Introduction to Physical Science Chapters 1 + 2

Outline

-SI units, derived units -sig figs and scientific notation

-metric conversion and conversion factors -functions and graphing

-lab setup -components of matter

-senses and technology -phases

-accuracy vs. precision -chemical vs. physical change

-% error, difference -elements and the periodic table, metals, nonmetals

Metric Units: Le Système International d'Unités (SI) SI Units Symbol Unit Unit Abbreviation p.34

Length L metre

Mass m kilogram (There are only 7 SI units)

Time t second

Temperature T kelvin

Amount of substance n mole

Electric current I ampere

Luminous intensity Iv candela

(Some derived units)

Area A square meter

(l x w)

Volume V cubic metre

(l x w x h)

Density D grams per cubic cm

(mass ÷ vol)

Molar mass M grams per mole (mass ÷ mol)

Concentration c moles per litre (mol ÷ vol)

SI Prefixes

Name Symbol Meaning p.35

tera 1012

1 000 000 000 000

giga 109

1 000 000 000

mega 106

1 000 000

kilo 103

1 000

hecto 102

100

deca 101

10

base 100

1

deci 10−1

0.1

centi 10−2

0.01

milli 10−3

0.001

micro 10−6

0.000 001

nano 10−9

0.000 000 001

pico 10−12

0.000 000 000 001

Metric Conversion and Conversion Factors p.40

To change the prefix from what you have to what you want, multiply it by a conversion factor to

cancel what you have, and give you what you want.

How many km in 30 000 m?

How tall are you in cm if you are 5'8" (68")?

How many seconds in 3.28 years? (Assume 365 days per year, no leap years)

Significant Figures (s.f. are numbers that count for precision purposes) p.46

• All ______________ numbers are significant. 615 has 3 sf

• ___________ zeros are never significant. 0.0032 has 2 sf

• ____________ zeros are always significant. 12 004 has 5 sf

• _____________ zeros are only significant if there is a decimal.

1 200 has 2 sf 60.00 has 4 sf

Exact numbers from conversion factors or from counting objects have an ____________ number

of significant figures. Rounded conversion factors are limited in significant figures

1000 m has an infinite number of sf 2.205 lbs has 4 sf (is rounded from

1 km 1 kg 2.204622…

For ____________ and ______________: add or subtract, then round off to the least

PRECISE POSITION

For ____________ and __________: multiply or divide, then round off to the least number of

SIGNIFICANT FIGURES.

Scientific Notation p.50

Move the decimal to the place after the first non-zero number. The number of places it is moved

becomes the exponent of 10, which is multiplied.

65 000 → 0.000 0400 →

Numbers >1 have ( ) exponents Numbers <1 have ( ) exponents

Note: number of sig figs is the same in both notations

Manipulating Variables

In order to calculate problems effectively, one must be able to rearrange a formulas to solve for

the desired variable. Examples – solve for x in each of the following.

1. y = abx 2. n = ax 3. w = n + yx2

by

FC1 Foundations of Chemistry 1 Chapters 7, 20 + 21

-ionic charges -hydrates

-polyatomic ions -writing equations

-oxidation numbers -Law of Conservation of Mass

-formulas from names -balancing equations

-names from formulas -types of reactions

Ionic Charges p.204

Compounds will tend to form by _________, losing, or _________ electrons so that the

___________ shell of each atom is satisfied. This is known as the ________ rule, and even

though there are many exceptions, it is a useful concept.

Sodium has 1 valence electron, and will lose that electron in a reaction to become ___. Notice

that Na+ has a full valence shell.

Chlorine has 7 valence electrons and will gain 1 electron to complete its octet → ____

Family Li Be B C N O F Ne

Group

Charge

Ions are named after their element, however anions have their suffix changed to "ide".

Ca2+

calcium O2−

oxide

Ag+ Cl

−

Many transition metals and metals to their right make more than one stable ion. In order to figure

out which one is being used, you need to see a formula, or be given the type in the name. For

example, copper makes 2 stable ions: Cu +

and Cu 2+

,

These ions are named and respectively.

Polyatomic Ions Ions can also be made of more than one atom: p.210

ammonium

sulfite

sulfate

chlorate

carbonate

nitrate

nitrite

phosphate

acetate (or C2H3O2−)

hydrogen carbonate (or bicarbonate)

hydroxide

Oxidation Numbers p.216

The distribution of electrons in a molecule can be described by oxidation numbers, which are

usually __________ to the charge an atom will make. Oxidation numbers differ from charges in

that they are not physically ________, but only a mathematical explanation of the bonding.

Some examples: _____________ Li+1

, Ba+2

, S–2

, Cu+1

, Br–1

_____________ Li+ , Ba

2+, S

2–, Cu

+, Br

–

Assigning Oxidation numbers

1. Atoms in pure elements have an Ox # of____. ex) Na, O2, P4, S8 all have Ox # of 0

2. Elements are assigned the Ox # equal to the charge they would have as an ____. The most and

least electronegative elements get their _______ charges. ex) NaOH Na is +1, O is −2 (H is +1)

3. Fluorine is always ____.

4. Oxygen is almost always ___. Exceptions H2O2 (O = −1), OF2 (O = +2)

5. Hydrogen is ___ when bonded to a more electronegative element.

6. Ox #s add up to _ in a neutral compound. In a polyatomic ion, they add up to equal the charge.

7. These rules can also be applied to _______ compounds. ex) NaCl Na is +1, Cl is −1.

Use these rules to assign Ox #s to each element in a compound:

NaF

H2SO4

PO43−

From Name to Formula p.206

Ionic compounds (Metal + Nonmetal; Polyatomic Ions)

Write the _____ from the names. Then, using subscripts to balance charges, derive a formula.

