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madam [21]
4 years ago
12

How many different values of l are possible in the third principal level?

Chemistry
1 answer:
KatRina [158]4 years ago
6 0

Answer:

0, 1, and 2.

Explanation:

According to the quantum theory, it's not possible to determine at the same time the velocity and the position of an electron (uncertainty principle), but, it's possible to determine the most probable space that the electron is, which is called the orbital. The electron is then characterized by 4 quantum numbers, which indicates the orbital it is.

The principal quantum number (n) represents the energic levels, or shells, that the electron is surrounding the nucleus. It is represented by letters (K, L, M, N, O, P, and Q), or by numbers (1 to 7).

The azimuthal quantum number (l) represents the energic sublevel, or subshell, inside the level, and it is represented by letters (s, p, d, and f), or by number (0 to 3). It varies from 0 to n-1.

The magnetic quantum number (m) represents the orbital, and it varies from -l to l passing by the 0. Each orbital can have at least 2 electrons.

The spin quantum number (s) represents the spin of the electron in the orbital, it is a measure of the rotation. Thus, one electron has spin -1/2 and the other +1/2.

So, for the third principal level (n = 3), the maximum possible value for l is 2 (3 - 1), so the possibles values are 0,1, and 2.

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What is the pressure of a mixture of 0.200 g of H2, 1.00 g of N2 , and 0.820 g of Ar in a container with a volume of 2.00 L at 2
alexdok [17]

Answer:

P(mixture) = 1.92 atm

Explanation:

Given data:

Mass of H₂ = 0.200 g

Mass of N₂ = 1.00 g

Mass of Ar = 0.820 g

Volume = 2 L

Temperature = 20°C

Pressure of mixture = ?

Solution:

Pressure of hydrogen:

Number of moles of hydrogen = mass / molar mass

Number of moles of hydrogen = 0.200 g / 2 g/mol

Number of moles of hydrogen = 0.1 mol

P = nRT / V

P = 0.1 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L

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Number of moles of nitrogen = mass / molar mass

Number of moles of nitrogen = 1 g / 28 g/mol

Number of moles of nitrogen = 0.04 mol

P = nRT / V

P = 0.04 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L

p = 0.96 atm. L /2 L

P = 0.48 atm

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Number of moles of argon = mass / molar mass

Number of moles of argon = 0.820 g / 40 g/mol

Number of moles of argon = 0.02 mol

P = nRT / V

P = 0.02 mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 293 K / 2L

p = 0.48 atm. L /2 L

P = 0.24 atm

Total pressure of mixture:

P(mixture)  = pressure of hydrogen + pressure of nitrogen + pressure of argon

P(mixture)  = 1.2 atm + 0.48 atm + 0.24 atm

P(mixture) = 1.92 atm

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