Download Chemistry 11 Final Examination Review

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Chemistry 11
Final Examination Review - Answers
Part A - True or False.
Indicate whether each of the following statements is true or false. Correct the false statements.
F
T
T
T
F
T
F
1. The mass of an electron is equal to the mass of a proton.
The mass of an electron is less than the mass of a proton.
2.
3.
4.
5.
The mass of a proton is approximately equal to the mass of a neutron.
The atomic number represents the number of protons in a nucleus.
The proton has a mass of approximately 1 u.
The difference in mass of isotopes of the same element is due to the different number of
protons in the nucleus.
The difference in mass of isotopes of the same element is due to the
different number of neutrons in the nucleus.
6. The isotope carbon-12 is used as the relative mass standard for the atomic mass scale.
7. The mass of the most common isotope of each element is listed on the periodic table.
The average mass of the naturally occurring isotopes of each element
is listed on the periodic table.
Part B - Multiple Choice
D
A
A
D
B
A
B
D
1. A(n) __ is used to represent a compound.
a) symbol
b) equation
c) subscript
d) formula
2. ___ atoms or groups of atoms are called ions.
a) charged
b) diatomic
c) neutral
d) monatomic
3. For the formula of a compound to be correct, the algebraic addition of the charges on the
atoms or ions in the compound must add up to __.
a) zero
b) one
c) two
d) four
4. Potassium bromide is an example of a(n) __ compound.
a) molecular
b) organic
c) polyatomic
d) ionic
5. The only common polyatomic ion that has a positive charge is the __ ion.
a) phosphate
b) ammonium
c) sulfate
d) nitrate
6. In the formula H2SO4, the number 4 would be called a(n) __.
a) subscript
b) oxidation number
c) coefficient
d) charge
7. Which subatomic particle contributes the least to the mass of an atom?
a) nucleon
b) electron
c) proton
d) neutron
8. In the free, or uncombined, state the number of protons in the nucleus of an element must
equal the __.
a) mass number
c) mass number - atomic number
b) number of neutrons in the nucleus
d) number of electrons present
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C
B
B
C
A
C
B
B
D
A
B
D
B
D
C
9. Which of the following ideas of the Bohr model is not retained in the modern theory of
atomic structure?
a) Electrons can absorb or emit energy only in whole numbers of photons.
b) Atoms have a central positively charged nucleus.
c) Electrons move around the nucleus as planets orbit the sun.
d) Most of the volume of an atom is empty space.
10. Which of the following orbitals is spherical in shape?
a) 3p
b) 2s
c) 4d
d) 5f
11. The third energy level of an atom may have __ electrons.
a) 2
b) 18
c) 8
d) 32
12. How many sublevels are possible at the fourth energy level?
a) 2
b) 3
c) 4
d) 18
13. Lustrous, malleable, ductile elements that are good conductors of electricity and heat are
classified as __.
a) metals
b) nonmetals
c) metalloids
d) noble gases
14. The electron configuration of a certain element ends with 3p5. Which of the following
describes its position in the periodic table?
a) period 5, group 13 b) period 3, group 15
c) period 3, group 17 d) period 5, group 15
15. Which of the following is an example of a metalloid?
a) iodine
b) boron
c) bromine
d) indium
16. The periodicity of the elements is basically a function of their __.
a) nuclear stability
b) atomic numbers
c) mass numbers
d) none of these
17. An element with seven electrons in the outer level would be a __.
a) metal
b) metalloid
c) noble gas
d) nonmetal
18. As the atomic number in a period increases, the degree of nonmetallic character __.
a) increases
c) increases then decreases
b) decreases
d) remains the same
19. Elements in a group have similar chemical properties because of their similar __.
a) nuclear configurations
c) mass numbers
b) outer electron configurations
d) names
20. The period number in the periodic table designates the __ for the row.
a) total nuclear charge
c) maximum number of outer electrons
b) maximum number of nucleons
d) highest energy level
21. The radii of the atoms become smaller from sodium to chlorine across period 3. This is
primarily a result of __.
a) the shielding effect
c) the increased number of electrons
b) increased nuclear charge
d) decreased metallic character
22. Compared to the stability of the original atom, the stability of its ion that resembles a noble gas
configuration would be __.
a) identical
b) sometimes less
c) less
d) greater
23. The formation of bonds between atoms depends on __.
a) the electron configurations of the atoms involved
b) the attraction the atoms have for electrons
c) both of the preceding factors
d) neither of the preceding factors
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D
A
A
D
A
D
B
D
C
A
A
C
D
A
B
24. The particle that results when two or more atoms form covalent bonds is a __.
a) single charged atom b) molecule
c) polyatomic ion
d) b or c
25. Compounds that have low melting points, are brittle, and do not conduct electricity are probably __.
a) covalent
c) ionic
b) metallic
d) compounds of polyatomic ions
26. The most active __ have the highest electronegativities.
a) nonmetals
b) metalloids
c) metals
d) noble gases
27. __ compounds have high melting points, conduct electricity in the molten phase, and tend to be
soluble in water.
a) hydrogen
b) metallic
c) covalent
d) ionic
28. The element in the following group that has the lowest electronegativity is __.
a) potassium
b) arsenic
c) bromine
d) chromium
29. If there are only two electron pairs in the outer energy level of an atom in a molecule, they will be
found __.
a) at 90º to one another
c) at 120º to one another
b) on the same side of the nucleus
d) on opposite side of the nucleus
30. The __ molecule has two bonding pairs and two unshared pairs of electrons.
a) CH4
b) H2O
c) NH3
d) HF
31. A certain atom contains 34 protons, 34 electrons, and 45 neutrons. This atom has a mass
number of __.
a) 34
b) 45
c) 68
d) 79
32. An example of a compound is
a) oxygen
b) mercury
c) salt
d) diamond
33. Carbon is classed as an element rather than as a compound because it
a) cannot be chemically decomposed into two or more substances
b) has been known for many centuries
c) is formed when wood is heated out of contact with air
d) combines with oxygen to form a gas
34. The positively charged particles in the nucleus of an atom are called
a) protons
b) neutrons
c) electrons
d) ions
35. The charged particles that are found outside the nucleus of an atom are called
a) protons
b) ions
c) electrons
d) mesons
36. A nonmetallic atom generally becomes a negative ion by
a) losing protons
b) losing electrons
c) gaining protons
d) gaining electrons
37. Which of the following subatomic particles has the smallest mass?
a) electron
b) neutron
c) proton
d) nucleus
38. Which of the following symbols represents an atom that contains the largest number of neutrons?
a)
C
C
235
92 U
b)
239
92 U
c) 239
93 Np
39. The nuclide symbol 16
8 O represents an oxygen atom with
d) 239
94 Pu
a) a mass of 8 u
c) a mass of 16 u
b) an atomic number of 16
d) 16 neutrons
40. If Z represents the atomic number of an element and A represents the mass number, then the
number of neutrons in one atom is
a) A
b) A + Z
c) A - Z
d) Z - A
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D
B
25
41. Which of the following statements about the elemental species 24
11 X and 12 Z is correct?
a) They are isotopes of the same element.
b) They are nonmetals.
c) They are members of the same chemical family.
d) They have the same number of neutrons per atom.
