Be sure to read over the comments on the test. I often comment on things even if you get credit for the question. Learn from the comments (as well as from your mistakes). “ATQ” (written on the test) means “answer the question”; it means that, in some way, your answer is not addressing the key issues.

If you found this rather easy, that is somewhat inherent in the first test. It will get harder. If you found this test hard, the key thing is to learn the areas that bothered you here. It is common to do better as chem proceeds -- if you commit to learning the important old material. If you are disappointed, one suggestion would be to turn in regular homework, for feedback. For example, I might suggest that you turn in 1-2 pages of homework each class period, for feedback. The content is your choice, depending on where you are.

A = 23; B = 17; C = 11. (Scores very near those boundaries are given a + or -. For example, the range 17-22.5 includes all B, from B- to B+.) 

mean = 23.6		median = 24		top score: 29.

Check my arithmetic. The circled number is the total for a question.
 
Test 2 notes. Handed out in class 9 (11/2); due the following class (11/9). It probably will cover Ch 6 through Ch 10. (This will be adjusted as needed, based on how far we actually get in class 8. Check the Updates page.) Chemistry is inherently cumulative, but the test will emphasize material from current chapters. Test form will be similar to Test 1. You may supply a PT plus one page of notes -- anything in your own handwriting.

•	A sample Test 2, from a previous class, is at the web site, along with an answer key. (A sample final is also there.)

A major portion of the test will be problems of the type on the written assignments for Ch 7 & 9. There will be some hard ones, where good dimensional analysis may well be the key to seeing a pathway to a solution. Among the other things you should be able to do: deal with charge neutrality in ionic compounds; balance equations (given the correct chemicals); recognize reactions as to type (Table 8.2) and figure out what the reaction would be for typical cases. I will assume that you know the most common polyatomic ions and acids (see Ch 6 handout; put them on your note page?).

Remember that I am available in class by ~6:30 for questions (test preparation, or turn-in questions). I will “guarantee” phone availability 7-9 pm on the evening before the due date.

Brief answers:

1. a. 58.08 mL (Will accept answers within 0.01 mL of that.)
b. 58.00 mL (The final zero shows that you estimated the last digit as a zero, just as you estimated the last digit in part a as an 8. You read the scale by the same procedure in both parts; after all, it is the same scale.)

c.
58.08 mL    1 L
         x   =  0.05808 L
           1000 mL

Careful. Be sure you show what cancels, and be sure the units you end up with are what you want. One of the points of dimensional analysis is that it helps protect you against silly errors.

2. a. Br. The question gives the atomic weight (mass), shown on the PT.
b. Br. The atomic weight is due to the nuclear particles, P & N.
c. K
d. As (or Se). (As is a semi-metal, but on the non-metal side. I did not intend this to be an issue here.)
e. O

3. Dinitrogen pentasulfide

4. all nonmetals. (Folks: H is not a metal!)

5. Your classification will depend on what you think is happening here. If you think that the CO2 is created by a chemical reaction (acid + rock  CO2 + ??), then it is a chemical change. If you think that the CO2 is in the rock, and is released by the action of the acid, then it is a physical change.

The former is what is actually happening; we will discuss this in Ch 8. But the question does not depend on you knowing what is happening; the purpose is to get you to offer a suggestion, and then to make the connection between a process and how we would classify it.

6. Path: distance  time; km  s; km  m  s.

4.0x104 km    1000 m        1 s
           x  x   =  0.13 s (2 SF)
               1 km      3.0x108 m

This is the time for one trip between ground and satellite. In a normal conversation, what we would sense is the time it takes to get a reply. That requires four such trips: your signal up to the satellite, and then down to the receiver; the reply up to the satellite, and then down to you. Thus the observed delay is about a half second -- which is very noticeable.

7. a. 
O-S-O O-S-O O-S-O O-S-O + O-O O-O  O-S-O O-S-O O-S-O O-S-O
                                                      
                                       O     O     O     O

Each side shows 4 S atoms and 12 O atoms. The atoms have moved around, to form new chemical substances. The atoms themselves are conserved. 

