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3 years ago

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.

Explanation:

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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To Earth, since it has the same radio and masses. This is what I believe.

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3 years ago

Reaction is

-Dominant- [34]

According to the task, you are proveded with patial pressure of CO2 and graphite, and here is complete solution for the task :
At first you have to find n1 =moles of CO2 and n2 which are moles of C
The you go :
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n1 n2 0
-x -x +2x
$n1-x n2-x 2x Kp=P^2 (CO)/Pco2$
After that you have to use the formula $PV=nRT$
$Kp=(2x)^2/n1-x$
Then you have to solve x, and for that you have to use RT/V
And to find total values: $P(total)=n(total)*R*T/V$
I am absolutely sure that this would be helpful for you.

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3 years ago

When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m

Evgen [1.6K]

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

$500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2$

$42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl$

Have a nice day!

#LearnwithBrainly

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3 years ago

What is the electron configuration of an electrically neutral alkaline earth metal? please explain

vlabodo [156]

Answer:

The elements in Group 2 (beryllium, magnesium, calcium, strontium, barium, and radium) are called the alkaline earth metals (see Figure below). These elements have two valence electrons, both of which reside in the outermost s sublevel. The general electron configuration of all alkaline earth metals is ns

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3 years ago

In going from room temperature (25 C) to 10 C above room temperature, the rate of reaction doubles. Calculate the activation ene

Sauron [17]

Answer:

Ea=5.29 × 10⁴ J/mol

Explanation:

In going from 25 °C (298 K) to 35 °C (308 K), the rate of the reaction doubles. Since the rate of the reaction depends on the rate constant (k), this implies that the rate constant doubles. We can find the activation energy (Ea) using the two-point form of the Arrhenius equation.

$ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} .(\frac{1}{T_{2}}-\frac{1}{T_{1}})\\ln\frac{2k_{1}}{k_{1}}=\frac{-Ea}{8.314J/K.mol}.(\frac{1}{308K}-\frac{1}{298K} )\\Ea=5.29 \times 10^{4} J/mol$

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3 years ago