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Transcript
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 2
The Components of Matter
2-1
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
MATTER
PURE SUBSTANCES
2-2
ELEMENTS
COMPOUNDS
Composed only
of atoms of the
same element.
Two or more
elements
chemically
combined.
MIXTURES
HOMOGENOUS
HETEROGENEOUS
(See definitions in slide
after next.)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Definitions for Components of Matter
Compound - a substance
composed of two or more elements
which are chemically combined.
Figure 2.1
Mixture - a group of two or more
elements and/or compounds which
are physically intermingled.
2-3
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 2.19
The distinction between mixtures and compounds.
MIXTURE
COMPOUND
S
Fe
Physically mixed therefore can be
separated by physical means; in
this case by a magnet.
2-4
Allowed to react chemically
therefore cannot be separated by
physical means.
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Mixtures
Heterogeneous mixtures : have one or more visible boundaries
between the components. Ex: water and oil
Homogeneous mixtures : have no visible boundaries because the
components are mixed as individual atoms, ions, and molecules.
Ex: water and dissolved salt.
Solutions : A homogeneous mixture is also called a solution.
Solutions in water are called aqueous solutions, and are very
important in chemistry. Although we normally think of solutions as
liquids, they can exist in all three physical states.
2-5
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Tools of the Laboratory
Basic Separation Techniques
Filtration : Separates components of a mixture based upon
differences in particle size. Normally separating a precipitate
from a solution, or particles from an air stream.
Crystallization : Separation is based upon differences in solubility of
components in a mixture.
Distillation : separation is based upon differences in volatility.
Extraction : Separation is based upon differences in solubility in
different solvents (major material).
Chromatography : Separation is based upon differences in solubility
in a solvent versus a stationary phase.
2-6
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Basic Separation Techniques
From full-size Silberberg text
Figure B2.3 Filtration
2-7
Figure B2.4 Crystallization
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Physical Properties: color, texture, odor,
density, BP, MP, etc.
Chemical Properties: how a substance reacts
with other substances
2-8
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Chemical
Physical
Properties*:
Physical properties: MP, BP, color, density, solubility
in water. (*Except Na reacts with water.)
2-9
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


2-10
Identify the type of matter in four ways:
element, compound, homogenous mixture,
or heterogeneous mixture
Know the three states of matter
Separate physical properties from chemical
properties
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Law of Mass Conservation:
The total mass of substances does not change during a
chemical reaction.
reactant 1
+
reactant 2
total mass
product
=
total mass
calcium oxide
+
carbon dioxide
calcium carbonate
CaO
+
CO2
CaCO3
56.08g
+
44.00g
100.08g
2-11
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 2.3 (4th ed.)
Law of Definite (or Constant) Composition:
No matter the source, a particular compound is
composed of the same elements in the same parts
(fractions) by mass.
Calcium carbonate
Analysis by Mass
(grams/20.0g)
8.0 g calcium
2.4 g carbon
9.6 g oxygen
20.0 g
2-12
Mass Fraction
(parts/1.00 part)
Percent by Mass
(parts/100 parts)
0.40 calcium
0.12 carbon
0.48 oxygen
40% calcium
12% carbon
48% oxygen
1.00 part by mass
100% by mass
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Law of Multiple Proportions:
If elements A and B react to form two compounds, the different
masses of B that combine with a fixed mass of A can be expressed
as a ratio of small whole numbers.
Example: There are two “Carbon Oxides,” A & B
Carbon Oxide I : 57.1% oxygen and 42.9% carbon
Carbon Oxide II : 72.7% oxygen and 27.3% carbon
Assume that you have 100 g of each compound. That means that Oxide I
has 57.1 g of oxygen and 42. 9 g of carbon; Oxide II has 72.7 g of oxygen
and 27.3 g of carbon.
gO
57.1
=
= 1.33
gC
42.9
gO
gC
72.7
=
27.3
= 2.66
2.66 g O/g C in II
1.33 g O/g C in I
2-13
=
2
1
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Dalton’s Atomic Theory (MEMORIZE!)
1. All matter consists of atoms.
2. Atoms of one element cannot be converted into
atoms of another element.
3. Atoms of an element are identical in mass and other
properties and are different from atoms of any other
element.
4. Compounds result from the chemical combination of
a specific ratio of atoms of different elements.
2-14
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Fig 2.4 Experiments to Determine the Properties of Cathode Rays
CONCLUSION
OBSERVATION
1. Ray bends in magnetic field.
2. Ray bends towards positive
plate in electric field.
consists of charged particles
3. Ray is identical for any cathode.
consists of negative particles
This led to the discovery of
Electrons.
2-15
particles found in all matter
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Figure 2.5
Millikan’s oil-drop experiment
for measuring an electron’s charge.
(1909)
2-16
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Millikan used his findings to also calculate the mass of an
electron.
mass of electron =
mass
determined by J.J. Thomson and
others
X
charge
charge
= (-5.686x10-12 kg/C) X (-1.602x10-19 C)
= 9.109x10-31 kg = 9.109x10-28 g
2-17
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At the beginning of the 20th Century: scientists
knew an atom contained both positive and
negative particles. One model was “plum
pudding” – like raisins in rice pudding.
Rutherford’s experiment showed that an alpha
particle (a Helium nucleus with 2 protons and
2 neutrons) aimed at gold foil was deflected
almost straight back. NOT WHAT HE
EXPECTED: HAD TO REVISE THEIR MODEL!!!
Led to modern nuclear atom model.
2-18
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Figure 2.6
2-19
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General features of the nuclear atom (Know these!)
Figure 2.7
Atom is electrically neutral, spherical & composed of a positively charged
central nucleus surrounded by one or more negatively charge electrons.
Nucleus consists of protons and neutrons.
2-20
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Table 2.2
Properties of the Three Key Subatomic Particles
Charge
Mass
Location
Name(Symbol) Relative Absolute(C)* Relative(amu)† Absolute(g) in the Atom
Proton (p+)
Neutron (n0)
Electron (e-)
1+ +1.60218x10-19
1.00727
0
0
1.00866
1-
-1.60218x10-19
0.00054858
* The coulomb (C) is the SI unit of charge.
†
2-21
1.67262x1024
Nucleus
1.67493x10-24 Nucleus
9.10939x10-28
Outside
Nucleus
You memorize what’s
boxed above.
The atomic mass unit (amu) equals 1.66054x10-24 g.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Atomic Symbols, Isotopes, Numbers
A
X
Z
The Symbol of the Atom or Isotope
X = Atomic symbol of the element
A = mass number; A = Z + N
Z = atomic number
(the number of protons in the nucleus)
N = number of neutrons in the nucleus
Isotope = atoms of an element with the same number of protons, but a
different number of neutrons
From Figure 2.8
2-22
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Sample Problem 2.4
Determining the Number of Subatomic
Particles in the Isotopes of an Element
28
29
PROBLEM: Silicon has three naturally occurring isotopes: Si, Si, and
30Si. Determine the number of protons, neutrons, and electrons
in each silicon isotope.
SOLUTION:
The atomic number of silicon is 14. Therefore
28Si
29Si
has 14p+, 14e- and 14n0 (28-14)
has 14p+, 14e- and 15n0 (29-14)
30Si
has 14p+, 14e- and 16n0 (30-14)
Now you try these uranium isotopes: 234, 235, 238, and 239.
2-23
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REMEMBER: In an atom of the same element
# protons never changes (except for nuclear
decay)
# neutrons can be different  ISOTOPES
# electrons can change  IONS
2-24
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Tools of the Laboratory
Figure B2.2
2.9
2-25
The Mass Spectrometer and Its Data
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Fig 2.9 is a simple schematic, but the mass spec determines the
actual mass of each isotope to many sig figs and its percent
abundance, usually to six or more sig figs.
