase change would infer an increase in "potential" as energy is being added to the compound by the system.
  • A phase change will display an increase in the kinetic energy at whatever point the compound is transforming from a high density to a low dense phase. The kinetic energy will decrease at whatever point the compound is transforming from a less dense to high dense phase.
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Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv
Triss [41]

Answer:

1. 3.57\times 10^{-4}was the K_a value calculated by the student.

2. 5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

Explanation:

1.

The pH value of Aspirin solution = 2.62

pH=-\log[H^+]

[H^+]=10^{-2.62}=0.00240 M

Moles of s asprin = \frac{2.00 g}{180 g/mol}=0.01111 mol

Volume of the solution = 0.600 L

The initial concentration of Aspirin  = c = \frac{0.01111 mol}{0.600 L}=0.0185 M

HAs\rightleftharpoons As^-+H^+

initially

c       0    0

At equilibrium

(c-x)      x   x

The expression of dissociation constant :

K_a=\frac{[As^-][H^+]}{[HAs]}:

K_a=\frac{x\times x }{(c-x)}

=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}

K_a=3.57\times 10^{-4}

3.57\times 10^{-4}was the K_a value calculated by the student.

2.

The pH value of ethylamine = 11.87

pH+pOH=14

pOH=14-11.87=2.13

pOH=-\log[OH^-]

[OH^-]=10^{-2.13}=0.00741 M

The initial concentration of ethylamine = c = 0.100 M

C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-

initially

c                    0    0

At equilibrium

(c-x)                x   x

The expression of dissociation constant :

K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}:

K_b=\frac{x\times x}{(c-x)}

=\frac{0.00741\times 0.00741}{(0.100-0.00741)}

K_b=5.93\times 10^{-4}

5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

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3 years ago
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