ch10 - 10 Liquids Solids and Changes of State 10.1 μ = Q x...

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Unformatted text preview: 10 Liquids, Solids, and Changes of State 10.1 μ = Q x r = (1.60 x 10 19 C)(92 x 10 12 m) • m C 10 x 3.336 D 1 30 _ = 4.41 D % ionic character for HF = D 4.41 D 1.82 x 100% = 41% HF has more ionic character than HCl. HCl has only 17% ionic character. 10.2 (a) SF 6 has polar covalent bonds but the molecule is symmetrical (octahedral). The individual bond polarities cancel, and the molecule has no dipole moment. (b) H 2 C=CH 2 can be assumed to have nonpolar C–H bonds. In addition, the molecule is symmetrical. The molecule has no dipole moment. (c) The C–Cl bonds in CHCl 3 are polar covalent bonds, and the molecule is polar. (d) The C–Cl bonds in CH 2 Cl 2 are polar covalent bonds, and the molecule is polar. 10.3 10.4 The N atom is electron rich (red) because of its high electronegativity. The H atoms are electron poor (blue) because they are less electronegative. 10.5 (a) Of the four substances, only HNO 3 has a net dipole moment. (b) Only HNO 3 can hydrogen bond. (c) Ar has fewer electrons than Cl 2 and CCl 4 , and has the smallest dispersion forces. 10.6 H 2 S dipole-dipole, dispersion CH 3 OH hydrogen bonding, dipole-dipole, dispersion 229 C 2 H 6 dispersion Ar dispersion Ar < C 2 H 6 < H 2 S < CH 3 OH 10.7 (a) CO 2 (s) → CO 2 (g), ΔS is positive. (b) H 2 O(g) → H 2 O(l), ΔS is negative. (c) ΔS is positive (more disorder). 10.8 ΔG = ΔH TΔS; at the boiling point (phase change), ΔG = 0. ΔH = TΔS; T = mol) kJ/(K 10 x 87.5 kJ/mol 29.2 = S H 3 _ vap vap • ∆ ∆ = 334 K 10.9 The boiling point is the temperature where the vapor pressure of a liquid equals the external pressure. P 1 = 760 mm Hg; P 2 = 260 mm Hg; T 1 = 80.1 o C ΔH vap = 30.8 kJ/mol ∆ T 1 _ T 1 R H + P ln = P ln 2 1 vap 1 2 T 1 _ T 1 = H R ) P ln _ P (ln 2 1 vap 1 2 ∆ Solve for T 2 (the boiling point for benzene at 260 mm Hg). T 1 = H R ) P ln _ P (ln _ T 1 2 vap 1 2 1 ∆ T 1 = J/mol 30,800 mol K J 8.3145 ln(760)] _ ) 260 [ln( _ K 353.2 1 2 • T 1 2 = 0.003 121 K 1 ; T 2 = 320 K = 47 o C (boiling point is lower at lower pressure) 10.10 ΔH vap = T 1 _ T 1 )(R) P ln _ P (ln 2 1 1 2 P 1 = 400 mm Hg; T 1 = 41.0 o C = 314.2 K P 2 = 760 mm Hg; T 2 = 331.9 K ΔH vap = • K 331.9 1 _ K 314.2 1 mol K J 8.3145 (400)] ln _ (760) [ln = 31,442 J/mol = 31.4 kJ/mol 10.11 (a) 1/8 atom at 8 corners and 1 atom at body center = 2 atoms (b) 1/8 atom at 8 corners and 1/2 atom at 6 faces = 4 atoms 230 10.12 For a simple cube, d = 2r; r = 2 pm 334 = 2 d = 167 pm 10.13 For a simple cube, there is one atom per unit cell....
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This note was uploaded on 08/27/2008 for the course CHEM 1280 taught by Professor Schmidt during the Fall '07 term at Toledo.

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ch10 - 10 Liquids Solids and Changes of State 10.1 μ = Q x...

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