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.
Answer:
Explanation:
Relation between ΔG₀ and K ( equilibrium constant ) is as follows .
lnK = - ΔG₀ / RT
The value of R and T are same for all reactions .
So higher the value of negative ΔG₀ , higher will be the value of K .
Mg(s) + N₂0(g) → MgO(s) + N₂(g)
has the ΔG₀ value of -673 kJ which is highest negative value . So this reaction will have highest value of equilibrium constant K .
This statment is false, the pen on a seismograph does not swing freely.
Answer:
In liquids, particles are quite close together and move with random motion throughout the container. Particles move rapidly in all directions but collide with each other more frequently than in gases due to shorter distances between particles. With an increase in temperature, the particles move faster as they gain kinetic energy, resulting in increased collision rates and an increased rate of diffusion.
Explanation:
In liquids, particles are quite close together and move with random motion throughout the container. Particles move rapidly in all directions but collide with each other more frequently than in gases due to shorter distances between particles. With an increase in temperature, the particles move faster as they gain kinetic energy, resulting in increased collision rates and an increased rate of diffusion.
Answer:
look at the picture above
