Name:				Total points:			Course grade:

Exam score is on the exam. 25 = A; 19 = B; 12 = C ; 6 = D. (Scores very close to cutoff have + or ‑.) Mean = 16.4; median = 11.8; hi = 31.

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 70 points total = A, 52 = B, 34 = C (again, ignoring +/-). Hi total = 84 (out of 88); mean = 61, median = 58. 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 Wednesday morning Dec 28.

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 Spring is on Monday nights in San Francisco, starting January 30. Hope to see some of you in X402. See the web site for lots of info! I encourage you to get a book 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 low enrollment is likely.

The following is a brief answer sheet. 

1. The rule of thumb is that ionic bonds form between a metal and a non-metal, whereas covalent bonds form between two non-metal atoms. So, choose a metal as the partner in part a, and a non-metal in part b. Many possible; examples would include Na2S and H2S.

Many simple ionic compounds have formulas that are entirely logical, so I expected your ionic compound to be “good”. Covalent compounds are often less predictable, which is why we use the prefixes in naming them. Many of you tried to use a sulfur oxide. Some knew that the common oxides of S are SO2 and SO3. Some wrote SO, which might seem reasonable; however, SO does not really exist, at least as a stable compound.

2. H gains electrons: H+ + e-  H0, or 2 H+ + 2 e-  H2. (The Al loses e, going from 0 to 3+.)

It is not necessary to write the net ionic equation here. Just look at the oxidation numbers (ON). Remember, free elements always have ON = 0, so when an element occurs free in a reaction, you can pretty well bet it is one of the redox components. In contrast, the Cl appears in the equation only as chloride, with ON = -1; it is a spectator.

3. We doubled n (number of moles). If we want to compensate by changing T, we must halve T -- on the Kelvin scale. The given T is 258 K. Half that is 129 K (which is -144 C).

If you start with pV = nRT n and T are inversely proportional. If n is doubled, T must be halved to result in no change of pV.

If your intuitive analysis leads you through pressure Doubling n would double p (at constant T). Now we need to compensate for the doubled P by lowering the T. You’ll end up in the same place.

4. a. pH = -log[H+] = 12

b. First, need to calculate [H+]. Kw = [H+][OH-]; 
[H+] = Kw/[OH-] = 10-14 M2/10-5 M = 10-9 M. Therefore, pH = 9.

c. The solution in part a is more basic. It has the lower [H+], or, equivalently, higher pH. 

5. a & b. H3PO4 (aq) + NO3- (aq)  HNO3 (aq) + H2PO4- (aq)
                acid 					             base		(one conjugate pair)
		               base                 acid				(one conjugate pair)

c. It will tend to go to the left; nitric acid is a familiar strong acid.

6. The M gives you mol/L; question is in grams; so you need a g/mol conversion. The chemical is FeBr2. A good path: g  mol  L.

8.47 g    mol           L
       x  x  = 0.291 L (or 291 mL) (3 SF)
         215.65 g   0.135 mol

7. a. +6. O is taken as -2; three O gives -6. Since the molecule is neutral, the S must be +6.
b. +4. Again, there is -6 worth of O, but this time the total is -2.

8. a. Ca(NO3)2 (aq) + 2 NaF (aq)  2 NaNO3 (aq) + CaF2 (s)

b. First, show all aqueous strong electrolytes in ionic form:
Ca2+ (aq) + 2 NO3- (aq) + 2 Na+ (aq) + 2 F- (aq)  2 Na+ (aq) + 2 NO3- (aq) + CaF2 (s)

Cancel terms that are identical on both sides (the spectators):
Ca2+ (aq) + 2 NO3- (aq) + 2 Na+ (aq) + 2 F- (aq)  2 Na+ (aq) + 2 NO3- (aq) + CaF2 (s)

Rewrite what is left, cleanly. This is the net ionic equation. It should come out balanced and looking reasonable. (If not, something is wrong, at some step.)
Ca2+ (aq) + 2 F- (aq)  CaF2 (s)

(Part a was entirely a review from previous test; Ch 8. Part b is one of the questions I specifically promised you.)

