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2 years ago

When an excited electron spontaneously emits a photon, which energy transformation takes place

2 answers:

7 0

When an excited electron spontaneously emits a photon, the energy transformation from potential to kinetic energy takes place.

If an atom is in an excited state means full of energy, it goes from higher energy level to lower energy level. The electron releasing energy in the form of a photon, which is emitted in a random direction.

When the electron returns to a low energy state, it releases the potential energy in the form of kinetic energy so we can conclude that the energy transformation from potential to kinetic energy takes place.

Learn more about excited electron here: brainly.com/question/81112

Learn more: brainly.com/question/25976890

notka56 [123]2 years ago

4 0

Given what we know, we can confirm that when an excited electron spontaneously emits a photon, the energy released is electromagnetic energy.

<h3>What is a Photon and what energy does it release when being emitted?</h3>

A photon is a particle.
This means that it is one of the smallest forms of matter that we can study.
Photons form electromagnetic fields.
Therefore, when being emitted by an electron, photons release electromagnetic energy.

Therefore, we can confirm that when an excited electron spontaneously emits a photon, the energy released is electromagnetic energy due to the properties of the photon being emitted.

To learn more about photons visit:

brainly.com/question/24309591?referrer=searchResults

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3 years ago

Read 2 more answers

Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

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The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

Molarity of CH4:

CH4 = $\frac{20}{125}$ = 0.16M

Molarity of H20:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

Molarity of H2:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

Molarity of CO:

CO = $\frac{0.144}{3}$ = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

Molarity of CH4:

CH4 = 0.16 - 0.048 = 0.112M

Molarity of H2O:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

$K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }$

$K_{c}$ = $\frac{0.048.0.144^{3} }{0.112.0.032}$

$K_{c}$ = 2. $10^{-2}$

The concentration equilibrium constant for the process is $K_{c}$ = 2. $10^{-2}$ .

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3 years ago