Answer:
Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]
Explanation:
According to Brönsted-Lowry acid-base theory:
- An acid is a substance that donates H⁺.
- A base is a substance that accepts H⁺.
When methylamine reacts with water, it behaves as a Brönsted-Lowry base, according to the following reaction.
CH₃NH₂(aq) + H₂O(l) ⇄ CH₃NH₃⁺(aq) + OH⁻(aq)
The basic equilibrium constant (Kb) is:
Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]
Answer: It is important to wet the filter paper in the Buchner funnel first with cold re crystallization solvent before the re crystallization mixture being filtered to minimize gaps around the edges of the filter paper which can prevent mechanical impurities from passing through. This gives better filtration where most impurities can be filtered. Furthermore, it provides good vacuum.
Answer:
No
Explanation:
In ideal solutions, the interactions between solute - solvent are approximately the same as those of solute - solute and solvent - solvent, that is the interactions are to be practically indistintiguishable after disolution.
The moment we have a release of energy (the solution feels warm) we are to conclude that there are strong interactions between the water and methanol molecules so we would expect the solution to be non ideal.
The reason for the interactions is the presence of hydrogen bonds between methanol and water.
The volume of O₂ produced: 84.6 L
<h3>Further explanation</h3>
Given
7.93 mol of dinitrogen pentoxide
T = 48 + 273 = 321 K
P = 125 kPa = 1,23365 atm
Required
Volume of O₂
Solution
Decomposition reaction of dinitrogen pentoxide
2N₂O₅(g)→4NO₂(g)+O₂ (g)
From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :
= 0.5 x mol N₂O₅
= 0.5 x 7.93
= 3.965 moles
The volume of O₂ :
Answer:
- <em>Oxidation half-reaction</em>:
Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻
- <em>Reduction half-reaction</em>:
Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)
Explanation:
The reaction that takes place is:
- Fe²⁺(aq) + Ce⁴⁺(aq) → Fe³⁺(aq) + Ce³⁺(aq)
The <em>oxidation half-reaction</em> is:
- Fe²⁺(aq) → Fe³⁺(aq) + 1e⁻
It is an oxidation because the oxidation state of Fe increases from 2+ to 3+.
The <em>reduction half-reaction</em> is:
- Ce⁴⁺(aq) + 1e⁻ → Ce³⁺(aq)
It is a reduction because the oxidation state of Ce decreases from 4+ to 3+.
