326 ❯ STEP 5. Build Your Test-Taking Confidence- A—The compound with the highest boiling
point has the strongest intermolecular forces.
This is only valid if the molar masses are similar. - C—The melting points of ionic materials depend
upon the lattice energy. The higher the lattice
energy, the higher the melting point is. Lattice
energies depend upon the sizes of the ions and
the magnitude of the charge. All three metal ions
are approximately the same size; therefore, the
size factor is minimal. This leaves the magnitude
of the charge, and lanthanum, with the largest
charge, should have the highest lattice energy. - A—The lattice energy depends upon both the
charge and the size of the ions involved. The
greater the charge is, the greater the lattice energy
will be (higher melting point). There is an inverse
relationship between the lattice energy and the
size of the ion. Therefore, the smaller the ion is,
the greater the lattice energy will be (higher melt-
ing point). - C—If the reaction is nonspontaneous at 1 atm
and 298 K, then ΔG > 0. For this reaction to
become spontaneous at a lower temperature the
reaction must be impeded by entropy (entropy
was negative). The enthalpy must be negative or
the reaction could never be spontaneous. - A—Enter the information into the Ka expression.
A pH of 2.0 means that [H+] = 10 - 2.0 M or 1.0 ×
10 -^2. Ka = [H+][A-]/[HA]; [H+] = [A-] = 1.0 ×
10 -^2 [HA] = 0.060 - 1.0 × 10 -^2 = 0.05; therefore,
Ka = (1.0 × 10 -^2 )/0.050 = 0.002 = 2 × 10 -^3. - C—A carbon atom with four single bonds
should be tetrahedral. Tetrahedral atoms have
an ideal bond angle of 109.5°. However, the
carbon atoms in cyclopropane are at the corners
of an equilateral triangle, where the ideal angle
is 60°. The discrepancy between the two ideal
bond angles leads to the relative instability of
cyclopropane. - B—The reaction is:
HgO(s) + 2 HCl(aq) → HgCl 2 (aq) + H 2 O(l)
When the last of the solid dissolves, we are
at the stoichiometric point indicated by this
reaction. No HgO remains and the chemist
stopped adding HCl at this point, so there is no
excess HCl present. The HgCl 2 must be soluble
because there is no solid remaining in the beaker.
Since the resultant solution is nonconducting,
there must be no electrolytes present; therefore,
the products must be molecular species. The
best answer should show only water molecules
and HgCl 2 molecules.- D—This answer shows the potassium ions and
bromide ions as separate ions in solution, which
is a property of strong electrolytes. The water
molecules are present in both the liquid and gas
states. Other diagrams incorrectly show water
dissociating, potassium bromide ion pairs, and
potassium bromide vaporizing. - B—There are two bromine isotopes with equal
abundances, which is why there are two nearly
equally intense peaks in the mass spectrum
of CH 3 Br. One of the peaks corresponds to
CH 379 Br and the other corresponds to CH 381 Br.
The separation between the peaks is two mass
units because the isotopic masses differ by two
units. In the case of CH 2 Br 2 , the possible com-
binations and masses are CH 279 Br^79 Br (172),
CH 279 Br^81 Br (174), CH 281 Br^79 Br (174), and
CH 281 Br^81 Br (176). The four possible combina-
tions are equally probably since the abundances
of the two bromine isotopes are nearly equal. If
they all had different masses, the mass spectrum
would consist of four equally intense peaks.
However, two of the combinations have the
same mass, which results in the 174 peak being
twice as intense as the other two peaks. - C—In diagram A, the forces to overcome are
ionic bonding (lattice energy) and hydrogen
bonding. In diagram B, the forces to overcome
are hydrogen bonding for both the solvent and
the solute. In diagram C, the forces to overcome
are London dispersion forces in both cases;
however, since the chlorine is a gas, the London
dispersion forces have already been overcome. In
diagram D, the forces to overcome are London
dispersion forces and hydrogen bonding. Since
London dispersion forces are normally weaker
than the other intermolecular forces, answer C
requires the least amount of energy.
20-Moore_PE01_p307-340.indd 326 31/05/18 1:58 pm