Recall: all compounds must have a net ionic charge of ________. The charges can be found by

the group on the periodic table, from roman numerals, or from memory (ions).

barium fluoride Ba

2+ F

− →

boron hydroxide B

3+ OH

− → (polyatomic ions need brackets if use more than 1)

uranium(VI) oxide U6+

O2−

→

Hydrates

Compounds that have ________ as part of their unit cell are called hydrates and are named with

a prefix showing the number of water molecules in front of "hydrate".

cobalt(II) chloride hexahydrate → CoCl2•

calcium sulfate hemihydrate → CaSO4•

½ - 6 -

1 - 7 - naming prefixes

2 - 8 -

3 - 9 -

4 - 10 -

5 -

Molecular compounds (Nonmetal + Nonmetal) p.211

Molecular compounds use the same prefixes to show structure. Simply turn the prefixes into

subscripts for that element. If there is no prefix, it is single. DO NOT _________ SUBSCRIPTS.

dinitrogen trioxide → iodine pentafluoride →

tetraphosphorous decoxide →

Give the formula for the following compounds:

1. sodium chloride 11. sodium chlorate 21. sulfur trioxide

2. calcium fluoride 12. calcium nitrate tetrahydrate 22. dinitrogen tetroxide

3. magnesium sulfide 13. aluminum sulfate 23. carbon tetraiodide

4. aluminum oxide 14. silver phosphate 24. diphosphorous pentachloride

5. zinc chloride monohydrate 15. ammonium chloride 25. carbon dioxide

6. chromium(II) oxide 16. iron(II) hydroxide 26. beryllium nitrate

7. copper(II) bromide 17. nickel(III) acetate 27. phosphorous pentafluoride

8. manganese(VII) oxide 18. titanium(IV) carbonate 28. cobalt(III) sulfate

9. tin(IV) iodide 19. sodium bicarbonate 29. dihydrogen monoxide

10. iron(III) oxide 20. uranium(III) oxide 30. potassium hydroxide

From Formula to Name p.206

Ionic Compounds (Metal + Nonmetal ; Polyatomic Ions)

Name using the names of the ions. Cation names are the same as the metal, monoatomic anions

end in "ide"

MgCl2 → Al2O3 →

If there are polyatomic ions, name them in full.

Na2CO3 → (NH4)2SO4 →

If there are metals that have more than one oxidation state, use _________ numerals to represent

which charge was used. To find this, work back from the known anion. We always know the

anion charge because the more electronegative element (on the right) gets its ________ charge.

Fe2O3 →

SnCl4 →

Cu(C2H3O2)2 →

MnO2 →

Molecular Compounds (nonmetals + nonmetals) p.211

Give prefixes to show number of each atom in the molecule.

P2O5 →

As2S3 →

Do not include “mono” on the first ion, but always on the second.

CO2 → N2O →

Name the following formulas:

1. LiCl 11. Ba(OH)2 21. SO2

2. K2S 12. Na2SO4•10H2O 22. CS2

3. Sr3N2 13. Cs3PO4 23. N2O4

4. Ga2O3 14. Sc(NO3)3 24. Cl2O

5. ZrF4 15. ZnSO4•7H2O 25. P4O10

6. NiBr2 16. Mn(CH3COO)2 26. Co(C2H3O2)2

7. FeO 17. Cu(ClO3)2 27. NH4HCO3

8. Co2S3 18. Ti2(SO4)3 28. Zr3P4

9. Mn3N2 19. NH4F 29. H2SO4

10. CrO3 20. Fe2(CO3)3 30. BF3

Writing Equations p.242

A chemical reaction is a change in which substances are converted into different substances.

magnesium and oxygen react to form magnesium oxide (word equation)

Chemical reactions are also represented by formula equations, which display the reactants and

products as symbols or formulas. They are separated by an arrow, which indicates a chemical

reaction has taken place, as well as its direction.

"reacts with" "to produce"

→ (formula equation)

reactants products

The (s) and (g) refer to the state of the substance. For additional symbols, see p.246

( ) solid ( ) gaseous

( ) liquid ( ) aqueous (dissolved in water)

Chemists often leave out the phases, as they don't affect the calculations. Also, recall that oxygen

is a diatomic element, which exists in pairs in nature. The following form polyatomic molecules,

which you need to remember:

ex) Write the equation: solid aluminum carbide reacts with water to produce methane gas and

solid aluminum hydroxide.

The Law of Conservation of Mass p.243

In normal chemical reactions, matter is neither created, nor destroyed and the total mass of the

reactants must equal the total mass of the products. In other words, the number of atoms that go

into a reaction has to equal the number that come out. As a result of this law, we must balance all

chemical equations using coefficients (large numbers in front of the symbol). You may not

change the formulas of the substances, just the ratio in which they react.

balanced (no "1" is used)

Coefficients multiply through the entire compound.

2NaCl

3MgCl2

2H2SO4

3Ca(NO3)2

5Mg(CH3COO)2

Balancing Equations p.250

• When balancing, look for polyatomic ions or elements that appear just once on each side

- balance those first.

• You may need to change coefficients several times, so use a pencil.

• If you run into an element that is difficult to balance, leave it and come back to it.

Fractions may be useful in some cases.

• Do not include "1" as a coefficient.

• Reduce coefficients to the lowest whole number ratio.

• Double check by recounting.

ex) balance these 3 equations:

1) ___ Zn + ___ HCl → ___ ZnCl2 + ___ H2

2) ___ Al4C3 + ___ H2O → ___ CH4 + ___ Al(OH)3

3) ___ C3H6 + ___ O2 → ___ CO2 + ___ H2O

Types of Chemical Reactions p.256

We will study 5 types of reactions: synthesis, decomposition, single replacement,

double replacement, and combustion.

1) Synthesis Reaction: where 2 or more simple substances combine to form a more

complex substance. A + B → AB

2) Decomposition Reaction: where a complex substance breaks down into 2 or more

simpler substances. AB → A + B

3) Single Replacement Reaction: (Single Displacement) where an uncombined element

replaces an element that is part of a compound. A + BX → AX + B

Note: single replacement reactions will only occur if the element doing the replacing

is more active (reactive) than the element being replaced. (activity series p.266)

Mg + 2HCl → Mg + NaCl →

4) Double Replacement Reaction: (Double Displacement) where different atoms in

2 different compounds replace each other, or when 2 compounds react to form 2

new compounds. Double replacement reactions will always occur to some extent.

