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olya-2409 [2.1K]
2 years ago
9

Dipole-dipole interactions are (weaker than, stronger than, equal to) hydrogen bonds.

Chemistry
1 answer:
kozerog [31]2 years ago
7 0

Answer and Explanation:

Dipole-Dipole interactions are <u>weaker than</u> hydrogen bonds.

Hydrogen bonds are a form of dipole-dipole interactions, being the strongest form of dipole-dipole interactions.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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The heat of vaporization for benzaldehyde is 48.8 kj/mol, and its normal boiling point is 451.0 k. use this information to deter
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The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

Explanation:

  • To solve this problem, we use Clausius Clapeyron equation: ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂).
  • The first case: P₁ = 1 atm = 760 torr and T₁ = 451.0 K.
  • The second case: P₂ = <em>??? needed to be calculated</em> and T₂ = 61.5 °C = 334.5 K.
  • ΔHvap = 48.8 KJ/mole = 48.8 x 10³ J/mole and R = 8.314 J/mole.K.
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  • ln(760 torr /P₂) = (48.8 x 10³ J/mole / 8.314 J/mole.K) (1/451 K - 1/334.5 K)
  • ln(760 torr /P₂) = (5869.62) (-7.722 x 10⁻⁴) = -4.53.
  • (760 torr /P₂) = 0.01075
  • Then, P₂ = (760 torr) / (0.01075) = 70691.73 torr.

So, The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

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