42. An element X has a mass number of 32 and an atomic number of 16. The most common ion
of element X is represented by
a) X+
b) X2c) Xd) X2+
Part C - Short Answer
1.
What is the maximum number of electrons that may occupy one orbital?
two
2. The Lewis electron dot diagram is used to represent only which electrons in the atom?
Valence electrons
3. What is the diagonal rule used to predict?
The order in which orbitals are filled
4. How many sublevels are possible at the third energy level?
three
5. How many orbitals are there in the f sublevel?
seven
6. What is the maximum number of electrons that can occupy a d sublevel?
ten
7. Which sublevel may contain a maximum of three pairs of electrons?
‘p’ sublevel
8. What must be true about the spins of two electrons occupying the same orbital?
They must have opposite spin
9. Write the electron configuration for each of the following elements:
a) lithium
1s22s1
b) radium
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s2
c) sodium
1s22s22p63s1
d) mercury
1s22s22p63s23p64s23d104p65s24d105p66s24f145d10
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10. Draw the energy level diagram for each of the following elements:
a) tin
6s
↑↓
5s
↑
↑↓
↑
5p
↑↓
4p
↑↓
↑↓
↑↓
3p
↑↓
↑↓
↑↓
2p
↑↓
↑↓
↑↓
4p
↑↓
↑↓
↑↓
3p
↑↓
↑↓
↑↓
2p
↑↓
↑↓
↑↓
↑↓
4d
↑↓
↑↓
↑↓
↑↓
↑↓
3d
↑↓
↑↓
↑↓
↑↓
↑↓
3d
↑↓
↑↓
4s
↑↓
3s
↑↓
2s
↑↓
1s
b) krypton
5s
↑↓
4s
↑↓
3s
↑↓
2s
↑↓
1s
↑↓
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c) gold
↑↓ ↑↓ ↑↓ ↑↓
5d
6p
↑↓
6s ↑↓ ↑↓
5p
↑↓
5s ↑↓ ↑↓
4p
↑↓
4s
↑↓ ↑↓
3p
↑↓
3s
↑↓ ↑↓
2p
↑↓
2s
↑
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4f
↑↓
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
4d
↑↓
↑↓ ↑↓ ↑↓ ↑↓ ↑↓
3d
↑↓
↑↓
↑↓
1s
d) potassium
↑
4s
3d
↑↓
↑↓
3p
↑↓
↑↓
↑↓
2p
↑↓
↑↓
3s
↑↓
2s
↑↓
1s
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11. Write the energy level population for each of the following elements:
a) calcium
2, 8, 8, 2
b) sulfur
2. 8. 6
c) scandium
2, 8, 9, 2
d) tungsten
2, 8, 18, 32, 12, 2
12. Element X has the following configuration: 1s22s22p63s22p64s23d104p65s1
a) What period is this element located in?
five
b) What group is this element in?
one (alkali metals)
c) Identify the element.
Rubidium (Rb)
d) Is the element a metal, nonmetal, or metalloid?
Metal
13. List five properties of metals and five properties of nonmetals.
Metals
-
good conductors of heat and electricity
lustre
malleable and ductile
high densities
high boiling points and melting points
resist stretching and twisting
solids at room temperature
Nonmetals
-
no lustre
poor conductors
may be solid, liquid or gas
low densities
low melting and boiling points
14. How are substances that are gases or soft solids at room temperature classified?
Substances that are gases or soft solids at room temperature are classified
as covalent compounds.
15. Would an element with two outer electrons be a metal or a nonmetal?
An element with two outer electrons would be a metal.
16. What Russian scientist designed the first periodic table?
Mendeleev designed the first periodic table.
17. Which group of elements has eight outer electrons?
The noble gases have eight outer electrons.
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18. For the transition elements, as the atomic number increases, to which sublevel are the electrons
being added?
For the transition metals, the electrons are being added to the “d”
sublevel.
19. According to the octet rule, how many pairs of outer electrons do the most stable atoms have?
According to the octet rule, the most stable atoms have four pairs of outer
electrons.
20. In the lanthanide series, as the atomic number increases, to which sublevel are electrons being
added?
In the lanthanide series, electrons are being added to the “f” sublevel.
21. Give examples of molecules with the following shapes:
a) linear
HCl, CO2, BeCl2
b) trigonal planer
BH3, BCl3, AlBr3
c) trigonal pyramidal
NH3, PH3
d) tetrahedral
CH4, CCl4
e) bent
H2O, H2S
22. Use electron dot formulas to show the complete balanced bonding reactions between the following
elements. Indicate the type of bond expected.
a) sodium and oxygen
b) nitrogen and hydrogen
c) zirconium and sulfur
d) magnesium and chlorine
*23.
Compare the boiling points of methane, ethane, propane, and butane. Use intermolecular bonding to
explain why they are different.
The boiling points, in order from least to greatest, are: methane, ethane,
propane, and butane. All four compounds contain only carbon and hydrogen.
All bonds are nonpolar, therefore all of the molecules are nonpolar. The only
type of intermolecular attraction is dispersion for all four molecules. As the size
of the atom increases (and the number of electrons) the amount of dispersion
increases.
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*24.
Compare the boiling points of propanoic acid, 1-butanol, diethyl ether, butanal, and pentane.
intermolecular bonding to explain why they are different.
The boiling points, in order from least to greatest, are:
diethyl ether, 1-butanol, and propanoic acid.
