2.6394*10^22 is the answer
Just multiply 151 x 0.0001
That will equal:
0.00151 km or if you want to round it then it would be 0.02km
Answer:
A) 6.48 g of OF₂ at the anode.
Explanation:
The gas OF₂ can be obtained through the oxidation of F⁻ (inverse reaction of the reduction presented). The standard potential of the oxidation is the opposite of the standard potential of the reduction.
H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻ E° = -2.15 V
Oxidation takes place in the anode.
We can establish the following relations:
- 1 Faraday is the charge corresponding to 1 mole of e⁻.
- 1 mole of OF₂ is produced when 4 moles of e⁻ circulate.
- The molar mass of OF₂ is 54.0 g/mol.
The mass of OF₂ produced when 0.480 F pass through an aqueous KF solution is:
Answer:
ΔG = 98.67 kJ/mol
Explanation:
Let' s consider the following reaction.
Hg₂Cl₂(s) → Hg₂²⁺(aq) + 2Cl⁻(aq)
The standard Gibbs free energy (ΔG°) for the reaction is:
ΔG° = 1 mol × ΔG°f(Hg₂²⁺(aq)) + 2 mol × ΔG°f(Cl⁻(aq)) - 1 mol × ΔG°f(Hg₂Cl₂(s))
where,
ΔG°f: standard Gibbs free energy of formation
ΔG° = 1 mol × (154.72 kJ/mol) + 2 mol × (-134.08 kJ/mol) - 1 mol × (-215.06 kJ/mol)
ΔG° = 101.62 kJ
This is standard Gibbs free energy change per mole of reaction.
The Gibbs free energy of the reaction (ΔG) can be calculated using the following expression.
ΔG = ΔG° - R.T.lnQ
where,
R: ideal gas constant
T: absolute temperature
Q: reaction quotient
ΔG = ΔG° - R.T.ln([Hg₂²⁺].[Cl⁻]²)
ΔG = 101.62 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) . (298.2 K) . ln [(0.926).(0.573)²]
ΔG = 98.67 kJ/mol