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MODULE FOR INTERNATIONAL STANDARD CLASS
TAMBUN SELATAN HIGH SCHOOL
Subject
Date – Time
Teacher
Module Code
Sub Theme
I.
:
:
:
:
:
Chemistry
……. x 45’
CH / BC 1.1 / 1
Atomic Structure
Standard Competence
To describe atomic structure, some periodic properties chemistry bond of molecule
structure and its characteristic.
II.
Basic Competence
1.1
III.
Indicator
1.
2.
3.
4.
5.
6.
7.
8.
9.
IV.
To define atomic, atomic structure, characteristic of elements atomic relatif
mass and periodic character from periodic table
To compare growth of periodic system
To determine group and period in periodic table
To determine atomic structure in periodic table
To determine valance electron elements from electron configurations and
periodic table
To determine the number of protons, neutrons, and electrons
To determine isotope, isoton, isobar
To determine characteristic of elements periodicity and atomic relatif mass (Ar)
from periodic table
To analyze table or graph of some periodic properties (atomic radius, ionization
energy, electron affinity, and electronegative)
To comparer the growth of atomic theory, to begin Dalton atomic Theory until
Niels Bohr
Teaching Materials
A. The growth of periodic system
 Elements consist of metal elements and non metal elements in ancient Greek
 Triade Doberenier Law by Johann Wolf Gang Dobereiner
Three elements base on increasing atomic mass so atomic mass in center is
same or nearly of
 Average mass element sum for left element and right element
Example :
Li
6,941
Na
Mass Na

K
39,10
= 6,941 + 39,10 = 23,02
2
Octav Law by JohnNew land
The elements base on increasing atomic mass so elements characteristic
appeared after eight element
Example :
1
2
3
4
5
6
7
H
Li
Be
B
C
N
O
8
9
10
11
12
13
14
F
Na
Mg
Al
Si
P
S

Mendeleev Periodic System
The periodic Law State’s that when elements are arranged in order of
increasing atomic numbers, elements with similar properties occur at periodic
(regularly recurring) intervals.

System of Modern Periodic
The periodic table consist of 7 periods and 8 groups. A period in the periodic
table is a horizontal row of elements.
Period
Period
Period
Period
Period
Period
Period
1
2
3
4
5
6
7
=
=
=
=
=
=
=
2 elements
8 elements
8 elements
18 elements
18 elements
32 elements
no complete
A group in the periodic table is a vertical column of elements. There are two
notations is use for designating individual periodic table groups. In the first
notation, which has been in use for many years, groups are designated using
roman numerals and the letters A and B.
Main Group
Another Name
Elements
IA
Alkali metals
H, Li, Na, K, R, Cs, Fr
II A
Alkalin Earth Metals
Be, Mg, Ca, Sr, Ba, Ra
Hologen
F, Cl, Br, I, At
Noble gases
He, Ne, Ar, Kr, Xe, Rn
A
VIII A
B. Interaction of electron configurations and periodic system
The place of elements in periodic system can be determined by electron
configurations, sum of shells are period and valency electron is group.
Example :
11 Na
=
Sum of shells =
2,
3,
8,
1
period = 3th
Cl of shells =
sum of shells =
2,
3,
8,
7
period = 3th
17
Exercise :
Determine period and group from :
a. 12 Mg
b. 14 Si
c. 16 S
d. 13 Ar
C. Subatomic particles = elctron, prototn, and neutrons
X = Atomic Symbol
Z = Atomic Number
A = Mass Number
A
Z
Atomic number
Mass number
Neutron amount
=
=
=
=
Number of protons = number of electrons
Number of protons + number of neutrons
Mass number – atomic number
A–Z
D. Isotope, Isobar, and Isoton
 Isotop are atoms that have the same number of protons and electrons but have
different numbers of neutrons.
Example :
16
17
1
2
and
,
and
8
8
1
1
 Isobars are atoms that have the same number of mass but have different
atomic number.
Example :
14
14
and
6
7
 Isotons are atoms that have the same number of neutrons.
Example :
16
15
O
O
C
N
O
N
H
H
8
7
E. Atomic relative ( Ar ) and molecule relative ( Mr )
Ar X =
Mass 1 X atomic
1/12 mass 12C
Mr
=
The sum of the atomic relative of atom X in a formula unit of
coumpound
Exercise :
Calculate the relative molecule of the following coumpounds, using a table of
atomic relative.
a) H2O
b) C6H12O6
Calculation is :
a) 2 x Ar of H = 2 x 1 = 2 amu
1 x Ar of O = 1 x 16 = 1 amu
Mr of H2O
b)
6
12
6
Mr
x Ar of C = 6 x 12 = 72 amu
x Ar of H = 12 x 1 = 12 amu
x Ar of O = 6 x 16 = 96 amu
of C6H12O6
180 amu
Exercise :
Calculate the relative molecule of the following coumpounds
a. H2SO4
b. CaCO3
c. CO (NH2)2
F. Growth of Atomic Theory
a. Dalton Theory
- All matter is made up of small particles called atoms
- All atoms of a given type are similar to each other and significantly
different from all other types
- The relative number and arrangement of the different types of atoms
found within o pure substance determines its identity
- Chemical change is a union, separation, or rearrangement of atoms to
give new substances
- Only whole atoms can participate in or result from any chemical change
because atoms are considered to be indestructible during these changes
b. Thomson Theory
The atom is like ball that has positive charge and separate negative charge
c. Ernest Rutherford Theory
The atoms consist of a nucleus with most of the mass and a positive charge,
arround which move enough electrons to make the atom elctrically neutral
d. Niels Bohr Theory
Electrons is an atom have energy lvels, and when an elctron in a higher
energy level drops (or undergous a transition) to a lower energy level, a
photon is emitted
Thomson
Rutherford
Bohr

G. Some Periodic Properties
1. Atomic Radius
- Within each period (horizontal row), the atomic radius tends to decrease
with increasing atomic number (nuclear charge). This the larges atom in
a period is a group IA and the smallest is a noble gas atom.
- within each group (vertical column), the atomic radius tends to increase
with the period number
2. Ionization energy
The first ionization energy or first ionization potential of an atom is the
minimum energy needed to remove the highest energy (that is, the
autormost) electron from the neutral atom is the gaseous state
3. Electron affinity
Electron affinity is the energy change for the process of adding an elctron to
an neutral in the gaseous state to form a negative ion
4. Electronegative
Electronegative is an atom tendency for receiving electrons
Exercise :
1. Determine the number of protons, neutrons, and electrons present in each of the
following atoms.
a. 235
b. 28
c. 27
d. 137
d. 31
Tn
Si
Al
Ba
P
90
14
13
56
15
2. How about are tendency of atomic radius, ionization energy, and electronegative in
periodic system.
3. Calculate the relative molecule for each of the following compounds
a. C2 H5 OH
b. Mg SO4
c. Na3 po4
d. HNO3
4. How are Dalton atomic theory, Rutherford theory, and Niels Bohr Theory.
MODULE FOR INTERNATIONAL STANDARD CLASS
TAMBUN SELATAN HIGH SCHOOL
Subject
Date – Time
Teacher
Module Code
Sub Theme
I.
:
:
:
:
:
Chemistry
……. x 45’
CH / BC 1.2 / 1
Chemical Bonding
Standard Competence
The describe atomic structure, some periodic properties, chemistry bond of molecule
structure and its characteristic.
II.
Basic Competence
1.2
To describe the term chemical bond of using periodic table
III. Indicator
1.
2.
3.
4.
5.
6.
7.
8.
To explain an element tendency for getting stable of another element bond
To describe valence electrons configurations of the nable gases
To explain the ionic bond model and give example it
To explain the terms single covalent bond, double covalent bond and triple
covalent bond
To research polar o some compounds and electronegative relationship in
experience
To explain the coordinate covalent bond on simple example
To explain the metal bond and interaction of metal physic characteristic
To predict various of bonds on compounds and compare physic characteristic
IV. Teaching Materials
1. The electron configurations of the noble gases.
2 He
=
2
10 Ne
=
2
8
18 Ar
=
2
8
8
36 Kr
=
2
8
18
8
54 Rn
=
2
8
18
18
8
2. The Octet Law
The tendency of atoms in molecules have eight electrons in their valence shells.
(Ne , Ar , Kr , Xe , Rn)
3. The Duplet Law
All of the noble gases except helium possess eight valence elctrons which is the
maximum number possible. Helium’s valence electron configuration is 152.
4. Covalent Bond
In 1916 Gilbert Newton Lewis proposed that the strong attractive force between
two atoms in a molecule result from a covalent bond, a chemical bond formed by
the sharing of a pair of electrons between atom.
Covalent bonds are formed between similar or even identical atoms.
Most often, two non metal atoms are involved.
a. A single – Covalent Bond
A single – Covalent Bonds is a bond where a single pair of electrons is shared
between two atoms.
Example :
H2

