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Lera25 [3.4K]
3 years ago
6

Use Lewis diagrams to show how electrons are shared to form covalent bonds in compounds wit the following atoms. Write the formu

las
a) two hydrogen atoms attached to a carbon atom that also has one oxygen atom attached to it 

Chemistry
1 answer:
bixtya [17]3 years ago
8 0
The Lewis structure/diagram for CH2O (aka Formaldehyde) can be written in either of the following ways shown in the picture.
The dots represent electrons in the valence shell of the atom (the outermost shell). The green dots are electrons that belong to the Oxygen atom, the blue belong to the Carbon atom, and the pink belong to the Hydrogen atoms.
Covalent bonds are bonds between atoms where atoms share electrons with each other. Atoms bond because they obey the octet rule ( the rule states that most atoms of main-group elements tend to want 8 electrons in their valence shells). 
Oxygen has 6 valence electrons, Carbon has 4, and Hydrogen has 1. H does not follow the octet rule, but C and O do, so the atoms are arranged in this way so that the O and C atoms have a full octet of electrons in their valence.

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Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the
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Answer : The equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

Solution : Given,

Equilibrium constant, K_c=1.8\times 10^{-5}

Initial concentration of HC_2H_3O_2 = 0.260 m

Let, the 'x' mol/L of H_3O^+ are formed and at same time 'x' mol/L of HC_2H_3O_2 are also formed.

The equilibrium reaction is,

                  HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)

Initially                0.260 m                       0                 0

At equilibrium    (0.260 - x)                     x                 x

The expression for equilibrium constant for a given reaction is,

K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

Now put all the given values in this expression, we get

1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}

By rearranging the terms, we get the value of 'x'.

x=2.154\times 10^{-3}m

Therefore, the equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

4 0
3 years ago
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