***C.4.A differentiate between physical and chemical changes and properties

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STAAR Chemistry Review Packet

Ms. Cole Chemistry

***
C.4.A differentiate betwee
n physical and chemical changes
and properties

Physical Properties of Matter

Definition: a property
that can be observed without changing the identity of the substance.

Chemical Properties

Definition: a property that can be observed only during a chemical change

Examples: reaction with acid, reaction with base, reaction with water, reaction with a
ir

Chan
ges of Matter:

Physi
cal Change
:

change in the form or state of a substance, for instance, from solid to liquid or liquid to gas or solid to
gas, without changing the chemical composition of the substance. As we will see later, chemical bonds are not broken

in
a physical change. Examples: Boiling of water and the melting of ice.

Chemical Change: the change of a substance into another substance, by reorganization of the atoms, i.e. by the making
and breaking of chemical bonds. In a chemical change chemical re
action takes place. The substance changes into a new
substance by rearranging the number and kinds of atoms.

Examples: Rusting of iron and the decomposition of water, induced by an electric current,

to gaseous hydrogen
and gaseous oxygen.

Recognizing a che
mical change (rules of thumb):

formation of a precipitate

change in color

effervescence (production of a gas)

change in energy (temperature)

change in sm
ell


+
C.4.B identify extensive and intensive properties

Extensive properties:

Definition: physical
property that depends on the amount of matter present or a property that changes when the size of
the sample changes.

examples: mass, volume, length

not very useful in identifying unknown substances


Intensive properties

Definition: physical property that
does not depend on the amount of matter present or doesn't change when you take
away some of the sample.

examples: density, boiling point, conductivity

can be very useful in identifying unknown substances


+
C.4.C compare solids, liquids, and gas
es in terms

of compressibility, structure, shape, and
volume

Particles in a:

gas are well separated with no regular arrangement.

liquid are close together with no regular arrangement.

solid are tightly packed, usually in a regular pattern. Particles in a:

gas vibrate

and move freely at high speeds.

liquid vibrate, move about, and slide past each other.

solid vibrate (jiggle) but generally do not move from place to place.

Liquids and solids are often referred to as condensed phases

because the particles

are very close

together.






***
C.4.D classify matter as pure substances or mixtures through investigation of their
properties

Mixtures:

two or more substances, combined in varying proportions
-

each retaining its own specific properties. The
components of a mixture can

be separated by physical means, i.e. without the making and breaking of chemical bonds.

Examples: Air, table salt thoroughly dissolved in water, milk, wood, and concrete.

characteristics

percentage composition varies from sample to sample

components are c
hemically different and retain properties in a mixture

do not melt/boil at a definite temperature

heterogeneous mixtures
(
not uniforml
y

mixed
)

uniformly mixed also called solution
s

Pure Substance:

a substance with constant composition. Can be classified an

either an element or as a compound.

Element: a substance that cannot be separated into two or more substances by ordinary chemical (or physical) means.
We use th
e
term ordinary chemical means to exclude nuclear reactions. Elements are composed of only one

kind of
atom.

Examples: Iron (Fe), copper (Cu), and oxygen (O2).

Compound contains two or more elements, in definite proportion by weight. The composition of a pure
compound will be invariant, regardless of the method of preparation. Compounds are compose
d of more than
one kind of atom. The term molecule is often used for

the smallest unit of a compound that still retains all of the
properties of the compound.

Examples: Table salt (sodium chloride, NaCl), sugar (sucrose, C12H22O11), and water


+
C.5.A
explain the use of chemical and physical properties in the historical development of
the Periodic Table

Mendeleev first arranged the elements in his periodic table according to atomic mass.

Mosley derived the modern periodic law stating that elements are a

function of their periodic number, thus the modern
periodic table is arranged by atomic number and not atomic mass.

A. Each element has a specific location on the periodic table, which indicates the elements physical and chemical
properties

1. Elements a
re arranged in the order of increasing atomic number from left to right across each horizontal row
-

a.k.a. period

2. Elements in each vertical column
-

a.k.a. family or group
-

all form compounds with similar chemical formulas
and properties because they
have the same number of valance electrons

3. Physical properties include density, hardness, conductivity, malleability, ductility and solubility

4. Chemical properties describe how an element behaves during a chemical reaction


***
C.5.B use the Periodic

Table to identify and explain the properties of chemical families,
including alkali metals, alkalin
e earth metals, halogens, noble
gases, and transition metals


Elements can be classified as metals, nonmetals, metalloids or noble gases

Metals


on the
left of the stairs (not including Hydrogen)

lose electrons to form positive ions (ionic radius is

smaller than atomic radius)

the greater the ability to lose electrons the more metallic character

have low ionization energy and electronegativity

are deform
able
-

malleable and ductile

good conductors of heat and electricity

mostly solids at room temperature

Nonmetals



on the right of the stairs & Hydrogens

gain electrons to form negative ions ionic radius is

larger than atomic radius)

the greater the abili
ty to gain electrons the more non
-
metallic properties


have high ionization energies and electronegativities

are brittle

are poor conductors of heat and electricity

tend to be gases or solids at room temperature

Metalloids



on the stairs

include the elemen
ts B, Si, Ge, As, Sb, and Te

(Al & Po are metals)

have properties of both metals and nonmetals

Noble Gases



last family on the right of the table

very stable

have complete valance shells

unreactive

gases at room temperature

extist as Monatomic molecules


***
C.5.C use the Periodic Table to identify and explain periodic trends, in
cluding atomic and
ionic radii,
electronegativity, and ionization energy

1. Atomic radius
-


a.) increases going down a family

b.) decreases going across a period

c.) factors
include nuclear charge, distance between nucleus and outer electrons, and amount of shielding

2. Ionic radius



a.) Cations
-

positive ions are smaller than the parent atom and the greater the positive charge in a period the
smaller the ionic size

b.) Ani
ons
-

negative ions are larger than their parent atom and the greater the negative charge across a period
the larger the ion

3. Electronegativity and ionization energy



a.)decreases as go down a family and increases across (L

R) a period

4. Metallic/ non
-
metallic properties


a.) metallic properties increase as go down a family and to the left across a period

b.) non
-
metallic properties increase as go up a family and towards the right across a period

5. Reactivity


a.) the reactivity of metals increases
down a family and decreases to the right across the table

b.) the reactivity of non
-
metals decreases down a family and increases to the right across a period

c.) noble gases are mostly unreactive

d
.) Fluorine is the most reactive non metal


+
C.6.A

Understa
nd the experimental design and conclusions used in the development of
modern atomic theory, including Dalton’s Postulates, Thomson’s discovery of electron
properties, Rutherford’s nuclear
atom, and Bohr’s nuclear atom

Atom
:

The smallest particle of an element that retains all of the chemical properties of the element.

T
he Theory & Evidence for John Dalton’s Atomic Theory:

Around 1805 an English chemist, John Dalton (176
6

1844), put forth a scientific
atomic theory. The major

difference between Dalton’s theory and that of others before him is that Dalton based his theory on evidence
rather than a belief. First, let us state his theory. We will then see what kind of evidence supported it.

Theory:



All matter consists of tiny par
ticles called atoms.



Atoms are indestructible and unchangeable. Atoms of an element cannot be created,



Destroyed
, broken into smaller parts or transformed into atoms of another element. Dalton based this
hypothesis on the law of conservation of mass and on

centuries of experimental evidence.



All atoms of the same element are identical. Elements are characterized by the mass of




Their

atoms. All atoms of the same element have identical weights,



Atoms combine in new ways during a chemical change. When elements

react, their



Atoms

combine in simple, whole
-
number ratios.



Evidence: Evidence for theories is often that they provided logical explanations for observed

observations. Two criteria
are
usually applied to any theory. First, does it agree with facts which
are already known? Second, does it predict new
relationships and stimulate additional observation and experimentation?


The Theory has been modified to account for new observations.

Atoms are indestructible and unchangeable. Atoms of an element cannot be
created, destroyed, broken into smaller
parts

or transformed into atoms of another element. We now know elements have subatomic particles and can be
transformed by nuclear change but not by physical or chemical change.

All atoms of the same element are ide
ntical. Elements are characterized by the mass of their atoms. All atoms of the
same element have identical weights. Atoms of the same element may have different numbers of neutrons but all atoms
of the same elements have the same number of protons.

J. J.
Thomson would have used. A cathode ray tube is a small glass tube with a cathode (a negatively charged metal
plate) and an anode (a positively charged metal plate) at opposite ends. By separating the cathode and anode by a short
distance, the cathode ray t
ube can generate what are known as "cathode rays"


rays of electricity that flowed from the
cathode to the anode, J. J. Thomson wanted to know what cathode rays were, where cathode rays came from and
whether cathode rays had any mass or charge.


Thomson’s

experiment also revealed the electron has a very large charge
-
to
-
mass ratio. In 1909, experiments conducted
by the American physicist Robert A. Millikan measured the charge of the electron. Scientists used this information and
the charge
-
to mass ratio of

the electron to determine that the mass of the electron is about one two
-
thousandth the
mass of the simplest type of hydrogen atom, which is the smallest atom known. More
-
accurate experiments conducted
since then indicate that the electron has a mass of 9
.109 × 10−31 kg, or 1/1837 the mass of the simplest type of
hydrogen atom. Based on what was learned about electrons, two other inferences were made about atomic structure.

Because atoms are electrically neutral, they must contain a positive charge to bala
nce the negative electrons.

Because electrons have so much less mass than atoms, atoms must contain other particles that account for most of their
mass

Final Conclusion
:

Plum Pudding model of the atom.

More detail of the atom’s structure was provided in 19
11 by New Zealander Ernest Rutherford and his associates Hans
Geiger and Ernest Marsden. The scientists bombarded a thin piece of gold foil with fast
-
moving alpha particles, which
are positively charged particles with about four times the mass of a hydroge
n atom.

Ex
periment:

Alpha particles are very dense positively charged radiation. The alpha particles are emitted from a
radioactive ore housed in a lead box with one opening for the fast traveling particles to leave the box. The alpha
particles are shoot a
t a piece of gold foil that Rutherford thought was not very dense because the model of the atom at
the time was the Plum pudding model. In the plum pudding model the mass is spread out over

the entire volume of the
atom. Rutherford thought shooting the alp
ha particles at the very this sheet

of gold foil was like shooting a 16 inch shell
at a piece of tissue paper. He thought the high speed very dense alpha particle would go through the gold leaf.

Hypothesis: If the plum pudding model of the atom is correct,

atoms have no concentration of mass or charge (atoms
are 'soft' targets)


Niels Bohr Observations, Conclusions and Model of the Hydrogen Atom

Observation: that light emitted from samples of atom exposed to energy (energizing the

atoms) when passed through

a
prism emit distinct wavelength
s
of light line spectrum instead of emitting all the colors of light. Each element emits
t
same line spectrum and are just as characteristic to that element as finger prints are to people.

Hypothesis: if energized atoms emit

only discrete wavelengths, maybe electrons can have only

Discrete

energies

from the nucleus called energy levels.