Looking ahead (Ch 8), this could be written in chemical equation form as: 
4 SO2 (g) + 2 O2 (g)  4 SO3 (g) (which of course could be simplified).

I did not worry about how you showed the atoms in each molecule (e.g., SO2) connected together, so long as you had the correct atoms. You will learn more about how to predict the atom order later. (Ch 11, but as a useful guideline, minimize O-O bonds.) The way I showed them above is about right, in showing the S connected to each O.

b. Mixture: two distinct chemical substances (SO2 and O2). (The RHS shows a compound, SO3.)

9. a. solid		b. gas

10. a. Path: mass  volume; g  mL. Density (given) is a CF between mass and volume.

2.0 g     mL
      x  =  0.089 mL (2 SF)
        22.59 g

If you use the density formula d = m/v, so v = m/d. Still, show a clear set-up, with clear units. Watch that the units work out correctly; this serves as a check on your algebra. It is also good to realize that you do not need to remember any formula to do this; you can do it entirely by following the units.

b. It is very small. The relevant comparison is with the dime; it is much smaller. It is about the size of 2-3 drops of water. Main idea: high density means there is a lot of mass in a unit of volume, so it takes a very small volume to have a certain amount of mass. In other words, mass and volume are inversely proportional. (If you decided that the Os bar was the size of a dime, or bigger, it should tell you that your answer is wrong.)

11. a. Atoms are neither created nor destroyed. But water is a molecule (of a compound), and its atoms can “come and go” to/from other chemical substances.

b. hydrogen gas + oxygen gas  water -- or the reverse. In symbols (and balanced, though I did not require it): 2 H2 (g) + O2 (g)  2 H2O (l).

12. a. Same. Iodine atoms are atoms that have 53 protons. If it did not have 53 protons, it would not be I.

b. Fewer. The 125 number is the mass number (P + N) for this isotope. The PT says that the average mass for I is 127. Isotopes differ (only) in the number of neutrons, so -- for isotopes of a particular element (i.e., for a given P) -- less mass means fewer neutrons. (You do not need to calculate the number of neutrons; it is sufficient to compare the mass number for this isotope with the average mass shown on the PT, since masses -- for the isotopes of one element -- differ only because of neutrons.)

13. a & b & c. electron.

B1. 1.59x1022 (3 SF).

This is an exercise to make sure you are learning the scientific notation features of your calculator. Procedures are different with different models, but I do want you to learn the procedures that are general with all models. These include use of the “enter exponent” key. You also need to be aware of the problem of entering a number such as 1035; does your calculator require a leading 1? Finally, the problem involves sig fig, which we will formally grade on after this test.

B2. #114. Ca nucleus has 20 protons, Pu nucleus has 94 protons. Add them together, and get 114 protons. 

This is indeed how element #114 was made. News story to accompany the announcement: N Rowley, Charting the shores of nuclear stability. Nature 400:209, 7/15/99.

This is a nuclear reaction. How and why it occurs is beyond our scope, but it does illustrate how atoms are made. The number of neutrons in the resulting atom is unpredictable, as neutrons tend to be “lost”. (In fact, they made the isotope of mass number 287; 3 neutrons “evaporated” in the process.)

B3. Path: The question is seconds  steps. How? You are given 2 CF which relate the given and wanted both to nm. Therefore: seconds  nm  steps.

10 s   2.5 nm     step
     x  x  =  3.1 steps (2 SF)
         s       8.1 nm

Why 2 SF? In context, seems to me rather clear that the 10 seconds is either intended as exact, or as 2 SF. In either case, this justifies 2 SF answer.

W Hua et al, Coupling of kinesin steps to ATP hydrolysis. Nature 388:390, 7/24/97.

B4. Metalloids: As, Antonin Scalia; Sb, Stephen Breyer. Transition metals: Ds, David Souter; Rg, Ruth Ginsburg.

pq1a (F2005-1)	10/10/05