Answer: 0.0220275 M
Explanation:
So, we are given the following data or parameters which are going to help in solving this particular Question/problem.
=> Averagely, we have the volume = 5.0 L of blood in human body .
=> Mass of sugar eaten = 37.7 g of sugar (sucrose, 342.30 g/mol).
Therefore, the molarity of the blood sugar change can be calculated as below:
The molarity of the blood sugar change = (1/ volume) × mass/molar mass.
Thus, the molarity of the blood sugar change = (1/5) × 37.7/342.30 = 0.0220275 M.
Each enzyme's active site is suitable for one specific type of substrate – just like a lock that has the right shape for only one specific key. Changing the shape of the active site of an enzyme will cause its reaction to slow down until the shape has changed so much that the substrate no longer fits.
Answer: Please see answer below
Explanation:
The steps of glycogen degradation is as follows from this order.
--->Hormonal signals trigger glycogen breakdown.
1. Glycogen is (de)branched by hydrolysis of α‑1,6‑glycosidic linkages.
2. Blocks consisting of three glucosyl residues are moved by remodeling of α‑1,4‑glycosidic linkages.
3.[Glucose 1‑phosphate is cleaved from the non reducing ends of glycogen and converted to glucose 6‑phosphate.
--->Glucose 6‑phosphate undergoes further metabolic processing
The degradation of Glycogen follows three steps:
(1) the release of glucose 1-phosphate from glycogen,
(2) the remodeling of the glycogen substrate to permit further degradation, and
(3) the conversion of glucose 1-phosphate into glucose 6-phosphate for further metabolism.
(https://www.ncbi.nlm.nih.gov/books/NBK21190)
Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7 ∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
= (2x10^-16)/(1x10^-7)^2
= 0.02 M
b) at PH =10when POH = 14- PH = 14-10 = 4
when POH = -㏒[OH-]
4 = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
= 2x10^-16 / (1x10^-4)^2
= 2x10^-8 Mc) at PH= 14
when POH = 14-PH
= 14 - 14
= 0
when POH = -㏒[OH]
0 = - ㏒[OH]
∴[OH] = 1 m
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
= (2x10^-16) / 1^2
= 2x10^-16 M
Answer:
wait i am kinda counfuesd
Explanation: