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NAME__ KEY ___________________
MIDTERM EXAM KEY
CHM 115 FALL 2002
DR. SCOTT TREMAIN
OCTOBER 18, 2002
THIS EXAM CONTAINS 5 PARTS WITH A TOTAL OF 28 QUESTIONS. PLEASE MAKE
SURE THIS EXAM PACKET CONTAINS 8 PAGES.
NOTE: SHOW ALL WORK IN ORDER TO RECEIVE FULL CREDIT. NO CREDIT WILL
BE GIVEN ON CALCULATION PROBLEMS EVEN THOUGH THE CORRECT ANSWER
(MAGICALLY?) APPEARS. NUMERICAL ANSWERS SHOULD BE GIVEN IN THE
CORRECT NUMBER OF SIGNIFICANT FIGURES AND UNITS. PUT YOUR ANSWERS
IN THE ANSWER SPACES IF PROVIDED. THE LAST PAGE CONTAINS A PERIODIC
TABLE AND USEFUL EQUATIONS AND CONSTANTS. I RESERVE THE RIGHT TO
ASK YOU TO REMOVE YOUR CALCULATOR BATTERIES IN ORDER TO CLEAR ITS
MEMORY.
THE ANSWER KEY TO THIS MIDTERM EXAM WILL BE POSTED ON THE COURSE
WEB PAGE LATER TODAY. GRADED MIDTERM EXAMS WILL BE RETURNED AS
SOON AS POSSIBLE.
GRADING
PART POINTS SCORE
I. 27 pts _______
II. 24 pts _______
III. 49 pts _______
IV. 30 pts _______
V. 45 pts _______
TOTAL 175 pts ______
Part I. (27 points) Terminology – Match the following terms with the MOST APPROPRIATE definition by placing the number of the term in the space provided. Not all terms will be used. Write only one number per blank.
- Acid 9. Isotope
- Amphoteric 10. Matter
- Anion 11. Mass number
- Base 12. Mixture
- Cation 13. Mole
- Compound 14. Oxidation
- Effusion 15. Reduction
- Hypothesis 16. Theory
__ 8 __ This is a tentative explanation for a set of observations (educated guess).
__ 6 __ This is a substance composed of two or more elements chemically united in fixed proportions.
__ 9 __ These are atoms which have the same atomic number but different mass numbers.
__ 11 _ This term identifies the total number of protons and neutrons in a nucleus.
__ 3 __ This is an ion with a net negative charge.
__ 13 _ This is the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon-12.
__ 2 __ This describes a species that can both accept a proton and donate a proton.
__ 14 _ This is the process (or half-reaction) in which there is a loss of electrons by a species.
__ 7 __ This is the process by which a gas flows through a small hole in a container.
Part III. (49 points) Multiple Choice. Circle the one correct answer.
- Convert 777 μg to kg. Express your answer in scientific notation.
a) 7.77 x 10^11 kg d) 7.77 x 10–7^ kg b) 7.77 x 10^7 kg e) 7.77 x 10–9^ kg c) 7.77 x 10–4^ kg
- A magnesium ion (Mg2+) has how many protons (p) and electrons (e)?
a) 12 p, 13 e b) 24 p, 26 e c) 12 p, 10 e d) 24 p, 22 e e) 12 p, 14 e
- Which of the following is a strong electrolyte?
a) sucrose b) CaCl 2 c) H 3 PO 4 d) NH 3 e) HC 2 H 3 O 2 (CH 3 COOH)
- Which of the following is NOT a strong base?
a) NaOH b) KOH c) Mg(OH) 2 d) Ca(OH) 2 e) Sr(OH) 2
- What is the oxidation number of N in the compound H 2 N 2 O 2?
a) –1 b) +1 c) +2 d) +4 e) +
- All of the following are postulates of the kinetic-molecular theory of gases EXCEPT
a) gases are composed of molecules whose size is negligible compared with the average distance between each gas molecule. b) gas molecules move randomly in straight lines in all directions at various speeds. c) the average kinetic energy of a molecule is proportional to the absolute temperature d) when gas molecules collide, they both gain kinetic energy. e) the forces of attraction or repulsion between gas molecules are very weak or negligible.
