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:
0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J
Answer:
The answer to your question is below
Explanation:
There are 4 types of chemical reactions:
- Synthesis is when two elements or compounds form only one compound.
- Decomposition is when 1 compound is broken into 2 or more products.
- Single replacement is when one element is replaced by another element.
- Double replacement is when the cations of two compounds are interchanged.
1.- Decomposition 2 Al₂O₃ ⇒ 4 Al + 3O₂
2.- Single replacement Mg + 2HNO₃ ⇒ Mg(NO₃)₂ + H₂
3.- Combustion 2C₆H₆ + 15O₂ ⇒ 12CO₂ + 6H₂O
4.- Synthesis 2Ag + S ⇒ Ag₂S
5.- Double replacement 3Ca(OH)₂ + 2H₃PO₄ ⇒ Ca₃(PO₄)₂ + 6 H₂O
Answer:
Atom 1 after gaing 2 electrons have -2 charge
And atom 2 after donating 2 electrons have +2 charge
Explanation:
