Answer:
i and ii
Explanation:
In the aerobic oxidation of glucose, the electrons formed are transferred to O2 after several others transfer reactions like passing through coenzymes NAD+ and FAD
Answer: mass- how much stuff" is in an object
Weight- there is a gravitational interaction between objects
Explanation:
Answer:
Explanation:
Hello,
In this case, the reaction is:
Thus, the law of mass action turns out:
![Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_3CH_2OH%5D_%7Beq%7D%7D%7B%5BH_2O%5D_%7Beq%7D%5BCH_2CH_2%5D_%7Beq%7D%7D)
Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change 
result:
![[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol](https://tex.z-dn.net/?f=%5BCH_2CH_2%5D_%7Beq%7D%3D29mol-x%3D16mol%5C%5Cx%3D29-16%3D13mol)
In such a way, the equilibrium constant is then:
Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:
Thus, the second change, 
finally result (solving by solver or quadratic equation):
Finally, such second change equals the moles of ethanol after equilibrium based on the stoichiometry:
Best regards.
Answer:
The cell wall surrounds the plasma membrane of plant cells and provides tensile strength and protection against mechanical and osmotic stress.
Answer:
The enthalphy change of the given reaction is -280 KJ.
Explanation:
X 2 + 3 Y 2 ⟶ 2 XY 3 Δ H 1 = − 380 kJ ....Eq-1
X 2 + 2 Z 2 ⟶ 2 XZ 2 Δ H 2 = − 130 kJ ....Eq-2
2 Y 2 + Z 2 ⟶ 2 Y 2 Z Δ H 3 = − 260 kJ ....Eq-3
4 XY 3 + 7 Z 2 ⟶ 6 Y 2 Z + 4 XZ 2 ΔH ....Eq-4
<em>By Hess law,</em>
<em> The heat of any reaction ΔH for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction:</em>
Eq-4 can be manipulated as,
Eq-4 = -2(Eq-1) +2(Eq-2) +3(Eq-3)
By Hess law, same happen with their enthalphy
Therefore,
ΔH = -2(ΔH1) + 2(ΔH2) + 3(ΔH3)
= -2 × -380 + 2 × -130 + 3 × -260
= -280 KJ
