Chemistry 140 Fall 2002
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
9th Edition
Contents
• Early Chemical Discoveries
• The Observations That Led to an Atomic View
of Matter
Chapter 2: Atoms and the Atomic Theory
• Electrons and the Nuclear Atom
• Chemical Elements
• Atomic Masses
• Introduction to the Periodic Table
• The Mole Concept
Early Discoveries
Law of conservation of mass - Lavoisier 1774 -
Law of constant composition (or Law of definite proportions)
proportions) - Proust
1799-
Important Laws that Govern Chemical Reactions
The Law of Conservation of Mass - There is no
measurable change in total mass during a chemical
reaction.
“Mass is neither created nor destroyed”
destroyed”
2 H 2O (l) electrolysis
100.0 g
2 H 2 (g) + O 2 (g)
88.8 g
11.2 g
Cu (s) + S (s)
∆
63.55 g
32.07 g
CuS (s)
95.62 g
The Law of Multiple Proportions
Atomic Theory -Dalton (1803-1808)-
1
Chemistry 140 Fall 2002
The Law of Constant Composition - The elemental
composition by mass of a given compound is the
same for ALL samples of that compound.
Or when elements combine to form compounds, they
do so in definite proportions by mass =
The Law of Definite Proportions
Na (s) + Cl 2 (g)
39.3 g
60.7 g
23.0 g
35.5 g
39.3 g
60.7 g
NaCl (s)
100.0 g
58.5 g
23.0g
=
35.5 g
= 0.647
Na2SO4 = 2 atoms of Sodium + 1 atom of Sulfur + 4 atoms of Oxygen
Percent of each Element
2 x Na = 2 x 22.99 = 45.98
1 x S = 1 x 32.07 = 32.07
4 x O = 4 x 16.00 = 64.00
142.05
Check
H
O
O/H
H 2O 2
11.2 g
178 g
15.9
H 2O
11.2 g
88.8 g
7.93
C
O
O/C
CO 2
42.9 g
114 g
2.66
CO
42.9 g
57.1 g
1.33
=2
=2
Mass Percent Composition of Na2 SO4
Elemental masses
The Law of Multiple Proportions - In different comcompounds containing the same elements, the masses of
one element combined with a fixed mass of the other
element are in the ratio of small whole numbers.
% Na = Mass Na / Total mass x 100%
% Na = (45.98 / 142.05) x 100% =32.37%
% S = Mass S / Total mass x 100%
% S = (32.07 / 142.05) x 100% = 22.58%
% O = Mass O / Total mass x 100%
% O = (64.00 / 142.05) x 100% = 45.05%
% Na + % S + % O = 100%
32.37% + 22.58% + 45.05% = 100.00%
Calculating the Mass of an Element
in a Compound Ammonium Nitrate
How much Nitrogen is in 455 kg of Ammonium Nitrate?
Ammonium Nitrate = NH4NO3 The Formula Mass of Cpd is:
4 x H = 4 x 1.008 = 4.032 g
2 x N = 2 X 14.01 = 28.02 g
Therefore gm Nitrogen/ gm Cpd
3 x O = 3 x 16.00 = 48.00 g
28.02 g Nitrogen
80.052 g
= 0.35002249 g N / g Cpd
80.052 g Cpd
455 kg x 1000g / kg = 455,000 g NH4NO3
455,000 g Cpd x 0.35002249 g N / g Cpd = 1.59 x 105 g Nitrogen
or: 455 kg NH4NO3
X 28.02 kg Nitrogen = 159 kg Nitrogen
80.052 kg NH4NO4
2
Chemistry 140 Fall 2002
Summary:
The Law of Conservation of Mass
The Law of Constant Composition
Dalton’s Atomic Theory
• Each element is composed of small particles called atoms.
The Law of Definite Proportions
The Law of Multiple Proportions
These laws that govern chemical reactions and com pound composition allow us to do stoichiometric
calculations. They were all well known by 1803 when
John Dalton put forth his atomic theory of matter.
This theory was able to “explain ” why these laws that
govern chemical reactions were true.
Electrons and Other Discoveries in Atomic
Physics
‚ Atoms are neither created nor destroyed in chemical reactions.
ƒ All atoms of a given element are identical (in mass and other
properties)
„ Compounds are formed when atoms of more than one element
combine in a simple numerical ratio.
Behavior of charges
• Some objects display a property called electric
charge, which can be either positive (+) or
negative (-).
• Like charges repel each other while opposite
charges attract each other
• All objects are made up of charged particles
• An object carrying equal numbers of positively
and negatively charged particles carries no net
charge and is electrically neutral.
3
Chemistry 140 Fall 2002
Cathode ray tube (CRT)
Properties of cathode rays
Michael Faraday
Cathode rays travel in staright lines and have properties
independent of the cathode material. Cathode rays subsequently
became known as electrons.
Mass (m) to Charge (e) Ratio = m/e
J.J. Thomson in 1897
For Electron,
m/e = -5.6857 x 10-9 g Coulomb -1
4
Chemistry 140 Fall 2002
Radioactivity
Charge on the electron
¶ From 1906-1914 Robert Millikan showed ionized oil drops
can be balanced against the pull of gravity by an electric field.
¶ The fall of a droplet in the electric field between condenser
plates is speeded up or slowed down, depending on the
magnitude and sign of the charge on the droplet.
¶The charge is an integral multiple of the electronic charge, e.
¶e= -1.6022 x 10-19 Coulomb
¶Mass of an electron= 9.1094 x 10-28 g
Radioactivity is the spontaneous
emission of radiation from a
substance.
¶ X-rays and γ-rays are highenergy light.
¶ α-particles are a stream of
helium ions, He2+.
¶ β -particles are a stream of
high speed electrons that
originate in the nucleus.
The nuclear atom
Geiger and Rutherford
1909
5
Chemistry 140 Fall 2002
The α-particle experiment
¶ Most of the mass and all of the
positive charge is concentrated in a
small region called the nucleus .
The nuclear atom
Rutherford
protons 1919
James Chadwick
neutrons 1932
Expected result according to the
¶ There are as many electrons outside
J. J.Thomson’s “ plum-pudding”
the nucleus as there are units of
model of the atom in which electrons
positive charge on the nucleus
are scattered in a cloud of protons
Nuclear Structure
Atomic Diameter 10-10 m= 1 Ao Nuclear diameter 10-14 m
Particle
Electron
Proton
Neutron
Mass
kg
9.1 09 x 10-31
1.673 x 10-27
1.6 75 x 10-27
amu
0.000548
1.00073
1.00087
Charge
Coulombs
–1.602 x 10-19
+1.602 x 10-19
0
(e)
–1
+1
0
6
Chemistry 140 Fall 2002
Atomic Definitions I: Symbols,
Isotopes,Numbers
Scale of Atoms
ÜThe heaviest atom has a mass of only 4.8 x 10-22 g
and a diameter of only 5 x 10 -10 m.
Biggest atom is 240 amu and is 50 Å across.
A
Z
X
The Nuclear Symbol of the Atom, or Isotope
X = Atomic symbol of the element, or element symbol
A = The Mass number; A = Z + N
Typical C-C bond length 154 pm (1.54 Å)
Molecular models are 1 Å /inch or about 0.4 Å /cm
Z = The Atomic Number, the Number of Protons in the Nucleus
N = The Number of Neutrons in the Nucleus
Isotopes = atoms of an element with the same number of protons,
but different numbers of Neutrons in the Nucleus
Neutral ATOMS
•
•
51
Cr = P + (24), e- (24),
N (27)
•
•
239
•
15
N = P+(7), e-(7), N(8)
•
•
56
Fe = P +(26), e-(26),
N (30)
•
•
235
Pu = P +(94), e- (94),
N (145)
U =P+(92), e-(92),
N (143)
7
Chemistry 140 Fall 2002
Atomic Definitions II: AMU, Dalton, 12 C Std.
Atomic mass Unit (AMU) = 1/12 the mass of a carbon - 12 atom
on this scale Hydrogen has a mass of 1.008 AMU.
Isotopic
Isotop
ic Masses:
Definition –
Isotopic Mass = The mass of an Isotope relative to the
Dalton (D) = The new name for the Atomic Mass Unit,
one dalton = one Atomic Mass Unit
on this scale, 12C has a mass of 12.00 daltons.
Isotopic Mass =The mass of an Isotope relative to the
Isotope 12C, the chosen standard.
Atomic Mass = “ Atomic Weight” of an element is the average of the
masses of its naturally occurring isotopes weighted
according to their abundances.
Isotope 12C , the chosen standard.
standard
The relative mass of an atom usually
expressed in relative atomic mass units ( amu or u).
u).
A single oxygenoxygen - 16 atom has a mass that is 1.332 9096
times that of a carbon - 12 atom.
(12 u)(1.332 9096) = 15.994 915 u
Isotopes of Hydrogen
•
•
•
1
H
1 Proton 0 Neutrons 99.985 % 1.00782503 amu
H (D) 1 Proton
1 Neutron
0.015 % 2.01410178 amu
3 H (T)
1 Proton 2 Neutrons
----------------1
The average mass of Hydrogen is 1.008 amu
1
2
Element #8 : Oxygen, Isotopes
•
16
O
8 Protons
99.759%
8 Neutrons
15.99491462 amu
•
17
O
8 Protons
0.037%
9 Neutrons
16.9997341 amu
•
18
O
8 Protons
0.204 %
10 Neutrons
17.999160 amu
8
1
• 3H is Radioactive with a half life of 12 years.
• H2O Normal water “light water “
•
mass = 18.0 g/mole , BP = 100.00000 0C
• D2O Heavy water
•
mass = 20.0 g/mole , BP = 101.42 0C
8
8
8
Chemistry 140 Fall 2002
Formation of a Positively Charged
Neon Particle in a Mass Spectrometer
Fig. 2.A
Fig. 2.B part A
Calculating the “Average” Atomic
Mass of an Element
Measuring atomic masses
Problem: Calculate the average atomic mass of Magnesium!
Magnesium Has three stable isotopes, 24Mg ( 78.7%);
25Mg (10.2%); 26Mg (11.1%).
24
Mg (78.7%)
Mg (10.2%)
26Mg (11.1%)
25
23.98504 amu x 0.787 = 18.876226 amu
24.98584 amu x 0.102 = 2.548556 amu
25.98636 amu x 0.111 = 2.884486 amu
24.309268 amu
With Significant Digits = 24.3 amu
9
Chemistry 140 Fall 2002
Calculate the Average Atomic Mass of
Zirconium, Element #40
Zirconium has five stable isotopes: 90Zr, 91Zr, 92Zr, 94Zr, 96Zr.
Problem: Calculate the abundance of the two Bromine isotopes:
79Br = 78.918336 g/mol and 81Br = 80.91629 g/mol , given that
the average mass of Bromine is 79.904 g/mol.
Plan: Let the abundance of
Solution:
Isotope (% abd.)
90
Zr
Zr
92Zr
94Zr
96Zr
91
(51.45%)
(11.27%)
(17.17%)
(17.33%)
(2.78%)
Mass (amu)
89.904703
90.905642
91.905037
93.906314
95.908274
amu
amu
amu
amu
amu
(%)
X
X
X
X
X
0.5145
0.1127
0.1717
0.1733
0.0278
Fractional Mass
=
=
=
=
=
46.2560 amu
10.2451 amu
15.7801 amu
16.2740 amu
2.6663 amu
91.2215 amu
With Significant Digits = 91.22 amu
Modern Reassessment of the Atomic Theory
79
Br = X and of 81Br = Y and X + Y = 1.0
X(78.918336) + Y(80.91629) = 79.904
X + Y = 1.00 therefore X = 1.00 - Y
(1.00 - Y)(78.918336) + Y(80.91629) = 79.904
78.918336 - 78.918336 Y + 80.91629 Y = 79.904
1.997954 Y = 0.985664
or Y = 0.4933
X = 1.00 - Y = 1.00 - 0.4933 = 0.5067
%X = % 79Br = 0.5067 x 100% = 50.67% = 79Br
%Y = % 81Br = 0.4933 x 100% = 49.33% = 81Br
Definitions
1. All matter is composed of atoms. Although atoms are composed
of smaller particles (electrons, protons, and neutrons), the atom
is the smallest body that retains the unique identity of the element.
• ELEMENT - A substance that cannot be separated
into simpler substances by chemical means
2. Atoms of one element cannot be converted into atoms of another
element in a chemical reaction. Elements can only be converted into
other elements in Nuclear reactions in which protons are changed.
• COMPOUND - A substance composed of atoms
of two or more elements chemically united in
fixed proportions
3. All atoms of an element have the same number of protons and
electrons, which determines the chemical behavior of the element.
Isotopes of an element differ in the number of neutrons, and thus
in mass number, but a sample of the element is treated as though
its atoms have an average mass.
• PERIODIC TABLE - “ MENDELEEV TABLE” A tabular arrangement of the elements, vertical
groups or families of elements based upon their
chemical properties - actually combining ratios
with oxygen
4. Compounds are formed by the chemical combination of two or more
elements in specific ratios, as originally stated by Dalton.
10
Chemistry 140 Fall 2002
Newly Discovered Elements
Fig. 2.16
1994
Revised
IUPAC Slate
Atomic
No.
ACS Slate
IUPAC Slate
104
105
Rutherfordium
Hahnium
Dubnium
Joliotium
Rutherfordium
Dubnium
106
107
Seaborgium
Neilsbohrium
Rutherfordium
Bohrium
Seaborgium
Bohrium
Hahnium
Hassium
108
Hassium
109
Meitnerium
Meitnerium
110
?
?
3 Labs
Meitnerium
111
?
?
GSI
112
?
?
GSI
Final Slate
9/12/97
The Periodic Table of the Elements
H
Li Be
B C N
Na Mg
Al Si P
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As
Rb Sr Y Zr NbMo Tc Ru Rh Pd Ag Cd In Sn Sb
Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi
Fr Ra Ac Rf Du Sg Bo Ha Me
He
O F Ne
S Cl Ar
Se Br Kr
Te I Xe
Po At Rn
Ce Pr Nd Pm SmEu Gd TbDy Ho ErTmYb Lu
Th Pa U Np PuAmCm Bk Cf Es FmMdNo Lr
Metals
Non Metals
Semi - metals
Metalloids
Fig. 2.17
11
Chemistry 140 Fall 2002
Groups in the Periodic Table
Main Group Elements (Vertical Groups)
Group IA - Alkali Metals
Group IIA - Alkaline Earth Metals
Group IIIA - Boron Family
Group IVA - Carbon Family
Group VA - Nitrogen Family
Group VIA - Oxygen Family ( Calcogens)
Group VIIA - Halogens
Group VIIIA - Noble Gases
Other Groups ( Vertical and Horizontal Groups)
Group IB - 8B - Transition Metals
Period 6 Group - Lanthanides (Rare Earth Elements)
Period 7 Group - Actinides
The Periodic Table of the Elements
H
Li Be
B C N
NaMg
Al Si P
K Ca Sc Ti V CrMn Fe Co Ni Cu Zn Ga Ge As
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb
Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi
Fr Ra Ac Rf Du Sg Bo Ha Me
He
O F Ne
S Cl Ar
Se Br Kr
Te I Xe
Po At Rn
Ce Pr Nd PmSmEu GdTb Dy Ho ErTmYb Lu
Th Pa U Np PuAmCm Bk Cf Es FmMd No Lr
Alkali Metals
The Periodic table
Alkaline Earths
Halogens
Noble Gases
Main Group
Transition Metals
Main Group
Lanthanides and Actinides
The Periodic Table of the Elements
H
Li Be
B C N
Na Mg
Al Si P
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As
Rb Sr Y Zr NbMo Tc R Rh Pd Ag Cd In Sn Sb
Cs Ba La Hf Ta W Re uOs Ir Pt Au Hg Tl Pb Bi
Fr Ra Ac Rf Du Sg Bo HaMe
He
O F Ne
S Cl Ar
Se Br Kr
Te I Xe
Po At Rn
Ce Pr Nd PmSmEu GdTb Dy Ho ErTmYb Lu
Th Pa U Np PuAmCmBk Cf Es FmMd No Lr
Boron family
Nitrogen family
The Alkali Metals
The Halogens
Carbon Family
Oxygen Family
The Alkaline
Earth Metals
The Noble Gases
12
Chemistry 140 Fall 2002
The Periodic Table of the Elements
H
He
Li Be
B C N O F Ne
Na Mg
Al Si P S Cl Ar
K Ca Sc Ti V CrMn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Rb Sr Y Zr NbMo Tc Ru Rh PdAg Cd In Sn Sb Te I Xe
Cs Ba La Hf Ta W Re Os Ir Pt A Hg Tl Pb Bi Po At Rn
Fr Ra Ac Rf Du Sg Bo Ha Me u
The Transition Metals
Ce Pr Nd PmSmEu GdTb DyHo ErTmYb Lu
Th Pa U Np PuAmCmBk Cf Es FmMd No Lr
Lanthanides: The
The Periodic Table of the Elements
Most Probable Oxidation State
+1
0
+3 +_4 - 3
H +2
Li Be
B C N
+1 + 2
Na Mg+3 +4 +5
Al Si P
K Ca Sc Ti V CrMn Fe Co Ni Cu Zn Ga Ge A
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn sSb
Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi
Fr Ra Ac Rf Du Sg Bo Ha Me
+3
The Actinides
+3
-2 -1
O
S
Se
Te
Po
He
F Ne
Cl Ar
Br Kr
I Xe
At Rn
Ce Pr Nd PmSmEu GdTb Dy Ho Er TmYb Lu
Th Pa U Np Pu AmCmBk Cf Es FmMd No Lr
Rare Earth Elements
•
•
•
•
The Periodic Table
The Mole
Read atomic masses.
Read the ions formed by main group elements.
Read the electron configuration.
Learn trends in physical and chemical properties.
• Physically counting atoms is impossible.
• We must be able to relate measured mass to
numbers of atoms.
– buying nails by the pound.
– using atoms by the gram
We will discuss these in detail in Chapter 10.
13
Chemistry 140 Fall 2002
Pair: A group containing 2 items or objects.
Dozen: A group containing 12 items or objects.
Molar Mass: The mass in grams of one mole of any
substance. The molar mass is numerically equal to the
formula mass.
Gross: A group containing 144 items or objects.
Substance
Mole: A group containing 6.0221 x 1023
items or objects.
One Mole = the amount of substance ( element or
compound ) that contains the same number of particles
as there are atoms in EXACTLY 12 g of 12C.
6.0221 x 1023 = Avogadro’
Avogadro’s Number
Formula Masses: The relative mass of a compound is
equal to the sum of the atomic masses of the elements
forming the compound.
Calculate the formula masses for NaCl, H2O, and
H 3PO 4 NaCl
Molar
Mass
2.015 88 g
Fe
55.847 u
55. 847 g
H 3PO 4
97.9952 u
97.9952 g
SO 3
80.06 u
80.06 g
NaCl
58.443 u
58.443 g
“ When in doubt, calculate moles!!”
moles!!”
# of Particles
Moles
Avogadro’’s Number
Avogadro
How many atoms of carbon are in 2.6 moles of C?
H 2O
1 Na @ 22.989 77 u = 22.989 77 u 2 H @ 1.007 94 u = 2.015 88 u
1 Cl @ 35.453 u
= 35.453 u
1 O @ 15.9994 u = 15.9994 u
1 NaCl
H2
Formula
Mass
2.015 88 u
= 58.443 u
1 H 2O
= 18.0153 u
H 3PO 4
3 H @ 1.007 94 u = 3.023 82 u
1 P @ 30.973 76 u = 30.973 76 u
4 O @ 15.9994 u = 63.9976 u
1 H 3PO 4
= 97.9952 u
(2.6 mol C) (6.0221 x 1023 atom C)
= 1.6 x 1024 atom C
mol C
How many moles of water are present when 2.673 x 1025
molecules of water are present?
(2.673 x 1025 molecules H2O)
(1 mol H2O)
(6.0221 x 1023 molecules H2O)
= 44.39 mol H2O
14
Chemistry 140 Fall 2002
“ When in doubt, calculate moles!!”
moles!!”
# of Particles
Moles
Avogadro’’s Number
Avogadro
“ When in doubt, calculate moles!!”
moles!!”
Mass
(36.5 g H2O)(1 mol H2O) (2 mol H) (6.0221 x 1023 atom H)
(18.015 g H2O) (mol H2O)
(mol H)
Molar Mass
How many moles of carbon are in 26.78 g of C?
(26.78 g C) (1 mol C)
= 2.230 mol C
(12.011 g C)
= 2.44 x 10 24 atom H
How many grams of Au atoms are present in 1.45 x
10 20 Au atoms?
How many grams of water are in 32.6 moles of water?
(32.6 mol H2O) (18.015 g H2O)
= 587 g H 2O
(1 mol H 2O )
(1.45 x 1020 atom Au)(1 mol Au)(196.967 g Au)
(6.0221 x 10 23 atom Au) (mol Au)
= 0.0474 g Au
How many moles of uranium are there in 1.23 kg U?
Molar Mass of U = 238.019 g/mol
“ When in doubt, calculate moles!!”
moles!!”
# of Particles
(1.23 kg U)(10
U)(10 3 g U) (1 mol U)
= 5.17 mol U
(1 kg U) (238.019 g U)
How many atoms of sulfur are there in 7.62 g S?
(7.62 g S)(1 mol S/32.066 g S)(6.0221 x
How many atoms of hydrogen are present in
36.5 g of water?
10 23 atom
S/mol S)
= 1.43 x 1023 atom S
What is the mass of a single Pt atom?
Moles
Avogadro’’s Number
Avogadro
Mass
Molar Mass
This allows you to do all types of calculations involving
the conversion of particles to mass, mass to particles,
mass to moles, moles to mass, particles to moles, or moles
to particles.
(1 Pt atom)(1 mol Pt/6.0221 x 1023 atom Pt) (195.08 g Pt)
= 3.239 x 10 - 22 g Pt
(mol Pt)
15
Chemistry 140 Fall 2002
Example 2-9
Combining Several Factors in a Calculation—Molar Mass, the
Avogadro Constant, Percent Abundance.
Potassium-40 (40K) is one of the few naturally occurring
radioactive isotopes of elements of low atomic number. Its
percent natural abundance among K isotopes is 0.012%. How
many 40K atoms do you ingest by drinking one cup of whole
milk containing 371 mg of K?
Want atoms of 40K, need atoms of K,
Want atoms of K, need moles of K,
Convert strategy to plan
and plan into action
Convert mass of K(mg K) into moles of K (mol K)
mK(mg) x (1g/1000mg) è mK (g) x 1/MK (mol/g) è nK(mol)
nK = (371 mg K) x (10-3 g/mg)x (1 mol K) / (39.10 g K)
= 9.49 x 10-3 mol K
Convert moles of K into atoms of 40K
nK(mol) x N A è atoms K x 0.012% è atoms 40K
atoms 40K = (9.49 x 10-3 mol K) x (6.022 x 1023 atoms K/mol K)
Want moles of K, need mass and M(K).
x (1.2 x 10-4 40K/K)
= 6.9 x 1017 40K atoms
The Atomic Masses of Elements:
The atomic masses shown on the Periodic Table are
average masses of naturally occurring samples of the
elements. These samples consist of mixtures of isotopes
with different atomic masses.
Isotope
Mass (u)
Fractional
Abundance
28Si
- 92.23 %
(Atomic mass - 27.976 93 u)
Atomic Mass
(u)
29Si
- 4.67 %
(Atomic mass - 28.976 49 u)
15.956 3 1
30Si
- 3.10 %
(Atomic mass - 29.973 76 u)
16O
15.994 915 x 0.997 587 =
17O
16.999 133 x 0.000 374
=
0.006 358
18O
17.999 16
=
0.036 700
x 0.002 039
Practice Problem: Given the following data from
the mass spectrometer, calculate the average
atomic mass for the element, silicon (Si).
Weighted Average = 15.9994 u
(27.976 93 u)( 92.23 u )
(100 u)
+
(28.976 49 u)(4.67
u)(4.67 u )
(100 u)
+ (29.973 76 u)( 3.10 u ) =
(100 u)
28.09 u
16
Chemistry 140 Fall 2002
Chapter 2 Questions
1,2(a,b,c) 4,77,9,10
1,2(a,b,c),3,
9,10,
14,16,18,21,22,
14,16,18,21
25,27,33, 55, 66
25,27
66.
17