Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :
where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:
Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
Answer:
C₂ = 1.26 mol/dm³
Explanation:
Given data:
Volume of HCl =V₁ = 15.7 cm³
Volume of LiOH = V₂ = 25 cm³
Concentration of HCl =C₁ = 2 mol/dm³
Concentration of LiOH =C₂= ?
Solution:
Chemical equation:
LiOH + HCl → LiCl + H₂O
Formula:
C₁V₁ = C₂V₂
by putting values,
15.7 cm₃× 2 mol/dm³ = C₂× 25 cm³
C₂ = 15.7 cm₃× 2 mol/dm³ / 25 cm³
C₂ = 31.4 cm₃.mol/dm³ / 25 cm³
C₂ = 1.26 mol/dm³
Answer:
b. E = 2,28V
Explanation:
The maximum work is the same than ΔG. As ΔG could be written as:
ΔG = nFE <em>(1)</em>
Where n is moles of electrons transferred, F is faraday constant (96485 J/Vmol) and E is the voltage of the cell.
For the reaction:
CH₃OH(l) + ³/₂O₂(g) → CO₂(g) + 2H₂O(l)
The oxidation state of C in CH₃OH is -2 but in CO₂ is +4, that means transferred electrons are +4 - -2 = <em>6e⁻</em>
Replacing in (1):
1320x10³ J = 6mol e⁻×96485J/Vmol×E
<em>E = 2,28V</em>
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I hope it helps!
