Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.
a) We need to find the major species from A to F.
Major Species at A:
1. 
Major Species at B:
1.
2. 
Major Species at C:
1.
Major Species at D:
1.
2. 
Major Species at E:
1.
Major Species at F:
1.
b) pH calculation:
At Halfway point B:
pH = pK
+ log[
]/[H
]
pH = pK = 6.35
Similarly, at halfway point D.
At point D,
pH = pK
+ log [H]/[H2]
pH = pK = 10.33
Answer:
Explanation:
percentage abundance of third isotope = 100 - ( 78.900 + 10.009)
= 11.091 %
Atomic mass
24.1687 x .789 + 25.4830 x .10009 + 24.305 x .11091
19.069 + 2.5506 + 2.69566
= 24.3153 amu
Answer:
4.2 g
Explanation:
The VOLUME of the ring is 4.2 - 4.0 = .2 ml = .2 cm^3
the MASS of the ring is this times the density
.2 cm^3 * 21 g/cm^3 = 4.2 g
Answer:
The answer is (e) : phosphoglucomutase, UDP-glucose pyrophosphorylase, glycogen synthase then amylo-(1,4-1,6)-transglycosylase.
Explanation:
Phosphoglucomutase: Convert glucose-6-phosphate to glucose-1-phosphate.
UDP-glucose pyrophosphorylase: Form UDP-glucose from glucose-1-phosphate.
Glycogen synthase: Add the new glucose from UDP-glucose to the growing glycogen chain.
Amylo-(1,4-1,6)-transglycosylase: This is a branching enzyme, it initiates formation of branches evolving from the main chain.
Answer:
1.829 meters is the answer
Explanation:
