Alcoholic fermentation is mainly used by various yeast species to make energy.
If there is no oxygen available, the yeasts have in the alcoholic fermentation another possibility of energy supply. But they can - as compared with cellular respiration - recover substantially less energy from glucose, in the form of adenosine triphosphate (ATP): by complete oxidation, a molecule of glucose provides 36 molecules of ATP, but by alcoholic fermentation only 2 molecules of ATP. These two molecules are obtained in glycolysis, the first step in the chain of reactions for both cellular respiration and fermentation.
The two additional steps of the fermentation, and thus the production of ethanol serve not to make energy, but the regeneration of the NAD + cofactor used by the enzymes of glycolysis. As NAD + is available in limited quantities, it is converted by the NADH reduced state fermentation enzymes to the NAD + oxidized state by reduction of acetaldehyde to ethanol.
Answer: Option (A) is the correct answer.
Explanation:
Plastic deformation is the change in shape of an object or metal caused by the load of excess stress.
Thus, metals experience plastic deformation when their crystal patterns have been disrupted by stress.
When stress is provided to the metal then their crystal pattern gets deformed resulting in change of shape of the metal. Plastic deformation is a permanent deformation.
Answer:
Explanation:The pi-molecular orbitals in propene (CH3-CH=CH2) are essentially the ... This central carbon thus provides two p-orbitals – one for each pi bond – and these two different p-orbitals have to be perpendicular, leading to a twisted structure as shown: ... It all comes down to where the location of the electron-deficient carbon
The formula for water is H2O so there would have to be two Hyrdogens and one oxygen. Therefore it would be 4g of Hydrogen and 16g of Oxygen leaving you with 20g.
The answer is D.
Hope this helps :) ~
Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after
is 0.00345 mol/L.
Explanation:

Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t = 
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after
is 0.00345 mol/L.