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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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What is the mass of HF produced by three reaction of 3.0 10 to the 23 molecules of H2 with excess F2
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Answer:

It is 20. g HF

Explanation:

H2 + F2 ==> 2HF  ...  balanced equation

Since the question is asking us to find the mass of product formed, we will want to first convert the molecules of H2 into moles of H2 (we could do this at the end of the calculations, but it's just as easy to do it now).

moles of H2 present (using Avogadro's number):  

3.0x1023 molecules H2 x 1 mole H2/6.02x1023 molecules = 0.498 moles H2

From the balanced equation, we see that 1 mole H2 produces 2 moles HF.  Therefore, we can now find the theoretical mass of HF produced from 0.498 moles H2:

0.498 moles H2 x 2 moles HF/1 mol H2 = 0.996 moles HF formed.

The molar mass of HF = 20.01 g/mole, thus...

0.996 moles HF x 20.01 g/mole = 19.93 g HF = 20. g HF formed (to 2 significant figures)

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The electron configuration for aluminum (Al) is shown below:
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A 100. 0 ml sample of 0. 10 m nh3 is titrated with 0. 10 m hno3. Determine the ph of the solution after the addition of 100. 0 m
DerKrebs [107]

The pH of the solution after the addition of 100. 0 ml of HNO3. The kb of NH3 is 1. 8 × 10-5. So, pH is 10.9 basic .

This neutralization occurs as the acid is added to the base:

NH3(aq)+ HNO3(aq)= NH4NO3(aq)+ H2O(l)

The initial moles of NH3present is given by,

nNH3= c×v= 0.10× 100/1000= 0.01m

The number of moles of

HNO3 added is given by:

nHNO3= c×v= 0.10× 100/1000= 0.001

It is clear from the equation that the acid and base react in a 1:1 molar ratio. So, the no. moles of NH3remaining will be 0.01 - 0.001= 0.009

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100.0+100.0= 200.0x cm3

The concentration of NH3 is given by,

[NH3]= c/v= 0.001/200/1000= 0.05x mol/l .

pOH= 12 (pKb− logb)

where b is the base's concentration.

Given that there is little dissociation, we can roughly compare this to the starting concentration.

pKb= −logKb= −log(1.8×10−5)= 4.744

pOH= 3.056

At 25∘xC, we know that, pH+ pOH=14

pH= 14− 3.056= 10.9

To know more about Equilbrium, visit-brainly.com/question/14366127

#SPJ4

8 0
1 year ago
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