Answer:
No
Explanation:
AS there are more that are sick in the HL+ group than in the control group.
We will use this two reaction equation:
H2SO3 + H2O ↔ H3O+ + HSO3- Ka1 = 1.3 x 10^-2
HSO3- + H2O ↔ H3O+ + SO3 2- Ka2= 6.3 x 10^-8
we will use the ICE table for the first equation:
H2SO3 + H2O ↔ H3O+ + HSO3-
initial 0.025 0 0
change -X +X +X
Equ (0.025-X) X X
Ka1 = [H3O+] [HSO3-] / [H2SO3]
1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X
∴ X = 0.0127
when [H3O+] = X
∴[H3O+] = 0.0127 M
and when [HSO3-] = X
∴[HSO3-] = 0.0127 M
and when [H2SO3] = 0.025 - X
∴[H2SO3] = 0.025 - 0.0127
= 0.0123 M
when Kw = [OH-][H3O+]
and Kw = 1.1 x 10^-14 / 0.0127
∴[OH-] = 1.1 x 10^-14 / 0.0127
= 8.66 x 10^-13 M
- by using the ICE table for the second equation:
HSO3- + H2O ↔ H3O+ + SO3 2-
initial 0.0127 0.0127 0
change -X +X +X
Equ (0.0127-X) (0.0127+X) X
when Ka2 = [SO32-] [H3O+] / [HSO3-]
by substitution:
6.3 x 10^-8 = X(0.0127+X) / (0.0127-X)
as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X
6.3 x 10^-8 = 0.0127X /0.0127
∴X = 6.3 x 10^-8
when [SO3 2-] = X
∴[SO32-] = 6.3 x 10^-8
<span>An element belonging to the halogen family would be expected to have a large ionization energy and a large electron affinity.
Flourine, Chlorine, Bromine, Iodine and astatine are the elements that belongs to the halogen family and mostly they have high values of ionization energy.
The amount of energy released when an electron is added to an atom or molecule to form a negative ion or anion is electron affinity.</span>Chlorine from this family has highest electron affinity.
2 hydrogen and 1 oxygen makes water
Answer:
c- dry cells can not be recharged
