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Mumz [18]
3 years ago
8

What are all of the possible values of l, ml, and ms for n = 4? (select all that apply.)?

Chemistry
1 answer:
8_murik_8 [283]3 years ago
5 0
For p<span>rincipal quantum number n=4:
1) a</span><span>zimuthal quantum number (l) can be l = 0...n-1:
l = 0, 1, 2, 3. 
</span><span>The azimuthal quantum number determines its orbital angular momentum and describes the shape of the orbital. </span><span>
2) m</span>agnetic quantum number (ml) can be ml = -l...+l.
ml = -3, -2, -1, 0, +1, +2, +3.
<span>Magnetic quantum number specify orientation of electrons in magnetic field and number of electron states (orbitals) in subshells.
3) t</span>he spin quantum number (ms), is the spin of the electron.
ms = +1/2, -1/2.
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6.023 × 10²³ atoms : 1 mole = x : 2.70 moles
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Expression for the v_{rms} speed is as follows.

            v_{rms} = \sqrt{\frac{3kT}{M}}

where,   v_{rms} = root mean square speed

                     k = Boltzmann constant

                    T = temperature

                    M = molecular mass

As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of v_{rms} as follows.

               v_{rms} = \sqrt{\frac{3kT}{M}}

                            = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}

                            = 498.5 m/s

Hence, v_{rms} for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its v_{rms} speed as follows.

                v_{rms} = \sqrt{\frac{3kT}{M}}

                              = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}

                              = 524.5 m/s

Therefore, v_{rms} speed for nitrogen is 524.5 m/s.

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