Use the data provided to calculate benzaldehyde's heat of vaporization smartwork5
1 answer:
This problem is incomplete. Luckily, I found a similar problem from another website shown in the attached picture. The data given can be made to use through the Clausius-Clapeyron equation:
ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁)
where
P₁ = 14 Torr * 101325 Pa/760 torr = 1866.51 Pa
T₁ = 345 K
P₂ = 567 Torr * 101325 Pa/760 torr = 75593.78 Pa
T₂ = 441 K
ln(75593.78 Pa/1866.51 Pa) = (-ΔHvap/8.314 J/mol·K)(1/441 K - 1/345 K)
Solving for ΔHvap,
<em>ΔHvap = 48769.82 Pa/mol or 48.77 kPa/mol</em>
ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁)
where
P₁ = 14 Torr * 101325 Pa/760 torr = 1866.51 Pa
T₁ = 345 K
P₂ = 567 Torr * 101325 Pa/760 torr = 75593.78 Pa
T₂ = 441 K
ln(75593.78 Pa/1866.51 Pa) = (-ΔHvap/8.314 J/mol·K)(1/441 K - 1/345 K)
Solving for ΔHvap,
<em>ΔHvap = 48769.82 Pa/mol or 48.77 kPa/mol</em>
You might be interested in
The final volume of the gas is 73.359 mL
<h3 />Given :
A sample gas has an initial volume of 72.0 mL
The work done = 141.2 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since, the process does 141.2 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm; we have
External Pressure P:
= 783 torr ×
= 1.03 atm
The work done W =
The change in volume ΔV=
ΔV =
ΔV =
ΔV = 0.001359 L
ΔV = 1.359 mL
The initial volume = 72.0 mL
The change in volume V is ΔV = V₂ - V₁
- V₂ = - ΔV - V₁
multiply both sides by (-), we have:
V₂ = ΔV + V₁
= 1.359 mL + 72.0 mL
= 73.359 mL
Therefore, the final volume of the gas is 73.359 mL .
Learn more about volume here:
#SPJ4
Answer:
<span>The main interactions between molecules of hydrogen chloride are examples of Dipole-Dipole Interactions.
Explanation:
The bond between Hydrogen and Chlorine is Polar Covalent Bond as the electronegativity difference between these two elements is 0.96 which is greater than 0.4. Chlorine being more electronegative attracts the electrons from Hydrogen making the Hydrogen partial positive and itself partial negative. The two poles on HCl makes it a dipole. Now, one HCl (a dipole) interacts with another HCl (another Dipole) through their opposite charges and creates Dipole-Dipole Interaction. The picture is as below, the green dashed lines are interactions,
</span>
<span>The main interactions between molecules of hydrogen chloride are examples of Dipole-Dipole Interactions.
Explanation:
The bond between Hydrogen and Chlorine is Polar Covalent Bond as the electronegativity difference between these two elements is 0.96 which is greater than 0.4. Chlorine being more electronegative attracts the electrons from Hydrogen making the Hydrogen partial positive and itself partial negative. The two poles on HCl makes it a dipole. Now, one HCl (a dipole) interacts with another HCl (another Dipole) through their opposite charges and creates Dipole-Dipole Interaction. The picture is as below, the green dashed lines are interactions,
</span>