Answer:
51:49 is the ratio of female participants to male participants.
Explanation:
Percentage of females in the race = 51 %
Percentage of males in the race = 100% - 51 % = 49%
Total participants in race = 400
Number of women participants = 
Number of men participants = 
Ratio of female participants to male participants :

Answer:
0.01836 M
Explanation:
Again the reaction equation is;
Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)
E°cell= 0.77 V
Ecell= 0.78 V
[Mn2+] = 0.040 M
[Fe2+] = the unknown
n=2
From Nernst's equation;
Ecell= E°cell- 0.0592/n log Q
0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]
0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]
0.01/ -0.0296= log [Fe2+] /[0.040]
-0.3378= log [Fe2+] /[0.040]
Antilog(-0.3378) = [Fe2+] /[0.040]
0.459= [Fe2+] /[0.040]
[Fe2+] = 0.459 × 0.040
[Fe2+] = 0.01836 M
Answer:
Option B is correct.
Explanation:
They live in less extreme environment than most archaebacteria.
Hoped this helped.
Answer:
= 19
ΔG° of the reaction forming glucose 6-phosphate = -7295.06 J
ΔG° of the reaction under cellular conditions = 10817.46 J
Explanation:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
Given that: at equilibrium, 95% glucose 6-phospate is present, that implies that we 5% for glucose 1-phosphate
So, the equilibrium constant
can be calculated as:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)


= 19
The formula for calculating ΔG° is shown below as:
ΔG° = - RTinK
ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k × 1n(19))
ΔG° = 7295.05957 J
ΔG°≅ - 7295.06 J
b)
Given that; the concentration for glucose 1-phosphate = 1.090 x 10⁻² M
the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M
Equilibrium constant
can be calculated as:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)

0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J