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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Answer:
Chlorine is limiting reactant
Explanation:
Based on the reaction:
Cl₂ + 2NaOH → NaClO + NaCl + H₂O
<em>1 mole of chlorine reacts with 2 moles of NaOH</em>
<em />
To find limiting reactant, we need to determine the moles of the reactants:
<em />
<em>Moles Cl₂ -Molar mass: 70.9g/mol-:</em>
800lb Cl₂ * (453.6g / 1lb) * (1mol / 70.90g) =
5118 moles Cl₂
<em>Moles NaOH -Molar mass: 40g/mol-:</em>
1200lb NaOH * (453.6g / 1lb) * (1mol / 40g) =
13608 moles NaOH
For a complete reaction of 13608 moles of NaOH you need:
13608 moles NaOH * (1mol Cl₂ / 2 moles NaOH) = 6804 moles of Cl₂
As the solution contains just 5118 moles of chlorine,
<h3>Chlorine is limiting reactant</h3>
Answer:
Work done = 600 J
Power used = 60 W
Explanation:
Given:
Force acting on the box is, 
Displacement of the box is, 
Time taken for the work, 
Now, we know that, work is said to be done by a force only when there is displacement caused by the force in its direction.
Here, the force acting on the box causes a displacement of 30 m in its direction. So, work done is equal to the product of force and displacement caused.
Therefore, work done on the box is given as:
Therefore, the work done is 600 J.
Now, we know that, power is given as work done per unit time.
So, power used is given as:
Therefore, the power used is 60 W.
Answer:
(a) Between methanol (CH₃OH) and glycerol (C₃H₅(OH)₃), the substance with the higher surface tension is glycerol (C₃H₅(OH)₃)
(b) Between tetrabromomethane (CBr₄) and chloroform (CHCl₃), the substance with the higher surface tension is chloroform (CHCl₃)
Explanation:
The surface tension of these substances at 20 °C given in mN/m, is as follows:
The surface tension of Methanol is 22.70
The surface tension of Tetrabromomethane is 26.95
The surface tension of Glycerol is 64.00
The surface tension of Chloroform is 27.50
(a) Between methanol (CH₃OH) and glycerol (C₃H₅(OH)₃), the substance with the higher surface tension is glycerol (C₃H₅(OH)₃)
(b) Between tetrabromomethane (CBr₄) and chloroform (CHCl₃), the substance with the higher surface tension is chloroform (CHCl₃)
