Question 1 (a)

(ii) Determine the value of AHS In for LiCl in kJ/molrxn. 1 mol LiCl
 10.0 gLiC1 x = 0.236 mol LiC1 42.39 g LiC1 -9.47 k.J 00.1 kJ/molrxn
 0.236 mol LiC1 1 point is earned for the number of moles of LiCl. 1
 point is earned for the correct Al-Isoln and the correct Sign.

Question 1 (c)

(c) Using principles of atomic structure, explain why the Na ion is
 larger than the Li+ ion. a ion are ina hi her rinci al ener level the
 valence 1 point is earned for a correct explanation elecfrons in the
 Li ion. Electrons inhi er based on occupied principal energy levels.
 principal energy levels are, on average farther from the nucleus.

Question 1 (d)

(d) Which salt, LiCl or NaCl, has the greater lattice enthalpy?
 Justify your answer. LiCl. Because the Li ion is smaller than the Na
 ion, ions in LiCl are stronger than in NaCl. This results in a greater
 lattice enthalpy. 1 point is earned for the correct choice and

Question 1 (f)

(f) The lattice enthalpy of LiCl is positive, indicating that it
 takes energy to break the ions apart. in LiCl. However, the
 dissolution of LiCl in water is an exothermic process. Identify all
 particle-particle interactions that contribute significantly to the
 exothermic dissolution process being exothermic. For each interaction,
 include the particles that interact and the specific type of
 intermolecular force between those particles. There are interactions
 between Li+ ions and polar water molecules and between Cl- ions and
 polar water molecules. These ar ion-dipole interactions. 1 point is
 earned for identifying the particles that interact. 1 point is earned
 for correctly identifying the type of interaction.

Positive ends of polar molecules are oriented toward negatively
 charged anion Negative ends of polar molecules are oriented toward
 positively charged cation

Solute—solu te (positive AH) Solvent—solvent (positive AH) Solute +
 solvent Solution Solven t—solute (negative AH) SO In Solute—solu te
 (positive AH) Solvent—solvent (positive AH) Solute + solvent (a)
 Negative AH so In Solven t—solute (negative AH) Solution - - (b)
 Positive AH soln

Question 2 (e)

The HCOg- ion has three carbon-to-oxygen bonds. Two of the
 carbon-to-oxygen bonds have the same length and the third
 carbon-to-oxygen bond is longer than the other two. The hydrogen atom
 is bonded to one of the oxygen atoms. In the box below, draw a Lewis
 electron-dot diagram (or diagrams) for the HC03- ion that is (are)
 consistent with the given information.

Question 2 (f)

(f) A student prepares a solution containing equimolar amounts of
 HC2H302 and NaC2H302. The pH of the solution is measured to be 4.7.
 The student adds two drops of 3.0 M HN03(aq) and stirs the sample,
 observing that the pH remains at 4.7. Write a balanced, net-ionic
 equation for the reaction between HN03(aq) and the chemical species in
 the sample that is responsible for the pH remaining at 4.7. C2H302- +
 H30+ HC2H302 + H20 OR C2H302- + H+ HC2H302 I point is earned for a
 correct equation.

Presence of HF counteracts addtion of base; pH increase is small
 After addition Initial buffered solution Presence of F- counteracts
 addition of add; pH decrease is small After addition of H+ Add H+ Add
 OH H20 + F HF -F OH A Figure 17.2 Buffer action. The pH of an HF/F-
 buffered solution changes by only a small amount in response to
 addition of an acid or base.

Question 3 (d)

(d) Explain why 12 is a solid at room temperature whereas Br2 is a
 liquid. Your explanation should clearly reference the types and
 relative strengths of the intermolecular forces present in each
 substance. Both Br2 and 12 molecules are nonpolar molecules, therefore
 the only possible intermolecular forces are London dispersion forces.
 The London dispersion forces are stronger in 12 because it is larger
 in size with electron cloud. The stronger London dispersion forces in
 12 result in a higher melting point,which makes 12 a solid at room
 temperature. 1 point is earned for identifying the forces in each
 substance as London dispersion forces. 1 point is earned for
 explaining why the forces are stronger in 12 than in Br2 .

Question 5 (b)

![Zeroth Order First Order Second Order Differential rate law Concentration vs. time Integrated rate law Straight-line plot to determine rate constant Relative rate vs. concentration Half-life Units of k, rate constant Rate = o o o o ALA] At Time Slope = —k Time Rate — o o ALA] = MA] At Time o o —kt or Slope = —k Time [A], M Rate, M/S o o o o 2 3 M Rate, M/s 2k 2 3 Rate = 2 3 A[A] = At Time Slope = k Time Rate, M/S 4 9 2 3 0.693 l/s ](./media/image22.png)

Order Rate Law Rate = Rate = kl Concentration - Time Equation = kot
 2.303 Half Life 2k o o .aga Graphical

Zero order 1st order 2nd Order (b) Plots from integrated rate
 equations: Zero order Conc. 1st order Conc. 2nd order (Conc.)2 3rd
 order 3rd order (Conc.)

Question 6 (a)

(a) Considering the value of K for the reaction, determine the
 concentration of Ba(EDTA)2-(aq) in the 100.0 mL of solution. Justify
 your answer. Based on the K value, the reaction goes essentially to
 Ba2+(aq) is the limiting reactant. The concentration of Ba2+ when the
 solutions are first mixed but before any reaction takes place is 0.20
 M/2 = 0.10 M. Thus the equilibrium concentration of Ba(EDTA)2-(aq) is
 0.10M. 1 point is earned for indicating that the equilibrium
 concentration of Ba(EDTA)2-(aq) is the same as the original
 concentration of Ba2+ when the solutions are mixed. 1 point is earned
 for the concentration with appropriate calculations.

Question 7 (a)

  • Since your buret is graduated to 0.1 mL, you will read your buret to 0.01 ml. The second decimal place is an estimate, but should be recorded.

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