Answer:
Michaelis constant is known as km which is the substrate concentration that encourages the compound to work at half maximum velocity represented by Vmax/2. Michaelis constant is inversely related to the substrate and the affinity of the enzyme.
Induced fit model: The premise of the purported induced fit hypothesis, which expresses that the attachment or association of a substrate or some other atom to an enzyme causes an adjustment to the enzyme in order to fit or restrain its activity.
In substrate, analog Km or Michaelis constant will be high as the substrate will stay because of analogs inhibit activity.
In the transitional state, analog Km will be in the middle of the substrate and product analogs. Progress state analogs are synthetic mixes with a structure catalyzed reaction that looks like the progressing condition of a substrate atom in a compound enzyme.
In item simple thus Km is the least.
0.0013 M = product ananlog,
0.025 M=Transition state, and
0.0045 M = Substrate analog
Il fait plus chaud à l'équateur et plus froid aux pôles car : 1°) les rayons du Soleil sont plus concentrés au niveau de l'équateur et plus diffus au niveau des pôles ; 2°) l'épaisseur d'air composant l'atmosphère, traversée par les rayons du Soleil est plus importante aux pôles qu'à l'équateur.
Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
