Answer:
5.7
Explanation:
(C₂H₅)₃NHCl dissociates according to the following equation.
(C₂H₅)₃NHCl ⇒ (C₂H₅)₃NH⁺ + Cl⁻
The molar ratio of (C₂H₅)₃NHCl to (C₂H₅)₃NH⁺ is 1:1. Then, the concentration of (C₂H₅)₃NH⁺ is Ca = 0.166 M.
(C₂H₅)₃NH⁺ is the conjugate acid of (C₂H₅)₃N. Given the Kb of (C₂H₅)₃N, we can calculate Ka for (C₂H₅)₃NH⁺ using the following expression.
Ka × Kb = Kw
Ka = Kw / Kb
Ka = 1.0 × 10⁻¹⁴ / 5.2 × 10⁻⁴
Ka = 1.9 × 10⁻¹¹
(C₂H₅)₃NH⁺ dissociates according to the following equation.
(C₂H₅)₃NH⁺ ⇄ (C₂H₅)₃N + H⁺
We can calculate [H⁺] using the following expression.
[H⁺] = √(Ca × Ka) = √(0.166 × 1.9 × 10⁻¹¹) = 1.8 × 10⁻⁶
The pH is:
pH = -log [H⁺] = -log 1.8 × 10⁻⁶ = 5.7
Answer:
D (hope I’m right, sorry if it’s not)
Answer: HF-dipole- dipole interaction
CH3OH- dipole-dipole interaction
CaCl2- ion-ion interaction
Explanation:
Both CH3OH and HF possess permanent dipoles which interact with water leading to the dissolution of the above named substances. Remember that water also possesses a permanent dipole. Which can interact with the dipoles on other polar molecules. CaCl2 is purely ionic and interacts with water via ion-dipole mechanism.
Hope the diagram below will help you. when a C=C is hydrogenated it gains 2 hydrogens turning it into a C-C.
Hope that helps
I gotchuuuuu it’s Isotopes