Electrons in an non
-
energized atom occupy the

lowest energy
orbit (closest to nucleus this is called

ground state.)

Electrons in energized at
oms absorb just enough energy to move
from a lower energy orbital to a higher energy orbit (further from nucleus) this is called excited state.

Electrons do not

remain in excited state and return to ground state releasing energy that corresponds to all the

possible difference in
energy of
the allowed orbits in the atom.


+
C.6.B

Understand the electromagnetic spectrum and the mathematical relationships
between energy, frequency, and wavelength of light.

Electromagnetic radiation (EMR) is a form of energy
that is produced by oscillating

electric and magnetic waves, or by
the movement of electrically

charged particles traveling through a vacuum or matter. This energy is then grouped into
categories based on its wavelength into the electromagnetic spe
ctrum.

G
eneral Properties of all electromagnetic radiation:

Electromagnetic radiation can travel through empty space. Most other types of waves must travel through some sort of
substance. For example, sound waves need either a gas, solid, or liquid to pass through

in order to be heard.

The speed of light is always a constant. c= (Speed of light: 3.00 X 10
8

m/s)

Waves and their Characteristics

Wavelength

(λ) is the distance of one full cycle of the oscillation. Wavelengths are measured between the distances
between
two consecutive points on a wave which can be between consecutive troughs and crests. It is usually
characterized by the Greek symbol
ʎ
.

Longer wavelength waves such as radio waves carry low energy; this is why we can listen to the radio without any
harmfu
l consequences. Shorter wavelength waves such as x
-
rays carry higher energy that can be hazardous to our
health. Consequently lead aprons are worn to protect our bodies from harmful ra
diation when we undergo x
-
rays.

Frequency
is defined as the number of cy
cles per second, and is expressed as sec
-
1 or

Hertz (Hz). Frequency is directly
proportional to energy (High frequency

=High energy) and can be express as:


Where E is energy, h is Planck's constant and v is frequency.

Planck's constant: h= 6.62607 x
10
-
34

J s


This wavelength frequently relationship is characterized by
:
C= λ
f

Where c is the speed of light, λ stands for the wavelength, and
f

stands for frequency. Shorter wavelength means greater
frequency, and greater frequency means higher ener
gy. Wavelengths are important in that they tell one what type of
wave one is dealing with.


In general as the energy increases the wavelength decreases and the frequency increases. The opposite is also true as
the energy decreases the wavelength increas
es
and the frequency decreases.


Thus, electromagnetic radiation is then grouped into categories based on its wavelength or frequency into the
electromagnetic spectrum. The different types of electromagnetic radiation shown in the electromagnetic spectrum
con
sists of radio waves, microwaves, infrared waves, visible light, ultraviolet radiation, X
-
rays, and gamma rays. The part
of the electromagnetic spectrum that we are able to see is the visible light spectrum.


The electromagnetic spectrum is more familiar t
o you than you might think. The microwave you use to heat your food
and the cell phones you use are part of the Electromagnetic Spectrum. The light that our eyes can see is also part of the
electromagnetic spectrum. This visible part of the electromagnetic

spectrum consists of the colors that we see in a
rainbow
-

from reds and oranges, through blues and purples.


+
C.6.C

Calculate the wavelength, frequency, and energy using Planck’s constant and the
speed of light.

Example 1:

A dental hygienist uses X
-
rays
( λ = 1.00 Å ) to take a series of dental radiographs while the patient listens’ to a radio
station (λ=325 cm) and looks out the window at the blue sky (λ=473nm). What is the frequency in (s
-
1
) of the
electromagnetic radiation from each source?

In each of
the calculations the wavelength is converted to meters first because the speed of light is 3.00 x 10
8

m/s. The
point to remember is the conversions of wavelength to m.

E =h
f
= 6.626 x 10
-
34

Js


x

4.69 x 10
14

1/s

=

3.11 x 10
-
19

J per photon




+
C.6.D use i
sotopic composition to calculate average atomic mass of an element

atomic number
(Z)
--
The number of protons in an atom.


All atoms of the same element have the same number of protons.

mass number
(A)
--
The sum of the number of neutrons and protons for an
atom. A different mass number does not
mean a different

element
--
just an isotope.

average atomic masses

atoms have masses of whole numbers, HOWEVER samples

of quadrillions of atoms have a few that are heavier or lighter [isotopes] due to different numbers

of neutrons present

percent abundance
--
percentage of atoms in a natural sample of the pure element represente
d

by

particula
r

isotope (number of atoms of a given isotope/total number of atoms of all isotopes of that element)

x 100%

C
alculating Average Atomic mass:

You can use % abundance of the isotope
s:

Or you can use relative abundance of the isotopes:


(
Atomic

mass of isotope × relative abundance) × (atomic mass of isotope × relative abundance) = average atomic mass…

To
calculate the average atomic weight, each exact atomic weight is multiplied by its percent abundance (expressed as a
decimal). Then, add the results together and round off to an appropriate number of significant figures.

Example:

Zinc

There are five natura
lly occurring isotopes of the element zinc. The relative abundance and mass of each are as follows:

Zinc
-
64 = 48.89%, 63.929 amu

Zinc
-
66 = 27.81%, 65.926 amu

Zinc
-
67 = 4.11%, 66.927 amu

Zinc
-
68 = 18.57%, 67.925 amu

Zinc
-
70 = 0.62%, 69.925 amu

Calculate the

average atomic mass of zinc.

(63.929 × .4889) + (65.926 × .2781) + (66.927 × .0411) = (67.925 × .1857) (69.925 ×

.0062) = 65.3868


***
C.6.E express the arrangement of electrons in atoms through electron configurations and
Lewis valence electron dot struct
ures

First some terms and more information about the structure of the atom:

Energy level is no longer an orbit but more like a boundary or maximum distance from the nucleus that electrons
occupy. 1, 2, 3

With
in each energy level (boundary or maximum distan
ce from the nucleus) there are sublevels describes the shape of
the orbital (region occupied by electrons are orbitals).

The s, p and d orbitals have the same shape but differ in their orientation (direction the orbitals are in in space around
an X, Y and
Z axis).


The
s

sublevel only has one orbital (possible orientation) for a total of 2 electrons [s=holds 1 to maximums of 2
electrons]

The
p
sublevel has three orbital (possible orientations) all same shape different orientations each holding 2 electrons
for
a maximum of 6 electrons to fill all 3 p sublevels [p ho
ld 1 to maximum of 6 electrons]

The
d

sublevel has five orbital (possible orientations) each holding 2 electrons for maximum of 10 electrons [d hold 1 to
maximum of 10 electrons] (each orbital hol
ds 1 to2 electrons)

The
f

sublevel has seven orbital (possible orientations) each holding 2 electrons for maximum of 14 electrons [f holds 1
to maximum of 14 electrons]. Each orbital holds 1 to 2 electrons or is unoccupied.

Each orbital can hold two electr
ons but the electrons in the same orbital have opposite magnetic fields called spin. The
electron’s like charges that repel are cancelled by their opposite spin which is an attractive force equal to the repulsion
of like electric charges. The important co
nsequence of spin is that each orbital can hold 2 electrons which have opposite
spins.

Electrons in a sublevel remain unpaired before the second electron is added to an orbital. (For example each of the 3 p
orbitals will each contain one electron before th
e second electron will occupy the orbital)


The energy levels do not contain the same number of sublevels. This makes sense because the energy levels further
from the nucleus are larger. You need to know the sublevels in each energy level, the number of orb
itals each sublevel
has and the maximum number of electrons a sublevel will contain when all the sublevel’s orbitals contain 2 electrons.

Electron Configurations
: Assigning each electron in an atom to the energy level and sublevel it occupies in the atom.

Neutral atoms have equal number of protons and electrons. The number of protons is indicated by the atomic number
on the periodic table if the atom does not have a charge it has the same number of protons as electrons. Electrons fill
one sublevels orbitals

before starting to fill another sublevel. The energy levels are not filled in numerical order. The
periodic table tells you what energy level and sublevel the last electron in that atom occupies and the order of filling.

This look at an electron configur
ation and determine what it tells us: 1s
2

2s
2
2p
5

means "2 electrons in the 1s subshell, 2
electrons in the 2s subshell, and 5 electrons in the 2p subshell"

1s
2
2s
2

2p
6

3s
2

3p
3

is an electron configuration with 15 electrons total; 2 electrons are in the 1s

subshell; 8 electrons are
in the energy level 2 with 2 electrons the 2s subshell and 6 electrons in the 2p subshell; and 5 electrons are in energy
level 3 with 2 electrons in the 3s subshell and 3 electrons in the 3p subshell.

The d sublevels are filled o
ne back from the outer energy level and the f sublevels are filled 2 back from the outer energy
level.
(n= outer energy level)

Writing electron configurations:

Each electron that is added to an atom is placed in the lowest
-
energy orbital that is available.

The orbitals are filled in
the order which is read off the periodic table when reading in order of atomic number:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s,

4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f

Each orbital can hold no more than two electrons. Two electrons in the sam
e orbital must have opposite spins
(Pauli’s exclusion principle).

If two or more orbitals are available at the same energy level and same sublevel (degenerate orbitals), one
electron is placed in each orbital until the available orbitals are occupied by on
e electron. Remember one
electron in each orbital before you double up.

Electron configurations are written as a list of sublevels which are occupied, followed by a superscript to indicate
how many electrons are in those sublevel’s orbitals.

Complete Elect
ron Configurations:

Determine the number of electrons in the atom from its atomic number. (See Below.)

Add electrons to the sublevels in the correct order of filling.

Add two electrons to each s sublevel, 6 to each p sublevel, 10 to each d sublevel, and 14

to each f sublevel.

To check your complete electron configuration, look to see whether the location of the last electron added corresponds
to the element’s position on the periodic table.

Write the electron configuration for Ni:


Ni atomic number is 28 so

it has 28 protons and 28 electrons because it does not have a charge.

1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
8

Electron 1 and 2 are in 1s, the 3rd and 4th electrons are in 2s, the

next six electrons which are the 5th to the 10th
electrons are in the 2p, the 11th and 12th el
ectrons are in the 3s, the next six electrons which are 13th to the 18th
electrons are in the 3p sublevels orbitals, the 19th and 20th electron occupy the 4s orbital, the last 8 electrons will
occupy the 3d orbitals but do not fill all the orbitals in the
3d sublevel.

1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
8

in order of filling which can be rearrange in numerical order of energy level

1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
8
4s
2

Add the electrons (superscripts to check answer): 2+2+6+2+6+2+8 = 28 If we add electrons to correct

orbitals we are
correct.

Here are a couple of electron configurations:

Chlorine (Cl):

Iron (Fe):

Yes there is a short cut: the short cut is sometimes called core notation: Using Ni as an example

Use the nearest noble
gas with a smaller atomic number than

the element you are writing the electron

core notation for: for Ni this would be
Ar with the atomic number 18. Write the symbol of the noble gas in [ ] with the number of electron of the noble gas as a
superscript. [Ar] (this is saying the first 18 elect
rons in Ni are in the same orbitals as the 18 electrons in Ar.

Add the rest
of the electrons reading from the periodic table. You will be assigning the 19th through 28th electron. Remember the
19th electron is in the same orbital for all atoms.

[Ar]

4s
2
3d
8




***
C.7.A name ionic compounds containing main group or transition metals, covalent
compounds, acids, and bases, using (IUPAC) nomenclature rules

Writing Names and Formulas for Ionic Compounds


those containing at least 1
metal

and 1
nonmetal
.

Naming
Ionic Compounds


2 naming systems currently acceptable

1.

IUPAC



International Union of Practical and Applied Chemistry is the newest system
-
this system uses Roman
numerals to give the charges or oxidation number of positive ions
ONLY

if the positive ion ha
s variable charges.

2.

“ous” and “ic” system



oldest system and still very commonly used. May be used ONLY if the positive ion has a
variable charge and exhibits only 2 oxidation numbers.

Review the 4 ions with which we may use this system.


Iron: Fer
rous, ferric, Copper: cuprous, cupric,
Tin:

stannous, stannic,
Lead: plumbous, plumbic

Writing Names for Binary Ionic Compounds
-

those containing only 1
metal

and 1
nonmetal
.

The correct full name of the cation (positive ion) is written first.

(Do not forget about the roman numerals on
some of the metal ions!)

The last syllable in the anion (negative ion) is dropped and

ide

is added.

Example: NaCl


Sodium Chloride

Example: CuS


Copper (II) Sulfide or Cupric Sulfide

Writing Names for Te
rnary Ionic Compounds
-

those containing
polyatomic

ions
. (It is imperative that you know the
correct names of these ions!!!!)

The correct full name of the cation (metal ion or polyatomic ion) is written first. (Do not forget about the roman
numerals on
some of the metal ions!)

The correct full name of the
anion

(polyatomic ion or nonmetal ion) is written second. If the anion is a
polyatomic ion do not change the ending. If the anion is a nonmetal ion then the ending is dropped and

ide

is
added.

Example: KNO
3

potassium nitrate

Example: NH
4
Cl ammonium chloride

***
C.7.B write the chemical formulas of common polyatomic ions, ionic compounds
containing main group or transition me
tals, covalent compounds, acids

Ionic compounds

Ions are
atoms or groups of atoms that have a charge. An ionic compound is a compound consisting of positive and
negative ions held together in an ionic bond.

metal or positive ion is the first part of the formula‐nonmetal or negative polyatomic io
n is the second p
art of formula

ALL COMPOUNDS ARE NEUTRAL AND WILL HAVE THE SMALLEST WHOLE NUMBER OF POSITIVE + IONS AND NEGATIVE


IONS THAT RESULT IN A NEUTRAL COMPOUND

Example: What is the formula for the ionic compound formed by sodium (Na) and chlorine (Cl)?

Symbols:
Na Cl

Combining capacities(charges of ions: Na (+1) Cl (

1)

Balanced charges: (+1) balances out (

1) so this ion pair is neutral.

There will be one sodium ion for each chlorine ion. The

formula is

NaCl.


Rules for Writing Chemical Formulas

Write the symbol

for the more metallic element before the non


metallic element. (Metals are

on the left side of the periodic table!)

Write the charge of the ions of the elements, using the periodic table.

The charge of the ions of the elements should “balance” to make a

neutral compound. To

balance the c.c. of 1 calcium ion (+2), you need 2 chloride ions (


1 plus


1).


Subscripts indicate the number of atoms each element has in the compound. If the element has only 1 atom, no
subscript is needed. The subscripts should
be the smallest possible whole numbers.

Example: 1 ion of calcium and 2 ions of chlorine is CaCl
2

Using the Cross


over method (the Macarena!!!)

Cross over the combining capacities from each of the ions so that they become the subscripts of the opposite

ions. Reduce the subscripts to lowest terms if necessary. Subscripts of 1 are not written. If the oxidation
numbers cannot be reduced, crisscross the numbers omitting the signs and write the formula for the compound.
Remember:

Write positive ion first

Dete
rmine the charge on each ion

Cris
s cross and reduce



Ex:
calcium chloride

= CaCl
2



Sodium chloride

= NaCl



Lithium oxide


= Li
2
O

***
C.7.C construct electron dot formulas to illustrate ionic and covalent bonds

Lewis Dot Diagrams:

A Lewis symbol (Lewis

dot diagram) is a symbol in which the electrons in the valence shell (outer energy level) of an
atom or simple ion are represented by dots placed around the letter symbol of the element. Each dot represents one
electron. Negative ions gain electros and po
sitive ions lose electron.

Rules:

o

Determine the atom symbol and the number of valence electrons

Group

1

2

13

14

15

16

17

18

Number valence electrons

1

2

3

4

5

6

7

8

o

Place one pair of electrons between each pair of bonded atoms. H is always a terminal atom.

ALWAYS connected
to only

one other atom!!

o

Subtract from the total the number of electron pairs you just used.

o

Place lone pairs about each terminal atom (EXCEPT H) to satisfy the octet rule

o

Left over pairs are assigned to the central atom. If the central a
tom is from the 3rd or higher period, it can
accommodate more than four electron pairs.

o

If the central atom is not yet surrounded by four electron pairs, convert one or more terminal atom lone pairs to pi
bonds pairs. NOT ALL ELEMENTS FORM DOUBLE BONDS!!
u
sually C, N, O, P, and S
!!

+
C.7.D
describe

the nature of metallic bonding and apply the theory to explain metallic
properties such as thermal and electrical conductivity, malleability, and ductility

Review the Three Types of Bonds and General
Characteristics:

IONIC
:

transfer of e


compounds
of cations of metals and ‐ ions non‐molecular/
the compound is crystalline

COVALENT


sharing e


nonmetals & nonmetals
forms molecular compounds and network solids which are non‐molecular

Nonpolar Covalent:
equal sharing of e


Polar Covalent: unequal sharing of e


METALLIC


bond: delocalized e



substances and metallic alloys

Valence sea model:

Key Concepts

A metal is a lattice of positive metal 'ions' in a
'sea' of delocalized electrons.

Metallic bonding refers to the interaction between the delocalized electrons and the metal nuclei.

The physical properties of metals are the result of the delocalization of the electrons involved in metallic bonding.

The phy
sical properties of solid metals are:

conduct heat &
electricity

generally high melting and boiling points

malleable (can be hammered or pressed out of shape without breaking)

ductile (able to be drawn into a wire)

metallic luster

opaque (reflect light)




Physical Properties of Metals Explained Conductivity

Solid and liquid metals conduct heat and electricity.

The delocalized electrons are free to move in the solid lattice. These

mobile electrons can act as charge carriers in the conduction of electricity o
r as energy

conduc
tors in the
conduction of heat.

Melting Points

In general, metals have high melting and boiling points because of the strength of the metallic bond.

The strength of the metallic bond depends on the number of electrons in the delocalized '
sea' of electrons.
(More delocalized electrons results in a stronger bond and a higher melting point.)

packing arrangement of the metal atoms. (The more closely packed the atoms are the stronger the bond is and
the higher the melting point.)

Group I metals

have relatively low melting points compared to other metals because they:

only have 1 electron to contribute to the delocalized 'sea' of electrons

are not forming as many metallic bonds as other metals because Group I atoms are inefficiently packed have
l
arge atomic radii so the delocalized electrons are further away from the nucleus resulting in a weaker

metallic bond

Malleable and Ductile

Metals are malleable and ductile.

The delocalized electrons in the 'sea' of electrons in the

metallic bond, enable th
e metal atoms to roll over each other when a stress is applied.

Optical Properties

Metals typically have a shiny, metallic luster.

Photons of light do not penetrate very far into the surface of a metal and are typically reflected, or bounced off,

the metal
lic surface.

+
C.7.E predict molecular structure for molecules with linear, trigonal planar, or tetrahedral
electron pair geometries using Valence Shell Electron Pair Repulsion (VSEPR) theory

VSEPR stands for Valence Shell Electron Pair Repulsion. It's a co
mplicated acronym, but it means something that's not
difficult to understand. Basically, the idea is that covalent bonds and lone pair electrons like to stay as far apart from
each other as possible under all conditions. This is because covalent bonds cons
ist of electrons, and electrons have the
same charge. Since the molecular geometry is determined by how many bonding and non‐bonding electron groups
surround the central atom, the first thing one needs to do is count how many of each there are. Doubl
e bond
s count as a
single bond.

B. Shapes

1.
Linear.



The bond angle is 180*



1
-
2 atoms bonded to the central atom

0

lone pairs



Write the Lewis structure for HCl

2.
Bent
.



The bond angle is 109.5*



2 atoms bonded to the central atom

1

lone pairs



Write the
Lewis structure for H
2
O

3.
Trigonal Planar


The bond angle is 120*



3 atoms bonded to the central atom

0

lone pairs

Write the Lewis structure for

BH
3

4.
Trigonal Pyramidal
.


The bond angle is 107*



3 atoms bonded to the central atom


1 lone pairs



Write
the Lewis structure for NH
3
.

5
. Tetrahedral
.


The bond angle is 109.5*



4 atoms bonded to the central atom


0 lone pairs



Write the Lewis structure for CH
4
.

+
C.8.A defin
e and use the concept of a mole

The mole (mol) is a unit for counting objects. It
is especially useful for counting tiny objects like atoms, molecules, ions,
and formula‐units.

1 mole of particles (atoms, ions or molecules)= 6.02 x 10
23

particles for any substance!

We can use the mole to count objects too small to see but we have one m
ore step. Because we know the mass of 1 mole
of a substance, we can weigh a sample made of atoms or molecules and convert to moles. We can also covert moles to
the mass of a substance that contains the desired number of moles of atoms or molecules.

1 mole
of atoms has a mass equal to the atomic weight in grams or molar mass.

1 mole of molecules has a mass equal to the molecular

weight in grams or molar mass.

How to determine the molar mass of an atom?

The molar mass of elements is found by looking at the atomic mass of the element on the periodic table. For
example, if you want to find the molar mass of carbon, you would find the atomic mass of carbon
on

the periodic table, and this is equal to the mola
r mass in grams per mole. So, in our example, carbon has a mola
r
mass of 12.01 grams per mole.

How to determine the molar mass of a molecule:

For any chemical compound that's not an element, we need to find the molar mass from the chemical formula.
To do t
his, we need to remember a few rules:

To find the molar mass of a compound:

Use the chemical formula to determine the number of each type of atom present in the compound.

Multiply the atomic weight (from the periodic table) of each element by the number of

atoms of that element
present in the compound.

Add it all together and put units of grams/mole after the number.

Example Molar mass of NaCl
= 58.5 grams/mole

If you have a subscript in a chemical formula, then you multiply the number of atoms of anything
next to that
subscript by the number of the subscript. For most compounds, this is easy. For example, in iron

(II) chloride, or FeCl
2
, you have one atom of iron and two atoms of chlorine. The molar mass will be equal to (1
atom x 56 grams/mole Fe) + (2
atoms x 35.5 grams/mole of chlorine) = 127 gr
ams/mole of iron (II) chloride.

