# 12.E: Homework Chapter 12 Answers

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$

Chemical Species/Chemical Interactions:

1. a.) Cu(s)                Atomic

b.) H2O(s)         Molecular

c.) NaCl(s)          Ionic

d.) Fe(s)           Atomic

3. This mixture would be miscible because both molecules are polar.

5. A solution that mixes completely, without any particles left unmixed. Homogenous mixtures are most commonly thought of as a mixture behaving as a single substance.

Intermolecular Forces:

7.  a.) CH4       Dispersion

b.) HF         H-Bond

c.) H2O       H-Bond

d.) CHCl3   Dipole-Dipole

9. a.) CH3CH2OH   Dispersion, Dipole-Dipole, H-Bond

b.) CCl4               Dispersion

c.) CHF3              Dispersion, Dipole-Dipole

11. When a Hydrogen atom is bonded directly to a Nitrogen, Fluorine, or Oxygen atom.

13. CH4 , CH3CHO2 , HF

15. a.) CO Dispersion, Dipole-Dipole

b.) HBr Dispersion, Dipole-Dipole

c.) CCl4 Dispersion

d.) KCl ion-ion

17. c.) H-Bond

19. a.) ammonia                 H-Bond

b.) silicon dioxide       Dispersion

c.) ethanol                     H-Bond

d.) acetic acid               H-Bond

21. b.) Dipole-Dipole

23. a.) CH4

25. a.) I2                   Dispersion

b.) F2                  Dispersion

c.) SO2                       Dipole-Dipole

d.) HCH3COO   H-Bond

Boiling/Melting Point:

27. The molecule with the higher boiling point will be the molecule with the strongest intermolecular force. If they all have the same intermolecular force, then you will need to determine which molecule is the longest.

29. *remember: Ionic bonding is stronger than Hydrogen bonding!*

NaCl > CH3CH2OH > CHCl3 > Ar

31. At room temperature, the molecule with the stronger intermolecular force will be a liquid. CO2 only possesses dispersion forces, while CH3OH possesses hydrogen-bonding forces. Therefore, CH3OH will be a liquid at room temperature.

33. NaCl will have the higher melting point because ionic bonding between sodium and chlorine is stronger than the dispersion forces in methane.

35. a.) NH4Cl          *Remember: NH4+ and Cl- form an ionic bond*

37. c.) CH4

39. b.) HCl

41. HI > HBr > HCl

43. Octane > Hexane > Butane

45. No, ethanol and propane will not have the same boiling point because we have to put the intermolecular forces into account when deciding boiling points—not just their molar mass. Because ethanol has hydrogen-bonding forces, ethanol will have the higher boiling point.

47. a.) propanol and methanol

b.) methane and pentane

c.) dihydrogen monoxide and dihydrogen monosulfide

49. Acetic acid would most likely be a liquid at room temperature because of the strong hydrogen-bonding forces.

51. CH3CH2COH will have the higher boiling point because of the strong hydrogen-bonding forces present in the molecule.

53. b.) CH4

Cumulative/Challenge Problems:

55. No, a hydrogen-bond would not be able to form by connecting two acetone molecules together because the hydrogen atom would not be covalently bonded to the oxygen atom in the structure. For a hydrogen-bond to form, the hydrogen atom must be covalently bonded to either Nitrogen, Fluorine, or Oxygen atoms.

57. b.) vinegar

59. a.) acetone   Dipole-Dipole

b.) butane          Dispersion

c.) iso-butane   Dispersion

12.E: Homework Chapter 12 Answers is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.