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:
44.2 mmHg
Explanation:
We have to apply the colligative property of lowering vapor pressure:
P° - P' = P° . Xm
where P' refers to vapor pressure of solution
P° refers to vapor pressure of pure solvent
Xm is the mole fraction of solute
Let's determine the mole fraction (moles of solute / Total moles)
Total moles = Moles of solute + Moles of solvent
Moles of solute → 15 g . 1 mol/18g = 0.833 moles
Moles of solvent → 100 g . 1mol / 46 g = 2.174 moles
Total moles = 0.833 + 2.174 = 3.007
Xm = 0.833 / 3.007 = 0.277
We replace data in the formula: 61.2 mmHg - P' = 61.2 mmHg . 0.277
P' = - (61.2 mmHg . 0.277 - 61.2 mmHg) → 44.2 mmHg
I believe that the answer would be D, both the speaker and the microphone
