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astraxan [27]
2 years ago
5

To show the bonding in a molecule or polyatomic ion, we can use a lewis structure, which shows bonding electron pairs as ______

and any nonbonding electrons as ______.
Chemistry
1 answer:
ArbitrLikvidat [17]2 years ago
3 0

Lewis structure, which shows bonding electron pairs as Lines and any nonbonding electrons as dots.

A chemical bond is  holds atoms collectively in molecules. Bonds get up from the electrostatic forces between definitely charged atomic nuclei and negatively charged electrons.

Lewis symbols to describe valence electron configurations of atoms and monatomic ions. A Lewis image consists of an elemental image surrounded by way of one dot for each of its valence electrons: parent 1 suggests the Lewis symbols for the factors of the 0.33 length of the periodic table.

Lone pairs are discovered in the outermost electron shell of atoms. They may be diagnosed with the aid of using a Lewis shape. Electron pairs are therefore taken into consideration lone pairs if two electrons are paired however aren't utilized in chemical bonding.

It is  representation of the valence shell electrons in a molecule. it's far used to expose how the electrons are arranged round individual atoms in a molecule. Electrons are shown as dots or for bonding electrons as a line between the two atoms.

Learn more about bonding in a molecule here:-brainly.com/question/2615067

#SPJ4

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Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088

by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
               = 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
       = -㏒ 0.088 = 1.06 
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Lubov Fominskaja [6]
I think it’s B. I’m not 100% sure but I believe it is B! Srry if this is late btw.
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Vadim26 [7]
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