Pentane
•
•
•
CH3
CH2
CH2
CH2
Use
pentane, butanal,
CH3
42 electrons, all nonpolar bonds ∴ nonpolar molecule
dispersion only
lowest boiling point
O
Butanal
•
•
CH3 CH2
CH
42 electrons ∴
approximately the same amount of dispersion nonpolar
C-C and C-H bonds
one polar C=O bond, ∴ dipole-dipole in addition to dispersion, causing a
higher boiling point than pentane
Diethyl ether CH3
•
•
•
CH2
CH2
O
CH2
CH3
42 electrons ∴ approximately the same amount of dispersion
nonpolar C-C and C-H bonds
two polar C-O bonds, ∴ dipole-dipole in addition to dispersion, causing a
higher boiling point than butanal because of the two polar bonds
OH
1-Butanol
•
•
•
•
CH2
CH2
CH2
CH3
42 electrons ∴ approximately the same amount of dispersion
nonpolar C-C and C-H bonds
one polar C-O bond and one polar O-H bond, ∴ dipole-dipole
Hydrogen bonding (O-H) in addition to the dispersion and dipole-dipole, ∴
a higher boiling point than diethyl ether
O
Propanoic acid
•
•
•
•
•
CH3 CH2
C
OH
40 electrons ∴ approximately the same amount of dispersion
nonpolar C-C and C-H bonds
two polar C-O bonds and one polar O-H bond, ∴ dipole-dipole
Hydrogen bonding (O-H) in addition to the dispersion and dipole-dipole
a higher boiling point than 1-butanol because of the additional polar bonds
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25. Write the correct names for the following chemical compounds.
a) HCl(aq)
hydrochloric acid
m) SO3
sulfur trioxide
b) KOH
n) NaC2H3O2 sodium acetate
potassium hydroxide
o) HFO2(aq) flourous acid
c) HgOH
mercury(I) hydroxide
d) FeCl3
p) Al(BrO4)3 aluminum perbromate
iron(III) chloride
e) Al2(SO4)3 aluminum sulfate
q) SbF3
antimony(III) fluoride
f) N2O5
dinitrogen pentoxide
r) Pd(CN)2
palladium(II) cyanide
g) HF
hydrofluoric acid
s) Ca(MnO3)2 calcium manganate
h) Pb(OH)2
lead(II) hydroxide
t) Be(NO4)2 beryllium pernitrate
i) NH4NO3 ammonium nitrate
u) NiSeO4
nickel(II) selenate
j) NaHCO3
sodium bicarbonate
v) H2SO3(aq) sulfurous acid
w) Ba(OH)2
k) Zn(NO2)2 zinc nitrite
barium hydroxide
x) PbS
l) H3PO4(aq) phosphoric acid
lead(II) sulfide
26. Write the correct chemical formula for each compound and balance the equation.
a) sodium carbonate + calcium hydroxide ¼ sodium hydroxide + calcium carbonate
Na2CO3 + Ca(OH)2 ¼ 2NaOH + CaCO3
b) carbon dioxide + water ¼ carbonic acid
CO2 + H2O ¼ H2CO3
c) phosphorus + oxygen ¼ phosphorus pentoxide
2P + 5O2 ¼ 2PO5
d) sodium + water ¼ sodium hydroxide + hydrogen
2Na + 2H2O ¼ 2NaOH + H2
e) zinc + sulfuric acid ¼ zinc sulfate + hydrogen
Zn + H2SO4 ¼ ZnSO4 + H2
f) aluminum sulfate + calcium hydroxide ¼ aluminum hydroxide + calcium sulfate
Al2(SO4)3 + 3Ca(OH)2 ¼ 2Al(OH)3 + 3CaSO4
g) calcium oxide + water ¼ calcium hydroxide
CaO + H2O ¼ Ca(OH)2
h) iron + copper(I) nitrate ¼ iron(II) nitrate + copper
Fe + 2CuNO3 ¼ Fe(NO3)2 + 2Cu
i)
iron(II) sulfide + hydrochloric acid ¼ hydrogen sulfide + iron(II) chloride
j)
potassium oxide + water ¼ potassium hydroxide
FeS + 2HCl ¼ H2S + FeCl2
K2O + H2O ¼ 2KOH
Page 11 of 28
k.
carbon + ferric oxide ¼ iron + carbon dioxide
3 C + 2 Fe2O3 ¼ 4 Fe + 3 CO2
l.
sulfur tetrafluoride + water ¼ sulfur dioxide + hydrofluoric acid
SF4 + 2 H2O ¼ SO2 + 4 HF
m. calcium hydroxide + phosphoric acid ¼ calcium phosphate + water
3 Ca(OH)2 + 2 H3PO4 ¼ Ca3(PO4)2 + 6 H2O
n.
ethane + oxygen ¼ carbon dioxide + water
2 C2H6 +
o.
7 O2 ¼ 4 CO2 + 6 H2O
aluminum sulfate + ammonia + water ¼ aluminum hydroxide + ammonium sulfate
Al2(SO4)3 + 6 NH3 + 6 H2O ¼ 2 Al(OH)3 + 3 (NH4)2SO4
27. Identify the type of reaction. Write the correct chemical formulas of the compounds, complete and
balance the equations and name the products formed.
