CHAPTER ONE The Foundations of Chemistry. 2 Why is Chemistry Important? Materials for our homes...
-
date post
20-Dec-2015 -
Category
Documents
-
view
217 -
download
0
Transcript of CHAPTER ONE The Foundations of Chemistry. 2 Why is Chemistry Important? Materials for our homes...
CHAPTER ONEThe Foundations of Chemistry
2
Why is Chemistry Important?
Materials for our homes
Components for computers and other electronic devices
Cooking Fuel Body functions
3
Some definitions / Vocabulary
Chemistry Science that describes matter – its properties,
the changes it undergoes, and the energy changes that accompany those processes
Matter Anything that has mass and occupies space.
(In other words: anything that has mass and volume)
Energy The capacity to do work or transfer heat.
Types of energy Kinetic and potential energy Heat energy, light energy,
chemical energy, mechanical energy
4
Natural Laws
The Law of Conservation of Mass During a chemical or physical change the mass
of the system remains constant
The Law of Conservation of Energy Energy cannot be created or destroyed in a
chemical reaction or in a physical change. It can only be converted from one form to another.
The Law of Conservationof Matter and Energy Read at home
5
States of Matter
Liquid Solid Gas
6
States of Matter
Change States heating cooling
Ice
Steam
Water
7
Substances
Substance matter all samples of which have
identical composition and properties
Examples water sulfuric acid
Properties physical properties – physical changes chemical properties – chemical
changes
8
Physical Properties
Physical properties changes of state density, color, solubility always involve only one substance
A substance cannot be broken down or purified by physical means!
9
Mixtures Mixture
a combination of two or more substances can be separated by physical means
Homogeneous mixtures have uniform properties throughout examples: salt water; air
Heterogeneous mixtures do not exhibit uniform properties throughout examples: iron+sulfur; water+sand
10
Chemical Properties
Chemical properties chemical reactions always involve changes in composition always involve more than one
substance
Examples burning of methane rusting of iron oxidation of sugar
11
Decomposition of Water
Element
Element
Compound
oxygen
hydrogen
water
12
Compounds and Elements Compounds
If a substance can be decomposed into simpler substances through chemical changes, it is called a compound
Elements If a substance cannot be decomposed
into simpler substances by chemical means, it is called an element
13
Important to remember both compounds and elements are substances a compound consists of 2 or more elements
Law of Definite Proportions different samples of any pure compound contain
the same elements in the same proportion by mass
Symbols of elements found on the periodic chart (learn Table 1-2) www.webelements.com
Compounds and Elements
14
Scientific Notation
Use it when dealing with very large or very small numbers:
42,800,000. =
0.00000005117 =
15
Measurements in Chemistry
QuantityQuantity UnitUnit SymbolSymbol length meter m mass kilogram kg time second s current ampere A temperature Kelvin K amt. substance mole mol
16
Metric Prefixes
NameName SymbolSymbol MultiplierMultiplier mega- M 106
kilo- k 103
deci- d 10-1
centi- c 10-2
milli- m 10-3
micro- 10-6
nano- n 10-9
pico- p 10-12
17
1000 m =
0.008 s =
30,000,000 g =
0.07 L =
Metric Prefixes: Examples
18
Use of Numbers
Exact numbers obtained from counting or by definition 1 dozen = 12 things for example
Measured numbers numbers obtained from measurements
are not exact every measurement involves an
estimate
19
Significant Figures Significant figures
digits believed to be correct by the person making the measurement
20
Significant Figures
Side B:
13.6 mm
>13.5 mm but <13.7 mm in my judgement!
13.6 mm
certain figures estimated figure
21
Significant Figures
13.6 mm
certain figures + estimated figure
significant figures
we always report only 1 estimated figure the estimated figure is always the last one
of the significant figures
22
1) Exact numbers (defined quantities) have an unlimited number of significant figures. We do not apply the rules of significant figures to them.
2) Leading zeroes are never significant: 0.000357 has three significant figures
3) Zeros between nonzero digits are always significant:
20.034 1509 1.0000005
Significant Figures - Rules
23
4) Zeroes at the end of a number that contains a decimal point are always significant: 35.7000 0.07200 40.0 41.0
5) Zeroes at the end of a number that does not contain a decimal point may or may not be significant (use scientific notation to remove doubt): 173,700 may have 4, 5, or 6 significant
figures
Significant Figures - RulesTrailing zeros
24
6) The position of the first doubtful digit dictates the last digit retained in the sum or difference.
Significant Figures - Rules
Addition/Subtraction Rule
Multiplication/Division Rule
7) In multiplication or division, an answer contains no more significant figures than the least number of significant figures used in the operation.
Study examples 1-1 & 1-2 in the book
25
The Unit Factor Method The basic idea of the method:
multiplication by unity (by 1) does not change the value of the expression
Principles: construct unit factors from any two
terms that describe identical quantity the reciprocal of a unit factor is also a
unit factor
Study examples 1-3 through 1-9 in the book
26
1 ft = 12 inUnit factors:
Example: Express 77.5 inches in feet
77.5 in = 77.5 in x
in 12ft 1
ft 1in 12
in 12
ft 1= 6.46 ft
The Unit Factor Method
See Table 1-7 for various conversion factors
27
More examples
9.32 yrd = ? mm
1. We use the following knowledge to build unit factors:
1 yrd = 3 ft 1 in = 2.54 cm1 ft = 12 in 1 cm = 10 mm
2. Multiply 9.32 yrd by unit factors to get the value expressed in mm:
3 ft1 yrd
12 in1 ft
x2.54 cm
1 inx9.32 yrd x10 mm1 cm
x = 8.52·103 mm
28
Density
density =mass
volume
tells us how heavy a unit volume of matter is
usually expressed as “g/ml” for liquids and solids and as “g/L” for gases
Table 1-8 lists densities of some common substances
29
Density: Example
Example: Calculate the density of a substance if 742 grams of it occupies 97.3 cm3.
mL 97.3 cm 97.3 mL 1 cm 1 33
g/mL 637mL 97.3g 742
.d
Learn examples 1-11 through 1-13 in the book
30
Specific Gravity
Sp. Gr. =d (substance)
d (water)
tells us how much heavier or lighter a substance is compared to water:
Sp. Gr. < 1 – lighter than waterSp. Gr. > 1 – heavier than water
specific gravity has no units – it is a dimensionless quantity
See example 1-14 in the book
31
Specific Gravity: Example
Example 1-15: Battery acid is 40% sulfuric acid, H2SO4, and 60% water by mass. Its specific gravity is 1.31. Calculate the mass of pure H2SO4 in 100.0 mL of battery acid.
What do we know?
1. The mass percentage of H2SO4 and H2O in the sample of battery acid.
2. Specific gravity of battery acid.3. Density of water (1.00 g/mL).
To find the mass of H2SO4, we need to know the mass of 100.0 mL of battery acid.
32
Specific Gravity: Example
311g/mL001
(bat.acid) O)(H
(bat.acid) .acid)Sp.Gr.(bat
2
..
