Answer:
CuSO4 cell will have the greatest amount of deposit among all three. The deposit will occur at the cathode
Explanation:
The valence of the elements in this case is as follows -
Cu - 2e-
Sn - 4e-
Cr - 3e-
CuSO4 cell will have the greatest amount of deposit among all three
The atoms of copper metal will deposit at the cathode. At the cathode, the least number of moles of electrons needed .
Hence, more amount of copper can be extracted out by the electrolyte
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>.
2Ca(OH)2(aq) + 2FeCl3(aq) on the dead locs
Tearing paper, Physical Change