Answer:
1. active site
2. substrate.
3. activity
4. cofactor
5. complex
6. coenzyme.
Explanation:
1. A substrate binds to an enzyme at the active site where the reaction occurs.
2. In a catalyzed reaction, a reactant is often called a substrate.
3 activity is a measure of how fast an enzyme can convert the reagent to the product.
4. An inorganic substance necessary for the function of some enzymes is called a cofactor.
5. When properly aligned, the enzyme and substrate form an enzyme-substrate (ES) complex.
6. A small organic molecule necessary for the function of some enzymes is called a coenzyme.
Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06
Answer:
138.96kJ is the maximum electrical work
Explanation:
The maximum electrical work that can be obtained from a cell is obtained from the equation:
W = -nFE
<em>Where W is work in Joules,</em>
<em>n are moles of electrons = 2mol e- because half-reaction of Zn is:</em>
Zn(s) → Zn²⁺(aq) + 2e⁻
F is faraday constant = 96500Coulombs/mol
E is cell potential = 0.72V
Replacing:
W = -2mol*96500Coulombs/mol*0.72V
W = - 138960J =
<h3>138.96kJ is the maximum electrical work</h3>
<em />
In Thomson's experiment, he showed that an electrical current can be made to flow from a positive site to a negative site.
