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Helga [31]
2 years ago
9

The following combinations are not allowed. If n and ml are correct, change the l value to create an allowable combination:

Chemistry
1 answer:
ololo11 [35]2 years ago
7 0

The allowable combination for the atomic orbital is n=3, l=1 or 2, m_{l}=+1.

<h3>What are the three quantum numbers of an atomic orbital?</h3>

Three quantum numbers specify an atomic orbital:

- The principal quantum number, n, which is a positive integer, describes the relative size of the orbital and its distance from the nucleus.

- l is the angular momentum quantum number that is related to the shape of the orbital; l is an integer from 0 to n-1 (so n limits l ),

- $\boldsymbol{m}_{l}$ is the magnetic quantum number that prescribes the three-dimensional shape of the orbital around the nucleus; m_{l} values are integers from -l to =l(l limits ml)

For n = 3, l can have three values: 0, 1, and 2. Hence, the l value must be lower than 3. Because of m_{l}= +1, the l value must be higher than 0. Two l values are consistent with n and m_{l) values:

l= 1 or 2

Therefore, the allowable condition is n=7, l=1 or 2, m_{l}=+3.

To know more about quantum numbers, visit: brainly.com/question/16979660

#SPJ4

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Answer:

n = 243.605

Explanation:

Given

Pressure, P = 200\ atm

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Volume, V = 30L

Required

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We'll apply the following formula to solve this question

n = \frac{pV}{RT}

Where

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The above equation is an illustration of the ideal gas law

Substitute values for p, V, R and T in:

n = \frac{pV}{RT}

n = \frac{200 * 30}{0.0821 * 300}

n = \frac{6000}{24.63}

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Korvikt [17]

<u>Given:</u>

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