Why does the designed world change?
1 answer:
You might be interested in
Answer:
Check the explanation
Explanation:
Answer – Given, acid and there are three Ka values
The transformation of is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.
Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L
First we need to calculate moles of each
Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1
= 0.162 moles
Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1
= 0.225 moles
[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M
[HPO42-] = 0.225 moles / 1.00 L = 0.225 M
Now we need to calculate the pKa2
pKa2 = -log Ka
= -log 6.2x10-8
= 7.21
We know Henderson-Hasselbalch equation
pH = pKa + log [conjugate base] / [acid]
pH = 7.21 + log 0.225 / 0.162
= 7.35
The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35
Answer:
a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe
b. Glu-Ala-Phe + Gly-Ala-Tyr
Explanation:
In this case, we have to remember which peptidic bonds can break each protease:
-) <u>Trypsin</u>
It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.
-) <u>Chymotrypsin</u>
It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.
With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the <u>"Lis"</u> and <u>"Arg"</u> (See figure 1).
In "peptide b", the peptidic bond that would be broken is the one in the <u>"Phe"</u> (See figure 2). The second amino acid that can be broken is <u>tyrosine</u>, but this amino acid is placed in the <u>C terminal spot</u>, therefore will not be involved in the <u>hydrolysis</u>.
