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The heat of vaporizationof nitrogen is 6,028.25 J/mol.The normal boiling pointof nitrogen is 77.58 K.The table for the question is in the attachment. Using theClausius-Clapeyron Equationto estimate the heat of vaporization.ln (P₁/P₂) = ΔH/R × (1/T₂ - 1/T₁)whereP = the pressureP₁ = 130.5 torrP₂ = 289.5 torrΔH = the heat of vaporizationR = constant gas = 8.314 J/mol KT = the temperature (K)T₁ = 65 KT₂ = 70 Kln (130,5/289,5) = ΔH/R × (1/70 - 1/65)ln 0.45078 × 8.314 = ΔH × ((65 - 70)/4550)- 0.79678 × 8.314 = ΔH × (- 5/4550)- 6.6244 = ΔH × - 1/910ΔH = - 6.6244 × - 910ΔH = 6,028.2 J/molThenormal boiling pointreaches when pressureis 1 atm or 760 torr.P₁ = 760 torrP₂ = 1,718 torrT₁ = T KT₂ = 85 Kln (760/1,718) = (6,028.2/8.314) × (1/85 - 1/T)ln 0.4424 = 725.0662 × ((T - 85)/85T)-0.81554 ÷ 725.0662 = ((T - 85)/85T)- 0.001125 × 85T = T - 85- 0.0956T = T - 85- 0.0956T - T = 85- 1.0956T = - 85T = - 85 ÷ - 1.0956T = 77.58 KLearn more aboutClausius-Clapeyron Equationhere:brainly.com/question/29414397#SPJ4...

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