Defining moles

1 mole of particles (atoms, ions or molecules)= 6.02 x 10
23

particles for any substance!

1 mole of atoms has a mass equal to the atomic weight in grams. Example:

1m
ole of copper atoms is the number of atoms in 63.546 g of Cu

1 mole of molecules has a mass equal to the molecular weight in grams. Examples:

1 mole H
2
O is the number of molecules in 18.015 g H
2
O

1 mole H
2

is the number of molecules in 2.016 g H
2
.

Vocabula
ry:

molar mass is the mass of one mole of a substance

Avogadro's number is the number of molecule
s in one mole for any substance

***
C.8.B use the mole concept to calculate the number of atoms, ions, or molecules in a
sam
ple of material

Usage of the
term m
olecule and formula unit:

The term molecule refers to the smallest unit of a covalent compound that contains the actual number of each
type of atom in the smallest unit of the compound that retains the properties of the compound.


Ionic compounds only have

formulas that represent the smallest whole number ratio of atoms of an ionic
compound. Ionic compounds do not form molecules. Formula unit is best term to use for the smallest unit of an
ionic compound but molecules is commonly used.



6.02 x 10
23

atoms (
particles) = the atomic mass in grams

6.02 x 10
23

atoms (particles)= 1 mole

1 mole = the atomic mass in grams


How many molecules are in 26.9 g of C
12
H
22
O
11
?


26.9 g/1 X 1 mol/342.0g X 6.02 X 10
23

molecules/mol =
4.74 X 10
22

molecules


+
C.8.C
calculate percent composition and empirical and molecular formulas

We will being using are mole conve
rsion tools. Let’s get digging!

Percent Composition

The
percent composition

is a list of the mass percent of each element in
a compound. You determine the %
composition
of each element

Calculate the percent composition of Mg(NO
3
)
2
.

1. Determine the mass of each element in one mole of the compound. (number of element in a formula unit ×
molar mass). Determine the molar mass of th
e compound (= the sum of the molar masses of the elements in

one mole of the formula unit).

1Mg = 1 x 24 = 24

2N = 2 x 14 = 28

6O = 6 x 16 = 96

148g/mole




24 g Mg
/148

×100 = 16.2 % of Mg

28 g N
/148

×100 = 18.9 % of N

96 g O
/148

×100 = 64.0 % of O

Empir
ical formula

Simplest formula for a compound; it is the smallest whole
-
number ratio of the atoms present.

Here's a rhyme: Percent or data to mass

Mass to mole

Divide by smallest

Multiply 'til whole

Example
:
Calculate the empirical formula for a compound c
omposed of 26.6% potassium, 35
.4% chromium, and 38.1%
oxygen.

Percent or data to mass: First, assume that you are given a 100
-
g sample, which would mean that the mass of each
e
lement in grams is equal to the
percent composition value. If given data in
grams you will need to calculate

grams of each element using the data.

From grams we find moles of each element in the formula.

Mass to mole

Find moles potassium.

26.6 g K
/ 39.1g/ mol =
0.06803 mole K

Next find moles of chromium.

35.4 g Cr
/52.0g/

mole Cr
=
0.6808 mole Cr

Next find moles of oxygen.

381 g
O / 16.0 g/mol = 2.381 mole O

Divide by smallest Next, set up a mole ratio that relates the moles of each element to the moles of the element that is
least present.

K 0.6803 /
0.6803

= 1

Cr 0.6808
/
0.6803

= 1

O 2.381/ 0.6803 = 3.5

Multiply 'til whole Because a chemical formula must have only whole numbers we multiply by 2 to yield: K
2
Cr
2
O
7

as the
empirical formula.




To find Molecular Formula:

Calculate empirical formula if not given

Divide the m
olar mass by the empirical formula mass.

Multiply all the subscripts in the empirical formula by the number you get when you divide the molar mass by th
e mass
of the empirical formula

Determine the molecular formula for a compound with the following percen
t composition 40.00% C, 6.72% H,
53.29% O. The molecular weight of this compound is 180 g/mol.

Calculate empirical formula:

Percent or data to mass 40.00% c = 40.00g C 6.72 & H = 6.72 g H

53.29% O


53.29g O


Multiply 'til whole already whole so the form
u
la is CH
2
O = empirical formula

Divide the molar mass by the empirical formula mass to find a multiple: Empirical mass :12.01g C + (2 x 1.008g H)
+ 16.00g O = 30g empirical formula mass

The molecular formula is a multiple of 6 times the empirical formula: C
(1 x 6) H(2 x 6) O(1 x 6) which becomes
C
6
H
12
O
6

***
C.8.D use the law of conservation of mass to write and balance chemical equations

Equations must be balanced because:

Law of Conservation of Matter:

Atoms can be neither created nor destroyed in an
ordinary chemical reaction,

so there must be the same number of

atoms on both sides of the equation. The mass of all the reactants (the substances going into a reaction)

must equal the mass of the products (the substances produced by the reaction).

Two typ
es of numbers are found in a chemical equation:

Subscripts:

H
2
O



the small numbers to the lower right of chemical symbols. Subscripts represent the number of atoms of each element in
the molecule.

Coefficients:

2
C
2
H
6
(g) +
7
O
2
(g)



4

CO
2
(g) +

6

H
2
O(g)

the large numbers in front of chemical formulas. Coefficients represent the number of mo
lecules of the substance in the
reaction.

Balancing TIPS:

1. Check for Diatomic Molecules
-

H
2

-

N
2

-

O
2

-

F
2

-

Cl
2

-

Br
2

-

I
2

If these elements appear by themselves i
n

an equation,
they must be written with the subscript 2

2. Balance atoms other than Hydrogen and Oxygen

3. Balance Hydrogen

5. Balance Oxygen

6. Recount All Atoms

Balance equations by changing

coefficients never by changing subscripts

7. If every coeffici
ent will reduce, rewrite in t
he simplest whole
-
number ratio.

An equation is not properly balanced if the coefficients are not written in t
heir lowest whole
-
number ratio.


+
C.8.E perform stoichiometric calculations, including determination of mass relations
hips
between reactants and products, calculation of limiting reagents, and percent yield

II. Solving Stoichiometry Problems

Setting the problem up: (mass
-
mass)

1. Write and balance the equation..

2. Identify the given and unknown

3. Identify the mole
ratio (coefficients from the balanced equation for the given and unknown).

4. Calculate the molar mass for the given and unknown.

Solving the Problem:

1. Convert grams of given to moles of given (Divide by molar mass of given)


2. Convert moles of given to
moles of unknown (Multiply by the mole ratio
-

coefficient of unknown/coefficient of given)

3. Convert moles of unknown to grams of unknown(Multiply by the molar mass of unknown)



Given Grams A X

1 mol A



X

coefficient B

X

molar m
ass B



1

Molar mass A



coefficient A

1 mol B


EXAMPLE: How many grams of silver chloride is produced from 5.0 g of silver nitrate reacting with an excess of barium
chloride?



2AgNO
3

+ BaCl
2

--
> 2AgCl + Ba(NO
3
)
2



5.0g AgNO
3

X 1 mol AgNO
3
/169.9 g AgNO
3

X 2 mol AgCl/2 mol AgNO
3

X 143.4g AgCl/mol = 4.2 g AgCl

LIMITING REACTANT

Is the reagent that is totally consumed during the reaction and therefore determines or limits the amo
unt of product
formed.

Steps: Do stoichiometry problems for ALL reactants (a separate calculation for each)

The answer that is the smallest amount came from your limiting reactant

PERCENT YIELD PROBLEMS:

Theoretical yield



this is what you

calculate from
stoichiometry

Actual yield



what you get in a lab in the real

world

% yield

= actual yield / theoretical yield

Suppose in the example where the theoretical yield was calculated to be 58.6g MgCl
2

discussed a chemist actually
obtained 55.4 g of MgCl
2
. Thi
s is called the actual yield and would be given to you in the problem.


To calculate the percent yield:

54g MgCl2 Actual Yield

× 100 = 94.5 % 58.6g MgCl2 Theoretical Yield


***
C.9.A describe and calculate the relations between volume, pressure, number
of moles,
and temperature for an ideal gas as described by Boyle’s law, Charles’ law, Avogadro’s law,
Dalton’s law of partial pressure, and the ideal gas law

Units:

STP is 1 atm and 0

C




K = 273 +

C (Change ALL temperatures to Kelvin!!!!)

1 atm = 760 m
mHg or 760 torr



1 atm = 101.3 kPa




Molar Volume of a Gas at STP 22.4 L/mol

Daltons Law

P
T

= P
1

+ P
2

+ P
3

+ …….

P
T

= total pressure

P
#

= the partial pressures of the individual gases

Total Pressure of a Gas = (Sum of the partial pressures of the compone
nt gases)

Boyle’s Law

:
V
1
P
1

= V
2
P
2






V
1

= initial volume



V
2

= final volume





P
1

= initial pressure





P
2

= final pressure

Charles’s
Law
:

V
1

/T
1

= V
2
/T
2






T
1

= initial temperature (in Kelvin)





T
2

= final temperature (in Kelvin)

Moles and Volume Law:
V
1

/n
1

= V
2
/n
2





Ideal Gas Law
:
PV = nRT




P = pressure in atm, kPa, or mmHg (Make sure you pick correct R!)


V = volume in liters


n = number of moles


T = temperature in Kelvin


Ideal Gas Constant =
R = 0.0821
L • atm
/

mol •
K

= 8.31
L • kPa

/

mol • K

= 62.4
L • mmHg

/

mol • K

+
C.9.B perform stoichiometric calculations, including determination of mass and volume
relationships between reactants and products for reac
tions involving gases

You will be learn
ing to solve problems involving Gas Laws and Stoichiometry.

You can recognize stoichiometry/ gas law problems because a chemical reaction is involved, gases are involved and you
are given information about one substance and asked about another, amount give
n or ask for may be a volume.

For calculations at STP use 22.4 L gas = 1 mole gas

Question is AT STP: Example 1

Potassium chlorate is used in the lab to make oxygen gas by the following (unbalanced) reaction:

2KCl
O
3
(s) →
2KCl(s) + 3O
2
(g)

How liters of oxy
gen at STP
may be made from reacting 1.226
grams of Potassium chlorate?

1.226 g KClO
3

X 1 mol KClO
3
/
122.6g X 3 mol O
2
/2 mol KClO
3

X 22.4L/1mol = .336 L O
2

+
C.9.C describe the postulates of kinetic molecular theory

A Molecular Model of Gases

observat
ion

hypothesis

Gases are easy to expand

gas molecules don't strongly attract each other

Gases are easy to compress

gas molecules don't strongly repel each other

Gases have densities that are about 1/1000 of solid or
liquid densities

molecules are much
farther apart in gases than in
liquids and solids

Gases completely fill their containers

gas molecules are in constant motion

Hot gases leak through holes faster than cold gases

the hotter the gas, the faster the molecules are
moving


THE KINETIC
MOLECULAR THEORY

is used to explain the behavior of gases and is based upon the following postulates:

Gases are composed of a many particles that behave like hard spherical objects in a state of constant, random
motion.

These particles move in a straight
line until they collide with another particle or the walls of the container.

These particles are much smaller than the distance between particles, therefore the volume of a gas is mostly
empty space and the volume of the gas molecule themselves is negligib
le.

There is no force of attraction between gas particles or between the particles and the walls of the container.

Collisions between gas particles or collisions with the walls of the container are elastic. That is, none of the
energy of the gas particle i
s lost in a collision.

The average kinetic energy of a collection of gas particles is dependent only upon the temperature of the gas.

Two Assumptions made in Kinetic Molecular Theory of "Ideal" Gases

Gas particles are much smaller than the distance between

particles, therefore the volume of a gas is mostly
empty space and the volume of the gas molecules themselves is

negligible.

There is no force of attraction between gas particles or between the particles and the walls of the

container.

The KMT Assumpt
ions

and the Gas Law Variables:

We can connect these assumptions with the four variable
s from the individual gas laws.

P= Pressure is force per unit area. What we observe as the pressure of a gas is the force of collisions as the
particles strike the walls of
the container. If these collisions occur frequently, the gas pressure is high. If the
collisions don’t occur very often, the pressure is low. Any change in the conditions that results in more frequent
collisions will increase the pressure.

V= What we obser
ve as the volume of a gas is the empty space the particles travel through. The larger the

volume, the greater the distance between particles. Any change in the conditions that results in a longer
distance between particles is due to an increase in

volume.

n= What we observe as n, or number of moles, is the number of particles.

T=What we observe as temperature of a gas is the average speed of the particles. The hotter the gas, the faster
the particles are moving. The speeds of the individual gas particles

va
ry, but they form a statistical distribution of speeds that looks like the following graph:

Incandescent light bulbs produce light by heating a filament. Filling the bulb with an inert gas like argon makes the
filament last longer. The bulb shown has a vol
ume of 150 cm3 and contains a mass of 0.16 g of argon (atomic mass of
argon is 39.9 amu).

If neon were used in place of argon, what mass of neon would be contained in the bulb (atomic mass of

neon is 20.2 amu)? Assume that the bulb is filled to the same pr
essure. Explain your answer.

What happens to the gas particles inside the bulb when it is turned on? Explain your answer in terms o
f the kinetic
molecular theory.

+
C.10.A describe the unique role of water in chemical and biological systems

Water is
polar
:
In water the oxygen has a
-

charge and hydrogen’s have a + charge. Water has a bent shape.

These partial charges result in the ability of wat
er to exhibit what is called in
chemistry as an intermolecular force. This
force is called a hydrogen bond and is a

result of water being polar and is the force that causes in water to attracting
itself and other substance. Hydrogen bonds occur between water molecules.

Water molecules stick to each other

a force called
cohesion
. They also stick to other things

a force
called

adhesion.

-

Strong
Surface Tension
-

Charged ends of water molecules are attracted to other substances with charges.

Water molecules pull on other water molecules. This pulling, called capillary action, allows liquid water to move against
gravity.
Ca
pillary action

is very important in nature. It makes it possible for
water to travel from the roots

of plants to
the tips of their leaves

even to the tops of towering trees!

The stickiness of the hydrogen bonds also causes liquid water to form a kind of “s
kin”

called surface tension


where it
meets the air

High Heat Capacity

-

Water’s ability to absorb heat helps protect Earth from wild temperature swings from night to day
and summer to winter. It keeps ocean temperatures fairly constant, so creatures that
live there have a relatively stable
environment. Water’s ability to soak up heat as it changes from one form to another helps keep you cool too. When
your sweat evaporates, it draws heat from your body.

Good solvent
-

Scientists call water the universal sol
vent. Rivers and streams carry nitrogen and phosphorus, which
plants need to grow. They also carry pollutants, which can harm living things. Water carries nutrients from plants’ roots
to their leaves. It carries food through an animal’s bloodstream to nour
ish its cells.

Water’s ability to exist in all three forms of matter at temperatures common on Earth helps it move from one

place to another.

Ice float
s

on water. Unlike most substances the solid form of water is less dense than the liquid. When water free
zes the
molecules move part. If ice didn’t float, then in winter the water at the top of the lake, near the cold air, would freeze
and sink to the bottom.

***
C.10.B develop and use general rules regarding solubility through invest
igations with
aqueous so
lutions

Solubility Rules

Observations are used to determine which compounds are soluble in water. A precipitate is an insoluble product formed
during reaction. We can deduce which products of the following series of reactions are soluble and which is inso
luble.

Using solubility helps you make predictions and write net ionic equations. In the example below solubility rules will be
used to identify a group of unknowns.

Example Using Solubility Rules

There is a major problem in the chemical stockroom. Upon ar
riving back to school after the summer, a teacher found a
box containing a bunch of unlabeled bottles and 4 different labels at the bottom of the box. The labels must have fallen
off due to the humidity. By using your solubility rules, help us figure out w
hich numbered solution matches up with the
following chemicals: AgNO
3
, NaCl, KNO
3
, NaOH

You will be given a labeled solution of CuSO
4

to help you with this task.



Data Table


1

2

3

4

CuSO
4

No reaction

No reaction

Ppt formed

No reaction

1

X

No reaction

Ppt formed

Ppt formed

2

No reaction

X

No reaction

No reaction

3

Ppt formed

No reaction

X

Ppt formed

4

Ppt formed

No reaction

No reaction

X


Sample Conclusion

Here is a brief description of how the students should come up with the identity of the
unknowns: Solution 3 is NaOH
because it is the only one of the four that will form a precipitate with CuSO4. Solution 2 is KNO3 because it does not
form a precipitate with anything.

Solution 1 is AgNO
3

because it forms a precipitate with Solution 3, which
we have already determined is NaOH.

Solution 4 is NaCl because it forms a precipitate with AgNO
3
, which is Solution 1.


How do you write a balanced net ionic equation?

A
net ionic equation

shows only those particles involved in the reaction and is balanced

with respect to both mass and
charge. Consider the equation for the reaction of lead with silver nitrate.

Pb(s) + AgNO
3
(aq)


Ag(s) + Pb(NO
3
)
2
(aq)

The nitrate ion is a spectator ion in this reaction. Spectator ions appear on both sides of an equation but
are not

directly involved in the reaction. The net ionic equation is as follows: Pb(s) + Ag
+
(aq)


Ag(s) + Pb
2+
(aq)

Why is this equation unbalanced? Notice that a single unit of positive charge is on the reactant side of the equation. Two
units of positive

charge are on the product side. Placing the coefficient 2 in front of Ag
+
(aq) balances the charge. A
coefficient of 2 in front of Ag(s) also rebalance

Pb(s) + 2Ag
+
(aq)


2Ag(s) + Pb
2+
(aq)

+
C.10.C calculate the concentration of solutions in units of molarity

Molarity is the most useful concentration for chemical reaction in solution because it directly relates moles of

solute to volume of solution. The definition of molarity (M) is
𝑀
=
mol
of solute/ liter of solution

A Few Solution Terms

Solution
-

a homogeneous mixture of two or more substances in a single phase.

solute
--
component in lesser concentration; dissolve (being dissolved)

solvent
--
component in greater concentration; dissolver (doi
ng the dissolving)

solubility
--
maximum amount of material that will dissolve in a given amount of solvent at a given temperature to
produce a stable solution.

Example 1:

How many moles of NaCl are dissolved in 250 mL of solution if the solution concentrati
on is 0.150 M?

0.15 M = x mol/.250L x = .0375 mol


Example 2
:
Calculate the molarity of a solution containing 8.61 g potassium hydroxide, dissolved in enough water to
form 750 mL of solution.

[8.61 g KOH X 1mol KOH/56.1 g KOH] / .750L = 0.205
M

+
C.10.D use molarity to calculate the dilutions of solutions

M
1
V
1
=M
2
V
2

Example:
What volume
of 6.0M HCl
is needed to
make 2.00 L of 1.00 M solution?

V
1

=

x mL

M
1

= 6.0M

V
2

= 2.00L

M
2

= 1.00

(6.0M)(xmL) = (1.00)(2.00L) =
.33 L HCl




***
C.10.E
distinguish between types of solutions such as electrolytes and nonelectrolytes
and unsaturated, saturated, and supersaturated solutions

Solubility is defined as the maximum amount of solute dissolved by a given amount of solvent at a specified
temperatur
e. The solubility of the given substances in a given solute is temperature
-
dependent.

Unsaturated:
If the solution contains less quantity of solute than what can be dissolved, at a specified temperature it is
unsaturated. When more solute is added into
solution the solute dissolves. An unsaturated solution is one in which
more of the solute could dissolve at the same temperature.

Saturated:
A saturated solution is one in which no more of the solute will dissolve at a specific temperature. Solution
conta
ining the maximum amount of solute at a specified temperature is saturated. When more solute is added into the
solution the solute will no longer dissolve.

Supersat
urated:
A supersaturated solution is when a solution which contains more solute than would n
ormally dissolve
at a certain temperature. The solution becomes supersaturated by elevating the temperature and the solution contains
the maximum amount of solute at an elevated temperature is carefully cooled to a lower temperature and is now
supersaturat
ed. When more solute is added into the solution, crystals will form.

W
hen a solid is added to a solvent in which it is soluble, solute particles leave the surface of the solid and become
solvated by the solvent, initially forming an unsaturated solution.

W
hen the maximum possible amount of solute has dissolved, the solution becomes saturated. If excess solute is
present, the rate at which solute particles leave the surface of the solid equals the rate at which they return to the
surface of the solid.

A supe
rsaturated solution can usually be formed from a saturated solution by filtering off the excess solut
e and lowering
the temperature. W
hen a seed crystal of the solute is added to a supersaturated solution, solute particles leave the
solution and form a cry
stalline precipitate.

***
C.10.F investigate factors that influence solubilities and rates of dissolution such as
temperature, agitation, and surface area

Dissolution
, which is also called
solvation
, is the process by which one substance, the solute, dissol
ves into another, the
solvent. Dissolution is brought about by interactions at the particle level. Solute particles are surrounded by solvent
particles until both are spread evenly throughout the entire solution.

Ionic compound: separated ions surrounded (
hydrated) by water
.

Solubility
is the maximum amount of a substance that can be dissolved in a given amount of solvent. Solubility depends
on temperature and pressure, as well as on the type of solvent being used. Solubility is not affected by an increase
in the
surface area of the solute or by agitation of the solution although the rate of solution formation (dissolution) is affected.

Factors Affecting Solubility:

Temperature
affects solubility of solids differentl
y than the solubility of gases.

