Answer:
2 mole of C4H10 were used in this reaction.
Answer:
11.7 mL
Explanation:
Density of a substance is given by the mass of the substance divided by the volume of the substance .
Hence , d = m / V
V = volume
m = mass ,
d = density ,
From the question ,
The density of the metal = 10.5 g/cm³
The mass of the metal = 5.25 g
Hence , the volume can be calculated from the above formula , i.e. ,
d = m / V
V = m / d
V = 5.25 g / 10.5 g/cm³
V = 0.5 cm³
Since , The unit 1 mL = 1 cm³
V = 0.5 mL
The vessel has 11.2 mL of water ,
The new volume becomes ,
11.2 mL + 0.5 mL = 11.7 mL
Just multiple 19.3 by 15 because cm^3 will cancel outand you are left with g with represents mass
0.04350179862
Explanation:
Make sure to check for sig figs though. Basically, you have 2.54 mL and you multiply that by the 0.789 g/mL and mL cancels out and you are left with 2.00406g. There are 46.06844 grams per mol of ethanol, so to cancel out grams we multiply by 1/46.06844. And then we are left with 0.04350179862 mol
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