He normal boiling point of liquid butanol is 391 k. assuming that its molar heat of vaporization is constant at 42.3 kj/mol, the boiling point of c4h9oh when the external pressure is 1.39 atm is

401 K
   The equation that expresses the vapor pressure of a liquid is the Clapeyron equation:
 Tb = (1/T0 - (R ln(P/P0))/Hvap)^(-1)
 where
 Tb = boiling point at pressure of interest
 R = Ideal gas constant
 P = vapor pressure of the liquid at the pressure of interest
 P0 = known pressure at T0
 Hvap = heat of vaporization
 T0 = boiling temperature
   So let's substitute the known values and calculate.
 Tb = (1/T0 - (R ln(P/P0))/Hvap)^(-1)
 Tb = (1/391K - (8.3144598x10^-3 kJ/(K*mol) ln(1.39atm/1atm))/42.3 kJ/mol)^(-1)
 Tb = (1/391K - (8.3144598x10^-3 kJ/(K*mol) ln(1.39))/42.3 kJ/mol)^(-1)
 Tb = (1/391K - (8.3144598x10^-3 kJ/(K*mol)*0.329303747)/42.3 kJ/mol)^(-1)
 Tb = (1/391K - (2.73798276760652x10^-3 kJ/(K*mol))/42.3 kJ/mol)^(-1)
 Tb = (1/391K - (6.47277250025181x10^-5 1/K))^(-1)
 Tb = (2.49281703203073x10^-3 1/K))^(-1) Tb = 401.152586471767 K Rounding to 3 significant figures gives a boiling point of 401 K

You use the atomic number to find out how many protons and neutrons there are in the atom. you use the atomic mass to find out how many protons and electrons there are in the atom. 

Idon’t know what the answer is i wish i could