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erastovalidia [21]
3 years ago
15

The number of energy levels to which an electron can jump depends on the

Chemistry
2 answers:
Kitty [74]3 years ago
6 0
The number of energy levels to which an electron can jump depends on the amount of energy the electron possesses. Each energy level has a specific amount of energy an electron needs to have before it can be in there. So, if an electron doesn't have enough energy to be in that energy level then it won't jump to that higher level.
Sever21 [200]3 years ago
4 0

Answer:

amount of the energy that the electron absorbs.

Explanation:

According to Bohr's postulates, electron present in the energy levels can jump to their higher energy levels on absorbing energy and also come back to their orbital by the radiating the absorbed energy.

<u>Thus, the number of the energy levels to which the electron can jump depends upon the amount of the energy that the electron absorbs.</u>

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The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
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