Answer:
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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:
1. Classical conditioning. Classical conditioning is a type of associative learning based on the association between a neutral stimulus with another that is significant for a person or an animal in order to generate a similar response.
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The conversion factor between moles of calcium ion and moles of 
in would be 3 to 1.
<h3>Mole fraction</h3>
if formed from 3 carbon atoms and 3 phosphate atoms according to the equation:
Thus, there are 3 moles of calcium in every 1 mole of .
The conversion factor between moles of calcium and moles of will. therefore, be 3 to 1 or simply 3:1.
More on mole fractions can be found here: brainly.com/question/8076655
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