Answer:
The correct answer is
e. NH3(aq) + H+(aq) --> NH4+(aq)
Explanation:
To solve this, we write out the indidual ionization reation for aqueous ammonia and nitric acid thus
For aqueous ammonia we have
NH₃(aq) + H₂O(l) ↔ NH₄⁺(aq) + OH⁻(aq)
Aqueous ammonia is a weak base and therefore undergoes partial ionization hence the reversible reaction sign
As the level of ionization will not be more than 5% OH⁻ cannot represebt the weak base
For nitric acid we have
HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)
a strong acid like nitric acid undergoes conplete ionization in the solution
The total equation is NH₃(aq) + HNO₃(aq) → NH₄NO₃(aq)
The sum of the ionic equation is
NH₃(aq) + H⁺(aq) + NO₃⁻(aq) → NH₄⁺(aq) + NO₃⁻(aq)
The ionic equation is
NH₃(aq) + H⁺(aq) → NH₄⁺(aq)
Answer:
The answer to your question is U-234
Explanation:
Data
Thorium-234
beta emission twice
Definition
Beta emission is when a beta particle (electron) is emitted from an atomic nucleus.
First beta emission
²³⁴₉₀Th ⇒ ²³⁴₉₁Pa + e⁻
Second beta emission
²³⁴₉₁ Pa ⇒ ²³⁴₉₂U + e⁻
The atom will be Uranium-324
Moles KClO₃ = 0.239
<h3>Further explanation</h3>
Given
Reaction
2KClO₃(s) ⇒2KCl(s) + 3O₂(g)
P water = 23.8 mmHg
P tot = 758 mmHg
V = 9.07 L
T = 25 + 273 = 298 K
Required
moles of KClO₃
Solution
P tot = P O₂ + P water
P O₂ = P tot - P water
P O₂ = 758 - 23.8
P O₂ = 734.2 mmHg = 0.966 atm
moles O₂ :
n = PV/RT
n = 0.966 x 9.07 / 0.082 x 298
n = 0.358
From equation, mol ratio KClO₃ : O₂ = 2 : 3, so mol KClO₃ :
= 2/3 x mol O₂
= 2/3 x 0.358
= 0.239
