What types of intermolecular forces exist between hydrogen fluoride molecules? L

Question

What types of intermolecular forces exist between hydrogen fluoride molecules? London forces; dipole-dipole interactions; hydrogen bonding; ion-dipole interactions I, II, and III II and III I only I and III III only The molar enthalpy of vaporization of hexane (C_6H_14) is 28.9 hj/mol, and its normal boiling point is 68.73 degree What is the vapor pressure of hexane at 22 degree C (R = 8.314 J/K mol)? 20 mmHg 656 mmHg 152 mmHg 790 mmHg 683 mmHg How much energy (heat) is required to convert 52.3 g of ice at -10.0 degree C to steam at 100 degree C? specific heat of ice: 2.09 J/g- degree C delta H_fus = 6.02 kJ/mol specific heat of water 4.18 J/g degree C delta H_vap = 40.7 kJ/mol specific heat of steam: 1.84 J/g degree C 1.59E2 kJ 1.24E2 kJ 1.15E2 kJ 1.13E2 kJ 7.99E1 kJ A certain solid metallic clement has a density 7.87 g/cm^3 and a molar mass of 55.85 g mol. It crystallizes with a cubic unit cell, with an edge length of 286.7 pm. Calculate the number of atoms per unit cell. 1 4 6 3 2

Explanation / Answer

1) Being the most electronegative element (F) and a bonded hydrogen with it, it will form hydrogen bond

dipole dipole interactions are present due to development of charge

London forces are present in all compound

2) the relation between enthalpy of vapourisation and boiling point temperature will be

we will use clausius clapyeron equation

ln(P2/P1) = delta H / R [ 1/T1 - 1/T2]

ln (P2 / 760) = 28.9 X 10^3 / 8.314 [1/295 - 1/ 341.73]

ln (760/ P2) = 3476.04 [ 0.00339 - 0.00292 ]

ln (760/ P2) = 1.633

taking antilog

(760 / P2) = 5.12

P2 = 148.4

approx 152

3) the energy required will be

a) energy required for heating ice from -10 to 0 C

Q1 = mass X specific heat X change in temperature = 52.3 X 2.09 X (10) = 1093.07 Joules = 1.093 KJ

Heat required for conversionof ice to water

Enthalpy of fusion = 6.02 / 18 KJ / g = 0.334 KJ / g

Q2 = mass X enthalpy of fusion = 52.3 X 0.334 = 17.47 KJ

Heat required for heating water to 100 C (from 0)

Q3 = Mass X specific heat X change in temperature = 52.3 X 4.18 X 100 = 21882.32 Joules = 21.882 KJ

Heat required to convert water to steam

Q4 = Mass X enthalpy of vapourisation

Ethalpy of vapourisation = 40.7 KJ / mole = 40.7 / 18 KJ /g = 2.26 KJ / g

Q4 = 52.3 X 2.26 = 118.198 KJ

total heat = Q1 + Q2 + Q3 + Q4

Total heat = 1.093 + 17.47 + 21.882 + 118.198 = 158.64 = 1.59 X 10^2 KJ

4) the formula is

density = Z X M / Na X a^3

Z = atoms per unit cell

Z = 7.87 X 6.023 X 10^23 X (286.7 X 10^-10)^3 / 55.85 = 2.00 X 10^7 X 10^23 X 10^-30 = 2

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