Download PRACTICE EXAM 1-C

PRACTICE EXAM 1-C
CHEMISTRY E-1ax
Problems from this exam will be discussed in sections
prior to the exam.
LAST NAME: ________________________________________
FIRST NAME: ________________________________________
You should give yourself 2 hours to complete this exam
NOTES:
1. This exam has 6 pages with 6 problems, plus the cover sheet, useful information,
periodic table, and scrap paper.
2. Note the point values of each exam question, and consider where you can best spend
your time.
3. Answer all questions on the exam sheets. Put your final answers in the boxes
provided. If you must use the back, then please indicate that clearly!
4. Do not use red pen!
5. For full credit, show all calculations; this also helps us award partial credit.
6. All numerical answers must include correct units for full credit.
1. ____________
2. ____________
3. ____________
4. ____________
5. ____________
6. ____________
TOTAL: ___________ / 100
1
1.
Potassium is an element that is biologically necessary for the proper functioning of nerve
and muscle cells in the human body. A shortage of potassium, called hypokalemia, can
result in abnormal heart rhythms and in some cases can be life-threatening.
a)
The most abundant natural isotope of potassium is 39
19 K . Determine the number of
protons (p), neutrons (n), and electrons (e) in one atom of this isotope of potassium. (3 pts)
# of p =
b)
# of n =
Most people are able to maintain normal potassium levels simply through a standard diet
because many different foods are rich in potassium. For example, one average banana
contains 422 mg of potassium. Determine the number of potassium atoms in one average
banana. (6 pts)
number of K atoms =
c)
atoms
Potassium levels in the blood are measured in milliequivalents (mEq) of potassium per
liter of blood, or “mEq/L”. A normal-range potassium level is around 4.3 mEq/L.
Determine the mass of potassium contained in 1.00 pint of blood with a potassium level
of 4.3 mEq/L. (7 pts)
Some Useful Conversions:
1 Eq = 1000 mEq
1 Eq of potassium = 1 mole of potassium
1 liter = 2.11 pints
mass of potassium =
1
# of e =
grams
( ______ / 16 pts)
2
2.
Compound X contains only the elements C, H, and O.
a)
When 6.00 grams of Compound X are completely combusted in excess oxygen,
8.80 g of CO2 and 3.60 g of H2O are collected. Determine the empirical formula of
Compound X. (6 pts)
empirical formula:
b)
When 6.00 grams of Compound X are titrated with NaOH, it is determined that
Compound X is a monoprotic acid, and 44.4 mL of 1.50-molar NaOH are required to
completely neutralize the sample. Determine the molecular formula of Compound X. (6 pts)
molecular formula:
c)
2
Write a complete, balanced equation for the combustion of liquid Compound X.
(Please include state symbols such as (s), (aq), etc.) (4 pts)
( ______ / 16 pts)
3
3.
A hydrated salt is an ionic substance that has a specific number of water molecules
enclosed within its crystal structure. One example of a hydrated salt is CoCl2·6H2O. As
indicated in this chemical formula, this hydrated salt consists of the ionic solid cobalt (II)
chloride with 6 molecules of water trapped in the crystal structure.
a)
When solid CoCl2·6H2O is heated, all of the water is driven off as gaseous water, and
pure solid cobalt (II) chloride (without any water of hydration) will remain. In the box
below, write a complete balanced equation for the complete dehydration of solid
CoCl2·6H2O upon heating. (Please include state symbols such as (s), (aq), etc.) (4 pts)
b)
You are given 3.196 grams of solid CoCl2·6H2O. How many moles of hydrated salt is
this? (Hint: Don’t forget to account for the mass of water in determining the molar
mass!) (4 pts)
number of moles =
c)
You heat this hydrated salt to drive off the water, but unfortunately you do not heat it
long enough, and the reaction does not go to completion. The final mass of solid
substances after the reaction is 1.854 grams. (Note that this includes the dehydrated
CoCl2 product as well as some hydrated salt that has not been completely dehydrated!)
Determine the number of moles of water that were removed from the original hydrated
salt. (4 pts)
moles of H2O =
d)
Determine the percent yield for this reaction by using a ratio of the moles of water
actually removed from the hydrated salt to the total moles of water originally present in
the hydrated salt. (4 pts)
percent yield =
3
( ______ / 16 pts)
4
4.
a)
Write the chemical formula for each of the following species. (1 pt each)
Barium Thiocyanate
Copper (I) Cyanide
Aluminum Sulfide
Sulfur Trioxide
b)
Write an acceptable chemical name for each of the following. (1 pt each)
FeCO3
NH3
NO
c)
Write the oxidation state of each atom in the box above the atom. (1 pt per box)
H2PO4
d)
–
MgH2
For parts (i) and (ii) below, consider the following balanced chemical reaction and circle the
best answer to each question:
Si + 2 Cl2 !
SiCl4
3 moles of Si are mixed with 4 moles of Cl2 and this reaction proceeds to completion.
i) How many moles of SiCl4 will be formed? (2 pts)
a) 2 moles
b) 3 moles
c) 4 moles
d) 5 moles
e) 7 moles
d) 2 moles
e) 3 moles
ii) How many moles of Si will remain unreacted? (2 pts)
a) 0 moles
e)
4
b) 1 mole
c) 1.5 moles
In the space below, write a balanced net-ionic equation for the complete neutralization of
sulfuric acid with potassium hydroxide. (You do not need to include state symbols such as
(s), (aq), etc.) (You may wish to write the complete reaction and do some work on scrap
paper, but we will only grade the net-ionic reaction written in the box below.) (2 pts)
( ______ / 18 pts)
5
5.
