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The student is now told that the four solids, in no particular order, are calcium bromide (CaBr2), sugar (C6H12O6), butanoic aci
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Answer : The equilibrium constant for the reaction is, 0.1133
Explanation :
First we have to calculate the concentration of .
Now we have to calculate the dissociated concentration of .
The balanced equilibrium reaction is,
Initial conc. 1.731 M 0
At eqm. conc. (1.731-x) (2x) M
As we are given,
The percent of dissociation of =
= 1.2 %
So, the dissociate concentration of =
The value of x = 0.2077 M
Now we have to calculate the concentration of at equilibrium.
Concentration of = 1.731 - x = 1.731 - 0.2077 = 1.5233 M
Concentration of = 2x = 2 × 0.2077 = 0.4154 M
Now we have to calculate the equilibrium constant for the reaction.
The expression of equilibrium constant for the reaction will be :
Now put all the values in this expression, we get :
Therefore, the equilibrium constant for the reaction is, 0.1133
I hope you understood
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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>.
