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NISA [10]
3 years ago
15

Why might adding inorganic phosphate to a reaction mixture where glycolysis is rapidly proceeding help sustain the metabolic pat

hway?
Chemistry
1 answer:
melamori03 [73]3 years ago
7 0

Answer:

The addition of inorganic phosphate helps to sustain the reaction by helping the reaction to occur in a thermodynamically unfavorable manner.

Explanation:

The reaction  of inorganic phosphate with  glucose is an endergonic reaction which is thermodynamically unfavorable as a result reaction need coupling of ATP hydrolysis to take place in biological system because ATP hydrolysis releases huge amount of free energy as a result the overall free energy change of this reaction becomes negative which helps the reaction to proceed in a thermodynamically favorable manner.

       That"s why inorganic phosphate can be used to sustain the rapidly proceeding glycolysis pathway. The addition of inorganic phosphate makes the reaction thermodynamically unvorable.

     

         

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Answer:

The structures are shown below.

Explanation:

When the acid reacts with water, it loses one proton (H⁺) and forms a base, which is the conjugate base of its acid.

The formal charge of an atom can be calculated by:

FC = X - (Y + Z/2)

Where X is the valence electrons of the neutral atom, Y is the unshared electrons, and Z is the shared electrons in the molecule.

a) When HCl deprotonates, it forms Cl⁻ as the conjugate base. The neutral atom Cl has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1 The structure is shown below in figure a.

b) When Hbr deprotonates it forms Br- as the conjugate base. The neutral atom has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1. The structure is shown below in figure b.

c) When CH3COOH loses a proton, it forms the conjugate base CH3COO⁻. The carbon as 4 valence electrons, hydrogen has 1 valence electron and oxygen has 6 valence electrons. The first carbon make simple bonds with each hydrogen and with the second carbon, and so, all the electrons are shared, and it has FC = 4 - (0 + 8/2) = 0, as so the hydrogens have FC = 1 - (0 + 2/2) = 0.

The second carbon does 1 simple bond with the first carbon, a double bond with one oxygen, and a simple bond with the other oxygen, and so doesn't have unshared electrons, and FC = 4 - (0 + 8/2) = 0.

The first oxygen does a double bond with the carbon, and so it has 4 unshared electrons, so FC = 6 - (4 + 4/2) = 0. The second oxygen does a simple bond with the carbon, and so has 5 unshared electrons, so FC = 6 - (5 + 2/2) = 0.

The structure is shown in figure c.

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The product that is used on the natural nail prior to application to assist in adhesion and serves to chemically bond the enhancement product to the natural nail is known as nail primer.

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