AX + BY → AY + BX

5) Combustion Reaction: where a substance combines with oxygen and released energy in

the form of heat and light. CXHy + O2 → CO2 + H2O (for hydrocarbons)

FC2 The Mole p.81

The mole is a unit for counting particles such as atoms and molecules. This is similar to counting

eggs by the dozen, but much larger. A mole is defined as the number of atoms in 12 g of carbon-

12, and is 6.022 x 1023

. This number is called Avogadro's number in honour of Amedeo

Avogadro, who's ideas were crucial in understanding this relationship. We can use this equation:

where n = number of moles in mol

#p = number of particles

ex) How many atoms of gold are there in 3.25 mol of Au?

ex 2) How many atoms of oxygen are there in 6.29 mol of H2SO4?

ex 3) How many moles of Na atoms are there in 5.21 x 1022

atoms of Na?

Molar Mass p.81

Because of the definition of the mole, the atomic masses on the periodic table can also be viewed

as the mass in grams of 1 mole of that element, or molar mass. From this, we get this equation:

where n = number of moles in mol m = mass in g

M = molar mass in g/mol

*When taking molar masses of elements from the periodic table, round to 2 decimal places.*

ex) How many moles of atoms are there in 156 g of silver?

ex 2) What is the mass of 0.54 mol of aluminum?

Calculating Molar Mass of Compounds p.222

When calculating the molar mass of compounds, find the total mass of each element and add

them together.

*In this class, round all molar masses to 2 decimal places.*

ex) Find the molar mass of NaCl.

ex 2) Find the molar mass of NH3.

ex 3) Find the molar mass of H2SO4.

ex 4) Find the molar mass of oxygen.

Molar Volume of a Gas at STP p.312

1 mole of any gas at STP will occupy 22.4 L of space. STP stands for standard temperature and

pressure and is defined in your text as 0°C and 1 atm of pressure.

Note: 1 atm = 101.325 kPa = 760 mm Hg = 760 torr. STAP is used in thermodynamics and is

25°C and 1 atm. At STP, 22.414 L (22.4 L) of any gas contains 1 mole of gas particles.

ex 1) How many moles of N2 are found in 56.8 L of the gas at STP?

ex 2) How many litres of N2 are found in 4.77 mol of the gas at STP?

Multi-step Problems and Sig Figs

When using 2 equations to solve a problem, or using calculated numbers, do not round numbers

while solving the problem. Round the final answer to the correct number of sig figs.

ex) What is the mass of 4.59 x 1024

molecules of carbon dioxide?

Percent Composition p.226

The percent composition tells you what percentage of the compound's total mass comes from

each element. To find this, divide the mass of each component, by the total mass, or divide their

accumulated molar masses by the overall molar mass. Don't forget to multiply by 100.

ex) Find the percent composition of a 42 g sample that contains 32.8 g iron and 9.2 g of oxygen.

ex 2) Find the percent composition of methane, CH4.

First find the molar mass:

Empirical Formulas p.229

An empirical formula is the lowest whole number ratio of bonding atoms in a compound. For

example, C2H4, C5H10, and C100H200 all have the same empirical formula: CH2

To find an empirical formula, you must find the lowest whole number ratio of moles present.

ex) A compound is analyzed and is found to contain 47.9 g of carbon and 127.7 g of oxygen.

Find its empirical formula.

ex 2) Find the empirical formula of an iron-oxygen compound that is 70.0 % Fe and 30.0 % O by

mass.

ex 3) Find the empirical formula of a compound that contains 6.61 g of hydrogen, 105 g of sulfur

and 209 g of oxygen.

Molecular Formulas p.232

Once you have an empirical formula, you can determine the molecular formula, or the actual

formula for that compound. It will be a multiple of the empirical formula, so you can compare

the molar masses to find the multiple. (the multiple will be a whole number like 1, 2, 3, etc.)

ex) Find the molecular formula of a compound if its molar mass is 42.09 g/mol and its empirical

formula is CH2.

ex 2) Find the molecular formula for a hydrocarbon if a 20.15 g sample contains 16.66 g of

carbon, and its molar mass is 58.14 g/mol.

first find the empirical

Concentration p.412

Concentration describes how many moles of solute there are dissolved per litre of solution.

where: C = concentration in mol/L or M

n = number of moles in mol

V = volume in L

ex) Find the concentration of a solution made by dissolving 0.291 mol of H2SO4 in enough water

to make 451 mL of solution.

ex 2) What mass of sodium chloride would you add to water in order to make 500.0 mL of a

1.00 M solution?

FC3: Stoichiometry Chapter 9

Stoichiometry comes from the Greek words "stoicheion" (element) and "metron," (measure) and

is a way of using the mole ratio to calculate many unknowns from given information. For

example, if you are given 43 g of reactant, how many grams of product will you get? To solve

these problems, we use the ratio in which they react, but we must first convert to moles.

Moles to Moles p.280

Use the mole ratio to predict unknowns.

ex) When sodium reacts completely with 5.00 mol of chlorine, how many moles of sodium

chloride will you get?

ex 2) The complete combustion of 4.29 mol of octane would yield how many moles of carbon

dioxide?

Mass to Moles p.284

If you are given the mass of a reactant (or product), first find out how many moles that is, then

solve as before.

ex) Excess of calcium reacts with 127 g of oxygen to form how many moles of calcium oxide?

Mass to Mass p.286

Once you know the moles of one reactant or product, you can determine the mass of any other.

ex) How many grams of oxygen and sulfur are produced when 298 g of sulfur trioxide

decomposes?

Solutions

We can perform stoichiometric calculations involving solutions; finding the # of moles is key.

ex) What volume of 2.00 M AgNO3 is required to produce 4.00 L of 5.00 M Cu(NO3)2 when

reacted with copper?

