<span>Answer is: Van't Hoff factor
(i) for this solution is 1.051 .
Change in boiling point from pure solvent to solution: ΔT
=i · Kb · b.
Kb - </span><span>molal boiling point elevation constant</span><span> is 0.512°C/m.
b - molality, moles of solute per kilogram of solvent.
b = 1.26 m.
ΔT = 101.63°C - 100</span>°C = 1.63°C.
i = 1.63°C ÷ (0.512°C/m · 1.26 m).
i = 1.051.
Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
Answer:
Pb²⁺ (aq) + 2I⁻ (aq) → PbI₂ (s)
General Formulas and Concepts:
- Solubility Rules
- Reaction Prediction
Explanation:
<u>Step 1: RxN</u>
Pb(NO₃)₂ (aq) + KI (aq) → PbI₂ (s) + KNO₃ (aq)
<u>Step 2: Balance RxN</u>
Pb(NO₃)₂ (aq) + 2KI (aq) → PbI₂ (s) + 2KNO₃ (aq)
<u>Step 3: Ionic Equations</u>
Total Ionic Equation:
Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + 2I⁻ (aq) → PbI₂ (s) + 2K⁺ (aq) + 2NO₃⁻ (aq)
<em>Cancel out spectator ions.</em>
Net Ionic Equation:
Pb²⁺ (aq) + 2I⁻ (aq) → PbI₂ (s)