5 Steps to a 5 AP Chemistry 2019

AP Chemistry Practice Exam 2 ❮ 347

CH 3 CH 3

O H O

C C

O H O

CH 3 CH 3

O ClO

C C

O Cl O

23. A dimer consists of two closely associated mol-
    ecules. In the gas phase, acetic acid tends to form
    dimers as illustrated on the left in the above dia-
    gram. Acetyl chloride, on the right in the above
    diagram, is not very efficient in forming dimers.
    Why is acetic acid better able to form dimers
    than acetyl chloride?
    (A) The molecular mass of acetyl chloride is
    higher than that of acetic acid making
    it harder for the acetyl chloride to form
    dimers.
    (B) It is easier to form a covalent bond between
    acetic acid molecules than between acetyl
    chloride molecules.
    (C) Acetic acid can form strong hydrogen bonds
    but acetyl chloride can only form weaker
    dipole-dipole attractions.
    (D) Acetic acid is an acidic compound but
    acetyl chloride is a neutral compound.

C

trans-1,2-dichloroethene

C

Cl

H

H

Cl

C

cis-1,2-dichloroethene

C

Cl

H

Cl

H

24. Two compounds with the formula C 2 H 2 Cl 2
    appear in the above diagram. These two com-
    pounds are isomers. The molecules are planar
    and have the approximate structures shown
    in the diagram. The boiling point of trans-1,
    2-dichloroethene is 47.5°C and the boiling point
    of cis-1,2-dichloroethene is 60.3°C. Which of the
    following best explains why cis-1,2-dichloroethene
    has a higher boiling point than its isomer, trans-1,
    2-dichloroethene?
    (A) The higher boiling isomer is more polar than
    the other isomer.
    (B) The higher boiling isomer is better able to
    form hydrogen bonds than the other isomer.
    (C) The higher boiling isomer has a greater molar
    mass.
    (D) The higher boiling isomer has greater London
    dispersion forces than the other isomer.

Use the information on the following proposed

mechanism to answer questions 25–27.

Step 1: N 2 O 5 (g) → NO 3 (g) + NO 2 (g) Slow

Step 2: NO 3 (g) → NO(g) + O 2 (g) Fast

Step 3: NO(g) + N 2 O 5 (g) →

N 2 O 4 (g) + NO 2 (g) Fast

Step 4: N 2 O 4 (g) ^ 2 NO 2 (g) Fast

25. The above represents a proposed mechanism for
    the gaseous dinitrogen pentoxide, N 2 O 5. What
    are the overall products of the reaction?
    (A) N 2 O 4 (g) + O 2 (g)^
    (B) 2 NO 2 (g) + O 2 (g)
    (C) N 2 O 4 (g) + NO 2 (g) + O 2 (g)
    (D) 4 NO 2 (g) + O 2 (g)

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21-Moore_PE02_p341-370.indd 347 31/05/18 1:54 pm