Answer:
The lock-and-key model:
c. Enzyme active site has a rigid structure complementary
The induced-fit model:
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
Common to both The lock-and-key model and The induced-fit model:
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
d. Substrate binds to the enzyme through non-covalent interactions
Explanation:
Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.
The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.
The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.
fourth period
The third period is similar to the second, except the 3s and 3p sublevels are being filled. Because the 3d sublevel does not fill until after the 4s sublevel, the fourth period contains 18 elements, due to the 10 additional electrons that can be accommodated by the 3d orbitals.
When we balance this equation
SnO₂(s) + H₂(g) ⇄ Sn(s) + H₂O(g)
We will get
SnO₂(s) + 2H₂(g) ⇄ Sn(s) + 2H₂O(g)
Solution:
Balancing this equation
SnO₂(s) + H₂(g) ⇄ Sn(s) + H₂O(g)
We have to balance the number of O
SnO₂(s) + H₂(g) ⇄ Sn(s) + 2H₂O(g)
We have to balance the number of H
SnO₂(s) + 2H₂(g) ⇄ Sn(s) + 2H₂O(g)
We will get the balanced equation
SnO₂(s) + 2H₂(g) ⇄ Sn(s) + 2H₂O(g)
The reaction quotient will be
Qc = [product] / [reactant]
Qc = [Sn(s) + H₂O(g)] / [SnO₂(s) + H₂(g)]
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Answer: A. they have similar chemical properties
Explanation: How are elements in the same group in the periodic table alike? A. they have similar chemical properties
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