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Trava [24]
3 years ago
13

2. Ultraviolet radiation has a frequency of 6.8 x 10'5Hz. Calculate the energy, in

Chemistry
1 answer:
Gnoma [55]3 years ago
3 0

Answer:

E=4.5\times 10^{-18}\ J

Explanation:

It is given that, the frequency of Ultraviolet radiation is 6.8\times 10^{15}\ Hz

We need to find the energy of the photon. The formula for the energy of a photon is given by :

E=hf\\\\\text{Where h is Planck's constant}\\\\E=6.63\times 10^{-34}\times 6.8\times 10^{15}\\\\E=4.5\times 10^{-18}\ J

So, the energy of the photon is 4.5\times 10^{-18}\ J.

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Describe how electron arrangements are given using electron configurations.
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According to the principle, electrons fill orbitals starting at the lowest available energy states before filling higher states (e.g., 1s before 2s). The Madelung energy ordering rule: Order in which orbitals are arranged by increasing energy according to the Madelung Rule.

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3 years ago
Jeffrey is on his skateboard and starts skating down a ramp. Ten seconds later he is travelling at
AfilCa [17]

Answer: 2m/s^2

Explanation:

Velocity of skateboard = 20.0 m/s.

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4 years ago
How many atoms of oxygen are present in 7.51 grams of<br> glycine with formula C₂H5O2N?
Blizzard [7]

1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

Mass of oxygen in glycine = 32g/mol

Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine

Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

Learn more about number of atoms at: brainly.com/question/8834373

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2 years ago
What is the acceleration due to gravity of the moon<br>​
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Explanation:

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