Name:				Total points:			Course grade:

Exam score is on the exam. 26 = A; 19 = B. (Scores very close to cutoff have + or ‑.) 

In figuring your course grade, I look at several things. First, I average your three test letter grades. For most people, this will give a proper course grade. But I do look to see if one of your early tests was unusually low (and I may consider that, discounting the early low score a little). After that, I look over the whole picture, trying to make sure everything looks fair and consistent. 

It worked out that 67 points total = A, 49 = B (again, ignoring +/-; possible = 86). Your total test points is shown above; check it for accuracy of my records.

If there are any questions that might reasonably affect course grade, please contact me ASAP. It’s easier to fix problems before grades are sent in. I plan to send grades in to the office Thursday morning May 19.

Beyond that, I would be happy to talk with you about the exam, or whatever, anytime. I certainly encourage you to continue working on X11 material -- beyond what we covered or catching up on things where you got behind. I am happy to discuss it with you.

Additional feedback/suggestions always welcomed; e-mail is fine. If you haven’t already done so, look at the Supplemental Information page at the web site for some things you might comment on. (You can also send me anonymous messages by using the web-site message form.)

X402 for the Summer is on Thursday nights in San Francisco, starting June 9. Hope to see some of you in X402. See the web site for lots of info! I encourage you to get a book well before the course starts; you can probably get a much better deal by ordering one online yourself. See textbook info, in X402 syllabus, at web site. I also encourage you to register for the course in advance, or at least to be prepared to register at the first class; a small enrollment is likely.

(More calculator forms welcome! This is a chance for you to be sure you understand your calculator, and to contribute to a resource for future students. You can mail me the form you got. However, you can also go to the web page, and either download the DOC file, which you can return by e-mail, or use the web form there. It is usually necessary to follow-up with you to get everything clear; although that is easier in person, it is possible by e-mail.)

The following is a brief answer sheet. 

1. The key point is that water is a polar molecule, and methane is quite nonpolar. I’ll skip the Lewis structures here (see p 303, 304), but water has polar bonds and is asymmetric (lone pairs on the O), whereas methane has bonds that are only minimally polar (often called nonpolar), and is symmetric, so even the small bond polarities cancel out. Thus water molecules are attracted to each other much more strongly than methane molecules are attracted to each other.

The question is about intermolecular bonds: the attraction between molecules. Any discussion of the strength of covalent bonds, within molecules, is irrelevant.

2. a. nitric acid (a familiar strong acid)
b. NO3- and C2H3O2- (= Ac-). (The conjugate base is what’s left after the acid gives off its H+. Note that both conjugate bases here have a -1 charge. It is critical that you show correct charges. There is no such chemical as NO3 (which, if it existed, would be “nitrogen trioxide”) ‑‑ and certainly that is not what you would expect from loss of a proton = H+.)
c. Ac-. (Strong acids have weak conjugate bases.)
d. HNO3 (aq) + Ac- (aq)  NO3- (aq) + HAc (aq). This shows the strong acid and strong base reacting to leave the weaks. Be sure your equation is balanced, including charges.

3. The Mg and Fe are both 2+. The key points are to realize that both contain the same silicate ion, and that Mg is “predictable” (from PT). Thus your logic might be Mg is known to be 2+, therefore the silicate ion has charge -4, and therefore the Fe in the iron silicate is 2+.

Oxidation numbers are always given per atom.

Some tried to deal with the ON of the Si, although you really don’t need to here. It is 4+. 

4. Yes. Dissolving is (part of) a dynamic equilibrium. At saturation, NaCl is constantly dissolving and undissolving (crystallizing) -- at equal rates. (Fig 16.3, p 437)

How do you do what is suggested in the question? With a radioactive isotope. The chemical properties, including dissolving properties, are the same, but one can measure the amount of radioactivity in the solution.

5. electron, proton, chlorine atom (Cl -- many protons and electrons), chlorine molecule (Cl2), salt crystal (macroscopic; many many atoms), sun (mass = 2x1030 kg; someone asked).

6. a. strong electrolyte; non-electrolyte.
b. ethylene glycol molecules. (The question basically tells you that, by stating that it is not ionic, and does not form ions in solution.)

Ethylene glycol (EG) is indeed common antifreeze. It is very polar and extremely soluble in water. In that regard, it is very much like alcohol. In fact, EG is in the class of alcohols. So is sugar, another very polar covalent compound that is a non-electrolyte. The question assumed no knowledge of EG; all the info you need about it is stated in the question.

7. a. decreases. (The molecules have less energy. At constant V, they would exert less P. To maintain constant P, we would need to lower V.)
b. Key issues: Must use Kelvin temperatures; since the gas is cooled, the volume must get smaller (see part a).

