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A midterm exam key for the chm 115 course at the university of x, given by dr. Scott tremain in fall 2002. It includes instructions for the exam, points distribution, and questions covering various topics such as terminology, conversions, chemical equations, and calculation problems.
Typology: Exams
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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.
__ 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.
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) 12 p, 13 e b) 24 p, 26 e c) 12 p, 10 e d) 24 p, 22 e e) 12 p, 14 e
a) sucrose b) CaCl 2 c) H 3 PO 4 d) NH 3 e) HC 2 H 3 O 2 (CH 3 COOH)
a) NaOH b) KOH c) Mg(OH) 2 d) Ca(OH) 2 e) Sr(OH) 2
a) –1 b) +1 c) +2 d) +4 e) +
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.
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.
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)
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
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
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
molMgCl mol AgNO =
molCl ionsareleft unreacted molMgCl
molCl mol MgCl −
− ^ =
2
2
MCl left unreacted L
molCl L
mol M −
− = = = 0. 552
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
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
where s K mol
kg m R ⋅ ⋅
2