CHEMISTRY 31: Section 1: MTWR 8:40
SS 96 - Final
Name:
_______________________________________
TA:
____________________________
Student Number: __ __ __ __ __ __
Lab Section:
A B C
Show your work, include the formulas and appropriate units, and write the answer to a
reasonable number of significant digits. Calculators may be used. No books or notes may
be used -- except for your pre-approved notecard. No talking. Do your own work. Quickly
read all of the problems before starting the test; do the easy questions first!
(04)
1.
Convert the following:
(a)
797.0 cg to ng
(a)!!!797.0!cg!x!
10 –2 g
1!ng
!x! –9 !=!7.970!x!10 9 !ng
1!cg
10 g
(10 m )
(b)!!!!5500.0!dm !x!
–1
3
(09)
2.
(1!dm )
3
3
!=!5.5000!m 3
Balance the following chemical equations:
(a)
1 Ca(H2PO4)2 (s) + 2 NaHCO3 (s) --->
2 CO2 (g) + 1 CaHPO4
(b)
(14)
3.
5500.0 dm3 to m3
(b)
(s)
+ 2 H2O (g) + 1 Na2HPO4 (s)
2 CH3OH (l) ---> 1 C2H6O (g) + 1 H2O (l)
Complete the following table:
Symbol
Mass
Number
Atomic
Number
Protons
Electrons
Neutrons
Charge
78Se2-
78
34
34
36
44
2–
95Sr2+
95
38
38
36
57
2+
3H+
3
1
1
0
2
1+
(01)
4.
Wasn’t this page easy?
Page 1 of 8
CHEMISTRY 31: Section 1: MTWR 8:40
Name:
(20)
(20)
(01)
SS 96 - Final
_______________________________________
5.
6.
7.
Student Number: __ __ __ __ __ __
Write the correct chemical name for the following compounds:
cadmium disulfite
CdS2O5
hypobromus acid
HBrO (aq)
iron (III) cyanate
Fe(OCN)3
magnesium peroxide
MgO2
rubidium oxalate
Rb2C2O4
strontium dihydrogen phosphite
Sr(H2PO3)2
ammonium tartrate
(NH4)2C4H4O6
sulfuric acid
H2SO4(aq)
diphosphorous pentoxide
P2O5
lithium chlorate
LiClO3
Write the correct formula for the following compounds:
Cl2O7
dichlorine heptoxide
Cd(C2H3O2)2
cadmium acetate
Hg2Cr2O7
mercury (I) dichromate
HIO4 (aq)
periodic acid
CsOCl or CsClO
cesium hypochlorite
HF (aq)
hydrofluoric acid
AsCl5
arsenic pentachloride
HNO2 (aq)
nitrous acid
KMnO4
potassium permanganate
Sr(NO3)2
strontium nitrate
Don’t you wish that you had done a better job of learning the names and
symbols for the acids and polyatomic ions?
Page 2 of 8
CHEMISTRY 31: Section 1: MTWR 8:40
Name:
(10)
SS 96 - Final
_______________________________________
8.
Student Number: __ __ __ __ __ __
If 25.0 mL of a silver nitrate solution reacts with excess potassium chloride
solution to yield 0.642 g of precipitate, what is the molarity of the silver in
the original solution?
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
!
1!mol!AgCl
1!mol!AgNO3 $
#" 0.642!g!AgCl!x! 143.3212!g!AgCl !x! 1!mol!AgCl &%
M AgNO3 !=!
!=!0.179!M !AgNO3
1!L $
!
#" 25.0!mL!x! 3
&
10 !mL %
(09)
9.
During an experiment, a student collected the following data. Determine
water’s density and the experimental error for the student’s experiment.
Temperature: 92 oC
beaker + parafilm
beaker + parafilm + 25 mL H2O
beaker + parafilm + 50 mL H2O
beaker + parafilm + 75 mL H2O
Accepted water density: 963.9693 kg/m3
Mass (g)
37.8736
62.7739
87.7989
112.8223
Average mass of 25.00 mL H2O
Mass of Water Added (g)
----------------------------------------24.9003
25.0250
25.0234
24.9829
m 24.9829!g
g
!=!
!=!0.9993216!
V
25.00!mL
mL
exp erimental!!!accepted
Error!=!