Solid solu
bility: ↑ temperature
often (not always) ↑ solubility. Pressure
has no effect on solubility

of solids

Gas solubility: ↑ temperature ↓ solubility

Gas solubility: ↑ pressure ↑ solubility

Factors Affecting Rate of Solution Formation (dissolution) of Solids:



temperature ↑rate of dissolution

↑ agitation (stirring or shaking) ↑rate of dissolution

↑surface area (grinding or crushing a solid) ↑rate of dissolution

+
C.10.G define acids and bases and distinguish between Arrhenius and Bronsted‐Lowry
definitions and
predict products in acid‐base reaction
s that form water

Properties of Acids and Bases

Acids

Bases

Tastes Sour

Tastes bitter

pH Less than 7

pH more than 7

Turn litmus red

Turns litmus blue

Acids are electrolytes

Bases are electrolytes

Acids react with

metals to produce hydrogen

Bases feel slippery




ARRHENIUS DEFINITION

BRØNSTED
-
LOWRY DEFINITION

An acid is a substance that contains a hydrogen and

ionizes to produce hydrogen ions in aqueous solutions.

Acid
-
produces a proton (H+) in water

BrØnsted
-
Lowr
y acid is defined as a hydrogen
-
ion

donor

Acid
-
donates ( gives up)a proton (H+)


A base is a substance that contains a hydroxide and
dissociates to produce hydroxide ions in water (OH
-
)
Base
-
produces a hydroxide ion in water (OH
-
)

BrØnsted
-
Lowry base is
defined as a hydrogen
-
ion

acceptor.

Base
-
accepts (bonds to H+) a proton (H+)


require acid
-
base reactions to occur in aqueous

solutions


limited to substances with those

"parts"; ammonia is a
MAJOR exception!

Brønsted
-
Lowry definition includes all
Arrhenius

bases that produce hydroxide ions that

can accept a proton. But the Brønsted
-
Lowry
definition also covers other bases, such as ammonia,
NH3, that do not contain hydroxide ions.

A neutralization reaction which has been put into a word equation: a
cid + base → salt + water

Examples:

HCl + NaOH → H
2
O + NaCl

H
2
SO
4

+ 2KOH → 2H
2
O + K
2
SO
4


***
C.10.H understand and differentiate among acid‐base reactions, precipitation reactions,
and oxidationreduction reaction

Recognizing Neutralization, Precipitat
ion and Oxidation and Reduction Reactions

Acid
-
base reactions are reactions that occur between

an acid and a base in which a proton is transferred.
The product includes a salt and, often, water.

Reactant an acid (formu
la starts with an H) and a
base
(formula end in OH or is NH
3
). If the base ends
in OH the products are a salt (first element a metal)
and water. Salt has the + ion of the metal and
-

ion of
the base. If the base is NH
3

product will only be a salt
that has a formula that starts with NH
4

a
nd the last
ion in the formula will be the anion in the acid.

Precipitation reactions are reactions that occur when

two aqueous solutions react and produce a solid
precipitate.

Use solubility chart: If one product is insoluble the

reaction is a precipitat
ion reaction.

An oxidation
-
reduction (redox) reaction, also called a

redox reaction, is any reaction that involves the
exchange of an electron or oxygen. In general, when
the oxidation number of a certain atom is different in
one of the reactants than it
is in one of the products,
it means that at least one electron(s) were
exchanged.

If the reaction if not an acid base or precipitation

reaction it will most likely be a redox reaction.
Assign oxidation numbers to all the elements in the
equation if an
elements oxidation number changes
the reaction is a redox reaction.


Oxidation Number Rules

The oxidation number for an atom in its elemental form is always zero.

The oxidation number of a monoatomic ion = charge of the monatomic ion.

Examples:

Oxidation
number of S
2
-

is
-
2.

Oxidation number of Al
3+

is +3.

The oxidation number of all Group 1A metals = +1 (unless elemental) and The oxidation number of all Group 2A
metals = +2 (unless elemental).

In ionic binary compounds the oxidation number is the same as
the oxidation number used to write the


formula of the compound

Hydrogen (H) has
two possible oxidation numbers: +1 when bonded to a nonmetal
-
1 when bonded to a metal

Oxygen (O) oxidation numbers is
-
2:

The oxidation number of fluorine (F) is always
-
1.

Th
e sum of the oxidation numbers of all atoms (or ions) in a neutral compound = 0.

The sum of the oxidation numbers of all atoms in a polyatomic ion
= charge on the polyatomic ion.

Example
: Determine the oxidation number of each element in Ba(NO3)2

Charge
each element

+2

+5

-
2


Ba

(NO3)2

(NO3)2

Total charge

2

2+ 2N +
-
12 =0

10

-
2 × (2 x 3) =
-
12

Explanation:

Ba(NO
3
)
2

Is the substance ionic?

Yes, metal + non
-
metal.

Barium ion = Ba
2+

Oxidation # for Ba = +2

Which elements have specific rules?

Oxygen has a
rule....
-
2 in most compounds

Oxidation # for O =
-
2

Which element does not have a specific rule?

N does not have a specific rule.

Use rule 8 to find the oxidation # of N

Let N = Oxidation # for nitrogen

(# Ba) (Oxid. # of Ba) + (# N) (Oxid. # N) + (# of O)

(Oxid. # of O) = 0

1(+2) + 2(N) + 6(
-
2) = 0

N = +5

+
C.10.I define pH and use the hydrogen or hydroxide ion concentrations to calculate the pH
of a solution

What is pH?

The strength of an acid depends on the concentration of hydrogen ions (H+) in a
solution, which can be

expressed in terms of molarity. For example, the concentration of hydrogen ions in water is 1.0 × 10

7

M. This means that for each liter of water, there is 0.0000001 mole of hydrogen ions. A much easier way to express the
strength
of an acid is by its pH, which is the negative logarithm of the hydrogen
-
ion concentration. pH = −log[H+]

The pH of water is the negative of the log of 1.0 × 10

7

M, which is 7. Water has an equal amount of hydrogen ions and
hydroxide ions, so it is consid
ered neutral. Solutions with a pH lower than 7 are considered acidic, and solutions with a
pH higher than 7 are considered basic.


How is the hydrogen ion concentration used to calculate the pH of a solution?

Consider an aqueous solution with a hydrogen i
on (H+) concentration of 1.0 × 10

5

M. The pH is c
alculated

using the
negative of the log of the hydrogen ion concentration.

pH = −log (1.0 × 10

5
) = −(−5) = 5

The hydrogen ion concentration of a solution can also be determined from its pH through a revers
e of the calculation.
The molar concentration of any substance is described using brackets, such as [H+]. [H+] = antilog (−pH)

(Note: Sometimes the antilog is referred to as the “inverse of the log” or as “log

1.” On some calculators, the button for
this f
unction is labeled “10 x.”) For example, to find the hydrogen ion concentration of a solution with a pH of 11, you
can calculate the antilog of −11. [H+]
= antilog (−11) = 1.0 × 10−11 M

Example 1: calculate the pH of a solution with a hydrogen ion concent
ration of 6.3 × 10

4

M

pH = −log (6.3 × 10

4
)

= −(−3.2) = 3.2




How is the hydroxide ion concentration used to calculate the pH of a solution?

The concentration of hydroxide ions is closely related to the pH of a solution. Recall that as water molecules
di
sassociate, equal amounts of hydroxide ions and hydrogen ions are formed.


pH + pOH = 14

This is why, in water, the pH and the pOH are both 7. If an acid is added to the water, the concentration of hydrogen
ions increases, lowering the pH. As the hydrogen
ion concentration increases, the hydroxide concentration decreases.
But the sum of the pH and the pOH is always 14. On the basis of this relationship, you can calculate the pOH of a solution
as long as you know its pH.


Example:

if a solution has a hy
droxide ion concentration of 1

× 10

10
, then you can calculate the pOH and find the pH.

pOH = −log (1

× 10

10
) =
10

Therefore, pH = 14


10 = 4

+
C.10.J distinguish between degrees of dissociation for strong and weak acids and bases

In general, strong acids

and strong bases dissociate, or ionize, completely in aqueous solution. In other words, almost
100 percent of a strong acid or strong base interacts with water and ionizes. A strong dissociation is described using one
arrow p
ointing to the right, as shown
.

HA + H
2
O → A
-

+ H
3
O
+

or

HA → A
-


+ H
+


hydroxide ions in water. For example, consider the dissociation of sodium hydroxide (NaOH).

NaOH (aq) → Na
+

(aq) + OH


(aq)

Degree that weak acids and base dissociate:

Weak acids and weak bases ionize only slightly in aqueous solution.

+
C.11.A understand energy and its forms, including kinetic, potential,

chemical, and thermal
energies

The Concept of Energy

The Usual Definition of Energy: the ability to do work and to
transfer heat. work is moving an object against an opposing
force

work
= distance × opposing force w = d x f

SI unit of work or energy: the
joule
(J)

Two Basic Forms of Energy:

potential energy
: energy of position

examples: boulder on a ledge, chemical bon
ds

kinetic energy:

energy of motion

examples: pool balls, molecules

Forms of Energy:

Energy is found in different forms including light, heat, chemical, and motion. There are many forms of energy,
but they can all be put into two categories: potential an
d kinetic.

Potential Energy

Potential energy is stored energy and the energy of
position


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Chemical Energy is energy stored in the bonds of

atoms and molecules. Batteries, biomass, petroleum,
natural gas, and coal are examples of stored
chemical energy. Chemical energy is converted to
thermal energy when we burn wood in a fireplace or
burn gasoline in a car's engine.

Mechanical Energy is energy stored in objects by
tension. Compressed springs and stretched rubber
bands are examples of stored mechanical energy.
Nuclear Energy is energy stored in the nucleus of an
atom


the energy that h
olds the nucleus together.
Gravitational Energy is energy stored in an object's
height. The higher and heavier the object, the more
gravitational energy is stored.

Radiant Energy is electromagnetic energy that travels

in transverse waves. Radiant energy in
cludes visible
light, x
-
rays, gamma rays and radio waves. Light is
one type of radiant energy

Thermal Energy is the vibration and movement of
the atoms and molecules within substances. As an
object is heated up, its atoms and molecules move
and collide fas
ter.

Motion Energy is energy stored in the movement of
objects. The faster they move, the more energy is
stored. It takes energy to get an object moving, and
energy is released when an object slows down.

Sound is the movement of energy through
substances i
n longitudinal (compression/rarefaction)
waves. Sound is produced when a force causes an
object or substance to vibrate


the energy is
transferred through the substance in a wave.

Electrical Energy is delivered by tiny charged
particles called electrons,
typically moving through a
wire. Lightning is an example of electrical energy in
nature, so powerful that it is not confined to a wire.


Thermal Energy, Tempe
rature & Heat: Comparison Chart


Thermal Energy

Temperature

Heat

Definition:

Thermal energy is
the sum of the
kinetic energy of the particles
(atoms/molecules) of an object
and is measured in joules.

Temperature is the average
kinetic energy of the
atoms/molecules of a substance.
It measures the hotness or
coldness expressed in terms of
any of sever
al arbitrary scales
like Celsius and Fahrenheit.

Heat is an energy
transfer that occurs
because of a difference
in temperature.

Symbol:

Q or q (often referred to as heat
instead of thermal energy)

T

Often reported as Q or
q

Unit:

Joules

Kelvin, Celsius
or Fahrenheit

Joules


Keep it straight:

Temperature
: Temperature is the average kinetic energy within a given object and is measured by three scales of
measurement (Fahrenheit, Celsius, Kelvin)

Thermal Energy

: Thermal energy is defined as the total of a
ll kinetic energies within a

given system.

