Answer:
7.4 - 5.4.
Explanation:
The relationship between HIn that is the non-ionized form and the In^-; the ionized form is an equilibrium Reaction which is given below;
HIn(aq) + H2O(l) <--------> H3O^+(aq). + In^-(aq).
From the question, we are given that the ka is 4 ×10^−7, therefore, the mathematical relationship between HIn and In^- is given below;
Ka = [H3O^+] [In^-] / [HIn] = 4 ×10^−7.
The formula we are going to be using to solve this question is given below;
pH = pKa (+ or -) 1.
Recall that the relationship between the ka and pKa;
pKa = - log (ka).
pKa = - log ( 4 ×10^−7).
pKa = 6.4
Therefore, for it to have a distinct color change, the pH range will be;
pH = pKa (+ or -) 1.
pH= 6.4 + 1 = 7.4.
pH= 6.4 - 1 = 5.4.
Hence, the pH range = 7.4 - 5.4.
Answer:
The wavelength the student should use is 700 nm.
Explanation:
Attached below you can find the diagram I found for this question elsewhere.
Because the idea is to minimize the interference of the Co⁺²(aq) species, we should <u>choose a wavelength in which its absorbance is minimum</u>.
At 400 nm Co⁺²(aq) shows no absorbance, however neither does Cu⁺²(aq). While at 700 nm Co⁺²(aq) shows no absorbance and Cu⁺²(aq) does.
Answer: O2+6H12O6=CO2+ENERGY(ATP)
I DON'T THINK SHE IS CORRECT
Explanation:
