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NeTakaya
1 year ago
14

Type in the maximum number of electrons that can be present in each shell or subshell below.

Chemistry
2 answers:
diamong [38]1 year ago
7 0

<span>n = 5 shell=50</span>

<span>n = 2 shell=8</span>

<span>n = 2, l = 0 subshell=2</span>

<span>n = 2, l = 1 subshell=6</span>

<span><span>3d subshell=</span>10</span>

<span>2s subshell=2</span>

<span><span>5f subshell=14</span></span><span><span> sorry if I'm late I just did this! hope it helps..</span></span>
IRISSAK [1]1 year ago
4 0

Pauli's principle states that in an atom there cannot be more than one electron described by the same combination of the 4 quantum numbers. As an orbital is described by a particular combination of the first three quantum numbers (n, l, and ml), each orbital can only accommodate 2 electrons (one of ms = + 1/2 and the other of ms = -1 / 2). Thus the layer K (n = 1), for which there is only one possible combination of n, l, and ml (that is to say an orbital), can accommodate a maximum of 2 electrons.

That is to say, that contains 4 orbitals, can contain a maximum of 8 electrons.


The orbitals are named s (contains 2 electrons max) p (contains 6 electrons max) d (contains 10 electrons max) and f (contains 14 electrons max).


As you get into the orbitals they have an extra "l" number.


The maximum number of electrons that can contain a given electronic layer is 2 n^2.




Let's go back to the question, for n = 5 the corresponding layer is the "O" layer, as well as the lower layers, and they can contain 50 electrons.


For n = 2 the corresponding layer is the layer "L" and the lower layer "K", and they can contain 8 electrons (2 + 6 electrons).


For n = 3 L = 0 it corresponds to the first under layer with 2 electrons.


For n = 3 L = 1 it corresponds to the second 6-electron sub-layer.


For n = 3 d (corresponding to l = 2) the corresponding layer is the layer "M" and it can contain 10 electrons.


For n = 2 s (corresponding to l = 0) it corresponds to 2 electrons


For n = 5 f (corresponding to l = 3) it corresponds to 14 electrons

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2 years ago
How much energy is required to heat 0.24 KG lutetium from 296.2K to 373.5 K? The specific heat for lutetium is 0.154 J/g-K
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6 0
1 year ago
The K w for water at 0 ∘ C is 0.12 × 10 − 14 M 2 . Calculate the pH of a neutral aqueous solution at 0 ∘ C . p H = Is a pH = 7.2
labwork [276]

Answer:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

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2 years ago
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gulaghasi [49]

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C. 0.04 moles per cubic decimeter.

Explanation:

The molar mass of the Iodine is 253.809 grams per mole and a cubic decimeter equals 1000 cubic centimeters. The concentration of Iodine (c), measured in moles per cubic decimeter, can be determined by the following formula:

c = \frac{m}{M\cdot V} (1)

Where:

m - Mass of iodine, measured in grams.

M - Molar mass of iodine, measured in grams per mol.

V - Volume of solution, measured in cubic decimeters.

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c = \frac{2.54\,g}{\left(253.809\,\frac{g}{mol} \right)\cdot (0.25\,dm^{3})}

c = 0.04\,\frac{mol}{dm^{3}}

Hence, the correct answer is C.

3 0
1 year ago
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