Chem 12 final exam

Submitted by Gary J. Snyder (inactive) on Friday, 5/6/2011, at 11:41 AM

Thursday, May 12, 9:00a - 12:00n

A - L Merrill 1; M - Z Merrill 2

ONE note card allowed, as announced in class.

Finals week office hours

Submitted by Gary J. Snyder (inactive) on Friday, 5/6/2011, at 11:36 AM

Prof Snyder - Wed 10a - 1p

Prof Stencel Mon 2 - 4p

Prof Hebda Tues 9 - 11a; 2 - 4p

old Exam 2

Submitted by Gary J. Snyder (inactive) on Wednesday, 3/30/2011, at 2:35 PM
this is from last Fall's Chem 12 course. Enjoy. Solutions will NOT be provided, although hints will appear here around April 6 or 7, and we will certainly be willing to go over these problems with you or to look at your written solutions and provide feedback.

Exam 1 Solutions

Submitted by Catherine A. Stillerman on Wednesday, 3/9/2011, at 1:34 PM
Prob 4a concentrations of oxygen and carbon monoxide should have been reported as 0.00500 M and 0.0100 M, respectively, or, if refined iteratively, 0.00493 and 0.00987 M. (-1 for me for being sloppy with sig figs. Drat.)

Average = 107.3 (71.5%)

Approximate letter grade rade ranges, assuming all 250 HW/Q points by end of semester (as announced in class)
A: 130 -
B: 105 - 129
C: 80 - 104
D: 60ish - 79

Hints for old exam problems

Submitted by Gary J. Snyder (inactive) on Wednesday, 3/9/2011, at 3:10 PM

1 - the stronger the interactions between molecules, the more difficult it is for them to go from liquid to gas.  Hydrogen bonding interactions are especially strong.  Other things being approximately equal, greater "surface area" generally leads to more van der Waals (London dispersion) interactions.

2 - with an N–C=O unit, there is another reasonable (minor) all-octet resonance contributor possible with + on N and - on O, (i.e. +N=C–O-).  This is not possible for H2C=O.  So the O of the amide has more negative charge and is a better hyd bond acceptor.

3 - 1st is straightforward; note that the acid concentration in the second is sooooooo tiny that the pH must be 7, i.e. the amount of H+ contributed by the acid is insignificant compared to that from autoionization of water.  (If you do this with your brain switch "off", you'll calculate the same pH as for the hydroxide solution in part a - that CAN'T be right, can it???  So be sure the switch is "on" at all times.)

4 - I did two iterations to get a chlorine concentration of 0.0135 M in the first part; for the second, so a stoichiometric shift to product (K is large!), then fine-tune.  I got a final NO conc of 0.00598 M.

5 - those are ugly numbers... 10.3% dissociation; with nitric acid present, that drops to 0.92%.  Adding less than one equiv of hydroxide converts the weak acid to a buffer, so the easiest way to solve this is with the HH eqn.  At the equivalence point, we have a solution of the weak base in water.  The usual calc gives pH = 7.96.

6 - We covered this in class.

7 - Structure, structure, structure.  Don't try to "match clusters of letters with things in the tables (e.g. the fact that you see an "NH3" embedded within the first formula does NOT mean that this should behave like NH3 itself!  Structure, structure, STRUCTURE.  What is that structure?!  That's CH3–NH3+ (with the positive on N) - that's a weak acid!  "NH3" alone is ammonia, which has a neutral N with a lone pair, and is a weak BASE.  Everything hinges on the STRUCTURE.  a - e: one is acidic, two are basic, and two are neutral.