This dipeptide is designated as: there is a structure for h3nchcnhchco with ch2oh group attached to the first (from left to righ
t) carbon, two oxygen atoms attached to the second and fourth carbons by double bonds, and ch2sh group attached to the third carbon. nitrogen has a charge of 1 plus. oxygen atom, which is attached to the fourth carbon by a single bond, has a charge of 1 minus.
1 answer:
A dipeptide is formed when 2 amino acids are joined together by a peptide linkage.
Each amino acid molecule contains an amino group ( -NH₂) and a carboxyl ( -COOH) group. During peptide formation, a carboxyl group of one amino acid reacts with amino group of the other , resulting in the formation of a linkage known as "peptide linkage". A water molecule is also eliminated in this reaction.
A general peptide formation reaction can be written as
In the above reaction, we can see a peptide linkage ( -CONH-) getting formed between 2 amino acid molecules.
In the given molecule the peptide linkage is shown by drawing a box around it. Please refer to the attached image.
If we want to know the amino acid molecules from which this peptide is formed, we will simply break the CO-NH bond.
We will add -OH group to CO part and -H to NH part to get 2 amino acid molecules as shown in the picture.
The two amino acids thus formed are Serine & Cysteine.
Serine is designated as "Ser" and Cysteine is designated as "Cys"
Therefore the given dipeptide is designated as "Ser-Cys"
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Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.