= 1  (needs 1e)
= 1  1e
1H
1H
Lewis Formula :
H .x
Line Diagram
H
H - H
b. A Double – Covalent Bond
A double – Covalent bond is a bond where two atoms two pairs of electrons
Example :
O2  8O =
2
6  (needs 2e)
=
2
6  2e
8O
Lewis Formula :
xx
Line Diagram
. .
x
x
x.
x.
O
.
.
=
O
c. A triple – Covalent Bond
A triple – Covalent bond is a bond where two atoms share three pairs of
electrons.
Example :
N2

=
2
5
 (needs 3e)
7N
=
2
5
 3e
7N
Lewis Formula
x
x
x
N xx
.
.
.
Line Diagram
N Ξ N
N ..
5. Coordinate Covalent Bond
Coordinate covalent bond is a bond in which both electrons of shared pair come
from one of the two atoms involved in the bond
NH4+ 

NH3
NH3
+
H+
7N
=
=
=
=
2
1
1
1

H
1
H- N -H

H
1H
1H
1H
.
x
H . xN x. H + H +
xx
Electron donor
 3e
1e
1e
1e
5



+
6. A Polar covalent bond and a non polar covalent bond
- A Polar covalent bond is one in which there is uniquel of bonding electrons
Example :
H Cl

1H
17 Cl
=
=
1
2
8
Or
H
Cl
7
xx
H
. x Cl xx
xx
δ+
δCl atom is more electronegative than the H atom, the bonding electrons in
H – Cl are pulled toward Cl.
A non polar – covalent bond is one in which there is equal sharing of
electrons.
Example :
H2 , Cl2 , F2 , O2 , etc
CO2 , CH4 , CCl4
-
7. A Metal bond
The metal consists of metal atoms where has positive can in electrons ocean
     
     
     
Posistive Ion
Electron Ocean
Energy of metal nucleus attraction is connected by electrons ocean will orm of
metal bond.
V.
Competence Test
A. Multiple Choice
1. The stable element is …
a. 3P
b. 8Q
2.
c.
10R
d.
16S
33T
will be stable with tendency of …
a. Give one electron
d. Form electron pair
b. Give two electrons
e. bond three electrons
c. Receive one electron
20Ca
3. The element atom will form ion bond on 17 X is …
a. 6C
b. 2O
c. 11Na
d. 14Si
4.
e.
12A
a.
b.
c.
d.
e.
e.
16S
and 9B will form compound is …
Ion bond on chemistry formula AB2
Ion bond on chemistry formula AB
Ion bond on chemistry formula A2B
Covalent bond on chemistry formula AB2
Covalent bond on chemistry formula A2B
5. The compound has covalent bond, except …
( 1H, 8O , 6C , 12Mg , 17Cl )
a. H2O
b. CH4
c. Cl2
d. MgO
e. HCl
6. The following compound is polar covalent compound is ….(35Br)
a. CH4
b. O2
c. Cl2
d. H2
e. HBr
7. The elctron valency has oxygen atom is H2O molecul is ….
a. 3
b. 2
c. 4
d. 5
e. 6
8. This is ….
1
H
H

x
X
N xH
XX
H
2
The coordinate covalent bond is …
a. 1 and 2
b. 1 and 3
c. 4
d. 2
e. 2 and 4
3
4
9. The sum free electron on N atom in NH3 molecule (picture no. 8) is …
a. 1 pair
c. 3 pairs
e. only one
b. 2 pairs
d. 4 pairs
10. There are electronegative elements on H, F, Cl, Br and I :
2,1 ; 4,0 ; 3,5 ; 2,8 ; and 2,5. The very polar molecule is …
a. HF
b. HCl
c. Hbr
d. HI
e. IF
11. The compound on NH4Cl is ….
a. ion bond
d. Covalent and coordinate bond
b. covalent bond
e. Ion, covalent and coordinate bond
c. coordinate bond
12. The following metal characteristic connects to metal bond is …
a. Conductor
b. The element mass is very huge and hard
c. The metal can put off valency electron
d. It is easy to form ionic bond on non metal
e. The metal has critic boild and high melting point
B. Answer the following questions correctly !
1. Explain to process compound : (12Mg, 8O, 6C, 1H)
a. MgO
b. C2H4
2. Explain to form H3O+
Keys 1.2:
A.
1.
2.
3.
4.
C
B
C
4
5.
6.
7.
8.
D
E
C
C
9. A
10. A
11. E
12. A
Forming Bond in Compound

