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First exam
Exercises
First Exam/ Exercises
1- Prefixes giga and deci represent, respectively:
a) 10-9 and 10-1. b) 106 and 10-3. c) 103 and 10-3
d) 109 and 10-1.
2- If 586 g of bromine occupies 188 mL, what is the density of bromine in
g/mL?
a) 3.12g/ml
b) 0.321 g/ml
c) 3.63g/ml
d) 3.08g/ml
d = m/V
m = 586 g, V = 188 mL
d = 586 / 188 = 3.12 g/ml
First Exam/ Exercises
1- Prefixes giga and deci represent, respectively:
a) 10-9 and 10-1. b) 106 and 10-3. c) 103 and 10-3
d) 109 and 10-1.
2- If 586 g of bromine occupies 188 mL, what is the density of bromine in
g/mL?
a) 3.12g/ml
b) 0.321 g/ml
c) 3.63g/ml
d) 3.08g/ml
3- 3.5 nanometer contains how many micrometers?
a) 3.5x 10+3
b) 3.5x 10-3
c) 3.5x 10+8
d) 3.5x 10-8
First convert from nm to m :
3.5 X 10-9m
Then from m to um
3.5 x 10-9 x 10+6 = 3.5 x 10-3
First Exam/ Exercises
1- Prefixes giga and deci represent, respectively:
a) 10-9 and 10-1. b) 106 and 10-3. c) 103 and 10-3
d) 109 and 10-1.
2- If 586 g of bromine occupies 188 mL, what is the density of bromine in
g/mL?
a) 3.12g/ml
b) 0.321 g/ml
c) 3.63g/ml
d) 3.08g/ml
3- 3.5 nanometer contains how many micrometers?
a) 3.5x 10+3
b) 3.5x 10-3
c) 3.5x 10+8
d) 3.5x 10-8
4- The element in group 2A and period 3 is:
a) Ga
b) Be
c) Al
d) Mg
First Exam/ Exercises
5- Predict the formula of a compound formed from Sr and Cl?
a) Sr2Cl
b) Sr2Cl2
c) SrCl2
d) SrCl
Sr+2
SrCl2
Cl-
First Exam/ Exercises
5- Predict the formula of a compound formed from Sr and Cl?
a) Sr2Cl
b) Sr2Cl2
c) SrCl2
d) SrCl
6- Which compound has the same empirical formula as C6H12O6?
a) C12H20O4
b) C2H8O4
c) C6H3O6
d) C12H24O12
C6H12O6 divided by 6 = CH2O
a- C12H20O4 divided by 4 = C3H5O
b- C2H8O4 divided by 2 = CH4O2
c- C6H3O6 divided by 3 = C2HO2
d- C12H24O12 divided by 12 = CH2O
First Exam/ Exercises
5- Predict the formula of a compound formed from Sr and Cl?
a) Sr2Cl
b) Sr2Cl2
c) SrCl2
d) SrCl
6- Which compound has the same empirical formula as C6H12O6?
a) C12H20O4
b) C2H8O4
c) C6H3O6
d) C12H24O12
7- An atom is
a) Smallest unit of matter that maintains its chemical identity.
b) The smallest unit of a compound.
c) Always made of carbon
d) Smaller than electron
First Exam/ Exercises
8-The right answer is:
a)Compounds are: N2, NH3 , H2O and CH4
b)Elements are: N2, H2O, H2 and F2
c) Molecules are: N2, H2, F2 and Cl2 and Fe2SO4
d) Compounds are: NH3, O2, H2O and CH4
9- Rubidium (Rb) and cesium (Cs) are members of which of the following
categories?
