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Anettt [7]
3 years ago
7

Describe the movement of electrons when the atom colors by applying ground state and excitement state?

Chemistry
1 answer:
Sladkaya [172]3 years ago
5 0

Answer:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations.The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum

Explanation:

The electron is jumped into higher level and back into lower level by absorbing and releasing the energy.

The process is called excitation and de-excitation.

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.  For example if electron jumped from K to L it must absorbed the energy which is equal the energy difference of these two level. The excited electron thus move back to lower energy level which is K by releasing the energy because electron can not stay longer in higher energy level and comes to ground state.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum

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The molarity of the stock Mn²⁺ ions is 0.0288 M

Based on the dilution formula;

  • The molarity of A is 0.00144 M
  • The molarity of B is 0.0000576 M
  • The molarity of C is 0.000001152 M

<h3>What is the molarity of a solution?</h3>

The molarity of a solution is the number of moles of a solute dissolved in a given volume of solution in liters.

  • Molarity = number of moles/volume

The molarity of the stock solution is:

moles of Mn²⁺ ions = mass / molar mass

molar mass of  Mn²⁺ ions = 55.0 g/mol

moles of Mn²⁺ ions = 1.584 / 55

moles of Mn²⁺ ions = 0.0288 moles

molarity of Mn²⁺ ions = 0.0288 / 1

molarity of Mn²⁺ ions = 0.0288 M

The dilution formula is used to determine the molarities of A, B, and C.

C₁V₁ = C₂V₂

C₂ = C₁V₁ / V₂

Where;

  • C₁ = initial molarity
  • V₁ = initial volume
  • C₂ = final molarity
  • V₂ = final volume

Molarity of A = 50 * 0.0288 / 1000

Molarity of A = 0.00144 M

Molarity of B = 10 * 0.00144 / 250

Molarity of B = 0.0000576 M

Molarity of C = 10 * 0.0000576 / 500

Molarity of C = 0.000001152 M

Learn more about molarity at: brainly.com/question/17138838

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Answer:

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