Answer:
[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52
Explanation:
Kb of the reaction:
NH3 + H2O(l) ⇄ NH4+ + OH-
Is:
Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]
<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>
<em>[NH₄⁺] = [OH⁻] = X</em>
<em>And as </em>[NH₃] = 0.619M
1.8x10⁻⁵ = [X] [X] / [0.619M]
1.11x10⁻⁵ = X²
3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]
<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />
% ionization:
[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%
pH:
As pOH = -log [OH-]
pOH = 2.48
pH = 14 - pOH
<h3>pH = 11.52</h3>
Answer:
ionic bond, also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound. Such a bond forms when the valence (outermost) electrons of one atom are transferred permanently to another atom.
Explanation:
Answer is: molar mass
of compound is 154,58 g/mol.<span>
m(</span>naphthalene<span>) = 10 g = 0,01 kg.
m(unknown compound) = 1,00 g.
</span>Δ<span>T (solution) = 4,47 °C.
Kf(</span>naphthalene) = 6,91°C/m<span>; cryoscopic
constant.
M</span>(unknown compound) = Kf(naphthalene)· m(unknown compound) ÷
m(naphthalene)<span> · ΔT(solution).
M(xylene) = </span>6,91°C/m<span> · 1 g ÷ 0,01 kg · 4,47</span>°C<span>.
M(xylene) = 154,58 g/mol.</span>
Answer:
Showing respect
Explanation:
Being first at everything is immature, playing on a cell phone is being off task, and using a fire blanket to keep warm in class shows that you are doing nothing to contribute to the lab.
