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Mandarinka [93]
3 years ago
15

Consider these two cases. Case 1: An electron jumps from energy level 5 to energy level 2 in an atom. Case 2: An electron jumps

from energy level 5 to energy level 1 in an atom. For case 1, what happens when an electron jumps from energy level 5 to energy level 2 in an atom? A photon is emitted by the atom. A photon is absorbed by the atom. A proton is absorbed by the atom. A proton is emitted by the atom. Assuming that both cases describe Hydrogen‑like atoms with one electron, for which case is more energy emitted or absorbed? More energy is emitted or absorbed for case 1. It is impossible to tell. The energy is the same for both cases. More energy is emitted or absorbed for case 2.
Chemistry
1 answer:
xenn [34]3 years ago
8 0

Answer:

1.  A photon is emitted by the atom.

2. More energy is emitted or absorbed for case 2

Explanation:

The Bohr's model of the atom was based on quantum mechanics idea. He suggested that electrons move in specific spherical orbits round the nucleus.

The claims of his model emphasized on the premise that electrons only move in permissible orbits or energy levels round the nucleus.

The ground state is the lowest energy state available where n= 1.

The excited state is energy levels of n = 2,3,4 .....

Problem 1

For an electron to return to the ground state from a higher energy level, i.e from  5 to 2 in case 1, energy is emitted in form of photons.

Problem 2

More energy is emitted or absorbed when an electron moves from a higher energy level to a more lower one or jumps from a lower energy level to a more higher one.

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What does CoH12O6 tell you about the glucose molecule?
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Answer:

Glucose = C6H12O6

molecular mass = 6(12) + 12(1) + 6(16)

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= 180 g

Explanation:

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3 years ago
3. What is the energy of a photon whose frequency is 5.2 x 1015 Hz? Use the equation: E = hxv
anzhelika [568]

Answer:

3. 3.45×10¯¹⁸ J.

4. 1.25×10¹⁵ Hz.

Explanation:

3. Determination of the energy of the photon.

Frequency (v) = 5.2×10¹⁵ Hz

Planck's constant (h) = 6.626×10¯³⁴ Js

Energy (E) =?

The energy of the photon can be obtained by using the following formula:

E = hv

E = 6.626×10¯³⁴ × 5.2×10¹⁵

E = 3.45×10¯¹⁸ J

Thus, the energy of the photon is 3.45×10¯¹⁸ J

4. Determination of the frequency of the radiation.

Wavelength (λ) = 2.4×10¯⁵ cm

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

Next, we shall convert 2.4×10¯⁵ cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

2.4×10¯⁵ cm = 2.4×10¯⁵ cm × 1 m /100 cm

2.4×10¯⁵ cm = 2.4×10¯⁷ m

Thus, 2.4×10¯⁵ cm is equivalent to 2.4×10¯⁷ m

Finally, we shall determine the frequency of the radiation by using the following formula as illustrated below:

Wavelength (λ) = 2.4×10¯⁷ m

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

v = c / λ

v = 3×10⁸ / 2.4×10¯⁷

v = 1.25×10¹⁵ Hz

Thus, the frequency of the radiation is 1.25×10¹⁵ Hz.

8 0
2 years ago
What is the molarity of a solution of 58.7 grams of MgCl2 in 359 ml of solution?
jekas [21]

Answer:

1.72 M

Explanation:

Molarity is the molar concentration of a solution. It can be calculated using the formula a follows:

Molarity = number of moles (n? ÷ volume (V)

According to the information provided in this question, the solution has 58.7 grams of MgCl2 in 359 ml of solution.

Using mole = mass/molar mass

Molar mass of MgCl2 = 24 + 35.5(2)

= 24 + 71

= 95g/mol

mole = 58.7g ÷ 95g/mol

mole = 0.618mol

Volume of solution = 359ml = 359/1000 = 0.359L

Molarity = 0.618mol ÷ 0.359L

Molarity = 1.72 M

6 0
3 years ago
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