Limiting Reactants p.288

Up until now, we have examined ideal stoichiometric equations. In reality, one reactant is

usually used up before the other, and will limit how much of the products you get. This reactant

is called the limiting reactant, and the reactants that are left over are said to be in excess. It is

important to use the limiting reactant when calculating yields.

To determine which reactant is limiting, you must be given the amounts of all of them. Take one

reactant and predict how much of the other one(s) you would expect. If you have more of the

other reactant than you expect, it is in excess; if you have less than you expect, it is limiting.

ex) If 4.00 mol of Na react with 2.50 mol of Cl2, find the limiting reactant.

We can also determine the amount in excess:

We can also determine the amount in excess in g:

Another way to look at it is to take the moles and divide by the coefficients. The smallest ratio is

the limiting reactant.

ex 2) Which reactant is limiting if 41.5 g of antimony react with 134 g of iodine? How many

grams of antimony(III) iodide would you expect?

ex 3) When 58 g of zinc reacts with 640 mL of 1.0 M hydrochloric acid, what mass of zinc

chloride is formed. Find how many grams of the excess reactant is left over.

Percent Yield p.293

So far we have been dealing with the theoretical yield, or the most product possibly produced.

Usually, the actual yield is less than this ideal due to side reactions and product lost in transfer or

in purification. We can calculate the efficiency of a reaction using this equation:

% yield =

*If given the actual yield in the question, you find the theoretical through stoichiometry.

*When given an actual yield, do not put that into the stoichiometric equation.

ex) When 59.7 g of propene is burned with 45.2 g of oxygen, 39.4 g of carbon dioxide is actually

produced. Calculate the percent yield of CO2. (first find the theoretical yield)

PW1 – Property of Waves 1 A wave is a transfer of energy in the form of a disturbance without the translocation of matter.

Waves transfer energy, waves do not transfer matter.

Medium: the substance the wave travels through

Pulse: a single disturbance

Period Wave: a series of disturbances occurring at regular time intervals.

Two types of waves include transverse and longitudinal waves:

Other waves:

Surface waves: the displacement is both perpendicular and parallel – for example water waves

Electromagnetic waves: waves made of electrical and magnetic energy (light, radio, microwave,

etc.) These do not require a medium for travel.

Frequency – how often something happens f = unit cycles/second or Hz

Period – how long something takes or

how long it takes for one complete wave cycle T = unit seconds

This is called a reciprocal relationship

Sample problem: A pendulum completes 45 full cycles in 15 seconds. What are the frequency

and the period of the pendulum?

Ex 2) A tuning fork has a frequency of 512 Hz. What is the period of vibration? How many

cycles would the fork complete in 2.0 s?

Universal Wave Equation: Recall that f = 1/T and T = 1/f, and to calculate the speed of an object

we use v = d/t, then where υ = speed in m/s

λ = wavelength in m

f = frequency in Hz

Ex) The wavelength of a wave in a ripple tank is 0.080 m. If the frequency of the wave is 2.50

Hz, what is the speed of the wave?

Ex 2) The horizontal distance between a crest and a trough is 4.0 m and the wave travels 9.0 m in

4.5 s. What is the frequency of the wave?

Ex 3) The period of a sound wave is 1.18 x 10-3

s. If the speed of the wave in air is 3.40 x 103

m/s, what is the wavelength?

When a wave strikes a boundary, the portion of the wave that is transmitted into the new medium

will change speed.

Waves will travel SLOWER in shallower water

Waves will travel FASTER in deeper water

Because frequency remains the same, we can rearrange the universal wave equation to get:

Ex) Waves in the deep end of a pool have a speed of 15.0 cm/s and a wavelength of 20.0 cm. If

the wavelength in the shallow end decreases to 12.0 cm, what speed will the waves be travelling?

Ex 2) The speed and wavelength of a water wave in deep water are 18.0 m/s and 2.0 m

respectively. If the speed in shallow water is 10.0 m/s, what is the wavelength?

The speed of a wave depends on the properties of a medium

More dense media will act to slow the speed of a wave

Less dense media will act to increase the speed of a wave

Your spring investigations: 2 people close together: slower wave, 2 people far apart: faster wave

When a wave reaches a boundary (fixed or free end) it will reflect back

Waves that approach a boundary are called incident waves

Waves that are reflected are called reflected waves

Fixed End Reflections

What happens when a wave pulse interacts with a fixed end (an end that DOES NOT MOVE)

The reflected pulse will be inverted Wavelength of reflected same as incident

Speed of reflected same as incident Amplitude of reflected SMALLER than incident

Free End Reflections

What happens when a wave pulse interacts with a free end (an end that IS FREE TO MOVE)

The reflected pulse will be erect (upright) Wavelength of reflected same as incident

Speed of reflected same as incident Amplitude of reflected SMALLER than incident

Transmission of Waves

Wave energy is partially transmitted to new medium and partially reflected back in original

medium. The speed of a wave in a medium depends on the characteristics of that medium:

Depth of water (water waves) Thickness of rope…

Temperature (sound waves)

When a wave enters a new medium its speed will change!

Amplitude and wavelength will also change. Frequency will remain the same

Fast (less dense) Medium to Slow (more dense) Medium

Property Reflected Pulse Transmitted Pulse

Speed

Wavelength

Amplitude

Orientation

Partial Transition

Slow (more dense) Medium to Fast (less dense) Medium

Compared to erect incident pulse:

Property Reflected Pulse Transmitted Pulse

Speed

Wavelength

Amplitude

Orientation

Constructive interference

Interference is what occurs when two waves collide or interfere with each other’s motion.

Constructive interference occurs where waves interfering with each other have displacements in

the SAME DIRECTION! (ie. 2 crests or 2 troughs)

When the two waves collide, they will form a resultant wave

The resultant wave will have a larger amplitude than either of the original waves

Think “constructing or building up”

When done overlapping, the waves will continue unchanged on their merry way!

