5 Steps to a 5 AP Chemistry 2019

(Marvins-Underground-K-12) #1
Use the following information to answer questions
44-45.

METAL ION

IONIC RADIUS
(pm) (CUBIC
ENVIRONMENT)

MELTING
POINT OF
OXIDE (°C)
Sodium Na+ 132 1275
(sublimes)
Cadmium Cd^2 + 124 1500
Lanthanum La^3 + 130 2256


  1. Each of the ions in the table form stable oxides
    (Na 2 O, CdO, and La 2 O 3 ). Lanthanum oxide,
    La 2 O 3 , has a melting point significantly higher
    than that of the other oxides. Which of the fol-
    lowing is the best explanation of why this is true?
    (A) Lanthanum is a lanthanide element and the
    melting points of these elements are always
    high.
    (B) There is more oxygen in the formula La 2 O 3
    than in the other formulas.
    (C) Lanthanum had the highest charge; there-
    fore, it has the highest lattice energy.
    (D) Alkali metals like sodium and transition metals
    like cadmium tend to have low melting points.

  2. The lithium ion, Li+, is smaller than the sodium
    ion. How does the melting point of lithium
    oxide, Li 2 O, compare to that of sodium oxide?
    (A) It is higher because smaller ions have a
    higher lattice energy.
    (B) It is the same because the charges are the same.
    (C) It is lower because smaller ions have a
    smaller lattice energy.
    (D) It is impossible to predict because there is
    insufficient information in the problem.
    46. During the investigation of a chemical reaction
    by a chemistry student, she was able to determine
    that the reaction was nonspontaneous at 1 atm
    and 298 K. However, she learned that cooling
    the system with dry ice (- 78 °C) caused the reac-
    tion to become spontaneous. Which of the fol-
    lowing combinations must apply to this reaction?
    (A) ΔH < 0, ΔS < 0, and ΔG = 0
    (B) ΔH > 0, ΔS < 0, and ΔG > 0
    (C) ΔH < 0, ΔS < 0, and ΔG > 0
    (D) ΔH > 0, ΔS > 0, and ΔG > 0
    47. What is the ionization constant, Ka, for a weak
    monoprotic acid if a 0.060 molar solution has a
    pH of 2.0?
    (A) 2.0 × 10 -^3
    (B) 2.0 × 10 -^1
    (C) 1.7 × 10 -^1
    (D) 5.0 × 10 -^3


H

H

H
C

C C
H

H

H


  1. Cyclopropane, pictured above, is a relatively unsta-
    ble compound. As seen in the diagram, the carbon
    atoms form the corners of an equilateral triangle
    and each carbon atom has two hydrogen atoms
    attached to complete an octet of electrons around
    the carbon atoms. Based upon this structure, why
    is cyclopropane a relatively unstable compound?
    (A) Hydrocarbon compounds are relatively
    unstable in general.
    (B) Compounds that have identical atoms
    bonded to each other are relatively unstable.
    (C) The bonds do not match the angles.
    (D) There is no resonance to stabilize the
    compound.


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