Answer:
Most of free energy available from oxidation of the glucose remains in pyruvate.
Explanation:
The overall reaction of the process glycolysis is:
Glucose + 2 NAD⁺ + 2 ADP + 2 Pi ⇒ 2 Pyruvate + 2 NADH + 2 H⁺ + 2ATP
Glucose is oxidized to give 2 molecules of pyruvate and 2 molecules of NADH and ATP (Energy currency).
<u>Though the free energy of oxidation of glucose is high but only 2 NADH is formed because the most of the free energy that is being released from the oxidation of glucose remains in the pyruvate which is produced in the reaction and thus only 2 molecules are formed.</u>
Answer:
The correct option is C.
Explanation:
Nuclear and chemical reactions are two types of reactions that one usually encounter in chemistry. These two reactions differ from each other significantly. For instance, the nuclear reactions usually involve the nucleus of the involving atoms while chemical reactions has to do with the electrons that are located outside of the nucleus of the atoms. Also, it is only chemical reaction that are influenced by factors such as temperature, pressure, catalyst, etc. Such factors does not determine the rate of nuclear reactions.
S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
<u>Explanation:</u>
S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
To measure the rate of this reaction we must measure the rate of concentration change of one of the reactants or products. To do this, we will include (to the reacting S₂O₈
²⁻ and I⁻
i) a small amount of sodium thiosulfate, Na₂S₂O₃,
ii) some starch indicator.
The added Na₂S₂O₃ does not interfere with the rate of above reaction, but it does consume the I₂ as soon as it is formed.
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
This reaction is much faster than the previous, so the conversion of I2 back to I⁻ is essentially instantaneous.
![rate = \frac{dI2}{dt} = \frac{1/2 [S2O3^2^-]}{t}](https://tex.z-dn.net/?f=rate%20%3D%20%5Cfrac%7BdI2%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%2F2%20%5BS2O3%5E2%5E-%5D%7D%7Bt%7D)
Answer:
The most likely outcome is that carrier protein dysfunction will increase the gradient which will lead to disruption of cellular metabolism.
Other commonly used units include g/L (grams of solute per liter of solution) and m/L (moles of solute per liter of solution). Solubility units always express the maximum amount of solute that will dissolve in either a given amount of solvent, or a given amount of solution, at a specific temperature.