Chart B = the mass charge ratio, which determines the mass to
many sig figs (look at fluorine on per table)
Chart C = a count of each isotope which is reported as percent
abundance, again to many sig figs
By using a mass spectrometer, an atom’s isotopes can be
counted, establishing relative abundance of each isotope and
its exact mass. From this data we can calculate a weighted
average atomic mass:
Weighted average atomic mass = fract1*mass1 + fract2*mass2 + ...
2-26
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We put boron in a mass spectrometer, and find
there are just two isotopes, with these results:
19.91% Boron-10 with mass 10.0129 amu and
80.09% Boron-11 mass with11.0093 amu
Average atomic mass = (0.1991 * 10.0129)
+ (0.8009 * 11.0093)
= 1.994 + 8.817 = 10.811 (sig fig rules)
Do problem 33 for practice on this method.
2-27
(Magnesium has three naturally occurring isotopes.
24Mg has a mass of 23.9850 amu and 78.99%
abundance, 25Mg has a mass of 24.9868 amu and
10.00% abundance, and 26Mg has a mass of
25.9826 amu and 11.01% abundance.)
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You have just practiced determining the
weighted average atomic mass. How is that
different from mass number?
ATOMIC MASS IS NOT THE SAME AS MASS NUMBER!
ATOMIC MASS IS NOT THE SAME AS MASS NUMBER!
ATOMIC MASS IS NOT THE SAME AS MASS NUMBER!
Think you can remember this?
2-28
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The Modern Reassessment of the Atomic Theory
1. All matter is composed of atoms. The atom is the smallest body that
retains the unique identity of the element.
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.
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. A sample of the element is treated as though its
atoms have an average mass.
4. Compounds are formed by the chemical combination of two or more
elements in specific ratios.
2-29
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Figure 2.10Copyright ©The McGraw-Hill
The modern periodic table.
2-30
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Write down group names, locations for metal,
nonmetals and metalloids.
(Alkali metals, alkaline earth metals, halogens,
noble gases, metalloids along stairs, metals
to lower left, nonmetals to upper right)
What trends do you know?
(Fr is largest, He is smallest)
Where are the transition metals and the inner
transition metals?
(Be able to locate them.)
2-31
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
2-32
You need to memorize the first 36 elements’
symbols and names, plus ten more: Ag, Sn, I,
Ba, Pt, Au, Hg, Pb, Rn, and U.
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Figure 2.11 from 4th
ed.
Copper
Metals, metalloids, and nonmetals.
Cadmium
Lead
Chromium
Bismuth
Arsenic
Silicon
Antimony
Chlorine
Bromine
Sulfur
Iodine
Boron
Tellurium
Carbon
(graphite)
Stop here and go to chapter 7!
2-33
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Figure 2.14
from 4th ed.
2-34
The relationship between ions formed and
the nearest noble gas.
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Figure 2.13
Formation of a covalent bond between two H atoms.
Covalent bonds form when elements share electrons, which usually
occurs between nonmetals.
2-35
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

2-36
Use the lecture notes, the textbook AND the
lab manual’s Dry Lab on Nomenclature.
All the tables in the dry lab will be very
useful!
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The seven common diatomic elements:
H2, O2, N2, F2, Cl2, Br2, I2
(Memory aid: HON+Halogens)
2-37
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Types of Chemical Formulas
A chemical formula is comprised of element symbols and numerical
subscripts that show the type and number of each atom present in the
smallest unit of the substance.
An empirical formula indicates the relative number of atoms of
each element in the compound. It is the simplest type of formula.
The empirical formula for hydrogen peroxide is HO.
A molecular formula shows the actual number of atoms of
each element in a molecule of the compound.
The molecular formula for hydrogen peroxide is H2O2.
A structural formula shows the number of atoms and the
bonds between them, that is, the relative placement and
connections of atoms in the molecule.
The structural formula for hydrogen peroxide is H-O-O-H.
2-38
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Figure 2.16
Some common monatomic ions of the elements.
Can you see any patterns?
See large Periodic Table
handout for variety of
charges of metals.
Be2+
2-39
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Remember that metals like iron or tin with
more than one possible charge have to show
which one by using Roman numerals for the
charge in parentheses.
Sn2+ is tin (II) ion
Sn4+ is tin (IV) ion
2-40
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Naming binary ionic compounds: see rules in your packet
The name of the cation is written first, followed by that of the anion.
The name of the cation is the same as the name of the metal.
Many metal names end in -ium.
The name of the anion takes the root of the nonmetal name
and adds the suffix -ide.
Calcium ion and bromide ion form calcium bromide.
2-41
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EXAMPLE: Form an ionic compound of Al & Br
and then Al & O.
Al forms Al3+ and Br forms Br-, so the ratio will
be 1:3 or AlBr3 aluminum bromide.
Again, Al forms Al3+ and O forms O2-, so the
ratio will be 2:3 or Al2O3 aluminum oxide.
Special trick called the criss-cross rule:
Xa+ & Yb-  XbYa
but if b=a, reduce formula to 1:1 ratio (unless
mercury (I) ion is involved)
2-42
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Naming Binary Ionic Compounds (sample problems 2.7 & 2.8
PROBLEM: Write formulas and name the ionic compound formed from the
following pairs of elements:
(a) magnesium and nitrogen
(b) iodine and cadmium
(c) strontium and fluorine
(d) sulfur and cesium
SOLUTION:
(a) Mg2+ & N3-; three Mg2+(6+) & two N3-(6-); Mg3N2
magnesium nitride
(b) Cd2+ & I-; one Cd2+(2+) & two I-(2-); CdI2
cadmium iodide
(c) Sr2+ & F-; one Sr2+(2+) & two F-(2-); SrF2
strontium fluoride
(d) Cs+ & S2-; two Cs+(2+) and one S2- (2-); Cs2S
cesium sulfide
Now do the follow-up problems.
2-43
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Sample Problem 2.9
Determining Names and Formulas of Ionic
Compounds of Elements That Form More
Than One Ion
PROBLEM:
Give the systematic names for the formulas or the formulas for the
names of the following compounds:
SOLUTION:
(a) tin(II) fluoride
(b) CrI3
(c) Iron (III) oxide
(d) CoS
(a) Tin (II) is Sn2+; fluoride is F-; so the formula is SnF2.
(b) The anion I is iodide(I-); 3I- means that Cr(chromium) is +3.
CrI3 is chromium(III) iodide
(c) Iron (III) is the name for Fe3+; oxide is O2-, therefore the
formula is Fe2O3.
(d) Co is cobalt; the anion S is sulfide(2-) so cobalt must be +2;
the compound is cobalt (II) sulfide.
2-44
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Hydrates are ionic crystals of salts with water
molecules incorporated in their crystal
structures.
Write "formula unit name - dash - Greek prefix
(representing # of water molecules) hydrate"
BaCl2.2H2O is barium chloride dihydrate
You try:
Name CuSO4.5H2O:
Give the formula for sodium sulfate
decahydrate:
2-45
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Naming oxoanions
Figure 2.17
Root
Suffixes
per
root
ate
ClO4-
perchlorate
root
ate
ClO3-
chlorate
root
ite
ClO2-
chlorite
root
ite
ClO-
hypochlorite
No. of O atoms
Prefixes
hypo
Examples
Table 2.6
Numerical Prefixes for Hydrates and Binary Covalent Compounds
Number
Prefix
Number
Prefix
Number
1
mono
4
tetra
8
octa
2
di
5
penta
9
nona
3
tri
6
hexa
10
deca
7
hepta
2-46
Prefix
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 2.10
Determining Names and Formulas of Ionic
Compounds Containing Polyatomic Ions
PROBLEM: Give the systematic names or the formula or the formulas for the
names of the following compounds:
(a) Fe(ClO4)2
SOLUTION:
(b) sodium sulfite
(c) Ba(OH)2 8H2O
(a) ClO4- is perchlorate; iron must have a 2+ charge. This is
iron(II) perchlorate.
(b) The anion sulfite is SO32- therefore you need 2 sodiums per
sulfite. The formula is Na2SO3.