Destructive Interference

Occurs where waves interfering with each other have displacements in the OPPOSITE

DIRECTION!

Some (or all) of the amplitude of the original waves will be decreased when the waves overlap

Think “destruction or knocking down”

When done overlapping, the waves will continue unchanged on their merry way!

Total Destructive Interference

Occurs where destructive interference causes a complete cancellation of waves

A temporary condition where the two waves completely destroy each other

Again, when done overlapping, the waves will carry on their merry way!

Principle of Superposition

When 2 waves interfere, the resulting displacement of the medium at any location is the

algebraic sum of the displacements of the individual waves at that section…

Standing Waves

A special case of wave interference

Waves generated with constant amplitude and wavelength

Waves reflect off of a fixed or free end

Incident waves interfering with reflected waves creates a standing wave pattern

Only occurs at specific “harmonic” frequencies (other frequencies will cause irregular motion)

Points of no displacement are called NODES Think “NO DESplacement”

Points of maximum displacement are antinodes

The distance between nodes is 1/2 of a Wavelength (λ)

The diagram shows the standing waves created by different harmonic

frequencies (1st or fundamental, 2nd, 3rd, 4th etc)

Loops are regions bound by nodes that oscillate back and forth

Why do we care: You want to avoid resonance…When an outside source vibrates at a particular

frequency that matches a harmonic or resonant frequency of an object, it can create LARGE and

regular amplitudes.

Ex) If the distance between 2 nodes of a standing wave is 15 cm and the frequency of the wave is

35 Hz, what is the speed of the wave?

Ex 2) If the distance between 5 nodes of a standing wave is 3.0 m and the frequency of the wave

is 40.0 kHz, what is the speed of the wave?

Ex 3) A spring 96.0 cm long vibrates at its 3rd harmonic with a speed of 12.0 m/s. What is the

frequency of this wave? If the spring vibrated at its 4th harmonic, what would the new

frequency be?

PW2 Law of Reflection:

The angle of incidence is always equal to the angle of reflection.

Mirrors - opaque surfaces which reflect light.

Incident Rays - The ray approaching the mirror.

Reflected Ray - the ray reflected by the mirror.

Point of Incidence - where the incident ray meets the mirror.

Normal - Perpendicular to the mirror from the point of incidence.

Angle of Incidence - the angle between the incident ray and the normal.

Angle of Reflection - the angle between the reflected ray and the normal.

The angle of incidence is always equal to the angle of reflection.

Plane Mirror: when you see an image in a mirror, your eye

cannot tell that light has been reflected. The light appears as

though it is behind the mirror (this is referred to as a virtual

image). When you look at yourself in the mirror and raise

your right hand, the image appears to raise its left hand.

This is called lateral inversion.

Characteristics of Images in a Plane Mirror

Same size as object Virtual

Vertically erect Laterally inverted

The image is the same distance behind the mirror as the object is in front of the mirror.

Curved Mirrors

1) Concave mirror (converging mirror) - makes parallel light rays converge on a focal point.

2) Convex mirror (diverging mirror) - makes parallel light rays diverge apart.

Drawing a Ray Diagram

-Draw two rays starting from the tip of the arrow

and reflect them off the mirror.

-The image forms where the two rays cross

after reflecting off the mirror.

Ray Rules:

Rays parallel to the P.A. reflect through F.

Rays through F reflect parallel to the P.A.

Rays through C reflect back through C.

Object Beyond the Centre of Curvature: Object Inside the Focal Point

Rules For Rays in a Diverging Mirror

1. A ray that is parallel to the principle axis is reflected from the principle focus.

2. A ray that travels towards the principle focus is reflected parallel to the principle axis.

3. A ray that travels towards the center of curvature is reflected back along the same path.

Images formed by Diverging Mirrors

The principle focus (F) and the center of curvature (C) are virtual, behind the mirror

Diverging mirrors only create virtual images that are erect, smaller than the original object, and

located between the vertex and the principle focus.

The curved mirror equation is given as: Where f = the focal length

do = the object distance

di = the image distance

Sign Conventions

1. All distances are measured from the vertex.

2. Distances of real images are positive.

3. Distances of virtual images are negative.

4. When measured upwards, hi and ho are positive. They are negative when measured

downwards.

The Magnification Equation is:

The magnification is positive for an erect image and negative for an inverted image.

Sample Problems

1) An object is located 30.0 cm from a converging mirror with a focal length of 5.0 cm. (a) At

what distance from the mirror will the image be formed? (b) If the object is 4.0 cm tall, how tall

is the image?

2) A diverging mirror with a focal length of 5.0 cm produces an image 4.0 cm from the mirror.

(a) What is the distance of the object from the mirror? (b) What is the magnification?

Reflection with a Parabolic Surface

Spherical Aberration: The failure of all the parallel light rays striking a spherical mirror to pass

through the focal point. It causes images to be blurry. To solve this problem, we use a parabolic

mirror.

Image Characteristics Summary

PW3 The speed of light is constant in a vacuum and travels 299 972 458 m/s. (rounded to 3.00 x 10

8

m/s). The speed of light is given the symbol “c” from the word “celerity” meaning “swiftness of

motion”.

How far would light travel in exactly one minute?

If the circumference of the Earth is 40 075 km, how many times will light travel around it?

What is the wavelength of light that has a frequency of 4.0 × 1014

Hz? What colour is this light?

As light passes through transparent substances, it “slows down” due to interaction with the

atoms. It actually is delayed as it is repeatedly absorbed and re-emitted by the atoms, effectively

slowing it down. This is quantified using the Index of Refraction (n).

n = Index of Refraction

c = Speed of light in a vacuum (3.00 x 108 m/s)

v = the speed of light in the substance

Note: The higher the Index, the slower the speed.

The speed of light in water is 2.25564 x 108 m/s. Find its Index of Refraction.

The Index for Ethanol is 1.36. Find the speed of light in Ethanol.

