Answer:
The mass of PbSO4 formed 15.163 gram
Explanation:
mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625
mole of Na₂SO₄ = 2 x 0.025 = 0.05
Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃
( Mole/Stoichiometry )
= 0.0625 = 0.05
From (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.
Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄
= 0.05 x 303.26 g
= 15.163 g
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:
19.28 g/cm^3 to the nearest hundredth.
Explanation:
The volume of water displaced = the volume of the metal.
density = mass / volume
0.0694 kg = 0.0694 * 1000
= 69.4 g.
Density = 69.4 / 3.6
= 19.28 g/cm^3.