<h3>
Answer:</h3><h3>
</h3><h2>E. Coli </h2>
<h2>its a easy and faster answer to look at (;</h2>
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:
Order zero
Explanation:
Let's consider the decomposition of ammonia to nitrogen and hydrogen on a tungsten filament at 800°C.
2 NH₃(g) → N₂(g) + 3 H₂(g)
The generic rate law is:
rate = k × [NH₃]ⁿ
where,
rate: reaction rate
k: rate constant
n: reaction order
When n = 0, we get:
rate = k × [NH₃]⁰ = k
As we can see, when the reaction order with respect to ammonia is zero, the reaction rate is independent of the concentration of ammonia.
Answer:
E
Explanation:
Dry weather with Mostly Clear sky’s
Answer:
There are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.
Explanation:
We can calculate the number of moles (η) of BeO as follows:
Where:
m: is the mass = 250 g
M: is the molar mass = 25.0116 g/mol
Hence, the number of moles is:
Therefore, there are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.
I hope it helps you!