Interpretation:
The number of monomer units in the given sample of nylon is to be determined, the part of the nylon’s structure thatis similar to the structure of the polypeptide is to be identified, tripeptides that can be formed from the given amino acids are to be discussed, and the polypeptide that is found in the silk is to be explained.
Concept introduction:
Each monomer of
A monomer is a small molecule that combines with a similar molecule to give a large molecule.
The total mass of
The number of pairs can be calculated as:

Answer to Problem 96AP
Solution:
(a) The combined weight of two monomer units of
(b) An Amide bond is similar to the polypeptide’s structure.
(c) Three letters, A(Ala), G(Gly), and S(Ser) can be arranged in six ways to form a tripeptide. Silk consists of only one polypeptide arrangement, without having two ends of the
Explanation of Solution
a) The number of monomer units are there in the sample
The average molar mass of
Adipic acid and hexamethylene diamine are the two monomers that are used in the formation of
On the other side, hexamethylene diamine contains
Therefore, the total mass of the two monomer units is
So, an average of
b) The part of the nylon’s structure similar to a polypeptide’s structure.
When a carboxyl group of one amino acid is linked with the other amino acid by peptide, it bonds to form polypeptide chains. So, this chain containsan amide bond, which is also found in nylon’s structure.
c) Number of tripeptides can be formed from the amino acids Alanine (Ala), glycine (Gly), and serene (Ser).
Alanine (Ala), glycine (Gly), and serene (Ser) are the three amino acids that form tripeptides. These three letters can be arranged in six ways, if only alternate polymers are taken into consideration. There is only one polypeptide chain that can be arranged in silk, leaving the two ends of the polymer.
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Chapter 25 Solutions
BURDGE CHEMISTRY VALUE ED (LL)
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