!x!100
accepted
d!=!
(
Error!=!
(
10.
)
3
"
10 –2 !m
kg 10 3 !g
1!cm 3 %
! $ 963.9693! 3 !x!
!x!
!x!
'
m
1!kg
$#
(1!cm )3 1!mL '&
Density: 0.9993216 g/mL
(01)
)
3
"
10 –2 !m
g
kg 10 3 !g
1!cm 3 %
0.9993216!
!!! $ 963.9693! 3 !x!
!x!
!x!
'
mL $#
m
1!kg
(1!cm )3 1!mL '&
!x!100!=!3.66!%
Experimental Error: 3.66%
Isn’t this fun?
Page 3 of 8
CHEMISTRY 31: Section 1: MTWR 8:40
Name:
(12)
SS 96 - Final
_______________________________________
11.
Student Number: __ __ __ __ __ __
What mass of aluminum hydroxide must be dissolved to prepare 1750.00 mL
of a solution whose [OH⎯] = 0.562341 M?
1!L $
1!mol!Al(OH )3 78.003558!g!Al(OH )3
!
–
!x!
!=
#" 1750.00!mL!x! 3
&% 0.562341!M !OH !x!
10 !mL
3!mol!OH –
1!mol!Al(OH )3
(
)
25.5877!g!Al(OH )3
(15)
12.
Menthol (MM = 156.3), a strong-smelling substance used in cough drops, is a
compound of carbon, hydrogen, and oxygen. When 0.1595 g menthol was
subjected to combustion analysis, it produced 0.449 g CO2 and 0.184 g H2O.
What is menthol’s empirical formula?.
1!mol!CO2
1!mol!C
12.0107!g!C
!x!
!=!0.0102023!mol!C!x!
!=!0.122537!g!C
44.0095!g!CO2 1!mol!CO2
1!mol!C
1!mol!H 2O
2!mol!H
1.00794!g!H
0.184!g!H 2O!x!
!x!
!=!0.0204271!mol!H !x!
!=!0.020589!g!H
18.015288!g!H 2O 1!mol!H 2O
1!mol!H
1!mol!O
g!O!=!0.1595!g!menthol!–!(0.122537!g!C!+!0.020589!g!H )!=!0.016374!g!O!x!
!=!0.0010234!mol!O
15.9994!g!O
0.0102023!mol!C
# C!=!
!=!10
0.0010234!mol!O
0.0204271!mol!H
# H !=!
!=!20
0.0010234!mol!O
0.0010234!mol!O
# O!=!
!=!1!!!!!!!!!!!empirical! formula :!!C10 H 20O!!! formula!mass!=!156.265
0.0010234!mol!O
molecular! formula!=!C10 H 20O
0.449!g!CO2 !x!
(01)
13.
Are you tired yet?
Page 4 of 8
CHEMISTRY 31: Section 1: MTWR 8:40
Name:
(09)
SS 96 - Final
_______________________________________
14.
Student Number: __ __ __ __ __ __
Write complete and balance the molecular and net ionic (or No Reaction,
where appropriate) equations for the following aqueous reactions:
a.
nitric acid + silver sulfate --->
2 HNO3 (aq) + Ag2SO4 (aq)  2 AgNO3 (aq) + H2SO4 (aq)
NO Net Reaction
b.
lead (IV) nitrate + iron (III) chloride --->
3 Pb(NO3)4 (aq) + 4 FeCl3 (aq)  3 PbCl4 (aq) + 4 Fe(NO3)3 (aq)
No Net Reaction
c.
mercury (I) nitrate + sodium sulfate --->
Hg2(NO3)2 (aq) + Na2SO4 (aq)  Hg2SO4 (aq) + 2 NaNO3 (aq)
No Net Reaction
(10)
15.
A 112.0251 g irregularly shaped object is placed in a graduated cylinder that
contains 15.0 cm3 of water. What is the solid’s density if the new water level
was 47.5 cm3?
d!=!
m
V
d!=!
112.0251!g
g
!=!3.44! 3
3
3
(47.5!cm !–!15.0!cm )
cm
(01)
16.
How are you doing so far?
Page 5 of 8
CHEMISTRY 31: Section 1: MTWR 8:40
Name:
(04)
SS 96 - Final
_______________________________________
17.
