Answer:
a. C: +3 ; b. N: +5 ; c. S:+6 ; d. C: +4; e. Mn: +7 ; f. Cr: +6.
Explanation:
Global charges in molecules is 0
You sum all the oxidation states to determine the oxidation state for the compound.
Na₂C₂O₄ → Sodium oxalate → Global charge: 0
Oxidation state for C: +3
HNO₃ → Nitric acid → Global charge: 0
Oxidation state for N: +5
H₂SO₄ → Sulfuric acid → Global charge: 0
Oxidation state for S: +6
HCO₃⁻ → Bicarbonate → Global charge: -1, this is an anion
Oxidation state for C: +4
KMnO₄ → Potassium permanganate → Global charge: 0
Oxidation state for Mn: +7
Cr₂O₇⁻ → Anion dichromate → Global charge: -2
Oxidation state for Cr: +6
Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:

Rate factor in the presence of catalyst:

Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:

![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;

Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;



The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Carbocation
I guess pls tell me if it wrong
The amount the amount of space a population has to grow in would be a limiting factor.