Answer:
stay the same.
Explanation: Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):
Na: 1s2 2s2 2p6 3s1
Ar: s2 2s2 2p6 3s2 3p6
As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:
Na: 3s1
Ar: 3s2, 3p6
All the new electrons electrons filled the third energy level (3s and 3p). So the energy level does not change, just the orbitals.
The equilibrium constant of the reaction is represented by the symbol K. Thus, option C is the correct and accurate statement about the equilibrium constant.
<h3>What is the equilibrium constant?</h3>
The equilibrium constant is a representation of the concentration of the products and the reactants of the reaction that is raised to the powers through their stoichiometry coefficient.
Its value varies and changes at different temperatures and is not always less than 1. The equilibrium constant is the ratio of the coefficient of the products to reactants.
Therefore, option C. equilibrium constant is represented by K is true.
Learn more about the equilibrium constant here:
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B) False- It has seven
A hexagon would have 6.
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>.
