Answer:
The two molecules of acetyl-CoA that are produced from a molecule of glucose goes through two turn in the citric acid cycle, one for each molecule of acetyl-CoA.
Explanation:
Glycolysis the process by which a molecule of glucose is broken down in a series of steps to yield two molecules of pyruvate. The overall equation for the reactions of glycolsis is given below:
Glucose + 2NAD+ ----> 2 Pyruvate + 2NADH + 2H⁺
Each of the two pyruvate molecules produced from glucose breakdown is further oxidized to two molecules of acetyl-CoA and CO₂ each.
2 Pyruvate ----> 2 AcetylCoA + 2CO₂
Each of the acetyl-CoA molecule then enters the citric acid cycle for its oxidation. In each turn of the cycle, one acetyl group enters as acetyl-CoA and two molecules of CO₂ leave.
Explanation:
The given reaction is as follows.
Value of equilibrium constant is given as 
= 4.3 \times 10^{6}[/tex].
Concentration of given species is ![[SO_2]](https://tex.z-dn.net/?f=%5BSO_2%5D)
= 0.010 M; ![[SO_3]](https://tex.z-dn.net/?f=%5BSO_3%5D)
= 10.M; ![[O_2]](https://tex.z-dn.net/?f=%5BO_2%5D)
= 0.010 M.
Formula for experimental value of equilibrium constant (Q) is as follows.
Q = ![\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO_%7B3%7D%5D%5E%7B2%7D%7D%7B%5BSO_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D)
Putting the given concentration as follows.
Q =
Q = 
Q = 
It is known that when Q > 
, then reaction moves in the backward direction.
When Q < , then reaction moves in the forward direction.
When Q = , then reaction is at equilibrium.
As, for the given reaction Q > then it means reaction moves in the backward direction.
Thus, we can conclude that the reaction is moving in the backward direction, that is, right to left to reach the equilibrium.
The answer would be B.
U-238 has a n to p ration of 1.6:1. 146 neutrons and 92 protons.
It is actually the most commonly used isotope is reactors.
C-14 is also a radioactive isotope with 8 neutrons and 6 protons.
The usual and ideal n to p ratio is 1:1 such as C-12 or Mg-24
Answer:
The traditional electrolyte for aluminium electrolysis is based on molten cryolite (Na3AlF6), acting as solvent for the raw material, alumina (Al2O3).Metals are found in ores combined with other elements. Electrolysis can be used to extract a more reactive metal from the ore.
Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions. As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere. In an aqueous solution, if the potential is high enough, OH− and O2− are generated at the anode, which can then react with the aluminum to produce aluminum oxide. Al^3+ can also be generated directly. The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the Al^3+ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode. In this way, coatings thicker than the normal passivation in air can be produced. However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made). This is the normal behaviour in aqueous solutions at near-neutral pH (5–7).
However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface. One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed OH−, leaving pores in the oxide layer through which the ions can travel and continue to react. These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.
