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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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When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular
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<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=1.03g

Mass of H_2O=0.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

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To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

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C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

Thus, the empirical and molecular formula for the given organic compound is C_3H_2O and C_6H_4O_2

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    calculation
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