Answer:
Anode: H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻
Cathode: 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq)
E° = 1.60 V
Explanation:
Let's consider the reaction taking place in a galvanic cell.
2 Fe⁺³(aq) + H₂(g) + 2 OH⁻(aq) → 2 Fe⁺²(aq) + 2 H₂O(l)
The corresponding half-reactions are:
Anode (oxidation): H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻ E°red = - 0.83 V
Cathode (reduction): 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq) E°red = 0.77 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.77 V - (-0.83 V) = 1.60 V
Answer is: <span> two samples have in common same amount of substance and same number of particles.
1) There are same amount of substance in both beakers:
n(Zn) = 1 mol.
n(ZnCl</span>₂) = 1 mol.
2) There are same number of particles (atoms, molecules, ions) in both beakers:
N(Zn) = n(Zn) · Na.
N(Zn) = 1 mol · 6.023·10²³ 1/mol = 6.023·10²³ atoms of zinc.
N(ZnCl₂) = n(ZnCl₂) · Na.
N(ZnCl₂) = 1 mol · 6.023·10²³ 1/mol = 6.023·10²³ molecules of zinc(II) chloride.
Na - Avogadro number.
a is the answer because all of the other answers are wrtong
1. Q=112.8 kJ
2. Q=5.01 kJ
<h3>Further explanation</h3>
The heat required for phase change :
Q = mLf
Lf=latent heat of fusion
- vaporization/condensation
Q = mLv
Lv=latent heat of vaporization
1.
m=50 g=0.05 kg
Lv (water) = 2256 kJ/kg
2.
m=15 g=0.015 kg
Lf for water = 334 kj/kg
