Answer is: Ka for propanoic acid is 6,57·10⁻⁵.
Chemical reaction: C₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).
n(C₂H₅COOH) = 0,04 mol.
V(C₂H₅COOH) = 750 mL = 0,75 L.
c(C₂H₅COOH) = 0,04 mol ÷ 0,75 L.
c(C₂H₅COOH) = 0,053 mol/L = 0,053 M.
[C₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [</span>C₂H₅COO⁻] ·
[H₃O⁺] / [C₂H₅COOH].
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10⁻⁵.
Answer:
H + ions are attracted to the cathode , gain electrons and form hydrogen gas. OH - ions are attracted to the anode , lose electrons and form oxygen gas.
Answer:
B
Explanation: Supersaturated
Answer:
kL, dL, mL, pL
Explanation:
pL stands for picoliters. This is equal to 1 × 10⁻¹² (0.000000000001) liters.
mL stands for milliliters. This is equal to 1 × 10⁻³ (0.001) liters.
dL stands for deciliters. This is equal to 1 × 10⁻¹ (0.1) liters.
kL stands for kiloliters. This is equal to 1 × 10³ (1000) liters.
Answer: option B. The number of the outer most energy level for elements in that period
Explanation:
