Chem II Homework Page, Exam 1 Material

Homework Page with Visible Answers

This page has all of the required homework for the material covered in the first exam of the second semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.

Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.

These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.

Intermolecular Forces, Liquids, and Solids (Chapters eleven and twelve)

  1. What is meant by a molecule that has a dipole?
    Answer
    The electrons have shifted to one side of the molecule.
  2. Which will have the strongest dipole (a) F2 or HF, (b) CH3Cl or CH3Br?
    Answer

    (a) HF

    (b) CH3Cl

  3. Which is more polarizable? (a) Cl2 or I2, (b) C2H6 or C10H22?
    Answer
    (a) I2

    (b) C10H22
  4. Ethanol, C2H5OH, has a boiling point of 78 °C while propane, C3H7OH, has a boiling point of 97 °C. Fully explain.
    Answer
    Propane is larger and will have more London Dispersion Forces giving it stonger intermolecular forces and requiring more energy to separate the molecules. Propane also has more mass and that also requires more energy to move them around and separate them. Both of these will contribute to a higher boiling point for propane.
  5. Propanol, CH3CH2OH, has a boiling point of 97 °C. Replacing the end CH3 with an OH gives ethylene glycol, HOCH2CH2OH, which has a boiling point of 197 °C. The masses are similar, so why is the boiling point of ethylene glycol so much higher than the boiling point of propanol?
    Answer
    The ethylene glycol can form hydrogen bonds on both ends of the molecule resulting in much stronger intermolecular forces and a higher boiling point.
  6. Which of the following materials is likely to have (a) no dipole-dipole forces, but the largest London dispersion forces, (b) the largest dipole-dipole intermolecular forces: I2, He, H2S, H2Te.
    Answer
    (a) I2, it is the largest nonpolar molecule.

    (b) H2S, S is more electronegative and will make the molecule more polar.
  7. What kind of attractive forces must be overcome to (a) sublime At2; (b) vaporize C2H5F; (c) boil hydrogen fluoride, HF; (d) melt LiBr? Explain.
    Answer
    (a) Nonpolar - London Dispersion Forces.

    (b) Polar - Dipole-dipole attractions (and dispersion forces).

    (c) Very polar - Hydrogen bonds between the molecules. Hydrogen bonds form when hydrogen is covalently bonded to N, O, or F.

    (d) No molecules, so there are no intermolecular forces - Ionic bonds.
  8. Rationalize the difference in boiling points between the members of the following pairs of substances (be sure to include all considerations):
  9. Which of the following materials is likely to have (a) no dipole-dipole forces, but the largest London dispersion forces, (b) the largest dipole-dipole intermolecular forces: I2, He, H2S, H2Te.
    Answer
    (a) I2, it is the largest nonpolar molecule.

    (b) H2S, S is more electronegative and will make the molecule more polar.
  10. Assume that liquid ammonia, NH3, has a specific heat of 4.75 J/g-C and that gaseous ammonia has a specific heat of 2.17 J/g-C. The heat of vaporization for ammonia is 23.35 kJ/mol at its boiling point of 33.4 C. (a) Draw the heating curve for converting 34 g of ammonia from a liquid at -40 °C to a gas at 0 °C. (b) Calculate the heat required to make the conversion described in part (a).
    Answer
  11. The normal melting and boiling points of benzene, C6H6, are 5.55 °C and 78.25 °C, respectively (at the normal melting and boiling point the pressure is 1 atm). Its triple point is at 5.51 °C and 0.047 atm. The critical point is at 288.95 °C and 47.9 atm. (a) Sketch the phase diagram for benzene (not necessarily in scale), showing the four points given above and indicating the area in which each phase is stable. (b) As it is heated, will solid benzene sublime or melt under a pressure of 2.5 atm? (c) Draw the heating curve for heating solid benzene until it is a gas at 1.0 atm.
    Answer
  12. Identify the following crystals (A-E) as ionic, polar-molecular, nonpolar-molecular, covalent (network), or metallic. Explain. There could be more than one or none of the crystal types in the chart.
    CrystalMelting Point (°C)Boiling Point (°C)Electrical Conductivity
      Solid          Liquid
    A-83.119.54NoNo
    B-259.14-252.5NoNo
    C15353000YesYes
    D6861330NoYes
    E-56.6-78.5NoNo
    F-182.48-164.8NoNo
    G35504827NoNo
    Answer

    A ⇒ good separation between melting and boiling points, relatively high m.p. for molecular, no conductivity - polar-molecular. (HF - polar molecule)

    B ⇒ close and low melting and boiling points, no condutivity - nonpolar-molecular. (H2 - nonpolar molecule)

    C ⇒ high mp and bp, good conductor - metallic. (Fe - metal)

    D ⇒ high melting and boiling points, conducts as a liquid - ionic. (KI - ionic compound)

    E ⇒ low and close together mp and bp, nonconductor - nonpolar-molecular. (CO2 - nonpolar molecule)

    F ⇒ low and close together mp and bp, nonconductor - nonpolar-molecular. (CH4 - nonpolar molecule)

    G ⇒ very high mp and bp, nonconductor - covalent (network). (Diamond - covalent network)

  13. Indicate the type of crystal (ionic, polar-molecular, nonpolar-moleculas, covalent, or metallic) and predict some of the properties (difference in melting and boiling points, conductivity, etc.) for each of the following substances upon solidification: HF, H2, Fe, KI, CO2, CH4, and diamond.
    Answer

    HF - polar molecule, good separation between melting and boiling points, relatively high m.p. for molecular, no conductivity - polar-molecular.

    H2 - nonpolar molecule, close and low melting and boiling points, no condutivity - nonpolar-molecular.

    Fe - metal, high or low mp and bp, good conductor - metallic.

    KI - ionic compound, high melting and boiling points, conducts as a liquid - ionic.

    CO2 - nonpolar molecule, low and close together mp and bp, nonconductor - nonpolar-molecular.

    CH4 - nonpolar molecule, low and close together mp and bp, nonconductor - nonpolar-molecular.

    Diamond - covalent network, very high mp and bp, nonconductor - covalent (network)

    CrystalMelting Point (°C)Boiling Point (°C)Electrical Conductivity
      Solid          Liquid
    HF-83.119.54NoNo
    H2-259.14-252.5NoNo
    Fe15353000YesYes
    KI6861330NoYes
    CO2-56.6-78.5NoNo
    CH4-182.48-164.8NoNo
    Diamond35504827NoNo
  14. Nickel metal has a face centered cubic unit cell with an edge length of 3.525 Å (3.525 x 10-8 cm). What is the density of Ni?
    Answer
    (58.7g/mole)(1 mole/6.02x10-23 atoms)(4 atoms/unit cell)(1 unit cell/[3.525x10-8]3 cm3) = 8.9 g/cm3
  15. Silver metal has a cubic unit cell with an edge length of 4.09 Å (4.09 x 10-8 cm) and a density of 10.5 g/cm3. Does silver (Ag) form simple, bcc, or fcc unit cells?
    Answer
    (6.03 x 1023 atoms/mole)(1 mole/107.9 g)(10.5 g/cm3)([4.09 x 10-8 cm]3/unit cell) = 4 atoms/unit cell ⇒ fcc.