(^30) › STEP 2. Determine Your Test Readiness
- C—The only ionic bonds are present in the sodium
compounds (eliminating B and D). The nitride ion
has no internal bonding (eliminating A), but the
nitrate ion has both s and p bonds. - C—Draw the Lewis structures. The number of
unshared pairs: (A) 1; (B) 0; (C) 3; (D) 0. - C—Draw the Lewis structure; the carbon on the
left in the formula is sp^3 , and the other is sp^2.
Chapter 12
- D—Both graphite and diamond are covalent
network solids. - A—Calcium is a metal, and answer A applies to
metallic bonding. - C—Calcium carbonate is an ionic compound.
- B—Sulfur dioxide consists of polar molecules.
- D—This is the definition of the critical point.
- D—This is a consequence of metallic bonding as
the atoms can easily move past each other with-
out breaking any bonds. - C—The carbonyl, C == O, and OH groups are
capable of participating in hydrogen bonds. - C—The more OH groups, the more hydrogen
bonding, and the more soluble in water (where
hydrogen bonding also occurs). - D—The solid begins to melt at A and finishes
melting at B. - B—The gas-liquid line always has a positive slope,
which eliminates A. Answer B negates C; there-
fore, both cannot be correct. The triple point is
not the same as the freezing point.
Chapter 13
- A—Molarity is moles per liter, and the moles
are already known; therefore, only the volume is
necessary to complete the determination. - D— (0.800 L)(0.50 mol Br-/L) =
0.40 mol needed.
(0.800 L)(0.20 mol Br-/L) =
0.16 mol present.
[(0.40 - 0.16) mol Br- to be added]
(1 mol MgBr 2 /2 mol Br-)
- D—(0.5000 L)(5.00 mol/L)(63.0 g/mol) = 158 g.
As always, estimate the answer by rounding the
values. - C—Equimolar gives a mole fraction of 0.5. 0.5 ×
480 mm Hg + 0.5 × 50 mm Hg = 265 mm Hg
(total vapor pressure) mole fraction ethyl ether =
(0.5 × 480 mm Hg)/265 mm Hg.
Chapter 14
- A—Add the equations and cancel anything that
appears on both sides of the reaction arrows. - B—This is the definition of the activation energy.
- C—The table shows second order in chlorine
dioxide (comparing experiments 1 and 2), because
doubling the ClO 2 concentration quadruples (2^2 )
the rate. The reaction is first order in the hydrox-
ide ion (comparing experiments 2 and 3), because
doubling the OH- concentration doubles (2^1 ) the
rate. When making this determination, make sure
there is only one concentration changing; i.e., do
not compare experiments 1 and 3.
Chapter 15
- D—To be an acid, the species must have an H+
to donate, and to be a base, the species must be
able to accept an H+. The carbonate ion has no
H+ to donate to be an acid. - B—Start with the acid with a pKa as near 8.5
(K = 10 - 8.5) as possible (H 2 PO 4 - ). To go to a
higher pH, add the acid (conjugate base) with the
smaller Ka (higher pKa). - B—This is an approximation. At pH = 5, [H+] =
10 -^5 M; therefore, Ka = (10-^5 )^2 /0.5. - A—HBr is a strong acid, and with equal con-
centrations and no base present, it will give the
lowest pH. - D—The weak acid and weak base give a nearly
neutral solution, as they will tend to neutralize
each other. - C—Only B and C are buffers. B is acidic (pH < 7)
and C is basic (pH > 7).