Student Number: __ __ __ __ __ __
For the following “basic” oxidation-reduction reaction:
Ni2+(aq) + Br2(l) ---> NiO(OH) (s) + Br⎯(aq)
Which substance acts as the reducing agent? Ni2+
Which substance acts as the oxidizing agent? Br2
(10)
18.
A student placed 50.00 mL of a basic solution of unknown concentration and
three drops of phenolphthalein into a beaker. After 64.37 mL of the 0.3450 M
acid was added, the phenolphthalein changed from pink to colorless. What
was the base’s [OH⎯])?
MaVa = MbVb
(64.37 mL)(0.3450 M) = (Mb)(50.00 mL)
Mb = 0.444 M OH–
(01)
19.
Wasn’t this page easy?
Page 6 of 8
CHEMISTRY 31: Section 1: MTWR 8:40
Name:
(20)
SS 96 - Final
_______________________________________
20.
Student Number: __ __ __ __ __ __
A chemical engineer determines the mass percent of iron in an ore sample by
converting the Fe to Fe2+ in acid and then titrating the Fe2+ with MnO4⎯. A
1.1081 g sample was dissolved in acid and titrated with 39.32 mL of 0.03190
M KMnO4. The reaction is:
8 H+(aq) + 5 Fe2+(aq) + MnO4⎯(aq) ---> 5 Fe3+(aq) + Mn2+(aq) + 8 H2O(aq)
1!L $
5!mol!Fe2 +
1!mol!Fe
55.845!g!Fe
!
–
39.32!mL!x!
0.03190!M
!MnO
!x!
!x!
!x!
#"
&%
4
3
–
2+
10 !mL
1!mol!MnO4 1!mol!Fe
1!mol!Fe
%!Fe!=!
!x!100
1.1081!g!sample
%!Fe!=!31.61!%
(
(01)
21.
)
Seven pages down, one more to go. How are you feeling?
Page 7 of 8
CHEMISTRY 31: Section 1: MTWR 8:40
Name:
(25)
SS 96 - Final
_______________________________________
22.
Student Number: __ __ __ __ __ __
What volume of oxygen (20.00 oC, 740.5 torr) is required for the complete
combustion of 51.350 L of toluene (C6H5CH3 , d = 0.8669 g/cm3) to produce
carbon dioxide and water vapor?
What is the percentage yield if 142.85 kL of carbon dioxide (395.0 oC,
1235 mB) is actually produced? What mass of carbon dioxide is actually
produced?
1 C6H5CH3 (l) + 9 O2 (g) ---> 7 CO2 (g) + 4 H2O (g)
!
10 3 !cm 3 $ !
g $
1!mol!C6 H 5CH 3
nC6 H 5 CH 3 !=! # 51.350!L!x!
!=!483.135!mol!C6 H 5CH 3
#" 0.8669! 3 &% !x!
&
1!L %
cm
92.13842!g!C6 H 5CH 3 !
"
!483.135!mol!C6 H 5CH 3 !x!
VO2 !=!
9!mol!O2
atm!L $
!
!x! # 0.082057!
& ( 20.00!+!273.15!K )!
1!mol!C6 H 5CH 3 "
mol!K %
!=!
1!atm $
!
#" 740.5!torr!x!
&
760!torr %
VO2 !=!1.0735!x!10 5 !L
ntheoretical,!CO2 !=!483.135!mol!C6 H 5CH 3 !x!
7!mol!CO2
!=!3381.945!mol!CO2 !theoretical
1!mol!C6 H 5CH 3
1!atm $ !
10 3 !L $
!
1235!mB!x!
142.85!kL!x!
#"
&
1013.25!mB % #"
kL &%
nactual,!CO2 !=!
!=!3175.711!mol!CO2 ! actual
atm!L $
!
#" 0.082057!
& ( 395.0!+!273.15!K )
mol!K %
44.0095!g!CO2
3175.711!mol!CO2 ! actual !x!
!=!1.3976!x!10 5 !g!CO2
1!mol!CO2
actual
3175.711!mol!CO2 ! actual
%!yield!=!
!x!100!=!
!x!100!=!93.90%
theoretical
3381.945!mol!CO2 !theoretical
(02)
23.
Yea! You’re done. Have a nice break. Aren’t you glad this is over with?
Page 8 of 8