I don't know man hahahaha
The molar concentration of the potassium hydroxide solution was 0.815 mol/L.
<em>Balanced equation</em>: 2KOH + H_2SO_4 → K_2SO_4 + 2H_2O
<em>Moles of H_2SO_4</em>:
15.1 mL H_2SO_4 × (1.50 mmol H_2SO_4 /1 mL H_2SO_4)
= 22.65 mmol H_2SO_4
<em>Moles of KOH</em>: 22.65 mmol H_2SO_4× (2 mmol KOH/2 mmol H_2SO_4)
= 45.30 mmol KOH
<em>Concentration of KOH</em>: c= "moles"/"litres" = 45.30 mmol/55.6 mL
= 0.815 mol/L
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: 4.0 g/mL
Explanation:
The volume increased by 5.0 mL. Recall that the number of significant figures is equal to the number of certain values you can read plus one. Here, the volume increased from 14.0 mL to 19.0 mL, so the volume of X is 5.0 mL.