Answer:
63.9%
Explanation:
moral mass of x (# x ions) -divided by- Moral mass of compound.
% of cl = (35.5g) (2 cl ions) -divided by- 111.1g × 100
%= 63.9%
Answer:
6.5x10⁻³M = [OH⁻]
Explanation:
The Kb of a Weak base as ethylamine is expressed as follows:
Kb = [OH⁻] [C₂H₅NH₃⁺] / [C₂H₅NH₂]
As the equilibrium of ethylenamine is:
C₂H₅NH₂(aq) + H₂O(l) ⇄ C₂H₅NH₃⁺(aq) + OH(aq)
The concentration of C₂H₅NH₃⁺(aq) + OH(aq) is the same because both ions comes from the same equilibrium. Thus, we can write:
Kb = [OH⁻] [C₂H₅NH₃⁺] / [C₂H₅NH₂]
6.4x10⁻⁴ = [X] [X] / [C₂H₅NH₂]
Also, we can assume the concentration of ethylamine doesn't decrease. Replacing:
6.4x10⁻⁴ = [X] [X] / [0.066M]
4.224x10⁻⁵ = X²
6.5x10⁻³M = X
<h3>6.5x10⁻³M = [OH⁻]</h3>
<span>Pb(NO3)2(aq) + K2SO4(aq) → PbSO4(s) + 2 KNO3(aq) is </span>
Answer :]
This is an oxidation-reduction (redox) reaction:
S0 + 2 e- → S-II (reduction)
2 K0 - 2 e- → 2 KI (oxidation)
S is an oxidizing agent, K is a reducing agent.