dd
d
Vm
d Vd m
Therefore,
g/mL 1.31 g/mL 001311 (bat.acid) ..d
g 131 mL 100.0 g/mL 311 (bat.acid) .m
g 52.4 100%40%
g 131 )SO(H 42 m
33
Heat and Temperature
Heat and Temperature are not the same thing: Heat is a form of energy T is a measure of the intensity of heat in a body
Heat always flows spontaneously from a hotter body to a colder body – never in the reverse direction
Body 1
T1
Body 2
T2
hotter T1 > T2 colder
Heat
34
Temperature Scales
3 common temperature scales
0ºF – freezing (salt+H2O)
30ºF – freezing H2O
90ºF – human body
0ºC – freezing H2O
100ºC – boiling H2O
0 K – absolute zero273.15 K – freezing H2O
Fahrenheit Celcius Kelvin
http://home.comcast.net/~igpl/Temperature.html
35
Temperature Scales & Water
Melting (MP) and boiling (MP) points of water on different temperature scales
MP BP Fahrenheit 32 oF 212 oF Celsius 0.0 oC 100 cC Kelvin 273 K 373 K
36
Temperature Conversion
degrees Kelvin degrees Celcius
? K = ?ºC + 273 ?ºC = ? K - 273
degrees Fahrenheit degrees Celcius
?ºF = (?ºC)·1.8 + 32 ?ºC = (?ºF – 32)/1.8
Examples 1-16 & 1-17 in the book
http://www.lenntech.com/unit-conversion-calculator/temperature.htm
37
Heat
Chemical and Physical changes: evolution of heat (exothermic processes) absorption of heat (endothermic processes)
Units of measurement: joule (J) – SI units calorie (cal) – conventional units 1 cal = 4.184 J
A “large calorie” (1 large cal = 1000 cal = 1 kcal) is used to express the energy content of foods
38
Specific Heat
The specific heat (Cp) of a substance: the amount of heat (Q) required to raise the
temperature of 1 g of the substance 1ºC (or 1 K)
Units of measurement:
Tmp Δ Q
C
KgJ
or Cg
J
39
Specific Heat: Example 1
Knowing specific heat, we can determine how much energy we need in order to raise the temperature of a substance by T = T2 – T1:
Calculate the amount of heat necessary to raise the temperature of 250 mL of water from 25 to 95ºC given the specific heat of water is 4.18 J·g-1 ·ºC-1.
What do we know? the temperature change the specific heat of water the volume of water the density of water
40
Specific Heat: Example 1
Examples 1-18 through 1-20 in the book
Tm heat) (specific heatTm Δ
heat heat specific
C70 C25 - C95 T
g 250 mL 250g/mL 001 .Vdm
J 107.3 J 73150 C70g 250Cg
J 4.18 heat 4
41
Specific Heat: Example 2
Given specific heats of two different substances, we can also calculate the heat transfer between them:
0.350 L of water at 74.0ºC is poured into an aluminum pot at room temperature (25.0ºC). The mass of the pot is 200 g. What will be the equilibrium temperature of water after it transfers part of its heat energy to the pot? The specific heats of aluminum and water are 0.900 and 4.18 J·g-1 ·ºC-1, respectively.
You might encounter this kind of problem at your first exam
42
What do we know? the pot and water come to equilibrium, that is
eventually they have the same temperature the specific heat of aluminum and water the mass of aluminum the volume of water the density of water finally, the Law of conservation of energy
which tells us that the amount of heat lost by water is the same as the amount of heat gained by the aluminum pot
Specific Heat: Example 2
43
Specific Heat: Example 2
Let’s denote the final temperature as Tf. Then the changes in temperature for water and aluminum are:
Tm heat) (specific heatTm Δ
heat heat specific
C025(Al) and - C047O)(H ff2 .TTT.T
g 350 L 1mL 1000
L 3500g/mL 001O)(H2 ..Vdm
Note that we used the unit factor method to convert L to mL
44
Specific Heat: Example 2
C)25.0-(g 200 Cg
J0.900)-C(74.0g 350
CgJ
4.18 ff
TT
Solving this equation with respect to Tf, we obtain
(Al) gain heat O)(H loss heat 2
Tm heat) (specific heat
Tf = 68.6ºC
Try to solve the equation yourself and analyze why the answer is given with 3 significant figures
45
Reading Assignment
Read Chapter 1 Learn Key Terms (pp. 40-41) Go through Chapter 2 notes
available on the class web site If you have time, read Chapter 2
46
Homework Assignment
Textbook problems (optional, Chp. 1): 11, 13, 15, 18, 27, 29, 30, 32, 36, 41, 43,
47, 49, 57, 62, 68, 80
OWL: Chapter 1 Exercises and Tutors – Optional Introductory math problems and Chapter
1 Homework problems – Required (homework #1; due by 9/13)