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 fuel value and the fuel density of pentane is, 49.09 kJ/g and 
respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:
Fuel value = 
Molar mass of pentane = 72 g/mol
Fuel value = 
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL =
Thus, the fuel density of pentane is
Answer:
250000 μL
Explanation:
If 1 L = 1000 mL
Then X L = 250 mL
X = (1 × 250) / 1000 = 0.25 L
Now we can calculate the number of microliters (μL) in 0.25 L:
if 1 μL = 10⁻⁶ L
then X μL = 0.25 L
X = (1 × 0.25) / 10⁻⁶ =250000 μL
Answer:
162 g Fe₂O₃
Explanation:
To find the mass of Fe₂O₃, you need to (1) convert grams C to moles C (via molar mass from periodic table), then (2) convert moles C to moles Fe₂O₃ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe₂O₃ to grams (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the given value.
Molar Mass (C): 12.011 g/mol
2 Fe₂O₃(s) + 3 C(s) ---> 4 Fe(s) + 3 CO₂(g)
Molar Mass (Fe₂O₃): 2(55.845 g/mol) + 3(15.998 g/mol)
Molar Mass (Fe₂O₃): 159.684 g/mol
18.3 g C 1 mole 2 moles Fe₂O₃ 159.684 g
-------------- x ---------------- x ------------------------- x ----------------- = 162 g Fe₂O₃
12.011 g 3 moles C 1 mole