calcium chloride + carbonic acid
+ H2CO3
a) calcium carbonate + hydrochloric acid ¼
CaCO3 + 2HCl ¼ CaCl2
double displacement
b) iron + sodium bromide ¼
Fe + NaBr ¼ no reaction b/c iron is less active than sodium
single displacement
ammonium chloride + iron(II) acetate
¼ 2NH4Cl + Fe(C2H3O2)2
c) ammonium acetate + iron(II)chloride ¼
2NH4C2H3O2 + FeCl2
double displacement
silver sulfate + ammonium bromide
2AgBr + (NH4)2SO4 ¼ Ag2SO4 + 2NH4Br
double displacement
d) silver bromide + ammonium sulfate ¼
e) zinc + sulfuric acid ¼
Zn + H2SO4 ¼ H2
single displacement
hydrogen + zinc sulfate
+ ZnSO4
f) neon + potassium ¼
Ne + K ¼ no reaction b/c noble gases are unreactive
combination
lead(II) chloride + water
+ 2H2O
g) lead(II )hydroxide + hydrochloric acid ¼
Pb(OH)2 + 2HCl ¼ PbCl2
double displacement
Page 12 of 28
iron(II) sulfide
Fe + S ¼ FeS
combination
h) iron + sulfur ¼
i)
potassium chloride + oxygen
potassium chlorate (heated) ¼
2KClO3 ¼ 2KCl + 3O2
decomposition
j)
calcium hydroxide
CaO + H2O ¼ Ca(OH)2
combination
calcium oxide + water ¼
k) dinitrogen pentoxide + water ¼ nitric acid
N2O5 + H2O ¼ 2HNO3
combination
l) carbon dioxide + water ¼ carbonic acid
CO2 + H2O ¼ H2CO3
combination
m) chlorine + chromium(III) bromide ¼ bromine + chromium(III) chloride
3 Cl2 + 2 CrBr3 ¼ 3 Br2 + 2 CrCl3
Single displacement
n) sulfur + oxygen ¼sulfur dioxide or sulfur trioxide
S + O2 ¼ SO2
or S + O2 ¼ SO3
combination
o) zinc + hydrochloric acid ¼ zinc chloride + hydrogen
Zn + 2 HCl ¼ ZnCl2 + H2
Single displacement
p) sodium + water ¼ sodium hydroxide + hydrogen
2 Na + 2 H2O ¼ 2 NaOH + H2
single displacement
q) magnesium + water ¼
Mg + H2O ¼no reaction – Mg is not active enough to replace H in water
Single displacement
r) copper + stannic nitrate ¼
Cu + Sn(NO3)4 ¼ no reaction – Cu is less active than Sn
Single displacement
s) aluminum + cupric sulfate ¼ copper + aluminum sulfate
2 Al + 3 CuSO4 ¼ 3 Cu + Al2(SO4)3
Double Displacement
Page 13 of 28
Part D – Calculations
Show all work. Express your answers using significant digits and include units with your answers.
1.
Calculate the average atomic mass of the following elements:
a) Mg-24
Mg-25
Mg-26
mass = 23.985 u
mass = 24.986 u
mass = 25.983 u
78.70%
10.13%
11.17%
(0.7870)(23.985) + (0.1013)(24.986) + (0.1117)(25.983) = 24.31 u
b) Ir-191
Ir-193
mass = 191.0 u
mass = 193.0 u
37.58%
62.42%
(0.3758)(191.0) + (0.6242)(193.0) = 192.2 u
2. Calculate the percentage of each of the isotopes of silver if silver-107 has a mass of 106.905 u and
silver-109 has a mass of 108.905 u and the average atomic mass of silver is 107.869 u
107
Ag = x,
109
Ag = y
x+y=1
106.905x + 108.905y = 107.869
-106.905x - 106.905y = -106.905
106.905x + 108.905y = 107.869
2y = 0.964
¼
107
use elimination or substitution
y = 0.482
¼
x = 0.518
109
Ag = 51.8%,
Ag = 48.2%
*3. Naturally, occurring silicon consists of three isotopes, 28Si, 29Si, and 30Si, whose atomic masses are 27.9769,
28.9765, and 29.9738, respectively. The most abundant isotope is 28Si, which accounts for 92.23 percent
of naturally occurring silicon. Given that the observed atomic mass of silicon is 28.0855, calculate the
percentages of 29Si and 30Si in nature.
Isotope
Mass
Percentage
28
Si
27.9769 g
92.23%
29
Si
28.9765 g
x
30
Si
29.9738 g
y
Average
28.0855
(eqn 1)
(eqn 1)
(eqn 1)
(0.9223)(27.9769) + (28.9765)(x) + (29.9738)(y) = 28.0855
25.8031 + 28.9765x + 29.9738y = 28.0855
28.9765x + 29.9738y = 2.2824
(eqn 2)
(eqn 2)
0.9223 + x + y = 1
x + y = 0.0777
(eqn 1)
(-28.9756)(eqn 2)