Refraction and Snell’s Law

As a light ray passes between media of different optical densities

(different speeds of light), the rays change direction due to refraction.

When a ray passes from air to glass, some of the light is reflected

while some is refracted. This is called partial reflection and partial

refraction.

Light bends towards the normal if traveling into something more dense (air to glass)

Light bends away from the normal if traveling into something less dense (glass to air)

If the angle of incidence is zero, there is no refraction.

Principle of Reversibility of Light - if a light ray is reversed, it travels back along the same path.

The angle the ray refracts to can be calculated using Snell’s Law:

Where: n = the index of refraction in medium 1 or 2.

θ = the angle of the ray in medium 1 or 2.

*All angles are measured from the normal*

A light beam in air (n=1.0003) strikes the surface of a pond at an angle of 32.0 ̊. Find the angle

the ray refracts to inside the water (n=1.33).

A ray of light travels from water (n=1.33) into an unknown substance. If the angle of incidence

is 25.0° and the angle of refraction is 13.4°, what is the index of refraction? What is the

substance?

Total Internal Reflection – Critical Angle

When the speed of light increases (glass to air) light is reflected to a greater extent than in the

cases where speed decreases.

As the angle of incidence increases, reflected rays become more intense. The angle of refraction

also increases to a maximum of 90°.

After this point, all light is reflected.

This point is called Total Internal Reflection.

This can only occur when light is speeding up.

(ie. going into a medium that is less dense).

Critical Angle - when the angle of refraction is 90.0°.

When calculating the critical angle, set the refractive angle to 90.0°.

What is the critical angle for light traveling from crown glass into air?

What is the critical angle for light traveling from diamond into water?

Lenses: There are 2 types

1) Converging Lenses (Convex) 2) Diverging Lenses (Concave)

Thickest in the middle Thinnest in the middle

Bring light rays together cause light rays to spread apart

Images in Lenses

Optical Center (O) – geometric center. From the center of the lens to the principle axis

Principle Axis (P.A.) – straight line perpendicular to the lens.

Principle Focus (F) – the point at which light rays are refracted through

Focal length (l) – distance between O and F. Measured along the principle axis.

Rules for Rays in a Converging Lens

1. A ray parallel to the P.A. is refracted through the principal focus.

2. A ray that passes through the secondary principal focus (F’) is refracted parallel to the P.A.

3. A ray that passes through the optical center (O) goes straight through (no bending)

Images Formed in Diverging Lenses

Parallel rays are refracted outwards

The principle focus in a converging lens is real, where in a diverging lens it is virtual

All images are ALWAYS virtual, smaller, and erect

Rules for Rays in Diverging Lenses

1. A ray parallel to the P.A. is refracted to pass through F.

2. A ray that passes through F’ is refracted parallel to the P.A.

3. A ray that passes through O goes straight through.

Thin Lens Equation

same as the MIRROR EQUATION

Sign Conventions

• All distances are measured from the O of the lens

• Distances of real images are positive

• Distances of virtual images are negative

• Object and image heights are positive when

measured upwards from the P.A. It is negative for any inverted image.

Magnification Equation

Examples

An object 8.0 cm high is 18 cm from a converging lens having a focal length of 10. cm. How far

from the image is the lens and how tall is it?

A diverging lens has a focal length of 6.0 cm. If an image is found 4.5 cm from the lens, how far

from the lens is the object? If the image is 0.012 m tall, what is the height of the object? What is

the magnification?

Lens Type Object

Location

Image

Location Orient

ation Size Type

Convex

(f is positive)

inside F

on F

between F &

2F

on 2F

beyond 2F

really far

beyond 2F

Concave (f is

negative) in front

The Human Eye

Cornea: - transparent covering of the eye

found in front of the pupil

-helps to focus the image

Lens: flexible and transparent. Converges light

and focuses images to the back of the retina

Lens is thin in the middle when the muscles

are relaxed.

Lens is thick in the middle when muscles contract.

This muscle adjustment is called ACCOMODATION.

Retina: a thin layer of light sensitive nerve cells that line the back of the eye - consists of rods

and cones rods are more numerous and do not detect colour. cones detect colour.

Colour Blindness - caused by lack of development of cones.

Blind Spot - Is the junction between the optic nerve and retina. There are no light sensitive cells

in this area.

Vitreous Humour – jelly-like part of the eye.

Ciliary Muscles – they contract and relax in order to change the shape of the lens. This changes

the focal length of your lens and allows us to focus.

Aqueous Humour – Liquid between the cornea and the lens.

Sclerotic – tough outer white wall.

Optic Nerve – is the junction between the retina and the brain. It sends signals to the brain.

The image is actually inverted on the retina but the brain straightens it out for you.

Iris (Diaphram) – located at the front of the eye.

Pupil – the hole in the center of the eye. The size of the hole is controlled by the iris.

Farsightedness (hypermetropia / hypermyopia)

Is a defect in the eye resulting in the inability to see nearby objects clearly.

Usually occurs because the distance between the lens and the retina is too small.

Can also occur if the cornea and the lens is too weak to focus the image on the retina.

Can be corrected by a converging lens.

Nearsightedness (myopia)

Is an inability to see far away objects clearly.

The distance between the lens and the retina is too great

or the lens-cornea combination is too strong.

This means the parallel light rays focus in front of the retina.

Can be corrected with a diverging lens.

Astigmatism

If the lens or cornea is curved more in one way than the other.

This causes a distorted image.

Corrected with lenses that have more curvature where the eye has less and vice-versa.

HT1 – Heat 1 The Kinetic Molecular Theory KMT is used to describe ____________________________, _________________, and

________________. KMT is based on 5 postulates:

1) Molecules are in ___________________________________________, travelling in straight

lines.

2) Motion is greatest in _________, less in ________________ and least in

___________________.

3) No energy is lost in _____________________ between molecules.

4) Molecules in motion have ____________________________ of motion.

