Calculate the vapor pressures of n-hexadecane at 15 °C (in bar and Pa) (1) by using the the boiling point and melting point Melting Point = 18 °C Boiling Point = 287°C (2) by using the Abraham solvation model
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Calculate the vapor pressures of n-hexadecane at 15 °C (in bar and Pa)
(1) by using the the boiling point and melting point
Melting Point = 18 °C
Boiling Point = 287°C
(2) by using the Abraham solvation model
Step by step
Solved in 4 steps with 4 images
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- Q.3 The elevation of boiling point and depression of freezing point of a certain solution of a monobasic acid HA in benzene are AT, and AT, respectively. HA exists usually as a monomer, but it dissociates into ions (with degree of dissociation a) at the boiling point T, of the solution, whereas at the freezing point T, of the solution, HA exists as (HA)n with a fixed value of n. The molal boiling point elevation constant and molal freezing point depression constant are K, and Kf, respectively. The correct expression for n is Ans X 1. n = [AT; K; (1+a)] / (AT, K, ) 2. n = [AT, K, (1+a)] / (AT; K; ) 3. n = (AT, K; ) / [AT, K, (1+a)) X 4. n = (AT, K, )/ [AT; K, (1+a)] %3DWhat is the molar fraction of glycine in an aqueous solution whose concentration is 0.140 molkg-1. Data: Mass H2O = 1 kg; M.M. H2O = 18.01 gmol-1.The generic structural formula for a 1-alkyl-3-methylimidazoliumcation is where R is a ¬CH21CH22nCH3 alkyl group. The meltingpoints of the salts that form between the 1-alkyl-3-methylimidazolium cation and the PF6- anion areas follows: R = CH2CH3 1m.p. = 60 °C2, R = CH2CH2CH31m.p. = 40 °C2, R = CH2CH2CH2CH3 1m.p. = 10 °C2, andR = CH2CH2CH2CH2CH2CH3 1m.p. = -61 °C2. Why doesthe melting point decrease as the length of alkyl groupincreases?
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