Heat:

It is important to remember that heat is caused by flow of thermal energy due to differences in temperature (heat
flows from object at higher temperature to object at lower

temperature), transferred through
conduction/convection/radiation. Additionally thermal

energy always flows fro
m warmer areas to cooler areas.

+
C.11.B understand the law of conservation of energy and the processes of heat transfer

Energy can be defined as the capacity to do work and to tra
nsfer heat.

Work
(w) is a force acting over a distance. W= f x d

Matter possesses energy both by virtue of its motion (kinetic energy) and by virtue of its position (potential energy).

Law of Conservation of Energy:

Energy can be
converted

from one form to
another but can neither be created nor destroyed. Both work & heat are
ways in which energy can be transferred. First Law of Thermodynamics:

The energy of the universe is constant.



Heat of Reaction

The heat of reaction, or the c
hange in enthalpy of reacti
on
is the heat exchanged between system and surroundings at
constant pressure, as the result of a chemical or physical change.

The system consists of those molecules which are reacting.

The surroundings are everything else; the rest of the universe.

Energy

and Chemical Reactions

All chemical reactions involve energy. Energy is used to break bonds in reactants, and energy is released when new
bonds form in products. Like the combustion reaction in a furnace, some chemical reactions require less energy to bre
ak
bonds in reactants than is released when bonds form in products. These reactions, called exothermic reactions, release
energy to the surroundings. In other chemical reactions, it takes more energy to break bonds in reactants than is
released when bonds
form in products. These reactions, called endothermic reactions, absor
b energy from the
surroundings.

Exothermic Reactions

In an exothermic reaction, it takes less energy to break bonds in the reactants than is released when new bonds are
formed when the p
roducts are formed. The word "exothermic" literally means "turning out heat." Energy, often in the
form of heat, is released as an exothermic reaction occurs. The general equation for an exothermic reaction is:

In many exothermic reactions, heat is rele
ased to the surroundings and as a result, the temperature of the surroundings
rises.

Endothermic Reactions

In an endothermic reaction, it takes more energy to break bonds in the reactants than is released when new bonds form
in the products. The word "end
othermic" literally means

"taking in heat." A constant input of energy, often in the form of heat, is needed in an

endothermic reaction. Not enough energy is released when products form to break more bonds in the reactants.
Additional energy is needed t
o keep the reaction going. The general equation for an endothermic reaction is:


In many endothermic reactions, heat is absorbed from the surroundings and as a result, the of the temperatu
re of the
surroundings drops..

Conservation of Energy

Whether a rea
ction absorbs energy or releases energy, there is no overall change in the amount of energy. Energy
cannot be created or destroyed. This is the law of conservation

of energy. Energy can change form


for example, from electricity to light


but the same

am
ount of energy always remains.

If energy cannot be destroyed, what happens to the energy that is absorbed in an endothermic reaction? The energy is
stored in the chemical bonds of the products. This form of energy is called chemical energy. In an
endothermic
reaction, the products have more stored chemical energy than the reactants. In an exothermic reaction, the opposite is
true. The products have less stored chemical energy than the reactants. The excess energy of the reactants is released
to th
e surroundings when the reaction occurs. The following graphs compare the energy changes in endothermic and
exothermic reactions.



EXOTHERMIC

ENDOTHERMIC

heat flows out of system

heat flows into system

heat is a product

heat is a reactant

system has
less energy

system has more energy

(Out of system)

system)

PE in bond


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Heat Flow


Sign of q

Type of Reaction


Products


Surroundings

from surroundings to
system


+ (positive)


endothermic


gain heat

lose heat (get
colder)

from system

to surroundings




(negative)


exothermic


lose heat

gain heat (get
hotter)


The
enthalpy change

or heat of reaction, ∆H, is equal to the heat gained or lost by the system under constant
-
pressure
conditions. The following sign conventions apply:

If ∆H > 0 the reaction is endothermic = +∆H

If ∆H < 0 the reaction is exothermic =
-
∆H

All chemical react
ions, even exothermic reactions, need a certain amount of energy to get started. This energy is called
activation energy. For example, activation energy is needed to start a car. Turning the key causes a spark that activates
the burning of gasoline in the
engine. The combustion of gas won’t occur without the spark of energy to begin the
reaction.

Why is
activation energy

needed? A reaction won’t occur unless atoms or molecules of reactants come together. This
happens only if the particles are moving, and mo
vement takes energy. Often, reactants have to overcome forces that
push them apart. This takes energy as well. Still more energy is needed to start breaking bonds in reactants. The graphs
below show the changes in energy in endothermic and exothermic react
ions. Both reactions need the same amount of
activation energy in order to begin.

f Products


Energy of Reactants

Net gain of energy (heat ) is absorbed

Net l
oss of energy (heat)released

Heat (thermal ener
gy) is released to surroundings

Heat
(thermal energy)
is absorb from the surroundings


Heat can be transferred between objects by three different processes: conduction, convection, or radiatio
n.

Three types of Heat Transfer:

Conduction:.

This method of heat transfer is most familiar to people
. If you

have ever burned yourself on a hot pan

because you touched it, you have experienced this first
-
hand. Conduction is

heat transfer through matter. Metals conduct heat well. Air is not as good a conductor of heat. This is a direct

contact
type of hea
t transfer.

Convection

is heat transfer by the movement of mass from one place to another. It can take place only in liquids and
gases.

Radiation

is the only way heat is transferred that can move through the relative emptiness of space. All other forms of
heat transfer require motion of molecules like air or water to move heat. The majority of our energy arrives in the form
of radiation from our Sun. Objects that are good absorbers of radiation are good radiators as well. (Microwave ovens,
sun, some electri
c heaters)

***
C.11.C use thermochemical equations to calculate energy changes that occur in chemical
reactions and classify reactions as exothermic
or endothermic


Heat Flow


Sign of q

Type of Reaction


Products


Surroundings

from surroundings to
system


+

(positive)



endothermic



gain heat


lose heat (get colder)

from system

to surroundings



(negative)


exothermic


lose heat

gain heat (get hotter)




Calculate Enthalpy of Reaction from Bond Energies

The bond enthalpy is the energy needed to break one mole of bonds in gaseous molecules under standard conditions
(298 K and 1 atm). The exact bond enthalpy of a particular chemical bond depends upon the molecular environment in
which the bond exists. There
fore, bond enthalpy values given in chemical
data books are averaged values.

Due to the principle of conservation of energy the total energy before and after the reaction dose not change. During a
reaction the bonds in the reactants are broken (requiring
energy) and new bonds are formed (releasing energy) to make
the products. Thus, the net energy that a reaction releases or absorbs must come from the difference in bond energies
of the products and the reactants. Note that D represents bond dissociation en
ergy, and H the entha
lpy of the reaction
as written.

ΔH = ΣnD

(
bonds broken)
-

ΣnD(bonds made)

Example 1

The bond energy (kJ) for H
2
, F
2
, and HF are 436, 158 and 568 kJ/mol respectively, calculate the enthalpy (energy)
of the reaction,

H
2
(g) + F
2
(g) = 2 HF

Solution

Based on t
he bond energies given, we have

H
2

→ 2H D = 436 kJ/mol

F
2

→ 2F D = 158 kJ/mol

2H + 2F → 2HF H =
-
568

2 kJ/mol

ΔH = ΣnD (bonds broken)
-

ΣnD(bonds made) ΔH = (436 kJ / mol + 158 k
J / mol)
-



H so the reaction is exothermic relea
sing energy to the surroundings

Standard Enthalpies of Formation

Molar Heat of Formation (∆Hf): is the amount of heat required / given off to make 1 mole of a compound from its
elemental components. Enthalpy
of Formation (also called Heat of Formation)

Molar Heat of Formation of ALL ELEMENTS is 0 kJ.

The state of the compound affects the magnitude of ∆Hf. (H
2
O (g) has ∆Hf = −241.8 kJ/mol; H
2
O (l) has ∆Hf = −285.8
kJ/mol)

Standard Molar Heat of Formation

(∆H°f)

is the heat of formation under standard conditions (1 atm and 25°C).

Enthalpies of formation for many chemical substances can be found in tables.

Example 1: Calculate the enthalpy of
reaction for decomposition of hydrogen peroxide into oxygen and water an
d determine if the reaction is exothermic or
endothermic.

2 H
2
O
2
(l) → O
2
(g) + 2 H
2
O(g)


Hf of reactants looked up in a table form in a textbook


Hf H
2
O
2
(l) =187.8 kJ/mol


Hf of products


Hf O
2
(g) = 0.0 kJ/mol


Hf H
2
O(g) =285.8 kJ/mol

Using the balanced
equation:


H = [ 0.0 + 2 mol (
-
285.8 kJ/mol) ]
-

[ 2 mol (
-
187.8 kJ/mol)] =
-
196 kJ/mol Because ∆H =
-
196 kJ/mol is negative,
the reaction is exothermic

+
C.11.D perform calculations involving heat, mass, temperature change, and specific
heat

Units of Energ
y:

calorie
--
amount of heat needed to raise the temp. of 1.00 gram of water 1.00 °C kilocalorie

1cal ==1000 Kcal the food
calorie with a capital C.

joule
--
SI
unit of energy; 1 cal = 4.184 J

Specific heat
,(s) also known as specific heat capacity (Cp)
(sometimes referred to as just heat capacity), is defined as the
amount of heat necessary to raise the temperature of 1 g of a substance by one degree. joules per gram per degree,
J/g°C or J/g K. The change in temperature ∆t =tfinal
-
tinitial is the same wh
ether you use oC or K.

Q=(mass) (∆t) (Cp)

The
heat capacity

of 1 mol of a substance is called its molar heat capacity. joules per mole per degree, J/mol°C or
J/molK.

Q= (∆t) (C)


Specific heat of water (liquid state) = 4.184 J/g°C ( or 1.00 cal/g°C)

Water h
as one of the highest specific heats known! That is why the earth stays at such an even temperature all
year round!

Heat(enthalpy) of fusion

is the amount of heat required to change a specified amount of a solid to a liquid or released
to change a given am
ount of a liquid to a solid at the melting or freezing point of the substance.

Heat(enthalpy) of vaporization

is the amount of heat to change a given amount of substance form an liquid to a gas or
released to change a given amount of a gas to a liquid

at t
he substance boiling point.

Heat of fusion

Q = m∆Hf

Heat of vaporization Q = m∆Hv

Heat of fusion and Vaporization of water:


Hf (H
2
O) = 80 kcal/kg

∆Hv (H
2
O) = 540 kcal/kg


Hf (H
2
O) = 6.01kJ/mol

∆Hv (H
2
O) = 40.67 kJ/mol

Calorimetry: is a method of measuring

heat flow between systems and surroundings. The device used to
measure heat flow between systems and surroundings is called a

Q=(mass) (∆t) (Cp) change in temperature

c)


There are
coffee cup calorimeters

and
bomb calorimeters
. You will be introduced to the bomb calorimeter in 11E
notes.

