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+.
Given:
1.50 L
62.5 grams
and the MM of MgO: 40.31 g/mol
Molarity: mol/L
First, find mol.
62.5 g x 1mole ÷ 40.31 g = 1.55 mol
then divide mol and the given liters
1.55mol ÷ 1.50 L= 1.03 M
Heat energy released : 167.2 kJ
<h3>Further explanation</h3>
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
m = 2000 g
c = 4.18 J/ g ° C
∆t = 20 ° C
Answer:
False.
Explanation:
From the balanced equation, 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
So for 0.40 moles of Mg(OH)2 we need 2*0.4 = 0.80 moles of HCl.
