In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. During this phase, the conversion of glyceraldehyde-3-phosphate to pyruvate and the coupled formation of ATP take place. Because Glucose is split to yield two molecules of D-Glyceraldehyde-3-phosphate, each step in the payoff phase occurs twice per molecule of glucose.
Glyceraldehyde 3-phosphate dehydrogenase Simultaneous oxidation and phosphorylation of G3P produce 1,3-bisphosphoglycerate (1,3-BPG) and nicotine adenine dinucleotide (NADH).
The divalent cation also affected the response of the enzyme from the endosperm and shoots to adenine nucleotides and inorganic pyrophosphate.
This phase is also called the glucose activation phase. In the preparatory phase of glycolysis, two molecules of ATP are invested and the hexose chain is cleaved into two triose phosphates. During this, the phosphorylation of glucose and its conversion to glyceraldehyde-3-phosphate take place. Steps 1, 2, 3, 4, and 5 together are called the preparatory phase.
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Sodium Oxide= Na2O
The formula mass of Na2O is (2x23) + 16 = 62g/mol
% Na= (46/62) x 100 = 74%
% O= (16/62) x 100 = 26%</span><span>
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Answer:
6 oxygen atoms
Explanation:
From the equation,
2Fe(OH)₃ → Fe₂O₃ + 3H₂O
From the reactant (left hand side) we have 2 moles of Fe(OH)₃ having (2 * 3 = 6) atoms of oxygen and decomposed to give Fe₂O₃ which contains 3 atoms of oxygen and 3 moles of water that also contains 3 atoms of oxygen.
Since the number of oxygen participating in the reaction is independent on the product (not a reversible reaction) then the total number of oxygen atoms participating in the reaction is 6
The chemical equation given is:
<span>2x(g) ⇄ y(g)+z(s)</span>
Answer: the higher the amount of x(g) the more the forward reacton will occur and the higher the amounts of products y(g) and z(s) will be obtained at equilibrium.
Justification:
As Le Chatellier's priciple states, any change in a system in equilibrium will be compensated to restablish the equilibrium.
The higher the amount, and so the concentration, of X(g), the more the forward reaction will proceed to deal witht he high concentration of X(g), leading to an increase on the concentration of the products y(g) and z (s).
