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Chapter 24 - Essay Example

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24.1 Carbon is able to form so many more compounds as compared to other elements because the carbon atom has a unique ability to form strong covalent bonds with other carbon atoms. This can result in long stable carbon chains and therefore a large number of compounds.
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Chapter 24
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Download file to see previous pages Whereas the aromatic compounds are those compounds which have a cyclic carbon chain with alternating double bonds. A very important group of aromatic compounds includes benzene which is a cyclic compound with 6 carbon atoms and alternate double bonds.
24.3 The hydrocarbons in which the adjacent carbon atoms have only one carbon-carbon bond are called saturated hydrocarbons. These are called saturated compounds because all the four carbon valences are satisfied and no more hydrogen can be attached to the carbon atom. These hydrocarbons are represented by the formula CnH2n+2. Example of saturated hydrocarbon is ethane (C2H6).
The hydrocarbons which contain one or more a double or triple bond in the compound are called unsaturated hydrocarbons. These compounds are called unsaturated because all the carbon valences are not satisfied by hydrogen atoms. The alkenes and alkynes are the examples of unsaturated compounds represented by the formula CnH2n and CnH2n-2. Example of unsaturated hydrocarbon is ethane (C2H4).
24.9 A carbon atom in a compound is called chiral, if the mirror image of this compound cannot be superimposed on itself. In a simplified way, if all the four substituent of a carbon atom are different, then such a carbon atom will be chiral. An example of such a compound is bromochlorofluoromethane.
24.25 CH3CH(NH2)COOH and CH2(OH)CH(NH2)COOH are chiral amino acids as all the four substituent of the central carbon atom are different, so the...
24.27 Structural formula of:
a) 3-methylhexane
b) 1,3,5-trichloro-cyclohexane
c) 2,3-dimethylpentane
d) 2-bromo-4-phenylpentane
e) 3,4,5-trimethyloctane
24.31 Structures are:
a) 1-bromo-3-methylbenzene
b) 1-chloro-2-propylbenzene
c) 1,2,4,5-tetramethylbenzene
24.34 Lewis Structure:
a) Alcohol:
b) Ether
c) Aldehyde
d) Ketone
e) Carboxylic acid
f) Ester
g) Amine
24.41 The products are:
a) CH3-CH2-COOH + H2O
b) H2-C=CH-CH3
c) Not Clear
24.43 The possible isomers of C7H7Cl with a benzene ring are:
24.49 Given:
Density of octane = 0.70 g/ml
Volume of octane = 1 lt = 1000 ml
Therefore; mass of octane is 700 g
As the molecular weight of octane is 114.23 g/mole; so the total moles of octane burnt are (700/114) = 6.13 moles
We know each mole of octane requires 12.5 moles of oxygen for complete combustion. Therefore for the combustion of 1 lt of octane, the oxygen required will be (12.5*6.13 =) 77.875 moles.
It is known that volume of 1 mole of a gas at 20oC is 24.04 lt. So the volume of oxygen required would be (77.875*24.04 =) 1872.2 lt.
Given that oxygen is 22% of the air. So the total volume of air required is (1872.2*100/22 =) 8510 lt.
Answer: 8510 lt. of air is required for complete combustion of 1 lt. of octane.
24.53 The structural isomers of C4H8Cl2 are:
1. 1,1-dichlorobutane
2. 1,2-dichlorobutane
This is a chiral structure.
3. 1,3-dichlorobutane
This structure has a chiral carbon.
4. 1,4-dichlorobutane
5. 2,3-dichlorobutane
This structure is chiral.
6. 1,1-dichloro-2-methyl-propane
7. 1,3-dichloro-2-methyl-propane
8. 1,2-dichloro-2-methyl-propane
24.59 The structure of:
a) Cyclopentane
b) ...Download file to see next pagesRead More
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