a) Halogens
b) Alkali metals
c) Alkaline earth metals d) Noble gases
10- What is the number of protons, electrons and neutrons in the atom of
a) 32 protons, 34 electrons, 16 neutrons
c) 16 protons, 18 electrons, 16 neutrons
b) 16 protons, 16 electrons, 16 neutrons
d) 18 protons, 16 electrons, 32 neutrons
32
16
S 2
First Exam/ Exercises
10- What is the number of protons, electrons and neutrons in the atom of 1632 S 2
a) 32 protons, 34 electrons, 16 neutrons
c) 16 protons, 18 electrons, 16 neutrons
P = 16, e = 16+2 = 18 , n = 32-16 = 16
b) 16 protons, 16 electrons, 16 neutrons
d) 18 protons, 16 electrons, 32 neutrons
First Exam/ Exercises
8-The right answer is:
a)Compounds are: N2, NH3 , H2O and CH4
b)Elements are: N2, H2O, H2 and F2
c) Molecules are: N2, H2, F2 and Cl2 and Fe2SO4
d) Compounds are: NH3, O2, H2O and CH4
9- Rubidium (Rb) and cesium (Cs) are members of which of the following
categories?
a) Halogens
b) Alkali metals
c) Alkaline earth metals d) Noble gases
10- What is the number of protons, electrons and neutrons in the atom of
a) 32 protons, 34 electrons, 16 neutrons
c) 16 protons, 18 electrons, 16 neutrons
b) 16 protons, 16 electrons, 16 neutrons
d) 18 protons, 16 electrons, 32 neutrons
32
16
S 2
First Exam/ Exercises
11- The correct systematic name for Cr2O3 is
a)Chromium (III) oxide.
b) Dichromium trioxide.
c) Chromium trioxide
d) Chromium oxide.
First Exam/ Exercises
compound
Summery of naming
Ionic
Molecular
Cation: metal or NH4+
Nonmetal + nonmetal
Anion: monotomic or polytomic
Nonmetal + metalloid
Cation has only
one charge
Cation has more than
one charge
•Alkali metal
Other metal
cations
•Alkaline earth metal
•Ag+, Al+3, Cd+2, Zn+2
•Name metal first
•Name metal first
•Specify charge of metal cation
with roman numeral (STOCK
SYSTEM)
•If monoatomic anion, add
ide to the anion
•If polyatomic anion use
name of anion from
previous table
•If monoatomic anion, add ide to
the anion
•If polyatomic anion use name of
anion from previous table
Pair Form one type
of compound
•Name first
element
• add ide to the
name of second
element
Pair Form more than
one type
of compound
•Name first
element
• add ide to the
name of second
element
•Add the prefix
(prefix mono
usually omitted
for the first
element
First Exam/ Exercises
11- The correct systematic name for Cr2O3 is
a)Chromium (III) oxide.
b) Dichromium trioxide.
c) Chromium trioxide
d) Chromium oxide.
12- The correct formula for calcium sulfate is
a) CaHSO4
b) CaSO4
c) CaS
d) Ca(HSO4)2
13- Gallium consists of 60.108% 69Ga with a mass of 68.9256 amu, and
39.892% 71Ga with a mass of 70.9247 amu, the average atomic mass of
gallium is:
a) 70.93 amu
b) 71.62 amu c) 68.93 amu
d) 69.723 amu
Average atomic mass = (atomic mass x abundenace)Ga-69+ (atomic mass x
abundenace)Ga-71
Average atomic mass = (68.9256 X (60.108 /100))Ga-69 + ( 70.9247 x (39.892
/100))Ga-71 = 69.723
First Exam/ Exercises
11- The correct systematic name for Cr2O3 is
a)Chromium (III) oxide.
b) Dichromium trioxide.
c) Chromium trioxide
d) Chromium oxide.