1. a. MgO
12Mg
=
=
8O
Mg2+
2-
xx



b. C2 H4
xx
6C
H
=
=
=
=
=
=
H
x
x
C xx


2
2
1
1
1
1
1
1
C
C
C
1
x
4
4
H


x

8
6
MgO
x
x
6C
1H
1H
1H
1H

2
2
1
H
H
2. The forming = H2O + H + 
1H = 1
1H = 1
6
8O = 2
H3O +
+
XX
H
 X
X
XX
H

XX
H-
- H
2

 2e
2e
MODULE FOR INTERNATIONAL STANDARD CLASS
TAMBUN SELATAN HIGH SCHOOL
Subject
Date – Time
Teacher
Module Code
Sub Theme
I.
:
:
:
:
:
Chemistry
……. x 45’
CH / BC 2.1 / 1
Element Symbol
Chemical Bonding
Balance the Equation
Standard Competence
To describe basic chemistry law and applied for starchiomtry
II.
Basic Competence
To write unorganic and simple organic on balance the equation
III. Indicator
1. To write binari molecule compounds and polyatomic from unorganic and organic
compound
2. To interpret balance the equation and give substance name reaction
IV. Teaching Materials






The using chemistry symbols in IUPAC ( International Union Pure and Applied
Chemistry.
The chemical equation is the symbolic representation of a chemical reaction in
term of chemical formulas.
The chemical formula of a substance is a notation that uses atomic symbols with
numerical subscript to convey the relative proportions of atom of the different
elements in the substance.
The chemical formula consist from molecule formula and empirical formula
The molecular formula is a chemical formula that gives the exact number of
different atoms of an element in a molecule.
Example : Compotition on water is 2 Hydrogen atoms and 1 oxygen atom
The empirical formula is a simple comparation sum on particel composition
substance and applied by element symbol.
Example : Glucose,
Molecule formula = C6 H12 O6
Empirical formula = ( CH2O )6
The name rules on compounds :
1. A Binary Covalent Compounds
a. When the two elements form only one compound, we name the compound by
giving the name of the first element, followed by the stem name of the second
element with the suffix – ide .
Example :
HCl
H2S
= Hydrogen Chloride
= Hydrogen Sulfide
b. When the two elements form more then one compound, we can distinguish between
these compounds by using prefix.
One = mono
Four = tetra
Seven = hepta
Two = di
Five = penta
eight = octa
Three = tri
Six
= hexa
Example :
NO
= nitrogen monooxida
NO2 = nitrogen dioxida
N2O3 = dinitrogen trioxida
2. A Binary ion compound
a. A metal with a monatomic ion + non metal + suffix ,-ide
Example :
NaCl = natrium chloride
MgF2 = magnesium fluoride
b. A monatomic cation is given the name of the elementif there is more than one
cation of that element, thecharge is denoted in the stock system of nomenclature by
a roman numeral in paren theses following the element name.
Example :
FeCl2 = ferrum (II) cloride
FeCl3 = ferrum (III) cloride
3. The acid name rules
The acid are substance that yield hydrogen ions, H + in aqueous solution
a. non oxoacid (with suffix –ide)
Example :
HCl = Cloride acid
HI
= Iodide acid
b. Oxoacid (with suffix –ate)
Example :
H2 SO4
=
Sulfuric acid
H2 SO3
=
Sulfurous acid
HCl O4
=
Perchloric acid
HClO
=
Hypochlorous acid
