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solid liquid

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York University
CHEM 1000
William Pietro

th Petrucci 10 ed Problem set #5: Structure Page 1 of 4 th Petrucci 10 Edition Problem Set #5: Solids and Liquids (Structural) Chapter 12 12-1 a) HCl is not a very heavy diatomic molecule; London forces are expected to only be moderately important. Hydrogen bonding is weak in the case of H-Cl bonds; Cl is not one of the three atoms (F, O, N) that from strong hydrogen bonds. Finally, because Cl is an electronegative atom, and H is only moderately electronegative, dipole-dipole interactions should be relatively strong. b)In Br2 neither hydrogen bonds nor dipole-dipole attractions are important; there are no H atoms in the molecule, and homonuclear molecules are nonpolar. London forces are more important than in HCl since Br 2 is heavier. c) In ICl there are no hydrogen bonds since there are no H atoms in the molecule. The London forces are as strong as in Br 2 since the two molecules have the same number of electrons. However, dipole-dipole interactions are important in ICl; the molecule is polar toward Cl. d) In HF London forces are not very important; the molecule has only 10 electrons. Hydrogen bonding is quite important and definitely overshadows even the strong dipole-dipole interactions. e) In CH 4, H bonds are not important; the H atoms are not bonded to F, O, or N. In addition the molecule is not polar, so there are no dipole-dipole interactions. Finally, London forces are quite weak since the molecule contains only 10 electrons, and this is why CH 4 is a gas with low critical temperature. 12-22 nacetonitrileRT -1 -1 nacetonitrile.00 atmx1.17 L)(0.08206 L atm mol K x354.75 K) = 0.0402 mol Therefore, H vap = (1.00 kJ 0.0402 mol) = 24.9 kJmol acetonitrile -1 -3 12-29 moles of Br 2 0.486159.8 g mol = 3.04 x 10 mol Now, -3 -1 -1 P = nRTV = (3.04 x 10 mol x 0.08206 L atm mol K x 298 K)0.2500L = 0.297 atm x (760 mmHg1 atm) = 226 mmHg 12-36 The final pressure of the gas mixture remains constant at 742 mmHg even though the gas mixture is now saturated with CCl so t4e partial pressure of N 2 must decrease. P totalP N2+ P CCl4 P N2 = Ptotal PCCl4= (742 - 481)mmHg = 261 mmHg Since the temperature does not change, P N2 totalV N2 total V totalV N2x P totalN2= 7.53 L x (742 mmHg481 mmHg) = 11.6 L
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