Answer:
HCl < CH₃COOH < NH₃ < NaOH
Explanation:
Given compounds:
Acetic acid: CH₃COOH
Ammonia; NH₃
Hydrochloric acid: HCl
Sodium hydroxide: NaOH
All the solutions are of the same molarity which is 0.1M. We need to see how these compounds dissociate to form solutions in order to establish their pH value:
For Acetic acid;
CH₃COOH + H₂O ⇄ H₃O⁺ + CH₃COO⁻
Acetic acid is a weak acid and it ionizes slightly in solutions. It would have a pH close to 7
For Ammonia;
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Ammonia is a weak base and it ionizes slightly in solutions. It sets up an equilibrium in the process. It's would be slightly above 7
For HCl:
HCl + H₂O → H₃O⁺ + Cl⁻
HCl is a strong acid and ionizes completely in solutions. It has a very low pH
For NaOH:
NaOH → Na⁺ + OH⁻
NaOH ionizes also completely in solutions and it breaks down into sodium and hydroxide ions. It is a strong base and it would have a high PH value.
HCl < CH₃COOH < NH₃ < NaOH
This is the trend of increasing pH
Answer:
A. Nonmetals have relatively high ionization energies.
Explanation:
Nonmetals have high ionization energies because they tend to gain electrons in order to fill their outer shell. Also, the electrons are closer to the nucleus and require more energy to remove them.
Hope that helps.
The Calcium atom has 2 valence electrons but a Calcium ion will have no electrons because it is a cation ion. Meaning that there are more protons than electrons. Also, the normal calcium symbol is Ca but the calcium ion is Ca²⁺
I hope this wasn't too late
When the first reaction equation is:
AgI(S) ↔ Ag+(Aq) + I-(Aq)
So, the Ksp expression = [Ag+][I-]
∴Ksp = [Ag+][I-] = 8.3 x 10^-17
Then the second reaction equation is:
Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+
So, Kf expression = [Ag(NH3)2+] / [Ag+] [NH3]^2
∴Kf = [Ag(NH3)2+] /[Ag+] [NH3]^2 = 1.7 x 10^7
by combining the two equations and solve for Ag+:
and by using ICE table:
AgI(aq) + 2NH3 ↔ Ag(NH3)2+ + I-
initial 2.5 0 0
change -2X +X +X
Equ (2.5-2X) X X
so K = [Ag(NH3)2+] [I-] / [NH3]^2
Kf * Ksp = X^2 / (2.5-2X)
8.3 x 10^-17 * 1.7 x10^7 = X^2 / (2.5-2X) by solving for X
∴ X = 5.9 x 10^-5
∴ the solubility of AgI = X = 5.9 x 10^-5 M