y = 0.0310,
use substitution or elimination
28.9765x + 29.9738y = 2.2824
-28.9765x – 28.9765y = -2.2515
0.9973y = 0.03093
x = 0.0467,
∴
29
Si = 3.10% and
y = 0.03093
0.9973
30
Si = 4.67%
Page 14 of 28
4. Complete the chart below:
Element
Atomic #
Mass #
Protons
Neutrons
Electrons
calcium-43
20
43
20
23
20
lead-211
82
211
82
129
82
plutonium-242
94
242
94
148
94
chromium-50
24
50
24
26
24
29
65
29
36
27
S2-
16
34
16
18
18
128 53 I
53
128
53
75
54
208
4+
82 Pb
82
208
82
126
78
65
Cu2+
34
5. Convert each of the following to moles.
a. 8.8 g of potassium carbonate
8.8 g K2CO3 ÷ 138.2052 g/mol = 0.0637 mol
b. 0.257 g of arsenic pentachloride
0.257 g AsCl5 ÷ 252.1866 g/mol = 1.02 x 10-3 mol
c. 12.5 L of carbon dioxide at STP
12.5 L CO2 ÷ 22.4 L/mol = 0.558 mol
d. 5.00 x 1013 atoms of iron
5.00 x 1013 atoms of Fe ÷ 6.022 x 1023 =
8.30 x 10-11 mol
e. 8.63 x 1028 molecules of water
8.63 x 1028 molecules of H2O ÷ 6.022 x 1023 =
1.43 x 105 mol
f. 450.0 mL helium at STP
0.4500 L He ÷ 22.4 L/mol = 0.0201 mol
g. 236.0 g of ammonium phosphate
236.0 g (NH4)3PO4 ÷ 149.086 74 g/mol = 1.58 mol
h. 15.0 g of butanoic acid
15.0 g CH3CH2CH2COOH
÷
88.106 32 g/mol = 0.170 mol
Page 15 of 28
6. Calculate the mass of each of the following.
a. 2.60 mol of sodium carbonate
(2.60 mol)(105.988 74 g/mol) = 276 g
b. five million atom of gold
(5 000 000 atoms) ÷ (6.022 x 1023) = 8.30 x 10-18 mol
(8.30 x 10-18 mol)(196.9665 g/mol) = 1.64 x 10-15 g
c. 25.0 mL of carbon dioxide at STP
(0.0250 L) ÷ (22.4 L/mol) = 0.001 12 mol
(0.001 12 mol)(44.0098 g/mol) = 0.0491 g
d. 4.50 x 1021 molecules of decanoic acid
CH3CH2CH2CH2CH2CH2CH2CH2CH2COOH or C10H202
4.50 × 10 21
= 0.00747 mol
6.022 × 10 23
g
(0.00747mol )(172.2676 mol
) = 1.29
g
7. Calculate the molarity of 825 mL of solution, which contains 30.0 g of acetic acid.
30.0 g CH 3 COOH
60.05256
g
mol
= 0.4996 g
0.4996 g
= 0.606
0.825 L
mol
L
8. What volume of solution can be made from 80.0 g of sodium hydroxide if a 2.00 M solution is required?
80.0 g NaOH
38.887 11
V =
n
C
g
mol
= 2.00 mol
2.00 mol
=
2.00
mol
L
= 1.00 L
9. What mass of calcium chloride is required to produce 750.0 mL of a 0.500 M solution?
n = C×V
=
(0.500
(0.375 mol)(110.984
mol
L
)(0.7500 L)
)
= 41.6 g
g
mol
= 0.375 mol
Page 16 of 28
10. What volume 0f 14.0 M nitric acid is required to produce 750.0 mL of a 0.250 M solution?
C V
V1 = 2 2
C1
C1 V1 = C2 V2
=
(0.250 )(0.7500L) = 0.0134 L
mol
L
mol
14.0
L
or
13.4 mL
11. What concentration results when 250.0 mL of a 0.125 M solution of hydrochloric acid is mixed with 125.0
mL of a 1.00 M solution?
C ab
C V + C b Vb
= a a
Va + Vb
(0.125
=
mol
L
mol
L
)(0.1250 L)
0.2500 L + 0.1250 L
0.03125 mol + 0.125 mol
=
)(0.2500 L) + (1.00
0.375 L
=
0.417
mol
L
12. What volume of 0.225 M sulfuric acid must be mixed with 500.0 mL of a 0.750 M solution in order to obtain
a 0.500 M solution?
C ab =
C a Va + C b Vb
Va + Vb
0.500
(0.500
0.500
(0.275
Va =
mol
L
mol
L
(0.225
=
)(V
mol
L
a
mol
L
)(0.5000 L)
Va + 0.5000L
=
(0.225
mol
L
)V
a
+ 0.375 mol
+ 0.250 mol
=
(0.225
mol
L
)V
a
+ 0.375 mol
+ 0.5000 L
)V
a
0.125 mol
0.275
)V + (0.750
)
a
mol
Va
L
mol
L
mol
L
=
0.125 mol
= 0.455 L
Page 17 of 28
E. STOICHIOMETRY - Begin each problem by writing a balanced chemical equation.
1.
Carbon dioxide is produced in the reaction between calcium carbonate and hydrochloric acid. How many
grams of calcium carbonate would be needed to react completely with 15.0 g of hydrochloric acid? How
many grams of calcium chloride would be formed?
CaCO3 + 2HCl ¼ H2O + CO2 + CaCl2
⎛ 15.0 g HCl ⎞⎛ 1 mol CaCO
⎜
⎟⎜
3
⎜
g ⎟⎜
⎜ 36.46094
⎟⎜
HCl
mol ⎠⎝ 2 mol
⎝
⎞
⎟
⎟ 100.0892
⎟
⎠
(
⎛ 15.0 g HCl ⎞⎛ 1 mol CaCl ⎞
⎜
⎟⎜
2 ⎟
⎜
⎟
⎜
⎟ 110.986
g
⎜ 36.460 94
⎟⎜
HCl ⎟
mol ⎠⎝ 2 mol
⎝
⎠
(
2.
g
mol
g
mol
) = 20.6 g
) = 22.8 g
CaCO 3
CaCl2
Sulfur dioxide may be catalytically oxidized to sulfur trioxide. How many grams of sulfur dioxide could be
converted by this process if 100.0 g of oxygen are available for the oxidation?
2SO2 + O2 ¼ 2SO3
⎛ 100.0 g O ⎞⎛ 2 mol SO ⎞
⎜
2 ⎟⎜
2 ⎟
⎜
⎟
⎜
⎟ 64.0648
g
⎜ 31.9988
⎟⎜ 1 mol O2 ⎟
mol ⎠⎝
⎝
⎠
(
g
mol
) = 400.4 g
SO2
3. Phosphoric acid is produced in the reaction between calcium phosphate and sulfuric acid. How much of the
phosphoric acid would be produced from 55.0 g of the calcium phosphate?
Ca3(PO4)2 + 3H2SO4 ¼ 3CaSO4 + 2H3PO4
⎛ 55.0 g Ca (PO )
⎜
3
4 2
⎜
g
310.183
⎜
mol
⎝
⎞⎛ 2 mol H PO
⎟⎜
3
4
⎟⎜
⎟⎜ 1 mol Ca3 (PO 4 )2
⎠⎝
⎞
⎟
⎟ 97.995
⎟
⎠
(
g
mol
) = 34.8 g
H3PO 4
4. How much magnesium sulfate is needed to completely react with 145 g of sodium chloride? How much
sodium sulfate could be produced by this reaction?
MgSO4 + 2NaCl ¼ Na2SO4 + MgCl2
⎛ 145 g NaCl ⎞⎛ 1 mol MgSO ⎞
⎜
⎟⎜
4 ⎟
⎜
⎟
⎜
⎟ 120.3686
g
⎜ 58.443
⎟⎜ 2 mol NaCl ⎟
mol ⎠⎝
⎝
⎠
(
⎛ 145 g NaCl ⎞⎛ 1 mol Na SO
⎜
⎟⎜
2
4
⎜
⎟
⎜
g
⎜ 58.443
⎟⎜
NaCl
mol ⎠⎝ 2 mol
⎝
⎞
⎟
⎟ 142.043
⎟
⎠
(
) = 149 g
MgSO 4
) = 176 g
Na2 SO 4
g
mol
g
mol
Page 18 of 28
5. Gold will dissolve in the acid mixture known as aqua regia according to the following reaction:
Au + HNO3 + 3 HCl Î AuCl3 + NO + 2 H2O
How much gold(III)chloride will be produced in this reaction when one starts with 5.0 mg of gold?
⎛ 5.0 mg Au ⎞⎛ 1 mol AuCl ⎞
3 ⎟
⎟⎜
⎜
⎟ 303.3255
⎜
mg ⎟⎜
⎜ 196.9665 mmol ⎟⎜ 1 mol Au ⎟
⎠
⎠⎝
⎝
(
mg
mmol
) = 7.7
mg
AuCl 3
How much hydrochloric acid must be added initially to dissolve the all this gold?