4. The base name rules
The base are substances that yield OH ions in equeous solution. The base consist from
metal and base valency.
Example :
NaOH  Na+
+
OH –
Base
metal ion
base valency
5. The Salts
The salts are ionic ompounds that contain only negative ion except hydroxide ion and
any positive ion except hydrogen ion.
The salts are joined together metal ion (base) and acid residue ion (acid)
Cation
Ca2+
K+
Anion
Cl CaCl2
KCl
2SO4
Ca SO4
K2 SO4
Example :
Ca Cl2 = Calcium chloride
KCl = Kalium Chloride
Ca SO4 = Calsium Sulfate
K2 SO4 = Kalium Sulfate
V.
Competence Test
A. Multiple Choice
1. The name of Cu S is ….
a. Cuprum ( I ) sulfide
b. Cuprum sulfide
c. Cuprum ( II ) sulfide
d. Cuprum
e. Cupro sulfide
2. The formula of diclorin oxide is ….
a. Cl O
b. Cl2 O
c. CO
d. CO2
e. Cl2 O5
3. The formula of Aluminium Sulfide is ...
a. Al2 S3
d. Al S
b. Al3 S2
e. Al2 S
c. Al S3
4. The following is base formula is ….
a. CH3 COOH
b. H2 SO4
c. HNO3
d. Mg ( OH )2
e. KNO3
5. The followis is arsenat acid is …
a. H2 SO4
b. H3 PO4
c. H2 A5 O3
d. HCl O3
e. H2 S
6. The formula on ferrum ( III ) sulfate is ….
a. Fe SO4
d. Fe2 S3
b. Fe2 (SO4)3
e. Fe S
c. Fe3 (SO4)2
B. Essay
1. Write the name from :
a. SO3
b. P2 O5
2. Write the formula from :
a. Magnesium Chloride
b. Sulfur heksa fluoride
3. Determine the forming compounds and give their names.
Cation
Anion
OH CO3 2PO43
Al 3+
Mg 2+
NH4+
Keys 2.1 :
A. Multiple Choice
1. D
2. B
3. A
4. D
5. C
6. B
B. Essay
1. a. SO3
b. P2 O5
= Sulfite
= dinitrogen penta oxide
2. a. Magnesium Chloride
b. Sulfur heksa fluoride
= Mg Cl2
= SF6
3.
Cation
Anion
OH CO3 2PO43
Al 3+
Mg 2+
NH4+
Al (OH)3
Al2 (CO3)3
Al PO4
Mg (OH)2
Mg CO3
Mg3 PO4
NH4 OH
(NH4)2 CO3
(NH4)3 PO4
Balancing Chemical Equations
Chemical equation : mA + nB
A and B
: reactan
C and D
: product
m, n, p, q
: coefficients
 Pc + q D
Example :
KCl O3 ( s ) 
Kalium chlorate
Atom
K
Cl
O
2
KCl O3 ( s ) + 3 O2 ( g )
Kalium chloride
Left
Right
2
2
2
2
6
6
2
Labels in chemical equation :
S
= Solid
l
= Liquid
g
= Gas
aq
= Aquaeous (water solution)
the method in balancing chemical equations :
1. a trial and error
2. algebra
3. redoks
Competence Test
1. Balance the equations and mentions it compounds !
a. Na OH (aq) + H3 PO4 (aq)  Na PO4 (aq) + H2O (l )
b. Al (s) + H2 SO4 (aq)  Al2 (SO4) aq + H2 (g)
c. C2H2 (g) + O2 (g)  CO2 (g) + H2O ( l )
2. Balance the following equation is
a.
b.
c.
d.
e.
Fe2 O3 (s) + 2 Al (s)  Al2 O3 (s) + Fe (s)
Al (s) + 3H2 SO4 (aq)  Al2 (SO4)3 (aq)
C2 H5 OH ( l ) + 3 O2 (s)  2 CO2 (g) + 3H2O (l )
Mg (OH)2 (s) + 2 HCl (aq)  mg Cl2 (aq) + H2O (l )
3Cu (s) + 6 HNO3 (aq)  3 Cu (NO3)2 (aq) + NO (g)
3. The correct coefficients for equation :
a Fe2 S3 + b H2O  c O2  d Fe (OH)9 + e S
The coefficients a, b, c, d, and e are :
a. 2, 3, 2, 2, 3
b. 2, 6, 4, 2, 3
c. 2, 6, 3, 4, 6
d. 4, 6, 3, 4, 12
e. 2, 6, 6, 4, 6
MODULE FOR INTERNATIONAL STANDARD CLASS
TAMBUN SELATAN HIGH SCHOOL
Subject
Date – Time
Teacher
Module Code
Sub Theme
I.
:
:
:
:
:
Chemistry
……. x 45’
CH / BC 2.2 / 1
Basic Chemistry Law