12- The correct formula for calcium sulfate is
a) CaHSO4
b) CaSO4
c) CaS
d) Ca(HSO4)2
13- Gallium consists of 60.108% 69Ga with a mass of 68.9256 amu, and
39.892% 71Ga with a mass of 70.9247 amu, the average atomic mass of
gallium is:
a) 70.93 amu
b) 71.62 amu c) 68.93 amu
d) 69.723 amu
14- Which pair of Atoms would be most likely to form an ionic compound?
a) C and N
b) K and Ca
c) P and Ar
d) Ba and S
First Exam/ Exercises
15- Two isotopes of the same element differ only in their
a) Number of neutron
b) Atomic number
c) Number of protons
d) Number of electron
16- What is the mass in grams of one atom of iron (Fe)?
a) 6.02  1023 g b) 1.66  10-24 g
c) 9.3  10-23 g
d) 55.85 g
Mass of one atom = atomic mass/ avogadro number
= 56 / 6.022 x 1023 = 9.3  10-23 g
First Exam/ Exercises
15- Two isotopes of the same element differ only in their
a) Number of neutron
b) Atomic number
c) Number of protons
d) Number of electron
16- What is the mass in grams of one atom of iron (Fe)?
a) 6.02  1023 g b) 1.66  10-24 g
c) 9.3  10-23 g
d) 55.85 g
17- A 4.691 g sample of MgCl2 is dissolved in enough water to give 750 mL of
solution. What is the molarity of this solution?
a) 3.70  10-2 M b) 1.05  10-2 M c) 6.58  10-2 M d) 4.93  10-2 M
M = n/V
n = mass /molar mass
= 4.691 / 95= 0.0493 mole
M = 0.0493 / (750 / 1000)= 0.0658 M = 6.58 X 10-2M
First Exam/ Exercises
15- Two isotopes of the same element differ only in their
a) Number of neutron
b) Atomic number
c) Number of protons
d) Number of electron
16- What is the mass in grams of one atom of iron (Fe)?
a) 6.02  1023 g b) 1.66  10-24 g
c) 9.3  10-23 g
d) 55.85 g
17- A 4.691 g sample of MgCl2 is dissolved in enough water to give 750 mL of
solution. What is the molarity of this solution?
a) 3.70  10-2 M b) 1.05  10-2 M c) 6.58  10-2 M d) 4.93  10-2 M
18- Calculate the percent composition by mass of C in picric acid (C6H3N3O7)?
a) 1.3 %
b) 18.3 %
c) 31.4 %
d) 48.9 %
First Exam/ Exercises
18- Calculate the percent composition by mass of C in picric acid (C6H3N3O7)?
a) 1.3 %
b) 18.3 %
c) 31.4 %
d) 48.9 %
n x molar mass of element
molar mass of compound
Molar mass of C6H3N3O7 =229.114g/mol
% C= 6x 12.01 x 100% = 31.4 %
229.11
x 100%
First Exam/ Exercises
15- Two isotopes of the same element differ only in their
a) Number of neutron
b) Atomic number
c) Number of protons
d) Number of electron
16- What is the mass in grams of one atom of iron (Fe)?
a) 6.02  1023 g b) 1.66  10-24 g
c) 9.3  10-23 g
d) 55.85 g
17- A 4.691 g sample of MgCl2 is dissolved in enough water to give 750 mL of
solution. What is the molarity of this solution?
a) 3.70  10-2 M b) 1.05  10-2 M c) 6.58  10-2 M d) 4.93  10-2 M
18- Calculate the percent composition by mass of C in picric acid (C6H3N3O7)?
a) 1.3 %
b) 18.3 %
c) 31.4 %
d) 48.9 %
First Exam/ Exercises
19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
a) 1.20 x 1022
b) 1.00 x 1022 c) 8.03 x 1021
d) 6.02 x 1021
Molar mass of ethane = 2 x 12 + 6 x 1 = 30 g/mole
First we should calculate the number of mole
Number of mole = mass / molar mass
Number of mole = 0.30 / 30 = 0.01 mole
We know that
Number of molecules = avogadro’s number x number of mole
= 6.022 x 1023 x 0.01
= 6.02 x 1021 molecules.
