A rectangular steel block of dimension 100cm × 50m × 20cm has a mass of 450kg. Calculate the density.
1 answer:
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The original concentration of the acid solution is 6.175 10^-4 mol / L.
<u>Explanation:</u>
Concentration is the ratio of solute in a solution to either solvent or total solution. It is expressed in terms of mass per unit volume
HBr + NaOH -----> NaBr + H2O
There is a 1:1 equivalence with acid and base.
Moles of NaOH = 72.90 10^-3
0.25
= 0.0182 mol.
[ HBr ] = moles of base / volume of a solution
= 0.0182 / 29.47
= 6.175 10^-4 mol / L.
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>.
