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Inessa [10]
3 years ago
9

What is the total number of electrons that can occupy the f sub level

Chemistry
2 answers:
Alisiya [41]3 years ago
7 0

<u>Answer:</u>

<em>14 electrons can occupy the f sub level</em>

<u>Explanation:</u>

Electronic configuration is the short representation which we use to represent the structure of an atom.

Atomic number of phosphorus is 15 so it has 15 electrons.

The electronic configuration of Phosphorus is given as 1s^2 2s^2 2p^6 3s^2 3p^3

Here the superscript number represents the electrons 2 + 2 + 6 + 2 + 3 = 15 electrons of Phosphorus

Orbital diagram too is used for this purpose.

A circle or a square is used to represent an orbital.

This is the orbital diagram of Radon (atomic number -86)  

Each orbital occupies 2 electron.

Shells are named using letters K, L, M, N and so on or using numbers 1, 2, 3, 4 etc.,

The number of electrons, a shell can accommodate is found using formula 2n^2 where n is the number of the shell.

Shell 1 contains subshell s

Shell 2 contains subshell s, p

Shell 3 contains subshell s, p, d

Shell 4 contains subshell s, p, d, f

Subshells or sub level are represented by letters s, p, d, f, g, h, i and so on.

s contains 1 orbital with 2 electrons

p subshell contains 3 orbital with 6 electrons

d with 5 orbitals and 10 electrons  

f with 7 orbitals and 14 electrons.

(Answer)

Zepler [3.9K]3 years ago
6 0

Answer:

14 Electrons

Explanation:

I know because I know

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A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
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Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

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it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

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