First Exam/ Exercises
19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
a) 1.20 x 1022
b) 1.00 x 1022 c) 8.03 x 1021
d) 6.02 x 1021
20- The empirical formula of an organic compound with 85.7% C and 14.3%
H is
a) CH
b) CH2
c) C2H
d) CH4
Divided by the smallest number
1- we change from % to g
of mole which is 7.14
85.7 g of C, 14.3 g of H
7.14
14.3
H:
=1
=2
2- change from g to mole using
C:
7.14
7.14
85.7
Thus the empirical formula is CH2
= 7.14 mol of C
nc =
12
14.3
= 14.3 mol of H
nH =
1
First Exam/ Exercises
19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
a) 1.20 x 1022
b) 1.00 x 1022 c) 8.03 x 1021
d) 6.02 x 1021
20- The empirical formula of an organic compound with 85.7% C and 14.3%
H is
a) CH
b) CH2
c) C2H
d) CH4
21- After balancing the following equation the coefficients are:
a C3H8 + b O2 → c CO2 + d H2O
a) a=1, b=5, c=2, d=4
b) a=2, b=5, c=3, d=4
c) a =1, b=3, c=3, d=4
d) a=1, b=5, c=3, d=4
First Exam/ Exercises
21- After balancing the following equation the coefficients are:
a C3H8 + b O2 → c CO2 + d H2O
a) a=1, b=5, c=2, d=4
b) a=2, b=5, c=3, d=4
c) a =1, b=3, c=3, d=4
d) a=1, b=5, c=3, d=4
C3H8 + O2 → CO2 + H2O
C3
C1
X3
C3H8 + O2 → 3 CO2 + H2O
H8
H2
X4
C3H8 + O2 → 3 CO2 + 4 H2O
O 2
O 6 + 4 =10 X 5
C3H8 +
5O2 → 3 CO2 + 4 H2O
First Exam/ Exercises
19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
a) 1.20 x 1022
b) 1.00 x 1022 c) 8.03 x 1021
d) 6.02 x 1021
20- The empirical formula of an organic compound with 85.7% C and 14.3%
H is
a) CH
b) CH2
c) C2H
d) CH4
21- After balancing the following equation the coefficients are:
a C3H8 + b O2 → c CO2 + d H2O
a) a=1, b=5, c=2, d=4
b) a=2, b=5, c=3, d=4
c) a =1, b=3, c=3, d=4
d) a=1, b=5, c=3, d=4
22- Calculate the number of O atoms in 2.50 g of sucrose (C12H22O11)
a) 3.87 x 1022
b) 4.84 x 1022
c) 5.81 x 1022
d) 6.78 x 1022
First Exam/ Exercises
22- Calculate the number of O atoms in 2.50 g of sucrose (C12H22O11)
a) 3.87 x 1022
b) 4.84 x 1022
c) 5.81 x 1022
d) 6.78 x 1022
First we calculate the number of mole
n = 2.5/ 342 = 0.007 mole
Number of molecules = Avogadro's number x number of mole
= 6.022 x 1023 x 0.007
= 4.4 x1022 molecules
From the chemical formula of sucrose C12H22O11
1 molecules of sucrose = 11 atom of O
4.4 x 1021 molecules = ? Atom of O
11 x 4.4 x 1021 = 4.84 x1021 atoms
First Exam/ Exercises
19- How many molecules of ethane (C2H6) are present in 0.30 g of C2H6?
