18 - C.Nitrogen Base (Includes Thymine, Guanine, Adenine,Cytosine)
19 - C.DNA
20 - C. Nucleus (Nucleus is where all the information is stored)
21 - A. Twisted ladder
22 - A. Punnet Square
23 - C. Phenotype (Physical Appearance)
24 - B. Genotype (Genetic Makeup)
25 - Cant do that one, I need a BB acc
26 - B. Hiding behind curly hair trait
And the last one about he flowers I also cant do because I need to log using BB
Answer:
The correct answer is: Serine, Histidine, Aspartate
Explanation:
The catalytic triad of an enzyme is composed of three aminoacid residues which are the most important for its catalytic activity. They are located in the catalytic site of the enzyme. In the case of chymotrypsin- a serine protease, the catalytic triad is composed by serine, histidine and aspartate (Ser-His-Asp). Serine proteases hydrolyse peptidic bonds in proteins and peptides. To do that, the histidine-which interacts with the aspartate by a hydrogen bond so its pKa increases- take a proton from the serine. Thus, deprotonated serine is able to attack the peptide bond and to perform hydrolysis.
Answer:
Explanation:
Given that;
The energy gap between the valence band and the conduction band in the widely-used semiconductor gallium arsenide (GaAs) is Δ = 1.424 eV.
So; that implies that:
Suppose that we consider a small piece of GaAs with 1020 available electrons, -- This is taking about the numbers of electrons used which is :
Temperature is given as:
Number of electrons can be calculated by using the formula;
Answer:
[H⁺] = 6.083x10⁻⁴ M, [C₆H₅OO⁻] = 6.083x10⁻⁴ M, [C₆H₅OOH] = 3.98x10⁻³M, pH = 3.22
Explanation:
Data: we have 0.56 gr of benzoic acid, disolved in 1Lt of water. Kₐ = 6.4x10⁻⁵
M (molar mass) of BA (Benzoic Acid) = 122 g/mol
Then, the inicial concentration is 0.56/122 = 4.59x10⁻³ M
We should consider the equation once it reaches the equilibrium:
C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺
C - x x x
And, for the Kₐ:
Kₐ = [H⁺][C₆H₅COO⁻]/[C₆H₅COOH] = x²/(C-x) , where C = 4.59x10⁻³
Then: x² + Kₐx - KₐC = 0
x² + 6.4x10⁻⁵ - 2.9x10⁻⁷ = 0
Resolving this cuadratic equation (remember to use Baskara equation), we obtain:
x = 6.083x10⁻⁴ M
Then: [H⁺] = [C₆H₅COO⁻] = 6.083x10⁻⁴ M
[C₆H₅COOH] = C - x = 3.98x10⁻³ M
pH = -Log [H⁺] = 3.22