6.8 L    213 K
       x  = 4.8 L (2 SF)
         303 K

8. a. N2; 10 VE. 
Lewis is :N:::N:	b. N3-; 8 VE. A single N with a full octet of electrons around it. That should then be shown in brackets, with a -3 charge at upper right (p 319).	

9. a.
650 mL   0.150 mol     1 L
       x  x  = 0.0975 mol (3 SF)
            L         1000 mL
b.
0.0975 mol      L
           x  = 0.122 L (3 SF)
             0.800 mol

If you used the dilution equation for part b -- and did it correctly -- fine; however, I designed the two parts to guide you through the logical two-step procedure. At least one person did try to use the dilution equation and mangled it, thus making my point that it is a dangerous equation.

The chemical is potassium hydrogen phosphate, K2HPO4. The “hydrogen phosphate” ion is HPO42- (Table 6.6, p 151). But note that you do not need any specific information about the chemical to do this question, since the problem is entirely in moles.

10. a. pH = 9 (pH = -log[H+] = -log(10-9) = -(-9) = 9)

b. First, need to calculate [H+]. Kw = [H+][OH-]; 
[H+] = Kw/[OH-] = 10-14 M2/10-11 M = 10-3 M. Therefore, pH = 3.

c. Solution b is more acidic; it has the lower pH (or the higher [H+]).

11. a. H (from +1 to 0)		b. Zn (from 0 to 2+)
c. Given complete equation: Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)

Show all aqueous strong electrolytes in ionic form:
Zn (s) + 2 H+ (aq) + 2 Cl- (aq)  Zn2+ (aq) + 2 Cl- (aq) + H2 (g)

Cancel terms that are identical on both sides (the spectators):
Zn (s) + 2 H+ (aq) + 2 Cl- (aq)  Zn2+ (aq) + 2 Cl- (aq) + H2 (g)		

Finally, rewrite what is left, clean:
Zn (s) + 2 H+ (aq)  Zn2+ (aq) + H2 (g)		

Check that this final net ionic equation is properly balanced (including charge), and makes sense. If you start with a balanced equation and do the intermediate steps correctly, the net ionic equation should come out balanced; if it does not, look for an error somewhere. As to “makes sense”, the net ionic equation shows the Zn replacing the H, and shows the electron transfers noted in parts a and b.

12. CaCl2. The two elements in this case are further apart on the PT, OR they have a greater difference in electronegativity, OR it consists of a metal and a non-metal. The three explanations above are more or less equivalent.

13. Given moles. Want amount of solution; given that the solution is described in %(w/w), this amount will come out in mass. [15.0%(w/w) means 15.0 g solute per 100 g solution.] 
Unit path: moles solute  mass solute  mass solution. (In the set-up, suc = sucrose.)

0.824 mol suc   342.30 g suc   100 g soln
              x  x  = 1.88x103 g soln (3 SF)
                  mol suc      15.0 g suc

This question is very similar to #3b in my set of Ch 16 questions, except that this one starts with moles as given.

14. C-F. Farthest part on PT; greatest difference in electronegativity (remember the diagonal trend line).

15. a. solid		b. gas

16. a. 26 valence electrons.
 ..  .. ..
:ClAsCl:
 ..    ..
   :Cl:
    ..	
b. As+-Cl-

c. Trigonal pyramid. The electron geometry is tetrahedral, but the As lone pair is “invisible” in the molecular shape. (If there were no lone pair, the molecule would be trigonal planar, as for BCl3.)	

“Trigonal” is not a valid term for describing shape. Valid shapes include trigonal planar and trigonal pyramid. Trigonal alone has no meaning.

17. Given: number of molecules, which can be converted to moles. Want: volume. Moles and volume are related by molarity, which is given. So the path will be something like: molecules  moles  volume. More specifically: molecules  moles  nanomoles  volume.

2000 molecules       mole          109 nmol      L
               x  x  x  =
                 6x1023 molecules   mol        500 nmol

= 6.7x10-15 L (which is 6.7 femtoliters = 6.7 fL)

SF? I didn’t worry about it, but I doubt the data is better than 2 SF, and it might be only 1.

Y Rondelez et al, Highly coupled ATP synthesis by F1-ATPase single molecules. Nature 433:773, 2/17/05.

18. Baseball. OH- = base.

A comic strip appeared last year making use of this idea. A couple of humorless literalists in the class suggested a sport involving water, such as water polo; oh, well.

Best of luck to each of you!

cq/pq3a	(Spring 2005)	 5/10/05