9. Covalent. Rule of thumb is that ionic compounds are solids at room temperature. This follows from the crystal lattice structure of ionic compounds. (It is not so much the strength of the individual bonds as the extensive lattice. If you are to compare bonds, it is the ionic bonds of the lattice vs the intermolecular bonds of the covalent compound. The covalent bonds per se are not relevant.)

As some of you noted, this question raises a contradiction. Another rule of thumb, that metal + non-metal gives an ionic compound (question #1), would suggest that the compound is ionic. But that is really a rather weak rule. The electronegativity (EN) of Ti (not shown in your book) is 1.5, thus the EN here is 1.5, a value that is borderline. This contradiction does not affect the question here; the question asked you to interpret a specific piece of evidence. If you found the evidence confusing, because of the contradiction, you should have asked about it, as indeed one person did.

10. How many moles do we need:
0.2 L    3.0 mol
       x  = 0.60 mol
           L
Where do those moles come from:
0.60 mol      L  
         x  = 0.033 L (2 SF)
           18 mol

If you used the dilution equation -- and did it correctly -- fine; however, I tried to aim you through the logical two-step procedure.

11. Na2SO4 (s)  2 Na+ (aq) + SO42- (aq)

12. sucrose molecules. (No ions are relevant. The information in the question tells you that.)

Given the way the question was worded, it was reasonable to also include H2O molecules. I didn’t really intend that, but if you noticed, good.

The difference between this question and #11 shows the difference between ionic and covalent compounds. (#12 here is essentially question #2 from the Ch 17 practice quiz, which I urged you to do.)

13. a. 
 ..        ..
H-O: .... H-O-H
           ..
  H
	Each O has an octet, including two lone pairs. Each H has a full shell of two.

The O is -, the H is +; hence the weak interaction between them.

The hydrogen bond, shown by ...., is long compared to a regular covalent bond, reflecting its relative weakness.	

b. None in the gas phase (that is the basic idea of a gas); four in the solid phase (maximum number possible). In the liquid phase, it will vary from 1-4.

c. It is “bent” (or “angular”). The electron geometry is tetrahedral, but molecular shape refers only to atom positions. So hide the two lone pairs, and the atoms show a “bent” shape.

14. Fr-F -- greatest difference in electronegativity (EN), using simply the main trend line for EN. (I allowed some room for alternative interpretation, so long as the reason was ok.)

15. Use hexane. Water and methanol are both polar solvents, but hexane is nonpolar. Since water did not work, it is likely that methanol will not work either, so try something of substantially different polarity. (Acid/base is irrelevant; none of these compounds are either acid or base. Remember, like dissolves like, referring to polarity. This question is like #14 from Ch 16 hw.)

16. Impossible. Multiply those together and you get 4x10-8 M2. That violates the Kw expression, which says that the product is always 10-14 M2.

17. In each case: What is the formula? How many VE? Then draw the Lewis.
a. a chlorine atom, Cl (7 VE)

 ..
:Cl:
  .
	b. a bromine molecule, Br2
(14 VE)

 .. ..
:Br:Br:
 .. ..	c. a fluoride ion, F- (8 VE)

  ..  -
: F :
  ..

(Hard to type, but proper way to show it is with brackets around the Lewis for the ion, with the charge shown outside the brackets.)	

18. You are asked for M = mol/L. You are given the L. Convert the amount, given in particles, to moles, using Avogadro’s number. (The chemical is PuO2, though you do not need the formula.)

1 formula unit        mol
  x  = 2x10-30 M (1 SF)
106 L             6x1023 f.u.

I encourage you to check my book list -- and to contribute to it.

Happy holidays, and best of luck to each of you!

Bob Bruner
<contact information>		

cq/pq3a	(Fall 2005)	 12/19/05