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ladessa [460]
3 years ago
14

I need you help in this homework

Chemistry
1 answer:
Mrrafil [7]3 years ago
4 0

Answer:

OPTION B ,COMPOUND CONTAINING AMMONIUM

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The photon of light that is emitted as an electron drops back to its original orbit is:
Alex73 [517]

Answer:

energy

Explanation:

The photon of light that is emitted as an electron drops back to its original orbit is energy and this energy is released during de-excitation process.

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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Sepuede compartir los vienes si tu mujer sefue con otro
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Si se Casaron por vienes separdos no tienes porque compartir.
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How does rock become soil?
Makovka662 [10]
I *think* through weathering and erosion. Hope this helped
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Explain why science is a continuous profession of study.
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3 0
3 years ago
A ball is dropped from a height of 1.20 m and hits the floor. The ball compresses and then reforms to spring upwards from the fl
diamong [38]

Answer:

\large \boxed{\text{98 m$\cdot$s}^{-2}}

Explanation:

The formula for the velocity of the ball is

v = \sqrt{2gh}

1. Velocity at time of impact

v = -\sqrt{2 \times 9.807 \times 1.20} = -\sqrt{23.54} = -\textbf{4.85 m/s}

2. Velocity on rebound

The ball has enough upward velocity to reach a height of 0.86 m.

v = \sqrt{2 \times 9.807 \times 0.86} = \sqrt{16.87} =\textbf{4.11 m/s}

3. Acceleration

a = \dfrac{\Delta v}{\Delta t} = \dfrac{4.11 - (-4.85)}{ 0.091} = \dfrac{8.96 }{0.091} =\textbf{98 m$\cdot$s}^{\mathbf{-2}}\\\\\text{The acceleration while the ball is in contact with the floor is $\large \boxed{\textbf{98 m$\cdot$s}^{\mathbf{-2}}}$}

5 0
3 years ago
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