⎛ 5.0 mg Au ⎞⎛ 3 mol
⎜
⎟⎜
⎜
mg ⎟⎜
⎜ 196.9665 mmol ⎟⎜ 1 mol
⎝
⎠⎝
HCl ⎞⎟
⎟ 36.461
Au ⎟
⎠
(
mg
mmol
) = 2.8
mg
HCl
g
H2 SO4
6. How many grams of sulfuric acid will react with 400.0 g of aluminum metal?
⎛ 400.0 g Al ⎞⎛ 3 mol H SO
2
4
⎜
⎟⎜
⎜
⎟
⎜
g
⎜ 26.98154 mol ⎟⎜ 2 mol Al
⎝
⎠⎝
⎞
⎟
⎟ 98.079
⎟
⎠
(
g
mol
) = 2181
7. It is desired to prepare 50.0 g of water by synthesis. How many litres of hydrogen must be used?
⎛ 50.0 g H O ⎞⎛ 2 mol H ⎞
2
2 ⎟
⎜
⎟⎜
⎜
⎟ 22.4
g ⎟⎜
⎜ 18.01528 mol ⎟⎜ 2 mol H 2 O ⎟
⎝
⎠⎝
⎠
(
8.
L
mol
) = 62.2
L
H2
An unknown amount of potassium chlorate was heated until no more oxygen was evolved.
potassium chloride remained in the test tube.
15.824 g of
2KClO3 ¼ 2KCl + 3O2
What mass of potassium chlorate had originally been placed in the tube?
⎛ 15.824 g
⎜
⎜
⎜ 74.5513
⎝
KCl ⎞⎟⎛⎜ 2 mol KClO3 ⎞⎟
⎟⎜
⎟ 122.5495
g
⎟
⎜
⎟
2
mol
KCl
mol
⎠⎝
⎠
(
g
mol
) = 26.0
g
KClO3
What volume of oxygen gas was evolved in the process?
⎛ 15.824 g
⎜
⎜
⎜ 74.5513
⎝
KCl ⎞⎟⎛⎜ 3 mol
⎟⎜
g
⎟⎜ 2 mol
mol
⎠⎝
O2 ⎞⎟
⎟ 22.4
KCl ⎟
⎠
(
L
mol
) = 7.13
L
O2
Page 19 of 28
9.
How many grams of carbon can be completely burned in 15.0 L of oxygen?
⎛ 15.0 L
⎜
⎜
⎜ 22.4
⎝
O2 ⎞⎟⎛⎜ 1 mol C ⎞⎟
⎟ 12.011
⎟⎜
L
⎟⎜ 1 mol O 2 ⎟
mol
⎠⎝
⎠
(
g
mol
) = 8.04
g
C
How many litres of carbon dioxide gas are produced in this reaction?
⎛ 15.0 L
⎜
⎜
⎜ 22.4
⎝
O2 ⎞⎟⎛⎜ 1 mol CO2 ⎞⎟
⎟⎜
⎟ 22.4
L
⎟
⎜
1 mol O 2 ⎟
mol
⎠⎝
⎠
(
L
mol
) = 15.0
L
CO2
10. How many grams of magnesium metal are required to liberate 250.0 mL of hydrogen gas from hydrochloric
acid?
⎛ 0.2500 L H ⎞⎛ 1 mol
2 ⎟⎜
⎜
⎜
⎟⎜
L
22.4 mol
⎜
⎟⎜ 1 mol
⎝
⎠⎝
Mg ⎞⎟
⎟ 24.305
H2 ⎟
⎠
(
g
mol
) = 0.271
g
Mg
Exactly how much acid would be used up in this reaction?
⎛ 0.2500 L H ⎞⎛ 2 mol HC lg ⎞
2 ⎟⎜
⎜
⎟
⎜
⎟⎜
⎟ 36.46094
L
22.4 mol
⎜
⎟⎜ 1 mol H 2 ⎟
⎝
⎠⎝
⎠
(
11.
g
mol
) = 0.8141
g
HCl
Will 30.0 L of fluorine gas completely react with 50.0 L of hydrogen gas? Which gas is in excess and by
how much? What volume of hydrogen fluoride is formed in this reaction?
⎛ 30.0 L
⎜
⎜
⎜ 22.4
⎝
F2 ⎞⎟⎛⎜ 2 mol
⎟⎜
L
⎟⎜ 1 mol
mol
⎠⎝
HF ⎞⎟
⎟ 22.4
F2 ⎟
⎠
⎛ 50.0 L
⎜
⎜
⎜ 22.4
⎝
H2 ⎞⎟⎛⎜ 2 mol
⎟⎜
L
⎟⎜ 1 mol
mol
⎠⎝
HF ⎞⎟
⎟ 22.4
H2 ⎟
⎠
(
(
L
mol
) = 60.0
L
mol
L
HF
) = 100. L
HF
60.0 L of hydrogen fluoride is produced.
Fluorine is the limiting reactant and hydrogen is in excess.
⎛ 30.0 L F ⎞⎛ 1 mol H ⎞
2 ⎟⎜
2 ⎟
⎜
L
⎟ 22.4 mol = 30.0 L HF needed
⎟⎜
⎜
L
⎜ 22.4 mol ⎟⎜ 1 mol O 2 ⎟
⎠
⎠⎝
⎝
50.0 L of hydrogen available - 30.0 L needed = 20.0 L in excess
(
)
Page 20 of 28
12. A mixture containing 100.0 g of H2 and 100.0 g of O2 is sparked so that water is formed. How much water
is formed?
⎛ 100.0 g H ⎞⎛ 2 mol H O ⎞
2 ⎟⎜
2
⎟
⎜
⎟ 18.01528
⎟⎜
⎜
g
⎜ 2.01588 mol ⎟⎜ 2 mol H 2 ⎟
⎠
⎠⎝
⎝
g
mol
) = 894
g
H2 O
⎛ 100.0 g O ⎞⎛ 2 mol H O ⎞
2 ⎟⎜
2
⎟
⎜
⎟ 18.01528
⎜
g ⎟⎜
⎜ 31.9988 mol ⎟⎜ 1 mol o 2 ⎟
⎠
⎠⎝
⎝
g
mol
) = 113
g
H2 O
(
(
Oxygen is the limiting reactant and hydrogen is in excess.
113 g of water are formed.
13. When copper is heated with sulfur, Cu2S is formed. How many grams of Cu2S could be produced if 100.0 g
of copper is heated with 50.0 g of sulfur?
