11/02/16
CHEM 101 Exam 2 (Ch. 4, 5, 6)
Name:
Open-End Questions and Problems
ANSWER KEY
Please read the following. To receive full credit for a question or a problem, in addition to the
correct answer, you must show a neat, complete, and logical method of solution where each
number is labeled with the appropriate unit and the final answer is rounded to the correct number
of significant digits. The correct answer without any work shown will generally get zero credit!
There are 4 questions for a total of 24 points. (Part of the exam with multiple choice questions is
worth 96 points. Both parts combined are 120 points total.
1. (6 points) Give two examples of each. Note: (i) do not use the same substance as an example
more than once; (ii) paper, plastic, wood, a stick, gasoline, milk, food are not pure chemical
substances and should not be used as examples.
Physical Change (a change that involves
Chemical Change (a change that involves
a specific pure substance)
a specific pure substance)
Example 1:
Example 1: ice (solid water) melts
potassium reacts with water to produce
potassium hydroxide and hydrogen gas
Example 2: sugar (sucrose) dissolves in
water
Example 2: sulfur burns in air to produce
sulfur dioxide
Physical Property (specific property of
Chemical Property (specific property of
a specific pure substance)
a specific pure substance)
Example 1: density of aluminum is 2.70 g/cm3 Example 1: bromine reacts with iron to
produce iron(III) bromide
Example 2: ammonia is gas at room
temperature and normal pressure
Example 2: neon does not react with other
chemical elements to form any stable
compounds
2. (6 points) In standardizing hydrochloric acid, 22.5 mL were required to neutralize 25.0 mL of
0.0500 M sodium carbonate solution. What is the molarity of the hydrochloric acid solution?
How much water must be added to 200 mL of it to make it 0.100 M?
Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)
25.0 mL × 0.0500 M = 1.25 mmol Na2CO3
1.25 mmol × 2 = 2.50 mmol HCl
(2.50 mmol HCl) / 22.5 mL = 0.111 M
dilution factor = (0.111 M) / (0.100 M) = 1.11
200 mL × 1.11 (final soln.) – 200 mL (initial soln.) = 22.0 mL water
3. (6 points) A chemist combines 60.0 mL of 0.322 M potassium iodide with 20.0 mL of 0.530
M lead(II) nitrate. (a) Write the full-formula, complete ionic, and net ionic equation (FFE,
CIE, and NIE) for the reaction that occurs. In each equation, indicate physical state of each
reactant and product: (s), (l), (g), (aq). (b) What precipitate forms? (c) What is the limiting
reactant? (d) How many grams of the precipitate form? (e) What is the concentration of each
ion that remains in solution?
Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq)
Pb2+(aq) + 2 NO3−(aq) +2 K+(aq) + 2 I−(aq) → PbI2(s) + 2 K+(aq) + 2 NO3−(aq)
Pb2+(aq) + 2 I−(aq) → PbI2(s)
mmol before the reaction
change in mmol
mmol after the reaction
Concentration in the final
solution (mol/L) *
Pb2+
NO3−
K+
I− (L.R.)
PbI2(s)
10.6
21.2
19.3
19.3
0
−9.66
0
0
−19.3
+9.66
0.94
21.2
19.3
0
9.66
0.0118
0.265
0.241
0
N/A
* Volume of the final solution: 20.0 mL + 60.0 mL = 80.0 mL
Mass of the precipitate: 0.00966 mol PbI2 × (461.0 g/mol) = 4.45 g PbI2
4.
(6 points) It is difficult to prepare many compounds directly from their elements, so ΔH°f
values for these compounds cannot be measured directly. For most organic compounds, it is
easier to measure the standard enthalpy of combustion by reaction of the compound with
excess O2(g) to form CO2(g) and H2O(l). The standard enthalpy of combustion of phenol,
C6H5OH, is −3053 kJ/mol at 25°C.
(a) Determine ΔH°f for the compound.
C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l); ΔH° = ─3053 kJ
ΔH°f, kJ/mol
X
0
─393.5
─285.8
6 mol × (─393.5 kJ/mol) + 3 mol × (─285.8 kJ/mol) ─ 1 mol × (X kJ/mol) = ─3053 kJ
X = 3053 ─ 6×393.5 ─ 3×285.8
X = ─165.4
ΔH°f[C6H5OH(s)] = ─165.4 kJ/mol
(b) Write the thermochemical equation for the reaction of formation (not of combustion) of
phenol.
6 C(graphite, s) + 3 H2(g) + ½ O2(g) → C6H5OH(s); ΔH° = ─165.4 kJ