Cliffs AP Chemistry, 3rd Edition

35.(A) The functional group of an amide is

The name of this amide is acetamide.

36.(B) The functional group of an amine is

The name of this amine is dimethylamine.

37. (D) The functional group of a thiol is
    S–H

The name of this thiol is ethanethiol.

38.(E) Do this problem by using the factor-label method (m.t. stands for metric tons, 1 m.t. =

103 kg).

...

...

...

...

m t alloy ...

m t alloy

mt Cu

mt Cu

m t Cu S

1

60 10

100

12 709

127 09

#^2 159 15 2

##

...

... ...

m t Cu S

m t ore m t ore

15 915

(^100) 60 10
2
##=^2
39.(C) Begin this problem by balancing the reaction.
2 C 2 H 6 (g) + 7 O 2 (g) →4 CO 2 (g) + 6 H 2 O(g)
Because ∆H°= Σ∆H°f products−Σ∆H°f reactants, you can substitute at this point.
∆H°= [4(−94.0) + 6(−60.0)] −2(20.0) = −696.0
However, remember that the question calls for the answer per mole of C 2 H 6. Thus, because
the balanced equation is written for 2 moles of C 2 H 6 , simply divide −696.0 by 2 and you
get the answer −348.0.
40.(B) Review the postulates of the kinetic molecular theory of gases, which are listed on
page 58.

41. (D) A principal quantum number of 4 tells you that you are in the fourth energy level. The
    fourth energy level contains electrons in the s, p, d, and forbitals. Counting the maximum
    numbers of electrons available in each of the four types of sublevels — 2 in the s, 6 in
    thep, 10 in the d, and 14 in the f — yields a total of 32.
    42.(C) When presented with a generic formula, such as AB 2 , the best way to answer the ques-
    tion is to use familiar examples that satisfy the conditions of the question. H 2 O would sat-
    isfy AB 2 as a bent molecule, and CO 2 would satisfy AB 2 as a linear molecule. In CO 2 , a
    linear molecule, the two dipoles cancel each other, resulting in a nonpolar molecule.
    However, in H 2 O, which also satisfies the AB 2 requirement, the two dipoles do not cancel
    each other out and result in a net dipole moment and a bent molecule. For both CO 2 and
    H 2 O, we have data on ionization energy, electron affinity, electronegativity, and bond en-
    ergy, but these are of no use, by themselves, in determining the geometry of the species.

N

CN

O

Part IV: AP Chemistry Practice Test