⎛ 100.0 g
⎜
⎜
⎜ 63.546
⎝
⎛ 50.0 g
⎜
⎜
⎜ 32.066
⎝
Cu ⎞⎟⎛⎜ 1 mol Cu2 S ⎞⎟
⎟ 159.158
⎟⎜
g
Cu ⎟
mol ⎟⎜ 2 mol
⎠
⎠⎝
(
S ⎞⎟⎛⎜ 1 mol Cu2 S ⎞⎟
⎟ 159.158
g ⎟⎜
1 mol S ⎟
mol ⎟⎜
⎠
⎠⎝
(
g
mol
g
mol
) = 125.2
) = 248
g
g
Cu2 S
Cu2 S
Copper is the limiting reactant and sulfur is in excess.
125.2 g of copper(I) sulfide is formed.
14. What volume (at STP) of carbon monoxide is required to produce 100.0 g of iron according to the equation:
(not balanced)
Fe2O3 + CO Î Fe + CO2
⎛ 100.0 g
⎜
⎜
⎜ 55.847
⎝
Fe ⎞⎟⎛⎜ 3 mol
⎟⎜
g
mol ⎟⎜ 2 mol
⎠⎝
CO ⎞⎟
⎟ 22.4
F2 ⎟
⎠
(
L
mol
) = 60.2
L
CO
Page 21 of 28
15. What is the molarity of a NaOH solution if 25.00 cm3 is required to completely neutralize 40.00 cm3 of a
1.50 M solution of H2SO4?
H2SO4 + 2NaOH ¼ Na2SO4 + 2H2O
C aVaR b
(1.50 mol L )(0.04000 L )(2) = 4.80 mol L NaOH
=
(0.02500 L )(1)
VbR a
Cb =
16. Calculate the volume of a 0.600 M solution of HNO3 necessary to neutralize 28.55 cm3 of a 0.450 M solution
of KOH.
HNO3 + KOH ¼ KNO3 + H2O
Va =
Cb VbR a (0.450 mol L )(0.02855 L )(1)
=
= 0.0214 L HNO3
(0.600 mol L )(1)
C aR b
17. A titration of 15.00 cm3 of household ammonia, NH4OH(aq), required 38.57 cm3 of 0.780 M HCl. Calculate
the molarity of the ammonia.
HCl + NH4OH ¼ NH4Cl + H2O
Cb =
C aVaR b
(0.780 mol L )(0.03857 L )(1) = 2.01 mol L NH OH
=
4
(0.01500 L )(1)
VbR a
18. What volume of 0.250 M H3PO4 is required to neutralize 30.00 cm3 of a 0.0500 M Ba(OH)2 solution?
2H3PO4 + 3Ba(OH)2 ¼ Ba3(PO4)2 + 6H2O
Va =
Cb VbR a (0.0500 mol L )(0.03000 L )(2)
=
= 0.00400 L H3PO 4
(0.250 mol L )(3)
C aR b
19. What mass of Ca(OH)2 would be required to completely neutralize 50.0 cm3 of 0.125 M HCl?
2HCl + Ca(OH)2 ¼ CaCl2 + H2O
(
)(
)
n = C ⋅ V = 0.125 mol L 0.0500 L = 0.00625 mol HCl
Ca(OH) ⎞⎟
(0.00625 mol HCl)⎛⎜⎜ 1 mol
(74.09268
2 mol HCl ⎟
2
⎝
⎠
)
g mol = 0.232 g Ca(OH)2
20. What mass of Mg(OH)2 would be required to completely neutralize 70.0 cm3 of 0.175 M HNO3?
(
2HNO3 + Mg(OH)2 ¼ Mg(NO3)2 + 2H2O
)(
)
n = C ⋅ V = 0.175 mol L 0.0700 L = 0.01225 mol HNO3
Mg(OH) ⎞⎟
(0.01225 mol HNO )⎛⎜⎜ 12mol
(58.31968
mol HNO ⎟
3
2
⎝
3
⎠
)
g mol = 0.357 g Mg(OH)2
Page 22 of 28
*21. Hydrazine is a nitrogen-hydrogen compound having the formula N2H4. It is an oily, colourless liquid that
freezes at 1.5°C and boils at 113.5°C. The principal use of hydrazine and certain compounds derived from it
is a rocket fuels, but it is also used in fuel cells, in the treatment of water in boilers to removed dissolved
oxygen gas, and in the plastics industry. One widely used method for the manufacture of hydrazine is the
Raschig process. The Raschig process involves three reaction steps. In the first step, sodium hydroxide is
reacted with chlorine to produce sodium hypochlorite, sodium chloride and water. The sodium hypochlorite
produced in the first step is reacted with ammonia in the second step to produce chloramine (NH2Cl) and
sodium hydroxide. The chloramine and sodium hydroxide produced in the second step is reacted with
ammonia in the third step to produce hydrazine, sodium chloride and water. A chemical plane using the
Raschig process obtains 0.299 kg of 98.0% hydrazine for every 1.00 kg of chlorine. What are the
theoretical, actual, and percent yields of pure hydrazine?
2NaOH + Cl2 ¼ NaOCl + NaCl + H2O
NaOCl + NH3 ¼ NH2Cl + NaOH
NH2Cl + NaOH + NH3 ¼ N2H4 + NaCl + H2O
⎛ 1000 g Cl ⎞⎛ 1 mol NaOCl ⎞⎛ 1 mol
2 ⎟⎜
⎟⎜
⎜
⎟⎜
⎟⎜
⎜
g
⎜ 70.906 mol ⎟⎜ 1 mol Cl2 ⎟⎜ 1 mol
⎠⎝
⎠⎝
⎝
= 452 g
N2 H 4
=
NH2 Cl ⎞⎟⎛⎜ 1 mol
⎟⎜
NaOCl ⎟⎜ 1 mol
⎠⎝
theoretica l yield
(
)
actual yield = (98.0%) 299 g = 293 g = actual yield
percentage yield =
actual yield
× 100 = 64.8% yield
theoretica l yield
N 2 H 4 ⎞⎟
⎟ 32.04516
NH2 Cl ⎟
⎠
(
g
mol
)
Page 23 of 28
*22.
The characteristic odour of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen,
and oxygen. Combustion of 2.78 mg of ethyl butyrate produces 6.32 mg of carbon dioxide and 2.58 mg
of water. What is the empirical formula of the compound?
