Answer:
0.33 mol/kg NH₃
Explanation:
Data:
b(NH₃) = 0.33 mol/kg
b(Na₂SO₄) = 0.10 mol/ kg
Calculations:
The formula for the boiling point elevation ΔTb is
i is the van’t Hoff factor — the number of moles of particles you get from a solute.
(a) For NH₃,
The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.
1 mol NH₃ ⟶ 1 mol particles
i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water
(b) For Na₂SO₄,
Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)
1 mol Na₂SO₄ ⟶ 3 mol particles
i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water
The NH₃ has more moles of particles, so it has the higher boiling point.
Answer:
A. non renewable energy sources cannot be used over again and renewable energy sources can be used again
Answer:
2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)
Explanation:
The unbalanced equation is
Al(s) + Ni²⁺(aq) ⟶ Ni(s) + Al³⁺(aq)
(i) Half-reactions
Al(s) ⟶ Al³⁺(aq) + 3e⁻
Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)
(ii) Balance charges
2 × [Al(s) ⟶ Al³⁺(aq) + 3e⁻]
3 × [Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)]
gives
2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻
3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s)
(iii) Add equations
2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻
<u>3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s) </u>
2Al(s) +3Ni²⁺(aq) + <em>6e</em>⁻ ⟶ 2Al³⁺(aq) + 3Ni(s) + <em>6e⁻
</em>
Simplify (cancel electrons)
2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)
Answer:
YASSSSSSS
Explanation:
i think the answer is true
