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andreev551 [17]
4 years ago
12

Which electron transition represents a gain of energy

Chemistry
1 answer:
Helen [10]4 years ago
3 0

Answer:

The transition from lower energy level to higher energy level require a gain of energy.

Explanation:

When transition occur from lower energy level to higher energy level require a gain of energy. Electron could not jump unto higher energy level without gaining thew energy.

When electron jump into lower energy level from high energy level it loses the energy.

For example electron when jumped from 2nd to 3rd shell it gain energy and when in return back to 2nd shell from 3rd shell it loses 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.

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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Answer:

2000 Joules.

Explanation:

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) In the Deacon process for the manufacture of chlorine, HCl and O2 react to form Cl2 and H2O. Sufficient air (21 mole% O2, 79%
LiRa [457]

Answer:

Here's what I get.

Explanation:

1. Write the chemical equation

\rm 4HCl + O$_{2} \longrightarrow \,$ 2Cl$_{2}$ + 2H$_{2}$O

Assume that we start with 4 L of HCl

2. Calculate the theoretical volume of oxygen

\text{V}_{\text{O}_{2}}= \text{4 L HCl} \times \dfrac{\text{1 L O}_{2}}{\text{4 L HCl}} = \text{1 L O}_{2}}

3. Add 35% excess

\text{V}_{\text{O}_{2}}= \text{1 L O}_{2}} \times 1.35 = \text{1.35 L O}_{2}}

4. Calculate the theoretical volume of nitrogen

\text{V}_{\text{N}_{2}} = \text{1.35 L O}_{2}} \times \dfrac{\text{79 L N}_{2}}{\text{21 L O}_{2}}} = \text{5.08 L N}_{2}}

4. Calculate volumes of reactant used up

Only 85 % of the HCl is converted.

We can summarize the volumes in an ICE table

           4HCl     +       O₂    +    N₂   →    2Cl₂   +   2H₂O

I/L:          4               1.35         5.08         0              0

C/L:  -0.85(4)        -0.85(1)        0      +0.85(2)   +0.85(2)

E/L:     0.60             0.50        5.08       1.70          1.70

5. Calculate the mole fractions of each gas in the product stream

Total volume = (0.60 + 0.50 + 5.08 + 1.70 + 1.70) L = 9.58 L

\chi = \dfrac{\text{V}_{\text{component}}}{\text{V}_\text{total}} = \dfrac{\text{ V}_{\text{component}}}{\text{9.58}} = \text{0.1044V}_{\text{component}}\\\\\chi_{\text{HCl}} = 0.1044\times 0.60 = 0.063\\\\\chi_{\text{O}_{2}} = 0.1044\times 0.50 = 0.052\\\\\chi_{\text{N}_{2}} = 0.1044\times 5.08 = 0.530\\\\\chi_{\text{Cl}_{2}} = 0.1044\times 1.70 = 0.177\\\\\chi_{\text{H$_{2}${O}}} = 0.1044\times 1.70 = 0.177\\\\

8 0
3 years ago
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