5) Molecules have _____________________________________ due to their position or state.

(the farther apart the molecules are, the greater the potential energy)

SOLID LIQUID GAS

Thermal Energy – _________________________________ of the molecules of the object.

Ethermal = EKE + EPE

Temperature – A measure of the _______________________________ of the molecules

• not a measure of ______________________

• is a measure of how fast on ___________________________ are moving

Heat (Q) – thermal energy transferred from one object to another due to a

______________________________________. Q = ∆Ethermal

• transferred by _______________________________________________________

• heat always flows from _______________________________

How Does Heat Move From One Place To Another? 1) _________________________________

• the movement of heat through ____________________

• hot molecules vibrate _________________ than cold ones

• when hot molecules collide with cooler ones they _______________________________

causing the cooler ones to vibrate faster

• speed depends on _______________________________

2) _________________________________

• transfer of heat by molecules moving from _____________________________________

• only occurs in ________________________________

3) _________________________________

• transfer of energy by

___________________________________

• ie. Light, x-rays, microwaves

Linear Expansion of Solids • solids ___________________ when they are heated and

_____________________ when they are cooled

• ________________________ depends on 3 factors:

1)

2)

3)

∆L = linear expansion (m)

α = coefficient of linear expansion (°C

–1)

L = initial length (m)

∆T = change in temperature (°C)

Sample Problems

1) An unknown metal with a coefficient of linear expansion equal to 38 × 10-6

°C-1

is 75.1 m

long. What is the final length of the metal if the temperature changes from 12.0 °C to 48.0°C?

2) A piece of copper 38.500m long is placed in the sun and has a temperature of 28.0 °C.

Several days later the length of the copper has changed to a length of 38.450m. What is the new

temperature of the copper?

3) An unknown metal increases in length from 63.200 m to 63.266 m. This happened when the

temperature rose from -8.3°C to 54.9°C. What is the coefficient of linear expansion? What is the

metal?

4) What is the initial length of a piece of steel if it is heated from 38.2°C to 45.7°C and it

increases in length by 0.15cm?

Applications of Linear Expansion

1)

2)

3)

Volume Expansion of Liquids

∆V = change in volume (m3)

β = coefficient of volume expansion (oC

–1)

V = initial volume (m3)

∆T = change in temperature (oC)

Sample Problems

1) If the coefficient of volume expansion is 38 × 10-6

°C-1

in a liquid that was originally

45.00m3 and ended up 45.10m

3, what was the change in temperature? What was the final

temperature if the original temperature was 12.0°C?

2) What is the coefficient in an unknown liquid if the temperature changes from 6.0°C to -

25.0°C and the liquid shrinks from 7.010m3 to 6.990m

3?

3) If β = 37 × 10-6

°C-1

and the liquid grows 0.10m3, what is the initial volume when it was

heated from -40.0°C to 80.0°C?

Applications of Volume Expansion 1)

2)

Measuring Quantities of Heat Specific Heat Capacity – ________________________________

• the heat required to raise the temperature of ____________________________________

Q = heat (J) lost or gained

m = mass (kg)

c = specific heat capacity (J / kg °C)

∆T = temperature change (°C)

Sample Problems

1. How much heat is required to warm 100.g of water from 10.0°°°°C to 60.5°°°°C?

2. A piece of lead is heated from -12.0°°°°C to 9.0°°°°C. If it required 3.75 kJ to do this, what is

the mass of lead?

3. An unknown metal has a mass of 3850g. When 14kJ of heat are applied, the temperature

rises from 6.2°°°°C to 21.5°°°°C. What is the metal?

Principle of Heat Exchange Heat lost by one object = Heat gained by another

- can be used to find ____________________________________ of an unknown

substance

Sample Problems

1) A 300.g chunk of iron is tossed into a bucket of methanol. If the original temperature of the

iron is 75.00°°°°C and the original temperature of the methanol is 4.50°°°°C, what would the mass of

the methanol present be if the final temperature is 4.80°°°°C?

2) A 200.g mass of an unknown metal at 90.0°C is placed in 400.g of water at 20.0°C. The final

temperature of the mixture is 23.5°C. What is the specific heat capacity for the metal? What is

the metal?

Conservation in Energy Transfer 3) A 0.500 kg sample of water is at 15.0°C in a calorimeter. A 0.0400 kg block of zinc at

115.0°C is placed in the water. Find the final temperature of the system.

4) A 0.250 kg piece of copper at 125.0°C is placed into 0.750 L of water at 18.0°C. Assuming

that no heat is lost to the surroundings, find the final temperature of the system.

Latent Heats - heat that causes _________________________________ with no change in temperature

- There are two types of latent heats:

Specific Latent Heat of Fusion (lf )

- the quantity of heat needed to ____________________________ 1kg of a solid at its

___________________________________ temperature

Specific Latent Heat of Vaporization (lv)

- the quantity of heat needed to __________________________________________ 1kg

of a substance at its _____________________________________________ temperature

Latent heat is calculated by:

OR

Q = heat , kJ

m = mass , kg

l = specific latent heat, kJ / kg

Sample Problems

1) How much heat is needed to vaporize 5.0 × 102 g of water when it has a temperature of

16.5 oC?

2) How much heat must be lost in order to freeze 150.0g of water from 45.10 °C to a

temperature of -6.80 °C?

3) 600.0 kJ of heat is put into a block of ice with a mass of 1500. g at a temperature of -15.0

°C. What is the final temperature of the water?

HT2 – Heat 2 Enthalpy (H) Enthalpy (H) is the thermodynamic potential of a system, and consists of the internal energy plus

the product of pressure and volume.

H = U + PV

Enthalpy cannot be directly measured and its unit is the joule (J).

