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.
<span>Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38
Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238
The total charge of all the vanadium ions would be +238.
Let x = number of of V+2
Then 100 – x = number of V+3
X(+2) + 100-x(+3) = +238
2x + 300 – 3x = 238
-x = 238-300 = -62
x = 62
Thus 62/100 are V+2
62/100 * 100 = 62%
</span>62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
Answer:
= C3H4N
Explanation:
We are given; 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen.
We first calculate the number of moles of each element.
Carbon = 90g/12 g/mol
= 7.5 moles
Hydrogen = 11 g/ 1 g/mol
= 11 moles
Nitrogen = 35 g/ 14 g/mol
= 2.5 moles
The we get the mole ratio of the elements;
= 7.5/2.5 : 11/2.5 : 2.5 /2.5
= 3 : 4.4 : 1
= 3 : 4 : 1
Therefore;
The empirical formula will be; C3H4N
Cation - Cr 4+
Anion - S ^ 2-
Chromium sulfide
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