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Viefleur [7K]
3 years ago
6

What is the maximum number of electrons that can occupy the hydrogen-like atomic orbitals defined by the quantum numbers: n-3, l

2?
(A) 2
(B) 6
(C) 10
(D) 14
Chemistry
1 answer:
denis23 [38]3 years ago
5 0

Answer:C

Explanation:

The orbital in question is the 3d orbital. The d-sublevel only contains a maximum of ten electrons in its five degenerate orbitals dxy,dyz,dxz,dz2 and dx2-y2. This is obtained the fact that each degenerate orbital holds only two electrons each.

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If kb for nx3 is 4.0×10−6, what is the poh of a 0.175 m aqueous solution of nx3?
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<h3><u>Answer;</u></h3>

pOH = 3.08

<h3><u>Explanation;</u></h3>

NX3 + H2O <----> NHX3+ + OH-  

Kb = 4.0 x 10^-6

Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).

c(NH₄⁺) = c(OH⁻) = x.

x² = Kb · c(NH₃)

x² = 4.0 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = c(OH⁻) = √(7.0 × 10⁻⁷)

    = 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

        =- log ( 8.367 × 10⁻⁴)

        <u>= 3.08</u>

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The ratio of atoms of potassium to ratio of atoms of oxygen is
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Select True or False: The equilibrium constant for the chemical equation 2NO(g) O2(g) 2NO2(g) is two times the equilibrium const
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Answer:

False

Explanation:

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It now follows that;

K'= K^2

Hence the statement in the question is false

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