a) 1.20 x 1022
b) 1.00 x 1022 c) 8.03 x 1021
d) 6.02 x 1021
20- The empirical formula of an organic compound with 85.7% C and 14.3%
H is
a) CH
b) CH2
c) C2H
d) CH4
21- After balancing the following equation the coefficients are:
a C3H8 + b O2 → c CO2 + d H2O
a) a=1, b=5, c=2, d=4
b) a=2, b=5, c=3, d=4
c) a =1, b=3, c=3, d=4
d) a=1, b=5, c=3, d=4
22- Calculate the number of O atoms in 2.50 g of sucrose (C12H22O11)
a) 3.87 x 1022
b) 4.84 x 1022
c) 5.81 x 1022
d) 6.78 x 1022
First Exam/ Exercises
23- An empirical formula of an organic compound is C3H4O2 , if the
molecular weight of the compound is (360 g/mol), the molecular formula
of the compound will be:
a) C6H8O4
b) C12H16O8
c) C9H12O6
d) C15H20O10
FIRST we calculate the molar mass of emperical formula
C3H4O2 = 72 g/mol
molar mass of compound
Ratio 
empirical molar mass
Ratio = 360 / 72 = 5
molecular formula = ratio x empirical formula
= 5 x C3H4O2 = C15H20O10
First Exam/ Exercises
23- An empirical formula of an organic compound is C3H4O2 , if the
molecular weight of the compound is (360 g/mol), the molecular formula
of the compound will be:
a) C6H8O4
b) C12H16O8
c) C9H12O6
d) C15H20O10
24- How many moles of FeSO4 are present in a 500 mg of this salt?
a) 1.64 x 10-3
b) 2.30 x 10-3
c) 3.29 x 10-3
d) 4.93 x 10-3
Mass = 500mg = 0.5 g
n= mass / molar mass
= 0.5 / 152
= 0.00329 = 3.29 x 10-3 g
First Exam/ Exercises
23- An empirical formula of an organic compound is C3H4O2 , if the
molecular weight of the compound is (360 g/mol), the molecular formula
of the compound will be:
a) C6H8O4
b) C12H16O8
c) C9H12O6
d) C15H20O10
24- How many moles of FeSO4 are present in a 500 mg of this salt?
a) 1.64 x 10-3
b) 2.30 x 10-3
c) 3.29 x 10-3
d) 4.93 x 10-3
25- How many milliliter of water must be added to 200 mL of 0.15M Na2CO3
to prepare 0.05 M Na2CO3?
a) 200 mL
b) 300 mL
c) 400 mL
d) 450 mL
M1 V 1 = M2 V 2
0.15 X 200 = 0.05 X V2
V2 = 0.15 X 200 / 0.05 = 600 ml
Add water = 600-200 = 400 ml
First Exam/ Exercises
23- An empirical formula of an organic compound is C3H4O2 , if the
molecular weight of the compound is (360 g/mol), the molecular formula
of the compound will be:
a) C6H8O4
b) C12H16O8
c) C9H12O6
d) C15H20O10
24- How many moles of FeSO4 are present in a 500 mg of this salt?
a) 1.64 x 10-3
b) 2.30 x 10-3
c) 3.29 x 10-3
d) 4.93 x 10-3
25- How many milliliter of water must be added to 200 mL of 0.15M Na2CO3
to prepare 0.05 M Na2CO3?
a) 200 mL
b) 300 mL
c) 400 mL
d) 450 mL
First Exam/ Exercises
26- What is the theoretical yield of chromium (Cr) that can be produced by
the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to
the chemical reaction?
2 Al + Cr2O3  Al2O3 + 2 Cr
a) 15.4 g
b) 17.33 g
c) 19.4 g
d) 13.48 g
First we have to determine the limiting reagent:
First start with Al
1-Convert to mole :
n = 8 / 27 = 0.296 mol
2- from equation
2mole Al ========= 2 mole Cr
0.296 mole Al =====? Mole Cr
2x 0.296 = 2 x ?
Mole of Cr = 0.296 mol
Mass = n x molar mass
= 0.296 x 52
=15.4g
second start with Cr2O3
1-Convert to mole :
n = 40 / 152 = 0.263mol
2- from equation
1mole Cr2O3 ========= 2 mole Cr
0.263mole Cr2O3 =====? Mole Cr
2 x 0.263 = 1 x ?
Mole of Cr = 0.526 mol
First Exam/ Exercises
26- What is the theoretical yield of chromium (Cr) that can be produced by
the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to
the chemical reaction?
