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Serggg [28]
3 years ago
13

When an excited electron in an atom moves to the ground state, the electron

Chemistry
2 answers:
kodGreya [7K]3 years ago
7 0

Answer is: (4) emits energy as it moves to a lower energy state.

Atom emits a characteristic set of discrete wavelengths, according to its electronic energy levels.

Emission spectrum of a chemical element is the spectrum of frequencies emitted due to an atom making a transition from a high energy state to a lower energy state.  

Each transition has a specific energy difference.  

Each element's emission spectrum is unique.


likoan [24]3 years ago
3 0

When an excited electron in an atom moves to the ground state, the electron  \boxed{\left( {\text{4}} \right){\text{ emits energy as it moves to a lower energy state}}}.

Further Explanation:

Electronic transition:

It is a process that occurs when an electron undergoes emission or absorption from one energy level to another energy level.

When an electron undergoes a transition from a lower energy level to a higher energy level then it requires energy to complete the process. This transition is an absorption process.

When an electron undergoes a transition from higher energy level to lower energy level then it emits energy to complete the process. This transition is an emission process.

An excited electron has higher energy than that present in the ground state of the atom. Energy is to be released when an excited electron returns to its ground state. Therefore the excited electron emits energy when it jumps from an excited state (higher energy state) to the ground state (lower energy state). Hence option (4) is the correct answer.

Learn more:

1. Which transition is associated with the greatest energy change? brainly.com/question/1594022

2. Describe the spectrum of elemental hydrogen gas: brainly.com/question/6255073

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: electronic transition, absorption, emission, lower, higher, energy level, excited state, ground state, emit, lower energy state.

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6 0
3 years ago
Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.
Rashid [163]

Answer:

1) The value of Kc:

C. remains the same.

2) The value of Qc:

A. is greater than Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium.

4) The concentration of N2 will:

B. decrease.

Explanation:

Hello,

In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:

1) The value of Kc:

C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

ln(K)=\frac{\Delta _RG}{RT}

2) The value of Qc:

A. is greater than Kc, since the reaction quotient is:

Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}

Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.

4) The concentration of N2 will:

B. decrease, since less reactant is forming the products.

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lisabon 2012 [21]

¿es demasiado tarde para ayudarte?

4 0
3 years ago
PLEASE SOMEONE HELP ME WITH THIS!
Bad White [126]

Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877  and 4) 1.04x10^{3\\}

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x  \frac{1 L}{1000 mL} = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x \frac{10 dg}{ 1 g} = 1.291 dg

For the next exercises, you need to follow some rules:

1.  All numbers  that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062  - 638.1848  = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.  

1.04x10^{3}

4 0
3 years ago
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