Answer:
650 mmol.
Explanation:
The equation for the fermentation of one mole of glucose is:
C₆H₁₂O₆ + 2 NAD⁺ + 2 ADP + 2 P i + 2 NADH → 2 EtOH + 2 ATP + 2 NADH + 2 NAD⁺
Since NAD⁺/NADH is used and regenerated, we can eliminate it from the equation:
C₆H₁₂O₆ + 2 ADP + 2 P i → 2 EtOH + 2 ATP
With the equation, we calculate the maximum amount of ethanol that could be obtained theoretically:
1000 mmol C₆H₁₂O₆ ------------ 2000 mmol EtOH
325 mmol C₆H₁₂O₆ ------------- x= 650 mmol EtOH
Therefore, the maximum amount of ethanol that could be produced is 650 mmol.
Answer:
<em>D is the answer</em>
Explanation:
First of all, conservation is when you conserve resources, or use them less, to be better for the environment. Using a gas powered car does not, as carbon monoxide and other toxic gases are sent into the atmosphere. Using a disposable bag is bad because the plastic is sent away to the environment every time you shop. Leaving water running when you brush your teeth wastes water and is bad for the environment. Recycling aluminium cans and glass bottles is good for the environment, and is conserving resources because the cans and bottles are reused, so instead of wasting, you are reusing.
Please leave brainliest! Thanks mate, and hope this helps!
Matter, air, people, gravity, protons, and space♀️
Answer:
The answer to your question is A. Decomposition
Explanation:
In the picture, we can observe that there is only one reactant that after the reaction are formed two products.
We observe that this is a decomposition reaction.
It is not a synthesis reaction, this reaction is the opposite of a decomposition reaction.
In a single-replacement and double-replacement reaction, there is an interchange of cations between the reactants, this does not happen in the picture.
Answer:
The percent ionization is 0,16%
Explanation:
The percent ionization is defined as the number of ions that exist in a substance.
![PI=\frac{[A-]}{[HA]} x100](https://tex.z-dn.net/?f=PI%3D%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D%20x100)
First, we find the [A-] using the ka equation
HA ⇄ 
[H+] = [A-]
![Ka=\frac{[H+][A-]}{[HA]}\\ \\](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%2B%5D%5BA-%5D%7D%7B%5BHA%5D%7D%5C%5C%20%5C%5C)
since the ionization constant is very small we can assume that the final concentration of [HA] is the same
![Ka=\frac{[H+]^{2} }{[HA]} \\\\](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%2B%5D%5E%7B2%7D%20%7D%7B%5BHA%5D%7D%20%5C%5C%5C%5C)
![[H+]=\sqrt[2]{Ka.[HA]} \\\\](https://tex.z-dn.net/?f=%5BH%2B%5D%3D%5Csqrt%5B2%5D%7BKa.%5BHA%5D%7D%20%5C%5C%5C%5C)
![[H+] =\sqrt{(2,610^{-7} )(0,1)} = 1,61210^{-4}](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%5Csqrt%7B%282%2C610%5E%7B-7%7D%20%29%280%2C1%29%7D%20%20%3D%201%2C61210%5E%7B-4%7D)
Now we calculate the percent ionization using these values
PI=0,16%
