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Kipish [7]
3 years ago
11

What is the maximum number of electrons that any 'p' sublevel can hold? 2 electrons 6 electrons 8 electrons 10 electrons

Chemistry
2 answers:
Varvara68 [4.7K]3 years ago
6 0
6 electrons... 's' can hold 2..... 'd' can hold 10 and 'f' can hold 14
BaLLatris [955]3 years ago
3 0

Answer:

6 electrons

Explanation:

With the quantum model of the atom, initiated by Bohr, the orbital (space where is more probably to find the electron) was divided into quantum numbers.

The principal number (n), is the number of the shell, it goes from 1 to 7 and is represented by the letter K, L, M, N, O, P, Q. The maximum number of electrons in the shells are, K = 2, L = 8, M = 18, N = 32, O = 32, P = 18, Q =8.

The secondary number (l) is the subshells, and it goes from 0 to 4 and is represented, in order, for the letter, s, p, d, f. The maximum number of electrons permitted in each subshell is s = 2, p = 6, d = 10, f = 14.

The magnetic number (ml), is represented by orbitals, and it goes from -l to +l, passing by 0. The maximum number of electrons is 2 for each one.

The spin number (ms) refers to the rotation of the electron, and it can be +1/2 or -1/2.

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kiruha [24]
<h3><u>Answer;</u></h3>

pH = 12.33

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

The equation of reaction is :

LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l)

Reactants left after the titrant is added;

Total Moles LiOH;

= 0.035L LiOH × (0.2moles/L)

= 0.007moles of LiOH

Moles of HCl;

= 0.023L HCl × (0.25moles/L)

= 0.00575moles HCl is the limiting reagent

Reacting amount of moles of LiOH;

= 0.0575 moles HCl *(1mole LiOH/1moles HCl)

=0.00575 moles LiOH (reacted)

Moles of LiOH left;

= 0.007moles total - 0.00575moles that react

= .00125 moles of LiOH (left)

LiOH is a strong base, which means that it ionizes completely.  

0.00125moles LiOH *(moles/0.058L) = 0.02155M of LiOH

LiOH(aq) --> Li+(aq) + OH-(aq)

[LiOH] = [OH-] = 0.02155 M

pOH = -log[OH-]

pOH = -log(0.02155)

pOH= 1.67

pH = 14 - pOH

pH = 14 - 1.67

pH = 12.33

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3 years ago
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VladimirAG [237]
I'm taking this lesson now, so imma help u ( if u need anything else ask me)

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Effusion Argon = 1/√40
Effusion Oxygen / Effusion Argon = √(40) / √(32)
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