- For a substance that remains a gas under the conditions listed below, deviation from the ideal gas law would be most pronounced at
a) 100°C and 1.0 atm d) –100°C and 10.0 atm b) 0°C and 2.0 atm e) 100°C and 0.1 atm c) –100°C and 1.0 atm
Part IV. (30 points) Chemical Equations.
- (15 points) Predict the products formed in the following precipitation reaction. Write the balanced molecular equation including phase labels on every species. ALSO write the complete ionic equation and net ionic equation.
MgCl 2 (aq) + Na 3 PO 4 (aq) ‡
a) Molecular Equation:
3 MgCl 2 (aq) + 2 Na 3 PO 4 (aq) ‡‡ Mg 3 (PO 4 ) 2 (s) + 6 NaCl(aq)
b) Complete Ionic Equation:
3 Mg2+^ + 6 Cl–^ + 6 Na+^ + 2 PO 4 3–^ ‡‡ Mg 3 (PO 4 ) 2 (s) + 6 Na+^ + 6 Cl–
c) Net Ionic Equation:
3 Mg2+^ + 2 PO 4 3–^ ‡‡ Mg 3 (PO 4 ) 2 (s)
- (15 points) Balance the following oxidation-reduction reaction (in acidic media) using the half-reaction method.
Zn(s) + ClO 3 – (aq) ‡ Zn2+(aq) + Cl–(aq)
Oxidiation ½ Reaction: Zn(s) ‡‡ Zn2+^ + 2 e–
Reduction ½ Reaction: 6 e–^ + 6H+^ + ClO 3 – (aq) ‡‡ Cl–(aq) + 3 H 2 O
Oxidiation ½ Reaction: 3 [Zn(s) ‡‡ 3 Zn2+^ + 2 e–]
Reduction ½ Reaction: 6 e–^ + 6H+^ + ClO 3 – (aq) ‡‡ Cl–(aq) + 3 H 2 O
6 H+^ + 3 Zn(s) + ClO 3 – (aq) ‡‡ 3 Zn2+(aq) + Cl–(aq) + 3 H 2 O
- (25 points) When a solution of AgNO 3 (aq) is added to a solution of MgCl 2 (aq), a white precipitate, AgCl(s), is formed. The balanced chemical equation for this precipitation reaction follows.
2 AgNO 3 (aq) + MgCl 2 (aq) ‡ 2 AgCl(s) + Mg(NO 3 ) 2 (aq)
a) (20 points) If a 250.0 mL solution of 0.500 M AgNO 3 (aq) is mixed with a 250.0 mL solution of 0.800 M MgCl 2 (aq), what is the theoretical yield (in grams) of AgCl(s). You may use the following molar masses: AgNO 3 = 169.9 g/mol; MgCl 2 = 92.2 g/mol; AgCl = 143.3 g/mol; and Mg(NO 3 ) 2 = 148.3 g/mol
Answer: _____ 17.9 g AgCl _______
Instead of providing the mass of each reactant, the molarity and volume of each reactant is provided. Calculate the number of moles of each reactant.
Decide which reactant is the limiting reactant by separately calculating the possible yield of AgCl(s).
Since 0.125 mol AgNO 3 yields 17.9 g AgCl, AgNO 3 is the limiting reactant and MgCl 2 is the excess reactant. Thus 17.9 g AgCl is the theoretical yield.
b) (5 points) What is the concentration (molarity) of the Cl–^ ions that remain unreacted in the combined solution above after the precipitation reaction is complete?
