Cliffs AP Chemistry, 3rd Edition

The Keqis needed for the following equation:

.

.

() () () ()

.

AgCl s I aq AgI s Cl aq K

I

Cl

L

mole

mole

16 10 L

10

20 10

80

eq

2

10 2

17 2

2

6

"

#

#

#

= #

++ =

--

7

7

A

A

70. (D)

Bond breaking (∆H 1 ) = HH + FF = 103 kcal ⋅mole–1+ 33 kcal ⋅mole–1

= 136 kcal ⋅mole–1

Bond forming (∆H 2 ) = 2 HF = 2(−135 kcal ⋅mole–1) = –270 kcal ⋅mole–1

∆H°= ∆H 1 + ∆H 2 = 136 kcal ⋅mole–1+ (−270 kcal ⋅mole–1) = −134 kcal/mole

71. (D) Whether you can answer this question depends on whether you are acquainted with
    what is known as the Maxwell-Boltzmann distribution. This distribution describes the way
    that molecular speeds or energies are shared among the molecules of a gas. If you missed
    this question, examine the following figure and refer to your textbook for a complete de-
    scription of the Maxwell-Boltzmann distribution.


72. (A) Because all choices have 18 electrons in their valence shell, you should pick the
    species with the fewest protons in the nucleus; this would result in the weakest electrosta-
    tic attraction. That species is sulfur.


73. (A) The central metal ion forms only two bonds to ligands, so the coordination number is 2.


74. (C) The body-centered cubic cell looks like this:

number ofmolecules

molecular speed

500 °C

0 °C

Answers and Explanations for the Practice Test