(c) Hydroxide is OH- and barium is a 2+ ion. When water is
included in the formula, we use the term “hydrate” and a prefix
which indicates the number of waters. So it is barium hydroxide
octahydrate.
2-47
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Naming Inorganic Acids
1) Binary acid solutions form when certain gaseous compounds
dissolve in water.
For example, when gaseous hydrogen chloride (HCl) dissolves in
water, it forms a solution called hydrochloric acid. Prefix hydro- +
anion nonmetal root + suffix -ic + the word acid - hydrochloric acid
Practice naming H2S(aq):
2) Oxoacid names are similar to those of the oxoanions, except for
two suffix changes:
Anion “-ate” suffix becomes an “-ic” suffix in the acid. Anion “-ite”
suffix becomes an “-ous” suffix in the acid.
The oxoanion prefixes “hypo-” and “per-” are retained. Thus, BrO4is perbromate, and HBrO4 is perbromic acid; IO2- is iodite, and
HIO2 is iodous acid.
Practice naming HClO3:
2-48
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Sample Problem 2.12
Determining Names and Formulas of Anions
and Acids
PROBLEM: Name the following anions and give the names and formulas of
the acids derived from them:
(a) Br -
(b) IO3 -
(c) CN -
(d) SO4 2-
(e) NO2 -
SOLUTION:
(a) The anion is bromide; the acid is hydrobromic acid, HBr.
(b) The anion is iodate; the acid is iodic acid, HIO3.
(c) The anion is cyanide; the acid is hydrocyanic acid, HCN.
(d) The anion is sulfate; the acid is sulfuric acid, H2SO4.
(e) The anion is nitrite; the acid is nitrous acid, HNO2.
2-49
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Very simple – you will know a compound is
covalent if it's two nonmetals. Indicate how
many of each atom in the compounds by
using Greek prefix (mono, di, tri, tetra, penta,
hexa, hepta, octa, nona, deca)
If first element is single, leave off mono. If first
element is hydrogen, leave off any prefix.
‘Prefix'element name - 'prefix'element root suffix 'ide'
Practice: CO, CO2, NO2, N2O4, P2O5, HF, H2S
2-50
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We still use old common names you'll have to
memorize:
water H2O
hydrogen peroxide H2O2
Ammonia NH3
Blood sugar
Table sugar
2-51
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Learning the rules for naming organic compounds
would take a month or more!
Just memorize these:
Methane
CH4
Ethane CH3CH3
Propane CH3CH2CH3
Butane CH3(CH2)2CH3
Octane CH3(CH2)6CH3 Benzene C6H6
Acetylene C2H2
Methanol CH3OH
Ethanol CH3CH2OH
1-Propanol CH3CH2CH2OH
Acetic Acid CH3COOH Formaldehyde HCHO
Glucose C6H12O6
Sucrose C12H22O11
2-52
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

2-53
Look up the molar mass on the Periodic
Table. Sum up all atoms in the compound.
(This part goes just before or with chapter 3.)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 2.15
PROBLEM:
Calculating the Molecular Mass of a Compound
Using the data in the periodic table, calculate the molecular (or
formula) mass of the following compounds:
(a) tetraphosphorous trisulfide
PLAN:
SOLUTION:
Write the formula and then multiply the number of atoms(in the
subscript) by the respective atomic masses. Add the masses for
the compound.
(a) P4S3
molecular = (4x atomic mass of P)
mass
+ (3x atomic mass of S)
(b) NH4NO3
molecular = (2x atomic mass of N)
mass
= (4x30.97 amu) + (3x32.07 amu)
= 220.09 amu
2-54
(b) ammonium nitrate
+ (4x atomic mass of H)
+ (3x atomic mass of O)
= (2x14.01 amu)+ (4x1.008 amu)
+
(3x16.00 amu)
= 80.05amu
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Do the follow-up problem for 2.15 in text:
Determine the formula and the molecular (or
formula) mass for each of these:
a. hydrogen peroxide
b. cesium chloride
c. sulfuric acid
d. potassium sulfate
2-55