Answer: _ 0.552 M (or 0.550 M) _
Find amount of MgCl 2 left unreacted by first calculating how much of the 0.200 mol MgCl 2 is used up in the reaction with 0.125 mol AgNO 3. (Other ways are possible to find amount of MgCl 2 left unreacted.)
(0.200 mol MgCl 2 to start with) – (0.0625 mol MgCl 2 used up) = 0.138 mol MgCl 2 left unreacted
MgCl 2 ‡ Mg2+^ + 2 Cl–
Volume of combined solution = 0.250 L + 0.250 L = 0.500 L
2 2
3 3 ( 0. 800 )( 0. 250 ) 0. 200
molMgCl M L M L mol MgCl
molAgNO M L M L molAgNO
L
mol M
g AgCl molAgCl
gAgCl molMgCl
molAgCl molMgCl
gAgCl molAgCl
gAgCl molAgNO
molAgCl molAgNO
2
2
3
3
^ =
2 3 3
2 3 2 0.^0625
- 125 molMgCl arerequiredtoreactwithallAgNO molAgNO
molMgCl mol AgNO =
molCl ionsareleft unreacted molMgCl
molCl mol MgCl −
− ^ =
2
2
MCl left unreacted L
molCl L
mol M −
− = = = 0. 552
- 500
2_
1 H
1 H
3 Li
11 Na
19 K
37 Rb
55 Cs
87 Fr (223)
4 Be
12 Mg
20 Ca
38 Sr
56 Ba
88 Ra (226)
21 Sc
39 Y
57 La*
89 Ac** (227)
22 Ti
40 Zr
72 Hf
104 Rf (261)
23 V
41 Nb
73 Ta
105 Db (262)
24 Cr
42 Mo
74 W
106 Sg (263)
25 Mn
43 Tc (98) 75 Re
107 Bh (262)
26 Fe
44 Ru
76 Os
108 Hs (265)
27 Co
45 Rh
77 Ir
109 Mt (266)
28 Ni
46 Pd
78 Pt
110 (269)
111 (272)
112 (277)
29 Cu
47 Ag
79 Au
30 Zn
48 Cd
80 Hg
5 B
13 Al
31 Ga
49 In
81 Tl
6 C
14 Si
32 Ge
50 Sn
82 Pb
7 N
15 P
33 As
51 Sb
83 Bi
8 O
16 S
34 Se
52 Te
84 Po (209)
9 F
17 Cl
35 Br
53 I
85 At (210)
2 He
10 Ne
18 Ar
36 Kr
54 Xe
86 Rn (222)
58 Ce
90 Th
59 Pr
91 Pa
60 Nd
92 U
61 Pm (145) 93 Np (237)
62 Sm
94 Pu (244)
63 Eu
95 Am (243)
64 Gd
96 Cm (247)
65 Tb
97 Bk (247)
66 Dy
98 Cf (251)
67 Ho
99 Es (252)
68 Er
100 Fm (257)
69 Tm
101 Md (258)
70 Yb
102 No (259)
71 Lu
103 Lr (262)
1 IA
1 2 3 4 5 6 7 2 IIA
3 IIIB
4 IVB
5 VB
6 VIB
7 VIIB
*Lanthanides
Metal
**Actinides
11 IB
12 IIB
13 IIIA
14 IVA
15 VA
16 VIA
17 VIIA
18 VIIIA
9 8 VIIIB 10
Transition Metals
Atomic number Symbol Atomic weight
Inner-Transition Metals
Main-Group Elements Main-Group Elements
Metalloid
Nonmetal
INFORMATION
volume
mass density = or V
m d = PV = nRT
NA = 6.022 x 10^23 things/mole mol K
L atm R ⋅
molar mass
mass moles = or MW
m n = Ptot = PA + PB + PC + …
liters solution
molessolute molarity = or L
mol M = Patm = Pgas + PH2O
M (^) 1 V 1 = M 2 V 2 u^ =^ MW
3 RT
where s K mol
kg m R ⋅ ⋅
2
- 314
