Answer:
A) E° = 4.40 V
B) ΔG° = -8.49 × 10⁵ J
Explanation:
Let's consider the following redox reaction.
2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)
We can write the corresponding half-reactions.
Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E°red = 1.36 V
Anode (oxidation): 2 Li(s) → 2 Li⁺(aq) + 2 e⁻ E°red = -3.04
<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>
The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V
<em>B) Calculate the free energy ΔG° of the reaction.</em>
We can calculate Gibbs free energy (ΔG°) using the following expression.
ΔG° = -n.F.E°
where,
n are the moles of electrons transferred
F is Faraday's constant
ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J
Answer:
The symbol of each element is, Ne, Na, Mg, and Al.
Explanation:
Below is the list of elements that has an atomic mass of less than 19.3 u.
The atomic mass of Neon is 20.1797 u and the atomic number is 10.
The atomic mass of Sodium is 22.989769 u and the atomic number is 11.
The atomic mass of Magnesium is 24.305 u and the atomic number is 12.
The atomic mass of Aluminium is 26.981539 u and the atomic number is 13.
Here, the symbol of each element is, Ne, Na, Mg, and Al.
I'm not so sure but I would say Answer Choice B
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
Answer:
Explanation:
Data:
Mass of NaCl = 4.6 g
Mass of water = 250 g
Calculations:
Mass of solution = mass of NaCl + mass of water = 4.6 g + 250 g = 254.6 g.
