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gregori [183]
3 years ago
7

Draw a lewis structure for ketene, c2h2o, which has a carbon–carbon double bond

Chemistry
1 answer:
MrMuchimi3 years ago
8 0

Visual representation of covalent bonding indicating the valence shell electrons in the molecule, lines represents the shared pair of electron and pair of electrons that are not involved in bonding are represented as dots(lone pairs) are known as Lewis structures.

Compound formation takes place in order to complete the octet of each element that is according to octet rule, each atom forms bond with other atom in order to complete their octet that is to get eight electrons in its valence shell and attain stability.

An organic compound of the form R^{'}R^{''}C=C=O is known as ketene.

The given ketene is C_2H_2O.

The number of valence electron of:

C = 4

H = 1

O = 6

The number of valence electrons in C_2H_2O = 2\times 4+2\times 1+1\times 6 =16

2 electrons are involved in each single bond between carbon and hydrogen and 4 electrons are involved in each double bond formed between carbon-carbon and carbon-oxygen. Hence, the total number of electrons involved in bond formation are 12 and rest 2 pair of electrons are present on oxygen as lone pair of electrons.

Therefore, the attached image is the Lewis structure of C_2H_2O .

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For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [
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Explanation:

The given data is as follows.

     E^{o} = 0.483,     [Co^{3+}] = 0.173 M,

     [Co^{2+}] = 0.433 M,     [Cl^{-}] = 0.306 M,

     P_{Cl_{2}} = 9.0 atm

According to the ideal gas equation, PV = nRT

or,             P = \frac{n}{V}RT    

Also, we know that

                Density = \frac{mass}{volume}

So,         P = MRT

and,          M = \frac{P}{RT}

                    = \frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}

                    = \frac{9.0}{24.436}

                    = 0.368 mol/L

Now, we will calculate the cell potential as follows.

          E = E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}

             = 0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}

             = 0.483 - 0.02955 log \frac{0.0689}{0.0162}

             = 0.483 - 0.02955 \times 0.628

             =  0.483 - 0.0185

             = 0.4645 V

Thus, we can conclude that the cell potential of given cell at 25^{o}C is 0.4645 V.

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