Standard Competence
To describe basic chemistry law and applied for stoichiometry.
II.
Basic Competence
To demonstrate and to inform about basic chemistry law on experiments
III. Indicator
1. To demonstrate Lavoisier Law (before and after on reaction in substance mass in
constant)
2. To demonstrate and to describe the data about two elements compounds mass
(Proust Law)
3. To demonstrate Dalton Law in some compounds
4. To use experiment data and to prove Gay Lussac Law
IV. Teaching Materials
A. Lavoisier Law
Before and after on substance reaction is constant.
Example : H2 + O2  H2 O
If 4 gram H2 is reactioned with 32 gr O2 is 36 gr H2O
B. Proust Law
The elements mass comparation on compounds are constant
Exp. No
Cu Mass (gr)
S mass (gr)
Cu S (gr)
1
0,24
0,12
0,36
2
0,31
0,15
0,46
3
0,41
0,20
0,60
4
0,51
0,25
0,75
5
0,64
0,31
0,95
Cu : S = 2 : 1
Constant
C. Dalton Law (Multiple Comporation Law)
If two elements can form more than a compound, and if an element mass is
constant so another element compound comporation is rounding and simple
amount.
Example :
Nitrogen and oxygen can form NO, N2 O, N2 O3, and N2 O4 on the following mass
composition.
Nitrogen mass
Oxygen mass
Comparation
Compound
(gr)
(gr)
N=O
N2 O
28
16
7 : 4
NO
14
16
7 : 8
N2 O3
28
48
7 : 12
N2 O4
28
64
7 : 16
So oxygen mass comparation in N2O : NO : N2O3 : N2O4 = 4 : 8 : 12 : 16
D. Gay Lussac Law
Gases volume on reaction and gases volume on product can measured on
temperature and pressure standard is rounding and simple amount.
V.
Vol H2
1
= Vol Cl2
=
1
= Vol Hcl
=
2
Vol H2
2
= Vol O2
=
1
= Vol H2O
=
2
Competence Test
1. If 18 gr glucose, C6H12 O6 is burned by oxygen so product 26,4 gr CO2 and 10,8
gr H2O.
How many grams of oxygen are needed on ombustion ?
2. The comparation on oxygen mass and hydrogen mass in compound is 8 : 1. If
100 gr oxygen can form H2O, how many grams of H2O is produced ?
3. A and B elements can form 2 compounds and are consist of 50 % and 60 % A.
Determine comparation of B element mass on A constant ?
4. 4 l C2H6 is burned in balance the following equation is :
C S 2(g) + 3 O2 (g)  CO (g) + 2 SO2 (g)
How many volumes for CO2, H2O are producted in STP.
Keys 2.2 :
1. C H12 O6 + 6 O2  6 CO2 + 6 H2O
18 gr
26,4
10,8
(26,4 + 10,8) – 18 = 37,2 – 18
= 19,2 gr O2
2. O : H
100 gr : 3 gr
O :
8 :
100 :
24 :
3. (A B)
A = 50%
B = 50%
H2 O is producted 27 gr
(A B)
A = 60%
B = 40%
If B = 50gr
A = 50gr
Comparation A1
= 50
= 10
= 1
= 2
H
1
3
3
If B = 50gr
A = 60 x 50
40
= 75gr
:
:
:
:
:
A2,
75
15
1,5
3
B is constant (50gr)
4. 2 C2 H6 + 7 O2  4 CO2 + 6 H2O
4 l
Vol O2 is needed
Vol CO2 is producted
Vol H2O is producted
= 7/2 x 4
= 4/2 x 4
= 6/2 x 4
= 14 l
= 8l
= 12 l
REFERENCES
1. Ebbing, 1984. General Chemistry. Houghton Mifflin
Company Boston
2. Purba, Michael 2004. Kimia SMA X 1 A
Erlangga Jakarta
3. Sudarmo, Unggul 2004. Kimia SMA X
Erlangga, Jakarta
4. Stoker, H. Stephen and Edward B Walker. 1939
Fundamentals of Chemistry. Allyn and Bacon. Inc
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