The following two solutions are mixed together:
450.0 mL of 0.200-molar silver nitrate, AgNO3
350.0 mL of 0.200-molar sodium phosphate, Na3PO4
A yellow precipitate of silver phosphate, Ag3PO4, is formed in a double displacement reaction.
a)
Write a complete, balanced equation, and then a net ionic equation for this chemical
reaction. (Please include state symbols such as (s), (aq), etc.) (6 pts)
Complete:
Net Ionic:
b)
Calculate the mass of silver phosphate that would be formed, assuming the reaction goes
to completion, and determine the molar concentration of all ions in the resulting solution.
(Note: Assuming that the reaction goes to completion, some of the concentrations may be
effectively zero.) (12 pts)
mass of Ag3PO4 formed =
[Ag+] =
[Na+] =
5
[NO3–] =
[PO43–] =
( ______ / 18 pts)
6
6.
You are given a mixture of barium hydroxide, Ba(OH)2, and strontium hydroxide,
Sr(OH)2. You dissolve this mixture in water and titrate it with hydrochloric acid.
Complete neutralization of the mixture requires 96.0 mL of 1.50 M HCl. (Note that
Sr(OH)2 is a strong base that reacts with HCl in the same manner as Ba(OH)2.) After
performing the titration, you add excess sodium sulfate to the resulting solution,
completely precipitating solid barium sulfate, BaSO4, and solid strontium sulfate, SrSO4.
A total of 15.31 grams of precipitate is collected. Determine the number of moles of
Ba(OH)2 and Sr(OH)2 that were in the original mixture. (16 pts)
number of moles of Ba(OH)2 =
moles
number of moles of Sr(OH)2 =
moles
6
( ______ / 16 pts)
Scrap Paper
Nothing on this page will be graded unless you clearly indicate
on a specific problem that additional work is located here.
NAME:_____________________________
Useful Information
This page will NOT be collected after the exam.
We will NOT grade anything written on this page.
Avogadro’s Number = 6.02 ! 1023
Density of water at 25°C = 1.00 g/mL
Unit Conversions:
1 mL = 1 cm3
1 foot = 12 inches
1 inch = 2.54 cm
1 pound = 453.6 grams
27
Co
28
Ni
29
Cu
30
Zn
Fr
Cs
227.03
Ac
138.91
89
La
Y
88.91
57
(261)
Rf
178.49
104
Hf
Zr
91.22
72
47.88
40
Actinide series
Lanthanide series
Ra
Ba
Sr
44.96
39
231.04
Pa
232.04
Th
140.91
91
140.12
90
59
Pr
58
Ce
(263)
183.85
[106]
W
Mo
95.94
74
52.00
42
(262)
Ha
180.95
105
Ta
Nb
92.91
73
50.94
41
238.03
U
144.24
92
60
Nd
(262)
186.21
[107]
Re
Tc
(98)
75
54.94
43
237.05
Np
(145)
93
61
Pm
(265)
190.20
[108]
Os
Ru
101.07
76
55.85
44
(244)
Pu
150.36
94
62
Sm
(266)
192.22
[109]
Ir
Rh
102.91
77
58.93
45
(243)
Am
151.96
95
63
Eu
195.08
Pt
Pd
106.42
78
58.69
46
(247)
Cm
157.25
96
64
Gd
196.97
Au
Ag
107.87
79
63.55
47
(247)
Bk
158.93
97
65
Tb
200.59
Hg
Cd
112.41
80
65.39
48
(251)
Cf
162.50
98
66
Dy
204.38
Tl
In
114.82
81
69.72
49
Ga
(252)
Es
164.93
99
67
Ho
207.20
Pb
Sn
118.71
82
72.61
50
Ge
(257)
Fm
167.26
100
68
Er
208.98
Bi
Sb
121.76
83
74.92
51
As
(258)
Md
168.93
101
69
Tm
(209)
Po
Te
127.60
84
78.96
52
Se
(259)
No
173.04
102
70
Yb
(210)
At
I
126.91
85
79.90
53
Br
35.45
35
Cl
226.03
26
Fe
S
32.07
34
19.00
17
(223)
25
Mn
P
30.97
33
16.00
16
137.33
88
24
Cr
Si
28.09
32
14.01
15
132.91
87
V
23
Al
26.98
31
12.01
14
87.62
56
Rb
Ca
Mg
10.81
13
(260)
Lr
174.97
103
71
Lu
(222)
Rn
Xe
131.29
86
83.80
54
Kr
Ar
39.95
36
20.18
18
Ne
85.47
55
Ti
22
F
40.08
38
21
Sc
O
39.10
37
K
Na
Li
24.31
20
N
9
22.99
19
C
8
9.012
12
B
7
6.941
11
Be
6
4.003
10
5
4
1.008
3
2
He
H
1
PERIODIC TABLE OF THE ELEMENTS