_______is more useful and shows the amount of energy gained or lost in a process.

Exothermic Reactions _________ energy Endothermic Reactions _______ energy

CH4 + 2O2 →CO2 + 2H2O + 890.8 kJ 6C + 3H2 + 49.1 kJ → C6H6

Activation Energy (Ea) is the amount of energy required to initiate a reaction

Calculate

∆H = Ea =

∆Hrev = Ea`=

Ea`is the activation energy for the reverse reaction

Ea and Ea` are always positive

Heat of Reaction The heat of reaction is the amount of heat released or absorbed

in a chemical reaction. This is the difference in ________________________________ between

the reactants and products.

H2(g) + ½O2(g) → H2O(g) + 241.8 kJ or ∆H = – 241.8 kJ/mol

A negative ∆H means the reaction is ______________

H2O(g) + 241.8 kJ → H2(g) + ½O2(g) or ∆H = + 241.8 kJ/mol

A positive ∆H means the reaction is ______________

The reverse of an exothermic reaction is ______________.

2H2(g) + O2(g) → 2H2O(g) + 483.6 kJ or ∆H = – 483.6 kJ

The amount of heat produced depends on the moles of reactants. Use twice as much, produce

________________________.

Sample problem:

Given: C4H10 + 13/2O2 → 4CO2 + 5H2O ∆H = -2877.6 kJ/mol

How much energy is released when 780 g of butane is burned?

Heat of Formation The amount of heat absorbed or released to form 1 mole of a substance from its elements is

called the molar heat of formation. For example, ∆Hf of H2O(g) is ___________ kJ/mol.

Standard molar heat of formation occurs at room temperature (25°C/298.15 K) and 1 atm. The

standard states (°) at that temperature are used (water is a liquid at 25°C). For liquid water,

H2O(l), ∆H°f = -285.8 kJ/mol

Write the heat of formation equation for H2O(l) :

The more heat released in formation, the more ______________ the

compound. If the ∆H°f is ______________ (or only slightly negative), the

compound can decompose into its elements. The ∆H°f = 0 for elements.

Some examples:

Substance ∆H°f (kJ/mol)

Fe2O3(s) – 1118.4

C2H2(g) (acetylene) +228.2

C(s) 0

Heat of Combustion The molar heat of combustion is the heat released from the combustion of 1 mole of that

substance. For example, propane has a ∆H°c = of – 2219.2 kJ/mol.

Substance Formula ∆H°c (kJ/mol)

Methane CH4 -890.8

Ethane C2H

6 -1560.7

Propane C3H

8 -2219.2

Butane C4H

10 -2877.6

Pentane C5H

12 -3535.6

Hexane C6H

14 -4163.2

Heptane C7H

16 -4817.0

Octane C8H

18 -5470.5

Calculating the Heat of Reaction

Sample problem

Find the heat of reaction given the following heat of formations:

C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)

∆H°f :

C2H6(g) = -83.3 kJ/mol O2(g) = 0 kJ/mol

CO2(g) = -393.5 kJ/mol H2O(l) = -285.8 kJ/mol

Heat of Solution The heat of solution is the net amount of energy absorbed or released when a specific amount of

solute dissolves in a solvent.

It is measured in kJ/mol (kJ of energy per mole of solute)

Solute ∆Hsoln (kJ/mol)

NaOH -44.51

NH4NO3 +25.69

NaCl +3.88

LiCl -37.03

Sample Problem

Find the heat of solution for KCl if 23.89 g of this salt dissolved in water absorbs 5.518 kJ. Note

the ∆Hsoln is positive because it is endothermic

Because of the law of conservation of energy, the total energy of the system and surroundings

must remain constant.

Qsystem + Qsurroundings = ______

or

Qsystem = - Qsurroundings

In other words, the ∆Hsoln will always have an ______________ sign to the ∆T.

For exothermic dissolutions that get ______________

∆T = positive (+)

∆Hsoln = ______________

For endothermic dissolutions that get ______________

∆T = negative (-)

∆Hsoln = ______________

Calorimetry Recall specific heat capacity (c) in J/kg°C

For calorimeters, we can use ______________ (C) in J/°C to make calculations easier

i.e. Q = C∆T= mc∆T

For problems where you have to add the heat gained by the calorimeter, simply add the heat

needed to raise the temperature

Sample problem

When 15.3 g of sodium nitrate, NaNO3 was dissolved in water in a calorimeter, the temperature

fell from 25.00°C to 21.56°C. If the heat capacity of the solution plus the calorimeter is 1071

J/°C, what is the enthalpy change when one mole of NaNO3 dissolves?

Sample Problem 2

A 0.500 g sample of naphthalene (C10H8) is burned in a bomb calorimeter containing 0.650 kg of

water at 20.0°C. After the combustion, final temperature of the water is 26.4°C. If the heat

capacity of the calorimeter is 420 J/°C, find the heat of combustion of naphthalene in kJ/mol.

Hess's Law

Hess's law states that the overall enthalpy change in a reaction is equal to the

______________________________________ for the steps in the reaction.

Sometimes we cannot measure the heat of a reaction because it takes too long (like

rusting) or it is not possible to separate from other reactions (like burning C - forms CO2 and

CO). We can use known steps to find what we cannot measure.

Sample Problem

Find the heat of reaction for the following reaction:

NO(g) + ½O2(g) → NO2(g) ∆H = ?

Given

½N2(g) + ½O2(g) → NO(g) ∆H = +90.3 kJ/mol

½N2(g) + O2(g) → NO2(g) ∆H = +33.2 kJ/mol

Sample Problem 2

Find the heat of reaction for the following reaction:

C(s) + 2H2(g) → CH4(g) ∆H = ?

Given

C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ/mol

H2(g) + ½O2(g) → H2O(l) ∆H = -285.8 kJ/mol

CH4(g) + O2(g) → CO2(g) + 2H2O(l) ∆H = -890.8 kJ/mol

Sample Problem 3

Find the heat of reaction for the following reaction:

5C(s) + 6H2(g) → C5H12(g) ∆H = ?

Given

C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ/mol

H2(g) + ½O2(g) → H2O(l) ∆H = -285.8 kJ/mol

C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(l) ∆H = -3535.6 kJ/mol