The freezing point depression of water is 1.86°C/m.kg
<span>K2S dissociates: </span>
<span>K2S ↔ 2K+ + S 2- </span>
<span>1mol K2S will produce 3 mol of ions. </span>
<span>Therefore in the freezing point equation: i = 3 </span>
<span>Depression of freezing point = Kf * i* m </span>
<span>Depression of freezing point = 1.86*3*0.195 </span>
<span>Depression of freezing point = 1.088°C </span>
<span>The solution will freeze at - 1.088°C </span>
Answer:
Explanation:
1)<u><em> Ionization equilibrium equation: given</em></u>
- H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
2) <em><u>Ionization equilibrium constant, at 25°C, Kw: given</u></em>
<u>3) Stoichiometric mole ratio:</u>
As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:
- [H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M
- ⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷ × 1.0 × 10⁻⁷ = 1.0 × 10⁻¹⁴ M
<u><em>4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C </em></u><em><u>and you need to calculate what the [H₃O⁺(aq)] is.</u></em>
Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:
Then you can substitute the known values and solve for the unknown:
- 1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M
- ⇒ [H₃O⁺] = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ M
As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.
<h3><u>Answer;</u></h3>
NH3/NH4+
<h3><u>Explanation;</u></h3>
From the equation;
NH3(aq)+HNO3(aq)→NH4+(aq)+NO3−(aq)
NH3 is the base; while NH4+ is the conjugate acid
HNO3 is the acid; while NO3- is the conjugate base
- The conjugate base of a Brønsted-Lowry acid is species that is formed after an acid donates a proton while the conjugate acid of a Brønsted-Lowry base is the species formed after a base accepts a proton.
Answer:
electrons move around the nucleus in fixed orbits of equal amounts of energy
Explanation:
Quantum Theory
