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NeTakaya
2 years ago
14

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

Chemistry
2 answers:
diamong [38]2 years 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]2 years 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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Write a balanced equation for the reaction of NaCH3COO (also written as NaC2H3O2) and HCl.
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The balance chemical equation is:

NaCH₃COO + HCl → NaCl + HCH₃COO

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2 years ago
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Consider a process in which the entropy of a system increases by 125 J K−1 and the entropy of the surroundings decreases by 125
expeople1 [14]

Answer : The process is not spontaneous.

Explanation :

As, we know that:

Change in entropy = Change in entropy of system + Change in entropy of surrounding

As we are given in question, the entropy of surroundings decrease by the same amount as the entropy of the system increases.

For the given reaction to be spontaneous, the total change in entropy should be positive.

Given :

Entropy change of system = +125J/K

Entropy change of surroundings = -125J/K

Total change in entropy = Entropy change of system + Entropy change of surroundings

Total change in entropy = 125 J/K + (-125 J/K)

Total change in entropy = 0

The process is at equilibrium because the entropy change is equal to zero. So, the process is not spontaneous.

4 0
2 years ago
Silver nitrate and aluminum chloride react with each other by exchanging anions: 3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agc
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Silver nitrate and aluminum chloride react with each other by exchanging anions: 3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agcl (s) what mass in grams of agcl is produced when 4.22 g of agno3 react with 7.73 g of alcl3?
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Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
WARRIOR [948]
Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
5 0
2 years ago
A student assistant is cleaning up after a chemistry laboratory exercise and finds three one-liter bottles containing alcohol so
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The approximate alcohol content is 210 ml.

Explanation:

It can be deduced from the question that each bottle is of 1000ml or 1 litre.

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20/100*500

=100 ml

The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml

so it is 200ml having 30% alcohol

30/100*200

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100 ml.  having 50% of alcohol

50/100*100

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The alcohol content obtained from all these 3 litres is:

100+60+50

= 210 ml of alchohol is obtained from 800 ml of mixture.

3 0
2 years ago
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