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114. F F
26. E
27. D
28. C
29. B
30. E
31. C
32.
A
33.
C
34.
C
35.
C
36.
B
37.
E
38.
D
39.
A
40.
C
41.
E
42.
B
43.
E
44.
A
45.
C
46.
E
47.
C
48.
D
49.
B
50.
E
51.
A
52.
D
53.
D
54.
E
55.
D
56.
C
57.
C
58.
E
59.
D
60.
B
61.
A
62.
E
63.
D
64.
A
65.
D
66.
C
67.
C
68.
C
69.
B
70.
D
71.
C

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Chemistry Subject Test
297
ARCO
■ SAT II Subject Tests
w w w . p e t e r s o n s . c o m / a r c o
14. The correct answer is (C). In this reaction 2Br
–
ions are oxidized to Br
2
, while 2O
–
atoms are
reduced to 2O
–2
atoms.
15. The correct answer is (A). BCl
3
is a Lewis acid, and Cl
–
is a Lewis base.
16. The correct answer is (D). Group I metals have only one valence electron, which they give up very
easily to form a cation with a full outer shell.
17. The correct answer is (E). The Noble gases have completely filled outer shells that are stable; there-
fore, they rarely react with other elements.
18. The correct answer is (C). Because of the ease at which electrons can flow through these elements,
the best electrical conductors are metals. Those with outstanding conductivity are Cu, Ag, and Au.
19. The correct answer is (A). The halogens require only one electron to fill their outer shells; since a
full outer shell is stable, much binding energy can be released when halogens acquire one electron.
20. The correct answer is (E). This problem requires that one be able to identify compounds that are
acids and use the concepts of molarity and normality. By using the concept of normality, one knows
that a .02M H
2
SO
4
solution has a [H
3
+
O] = .04M, while a .02M HNO
3
solution has a [H
3
+
O] = .02M.
21. The correct answer is (D). The .02M Mg(OH)
2
solution contains .04M OH
–
ions; this is twice as
much as that contained in the .02M KOH solution.
22. The correct answer is (C). Knowledge of the colligative properties of solutions is required for this
question. The solution with the lowest number of particles per liter of solvent will have the lowest
boiling point elevation. All the acids and bases dissociate into two or three particles, whereas glucose
is a undissociating molecule.
23. The correct answer is (D).
24. The correct answer is (A). Given that T is constant, we know that for an ideal gas P
1
· V
1
= P
2
· V
2
.
The curve fitting this equation is (A).
25. The correct answer is (C). The rate equation for a first order reaction is 
−dS
dt
 = K
1
· S where K
1
is
the first order rate constant. The solution to this rate equation is St = So · e
–
K
1
t which gives the amount
of substrate remaining at any time value for t. Therefore, the amount of product at time = t
is So – So · e–K
1
t and curve C fits this equation.

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