The answer to the question is A.
<span>Answer
is: Ka for propinoic acid is 6,57·10</span>⁻⁵.<span>
Chemical reaction: C</span>₂H₅COOH(aq)
+ H₂O(l) ⇄ C₂H₅COO⁻(aq)
+ H₃O⁺(aq).<span>
n(C</span>₂H₅COOH) = 0,04 mol.<span>
V(C</span>₂H₅COOH) = 750 mL = 0,75 L.<span>
c(C</span>₂H₅COOH) = 0,04 mol ÷ 0,75 L.<span>
c(C</span>₂H₅COOH) = 0,053 mol/L = 0,053 M.<span>
[C</span>₂H₅COO⁻]
= [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [C</span>₂H₅COO⁻] · [H₃O⁺] /
[C₂H₅COOH].<span>
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10</span>⁻⁵.
The impact of speed and velocity at greater speeds affects the car much greater than at slower speeds due to impact and timing and velocity and speed.