__ CxHyOz
+ __ O2
2.78 mg
Î __ CO2
6.32 mg
+
__ H2O
2.58 mg
⎛ 12.011 g ⎞
⎟(0.00632 g ) = 0.001725 g of carbon
mass of carbon = ⎜⎜
⎟
⎝ 44.0098 g ⎠
⎛ (2)(1.00794 g) ⎞
⎟(0.00258 g) = 0.0002887 g of hydrogen
mass of hydrogen = ⎜⎜
⎟
18.01528
g
⎠
⎝
total mass − mass of carbon − mass of hydrogen = mass of oxygen
0.00278g − 0.001725g − 0.0002887 g = 0.0007665g of oxygen
Empirical Formula
0.001725 g of carbon
12.011 g mol
0.0001436 mol C
0.00004791
2.998 : 5.979 : 1
0.0002887 g of hydrogen
1.00794 g mol
0.0002864 mol H
0.00004791
=
3:6:1
0.0007665 g of oxygen
15.9994 g mol
0.00004791 mol Ol
0.00004791
∴ Empirical Formula = C3H6O
Page 24 of 28
*23.
Nicotine, a component of tobacco, is composed of carbon, hydrogen, and nitrogen. A 5.250 mg sample of
nicotine was combusted, producing 14.242 mg of carbon dioxide and 4.083 mg of water. What is the
empirical formula for nicotine?
⎛ 12.011 g ⎞
⎟⎟(0.014242 g) = 0.003887 g of carbon
mass of carbon = ⎜⎜
44.0098
g
⎠
⎝
⎛ (2)(1.00794 g) ⎞
⎟⎟(0.004083 g ) = 0.00045688 g of hydrogen
mass of hydrogen = ⎜⎜
18.01528
g
⎠
⎝
total mass − mass of carbon − mass of hydrogen = mass of nitrogen
0.005250g − 0.003887g − 0.00045688g = 0.0009062 g of nitrogen
Empirical Formula
0.003887 g of carbon
12.011 g mol
0.0003236 mol C
0.00006471
5.002 : 7.006 : 1
0.00045688 g of hydrogen
1.00794 g mol
0.0004533 mol H
0.00006471
=
5:7 :1
0.0009062 g of nitrogen
14.0067 g mol
0.00006471 mol N
0.00006471
∴ Empirical Formula = C5H7 N
Page 25 of 28
F - ORoANICCHEA,II5TRY
1. Drowthe structureof eachof the followingcompounds:
f ) 2-phenylpropone
o) 2,3-dinifrophenol
0l+
cH.-CH-c{u
,\ so.
I o}. NDz
/\
v
\?
1,3-dibromonophtholene
b) ?,3,4-trimelhyloctone
6r
I
c[l.-Ct'l;CH,* t$ ,- Ct|,
c,fl,a,tt.C{r'
ail. L,$,ctl,
./ \,/
\
tn\ol
\AA6V
h) 1,?-ethanediol
c) 1,3-cyclohexadiene
- t'*'
9H' b H
brt
proponol
d) 4,4-dimethyl-Z-pentene
0
ctlr
clf s- c*"-in
c H r - C $ = c -l lc-ifl r
Cil"
e) phenylethene
a
CA=C$
j)
pentonoicocid
0
cH,- C*,*c*r-(il, -t'- oFl
Review- Answers- Poge24of ?7
Chemistry11- FinolExominofion
Page 26 of 28
p) 3-methyl-3-pentonol
k) propylethonoote
a
CHr-,u,-
ll
Cilr't-o-(il"-cHr-CH,
?l{
,
cl'lL-cH=
f
cll t
l)
q) 4-methylphenol
proponone
OH
0
'
tl
(,'cH,
CL\-
.
t
./\
t \
I O I
LY- /
r) 2-omino-3-phenylproponol
m) l-propoxybutone
o
ll
CH,- chl - crl
I
Nt{'
U]3
0- Ct|r-Cilr'Cl#1
C[gC$t{r'Cfir-
0
n) ethylchloroethonoote
s) 3-methyl-3-pentonol
0
ll
oH
c br t ' c t | -l l ' Q H ' - c H s
ctt3
CHr-C-oC|r-cil,
c
)
I
gl
o ) iodobenzene
f)
ethylmethylefher
-t'
l.
CH,(v1r'o' ct'
Y
of 27
Review- Answers- Poga.?5
Chemistry11- FinolExominofion
Page 27 of 28
x) 3-ethyl-5-methyl-2-nophthol
u) butyl butanoote
cH,
CA;Cll
;CHr-C'o' L|v-cl|,-(H;
OH
ctlz-cl{3
V) 1,3,5-cyclohexanetriol
v) 3-ethylhexonoicacid
0
6il
-c-6H
,'f- cH,
cu{cHr'cttf
ll
*00,n
CH,
I
CH,
z) propyl2,3-dimethylpentonoote
w) nitromethone
o
clli- ct{r-qht-c,H,-{ - o- ctlr-ct{r-CHs
itlr c*r
c LL-No,
2 . Write o boloncedeguotionfor eachof the reoctionsbelow.
+ oxygen)
o) nophfholene
*tzos 4l0co)
@@Q
+ 4tl'o
C,oHg
b) 2-nophthol+ oxygen)
2 @SoH + 23ou
a ?o Coe,
* 9'tA'O
C,o&O
c) 3-methoxypentane+ oxygen)
- ctt; c*s
cht
ct{.-cHro
lnt
" 1O z
-+ Lffi,
I
ffi3
c6il1ao
Chemistry 11- Finol ExominotionReview- Answers- Page26 of 27
v7 d-O
Page 28 of 28
3. Name the following compounds.
a)
3-methyl-5-nitro-3-hexene
b)
1,3-dichlorocyclohexane
c)
1,3-diamino-5-nitrobenzene
d)
3,6-dibromo-2-naphthol
e)
diethyl ether or ethoxyethane
f)
1,4-cyclohexadiene
g)
2,3-dimethylbutanoic acid
h)
2,3,4-pentanetriol
i)
2-methyl-3-pentanone
j)
3-ethyl-2-methylpentanal
k)
methyl propanoate
l)
4,5-dimethyl-7-propyl-1-naphthol
m)
2,4-hexadiyne
n)
1,2-butadiene
o)
pentyl 3,3-dimethylbutanoate
p)
1,3,5-cyclohexanetriol
q)
1-amino-2-bromo-4-nitrocyclopentane
r)
2,3-dimethylpentanedioic acid
s)
butyl 2-methylpropanoate
t)
phenyl-3,4-diaminopentanoate