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:
= 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:
temperatures within the outer core range from 7,200 to 9,000 F. Pressure also increases in the outer core due in part to the weight of the crust and mantle above.The Earth's magnetic field is generated by the outer core.
Explanation:
D. It is the heat required to change a gram of substance from a liquid to a gas.
Explanation:
The heat of vaporization is the heat required to change a gram of substance from a liquid to a gas.
- It is also known as the enthalpy of vaporization.
- The heat of vaporization is the quantity of heat needed to change one gram of a substance from liquid to gas.
- This heat of vaporization is dependent on the pressure conditions the process is taking place.
- Different liquids have their heat of vaporization.
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Answer:
the answer is b mitosis in humans
The equation is L = m/M
First, covert 10. grams of AgNO3 to moles which is 0.059 moles.
Divide 0.059 moles by 0.25M which is 0.24 liters.