Coffee
-
cup calorimetry



in the lab this is how we experiment to find energy of a particular system. We use a Styrofoam
cup, reactants tha
t begin at the same temperature
and look for change in temperature. After all data is collect
ed (mass
or volume; initial and
final temperatu
res) we can use the specific formula to find the energy released or absorbed. We
refer to this process as constant pressure calorimetry.
∙∙

q = ∆H @ these conditions
.
∙∙

Coffee cup calorimetry

Cp: specific heat capacity measured at constant pressure. For
gases, this means the measurement was taken allowing
the gas to expand as it was heated.


Constant P
ressure Calorimetry/ Coffee Cup
assumed:

solutions have the density of water 1 g/mL

solutions have the specific heat of water 4.18 J/gCo

no heat lost to sur
roundings


Example
:

How much heat is required to raise 80.0 g of aluminum from 12°C to 46°? The specific heat capacity of
aluminum is 0.90kJ/kgoC. Q is heat so we are solving for Q. we can just plug into:

Qp = (m) (cp) (t), If we use 0.90kJ/kgoC we have

to use kg

80 g X 1 kg =0.080 kg.

1

1000g

Q= (0.080kg)(0.90kJ/kgoC)(46oC
-
1
2oC) = 2.488 kJ

+
C.11.E use calorimetry to calculate the heat of a chemical process

Tabulated ∆Hrxn values are derived from experimental data. Experiments that measure heat exchang
e are referred to
collectively as calorimetry experiments, and the measurement device is called a calorimeter. There are two important
types of calorimetry experiments: constant
-
pressure calorimetry (also known as coffee cup calorimetry) and constant
-

volu
me calorimetry (a
lso known as bomb calorimetry).

Sign Conventions in Constant
-
Pressure Calorimetry

Exothermic reactions

Endothermic reactions

Temperature increases (+∆T)

呥mp敲慴畲e⁤散 敡獥s

-
∆T)

⭑+獵srounT楮gs

-

儠獵srounT楮gs



儠sy獴敭

⬠儠sy獴敭


H⁩猠n敧慴eve


H⁩猠po獩瑩We






Constant Pressure Calorimetry/ Coffee Cup

Heat of Reaction is the heat liberated or absorbed when a chemical reaction takes place always reported as kJ/mol for a
specific reactant or product.

An exothermic reaction
liberates heat, temperature of the reaction mixture increases.

An endothermic reaction absorbs heat, temperature of the reaction mixture decreases.

The heat of reaction for a neutralization reaction is known as the heat of

neutralization. The heat of reac
tion for a solute dissolving in a solvent is known as the heat of solution. The heat of
reaction for a precipitation reaction is know
n as the heat of precipitation.

Measuring Heat of Reaction Experimentally

A known quantity of reactant is placed in a
well
-
insulated vessel (a styrofoam cup)

he initial temperature of this reactant is recorded, ti.

A known quantity of the second reactant is added, the vessel is sealed with a lid and the

reaction mixture stirred.

The final temperature of the reaction mixtu
re is recorded, tf.

The the heat released or absorbed, in joules, for the reaction is calculated:

Q=(mass) (∆t) (Cp) ∆t=tf
-
ti

he enthalpy change in kJ per mole of a given reactant for the reaction is calculated:

H = Q/1000 ÷ moles

Common assumptions for re
action mixtures made up of aqueous solutions:

density of aqueous solution assumed to be the same as for water, 1g/mL , 100mL of solution is said to have a mass of
100g

additivity of volumes of reactants is assumed, 100mL of reactant a + 200mL of reactant b

= 300mL of reaction mixture

specific heat capacity of the reaction mixture assumed to be the same as water, specific heat capacity = 4.184 J/goC

Heat is not lost to, or absorb
ed by, the surroundings.

Example: Heat (Enthalpy) of Solution

Ammonium chloride

is very soluble in water. When 4.50 g NH
4
Cl is dissolved in 53.00 g of water, the temperature of the
solution decreases from 20.40 °C to 15.20 °C. Calculate the enthalpy of dissolution of NH
4
Cl (in kJ/mol). (or enthalpy or
heat of solution) You are asked
to calculate the enthalpy change for the dissolution of ammonium chloride. You are
given the masses of the solid and water, and the temperature change that occurs when the two are combined.

First calculate the energy change for the surroundings (Q solutio
n) in the coffee cup calorimeter.

Q solution = m(solution) × Cp solution × ∆T

Cp solution = 4.18 J/g • °C ( always assume same as water for coffee cup calorimetry) m solution = 4.50 g + 53.00 g =
57.50 g


T = T final


T initial = 15.20 °C
-

20.40 °C =
-
5.
20 °C

Q solution = m(solution) × Cp solution × ∆T

Q = (57.50 g)(4.18 J/g • °C)(
-
5
.20 °C) =
-
1250 J The surroundings lost energy is negative

so the system gained energy and the reaction is endothermic and ∆H is positive

Q dissolution = 1250 J

Bomb Calorimet
ry

A bomb calorimeter is used to measure heat flows for gases
and high temperature reactions.

A bomb calorimeter works in the same manner as a coffee cup calorimeter, with one big difference. In a coffee cup
calorimeter, the reaction takes place in the wa
ter. In a bomb calorimeter, the reaction takes place in a sealed metal
container, which is placed in the water in an insulated container. Heat flow from the reaction crosses the walls of the
sealed container to the water. The temperature difference of the
water is measured, just as it was for a coffee cup
calorimeter. Analysis of the heat flow is a bit more complex than it was for the coffee cup calorimeter because the heat
flow into the metal parts of the calorimeter must be taken into account:


Bomb calor
imeter



weighed reactants are placed inside a steel container and ignited
.
( also referred to as constant
volume calorimetry) This is used by industry to determine number of food calories that we consume!

Cv: specific heat capacity measured at constant
volume. For gases, this means the measurement was made in a sealed
container, allowing the pressure to rise as the gas was heated.

Solve bomb Calorimetry problems with

Q = C

v bomb Х
t

You do not need mass because the mass of the calorimeter is always the

same.



Example:

The combustion of 1.010 g sucrose, C
12
H
22
O
11
, in a bomb calorimeter causes the temperature to rise from
24.92 to 28.33 °C The heat capacity of the calorimeter assembly is

ucrose expressed in kilojoules per mole of C
12
H
22
O
11
.

A combustion reaction is an exothermic reaction, which means that energy flows, in the form of heat, from the reaction
to the surroundings. Therefore, the Q for a combustion reaction is negative. The h
eat traveled from the reaction to the
solution causing the temperature to rise. =exothermic



Find kJ of heat released:

Q = C v bomb Х
t

=4.90

+
C.12.A describe the characteri
stics of alpha, beta, and gamma
radiation

Summary of Change in Nuclid
e
during Radioactive Processes:

Types of Radiation you need to know:

A= mass number which is the sum of the neutrons and protons

Z= at
omic number which is the number
of protons


PROCESS

CHANGE IN A

CHANGE IN Z

CHANGE IN
NEUTRON/PROTON RATIO

ALPHA EMISSION

‐4

‐2

䥎䍒䕁CN

䉅呁⁅M䥓卉低

0

1

M䕃剅䅓N

䝁MMA

0

0

X


Effects of Radiation:

2 types of damage:

Somatic damage = damage to the organism

Genetic damage = damage to the DNA and produces malfunctions in the off spring of the organism


Biological Effect of
Radiation Depends Upon:

Energy of the radiation Rad = 10‐2 J/Kg

Penetrating ability of the radiation

Ionizing ability of the radiation ~ Although gamma radiation is very penetrating it does not transform molecules in an
organism to ions. Alpha particles ar
e not very penetrating but do cause molecules in an organism to become ions.

Chemical properties of the radioactive isotope ~ Sr can be absorbed in the bone but Kr will pass through the body
quickly. REM is a unit to measure the amount of damage to living
tissue caused by different types of radiation and
different sources of radiation.

EFFECT OF SHORT TERM EXPOSURE TO RADIATION

Dose REM

Clinical Effect

0


25

Non‐detectable

25


50

Temporary decrease in white blood cell count

100


200

Strong decrease in

white blood cell count

500

Death of hal
f the population exposed for 30
days


***
C.12.B describe radioactive decay process in term
s of balanced nuclear equations

IV. Radioactive Decay
is the spontaneous disintegration of a nucleus into a slightly
lighter nucleus, accompanied by
emission of particles, elec
tromagnetic radiation, or both.

Types of Radioactive Decay

A. Alpha Emission


an alpha particle (

) is 2 protons and 2 neutrons (or a helium atom) bound together and is
emitted from the nucleus
during some kinds of radioactive decay.

210



206


4



Po






Pb


+


He


84



82


2


Clothes will s
hield you fro
m alpha particles.

B. Beta Emission



a beta particle (

) is an electron emitted from the nucleus when a neutron is converted
to a
proton.

14



14


0



C






N


+





6



7


-
1

Metal foil will
shield you from beta particles.

C. Positron Emission



a positron is a particle that is emitted from the nucleus when a proton is converted into
a neutron.

38



38


0



K






Ar


+





19



18


1


D. Gamma Emission



gamma rays (

) are high
-
energy electromagnetic waves emitted from a nucleus as it
changes from an excited state to a ground energy state. Very similar to light, but is much more dangerous.
Gamma e
mission usually occurs immediately following other types of decay. (Page 713 )

Lead or concrete will

protect you from gamma rays.

+C.12.C Compare fission and fusion reactions

Nuclear Fission and Fusion

Nuclear Fission



the splitting of a nucleus into smaller fragments (the splitting is caused by bombarding the nucleus
with neutrons). This process releases enormous amounts of energy. (page 717)

A nuclear chain reaction is a reaction in which the material that start
s the reaction (neutron) is also one of the products
and can be used to start another reaction.

1. Nuclear Reactors

use controlled


fission chain reactions to produce energy or radioactive nuclides.

2. Nuclear Power Plants

use heat from nuclear reactors

to produce electrical energy.

Nuclear Power Plants produce a great deal of energy, the current problems with nuclear power plants include
environmental requirements, safety of operation, plant construction costs, and storage and disposal of spent fuel a
nd
radioactive waste.

3. Atomic Bomb



fission reaction

Nuclear Fusion



light mass nuclei combine to form a heavier, more stable nucleus. Nuclear fusion releases even more
energy per gram of fuel then nuclear fission!!

1. Sun/Stars



four hydrogen nucl
ei combine at extremely high temperatures and pressures to form a helium
nucleus


this is a fusion reaction.

2. Hydrogen Bomb

(page 719)
-

uncontrolled fusion reactions of hydrogen are the source of energy for the
hydrogen bomb. Hydrogen bombs generate a

great deal more of energy then an atomic bomb.

3. Fusion as a Source of Energy
: because fusion reactions generate a great deal more energy and their products
are less harmful then fission reactions. Research is being done to try to use fusion instead

of fission, there are a
few problems: temperature of 10
8

Kelvin is required and no known material can withstand the temperature.