2 Al + Cr2O3  Al2O3 + 2 Cr
a) 15.4 g
b) 17.33 g
c) 19.4 g
d) 13.48 g
27- If the actual yield for the experiment in question (26) produced 13.0g,
what is the percentage yield?
a) 84.4%
b) 75.0%
c) 67%
d) 96.4%
% Yield 
Actual yield
X 100
Theoretica l yield
13
% YIELD 
X 100  84.4%
15.4
First Exam/ Exercises
26- What is the theoretical yield of chromium (Cr) that can be produced by
the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to
the chemical reaction?
2 Al + Cr2O3  Al2O3 + 2 Cr
a) 15.4 g
b) 17.33 g
c) 19.4 g
d) 13.48 g
27- If the actual yield for the experiment in question (26) produced 13.0g,
what is the percentage yield?
a) 84.4%
b) 75.0%
c) 67%
d) 96.4%
28- The mass of Na that will react with excess of water to produce 16.0 g
NaOH is
2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)
a) 5.75 g
b) 6.90 g
c) 8.05 g
d) 9.20 g
First Exam/ Exercises
28- The mass of Na that will react with excess of water to produce 16.0 g
NaOH is
2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)
a) 5.75 g
b) 6.90 g
c) 8.05 g
d) 9.20 g
1-First make sure the equation is balanced
2- g to mole
n = mass / molar mass
= 16 / 40
= 0.4 mol
From equation
2 mole Na =======2mole NaOH
?mole Na ======== 0.4 Mole NaOH
2 X 0.4 = 2 X ?
Mole of Na = 0.4 mole
Mass = n x molar mass
= 0.4 x 23 = 9.2 g
First Exam/ Exercises
26- What is the theoretical yield of chromium (Cr) that can be produced by
the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum (Al) according to
the chemical reaction?
2 Al + Cr2O3  Al2O3 + 2 Cr
a) 15.4 g
b) 17.33 g
c) 19.4 g
d) 13.48 g
27- If the actual yield for the experiment in question (26) produced 13.0g,
what is the percentage yield?
a) 84.4%
b) 75.0%
c) 67%
d) 96.4%
28- The mass of Na that will react with excess of water to produce 16.0 g
NaOH is
2Na(s) + 2H2O(l) → H2(g) + 2NaOH(aq)
a) 5.75 g
b) 6.90 g
c) 8.05 g
d) 9.20 g
First Exam/ Exercises
29- A 100.0 mL of 0.25 M HCl is diluted with water to a total volume of 200.0
mL. What is the final concentration in the resulting solution?
a) 0.125M
b) 0.083 M
c) 0.05 M
d) 0.0625 M
M1 V 1 = M2 V 2
0.25 X 100 = M2 X 200
M2= 0.25 X 100 / 200 = 0.125 M
First Exam/ Exercises
29- A 100.0 mL of 0.25 M HCl is diluted with water to a total volume of 200.0
mL. What is the final concentration in the resulting solution?
a) 0.125M
b) 0.083 M
c) 0.05 M
d) 0.0625 M
30- What is the mass of carbon in 10 g carbon dioxide (CO2)?
a) 4.1 g
b) 2.73 g
c) 6.82 g
d) 5.45 g
Mole of CO2 = mass / molar mass
= 10 / 44 = 0.227 mol
From the formula
1mole C ====== 1 mol CO2
? Mole C ====== 0.227 mole CO2
Mole of C = 0.227 mol
Mass of C = n x molar mass
= 0.227 x 12 = 2.73 g.
First Exam/ Exercises
29- A 100.0 mL of 0.25 M HCl is diluted with water to a total volume of 200.0
mL. What is the final concentration in the resulting solution?
a) 0.125M
b) 0.083 M
c) 0.05 M
d) 0.0625 M
30- What is the mass of carbon in 10 g carbon dioxide (CO2)?
a) 4.1 g
b) 2.73 g
c) 6.82 g
d) 5.45 g