In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
Reaction is NH4OH <-> NH4+ OH- (note this is reversible)
Draw up an ICE table
Let x be equilibrium conc of OH- assume init conc of OH is 0M and init conc of NH4+ is 0M also. Init conc of NH4OH is 0.1M so equilibrium conc will be 0.1-x.
%dissociation = x/0.1-x * 100%
1 = 100x/0.1-x
0.1-x = 100x
101x = 0.1
x = 0.0009901
pOH = -log(0.0009901) = 3.00
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.
For more information on phosphorylation click on the link below:
brainly.com/question/7465103
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Answer:
92.344mL
Explanation:
acording to boyle's law that PV=constant then P1V1=P2V2
The correct answer would be option 3. I order for a single displacement to occur there must be a single element and one compound. From then, the single element will then swap places with one of the elements in the compound (or in this case single circle and combined circles). Hope this helped!
