Transcript Week 6 Review 2014-15

KF Chemistry
1st 6 weeks Exam Review
Week 6 CCA
September 26, 2014
Pure Substances vs. Mixtures
• Pure substance: matter that has a fixed
(constant) composition and unique
properties. Contains only 1 type element or
compound; homogeneous
Mixture:
Contains at least 2 PHYSICALLY
combined compounds; can be homogeneous
or heterogeneous
Elements
• pure substance that cannot be separated
into simpler substance by physical or
chemical means.
Compounds
Pure substance composed of two or more different
elements joined by chemical bonds.
– Made of elements in a specific ratio
that is always the same
– Has a chemical formula
– Can only be separated by
chemical means, not physically
Mixtures
• A combination of two or more pure
substances that are not chemically
combined. Zn + Cu
• substances held together by physical forces,
not chemical
• No chemical change takes place
• Each item retains its properties
in the mixture
• They can be separated physically
3 classes of MIXTURES
Solution Colloid
homogenous
Examples
Particle Type
homogenous
Suspension
heterogenous
salt water, Soot, fog,
Muddy water,
air
mayonnaise
Italian dressing
ions, atoms Small Clusters Large Clusters
small
medium
large
Scatter Light?
(TYNDALL EFFECT)
No
yes
yes
Settle while standing?
No
No
yes
Separate by filtration?
No
No
yes
Particle Size
Mixtures vs. Compounds
http://www.bbc.co.uk/schools/ks3bitesize/science/chemistry/elements_com_mix_6.shtml
Properties of Matter (4D, Level 3)
•
•
•
Identify each of the following items as a mixture or a pure substance.
- If a mixture, identify as homogeneous or heterogeneous.
- If a pure substance, identify as a compound or an element.
Sample
Gold
Water
Italian Dressing
Milk
Cobalt
Coffee
Calcium Carbonate
Cake Batter
Type of Sample
Type of Mixture or Pure
Substance
Physical Properties
• Physical Property: Can be observed without
changing the substance composition
– Ex: Hardness, Color, Conductivity, Malleability, Melting
Point, Boiling Point
• Physical Change: Properties of the material may
change, but the COMPOSITION does not
Water –
MP is 0°C
Gallium – MP is 30°C
Chemical Properties
• Chemical Property: Can ONLY be observed by
changing the substance composition
– Ex: Burn, rot, rust, decompose, ferment, explode,
corrode
• Chemical Change: The COMPOSITION of matter
ALWAYS changes
• Also called a chemical reaction
Properties of Matter (4A, Level 2)
•
Identify the following properties as physical or chemical.
Property
Temperature
Bond Strength
Calorie Content
Mass
Density
Reactivity
Length
Melting Point
Physical Property
Chemical Property
Properties of Matter (4A, level 3)
•
Identify the following changes as physical or chemical.
Property
A rock is crushed
Condensation of water
vapor
Pancakes cook
Salt is dissolved in water
An apple is cut
Food is digested
Alcohol evaporates
Ice melts
Physical Change
Chemical Change
Extensive vs. Intensive Properties
• Extensive Properties: Depends on amount
• Ex: Volume
• Intensive Properties: Depends on type of matter
• Ex: Hardness
Properties of Matter (4B, Level 3)
• Identify the following properties as extensive or
intensive.
Property
Temperature
Bond Strength
Calorie Content
Mass
Density
Reactivity
Length
Melting Point
Extensive Property
Intensive Property
States of Matter
• Solid:
• Definite Shape
• Definite Volume
• Incompressible
• Liquid:
• Indefinite Shape
• Definite Volume
• Not Easily Compressed
• Gas:
• Indefinite Shape
• Indefinite Volume
• Easily compressed
States of Matter
Solid
Density
Viscosity
Compressibility
Structure
Shape
Volume
Movement
Draw a PictureNanoscopic Eyes
Liquid
Gases
The Law of Conservation of
Mass
Law of Conservation of Mass
During a chemical change,
matter is neither created nor
destroyed.
According to the law of conservation of
mass, how much zinc was present in
the zinc carbonate?
A 40 g
B 88 g
C 104 g
D 256 g
(All practice problems are from TEA released TAKS Tests)
In the procedure shown above, a calcium chloride
solution is mixed with a sodium sulfate solution to
create the products shown. Which of the following is
illustrated by this activity?
(A) The law of conservation of mass
(B) The theory of thermal equilibrium
(C) The law of conservation of momentum
(D) The theory of covalent bonding
Use SIGNIFICANT FIGURES to answer
•
•
•
•
•
All non-zero numbers are significant
EX: 23.4 is 3 sig.figs., 19 is 2 sig.figs., etc.
All zeros between numbers are significant—
EX: 2003 is 4s.f., 2.07 is 3 s.f., 1.009 is 4 s.f. (captive zeros)
All zeros to the right of numbers aren’t significant if there’s no decimal
EX: 1200 is 2s.f., 1000 is 1. (trailing zeros)
All zeros to the left of numbers aren’t significant if there’s no number in front
EX: 0.0045 is 2s.f. (Leading zeros)
All zeros after a number and a decimal are significant
EX: 1.00 is 3 s.f. 230.0 is 4 s.f.
When multiplying and dividing, use the least number of significant figures in your answer.
When adding and subtracting, use the least number of decimal places in your answer.
LOOK AT YOUR FORMULA SHEET FOR THE RULES! ! !
SIG FIGS-NOW YOU TRY IT
1. 22.4 x 1.2=
5. 561.25 + 105.2=
2. 1036 x 2.00=
6. 124.12 - 104.121=
3. 0.00686/19.00=
7. 343.2 x 510=
4. 1.200/13.686=
8. 100.25 + 68.750=
Scientific
Notation
(6.02E23)
Open (
Enter 6.02
2nd (Blue or Yellow button)
EE ( , button above 7)
Type in the exponent 23)
Your screen looks like:
(6.02E23)
0
6
2 3
.
change these numbers from Standard Notation
to Scientific Notation
1) 9872432
5) 356890
2) .0000345
6) 80345
3) .08376
7) 0.00000075
4) 5673
8) 1000
Use this example to help you learn how to find neutrons
Example: Carbon (C)
Atomic Number = 6
Mass Number = 12
#Protons = 6
#Electrons = 6
#Neutrons = (mass number) – (atomic number)
= (12) – (6)
=6
Element
Name
aluminum
Symbol
Atomic
Number
Al
13
Mass
Number
Number of
Electrons
Number of
Protons
Number of
Neutrons
13
13
14
bismuth
calcium
20
copper
iodine
129
82
8
50
30
Isotopes of Carbon
Naturally occurring carbon consists of three isotopes, 12C, 13C, and 14C. State the
number of protons, neutrons, and electrons in each of these carbon atoms.
12C
6
13C
14C
6
6
#P _______
_______
_______
#N _______
_______
_______
#E _______
_______
_______
Average Atomic Mass
Use this worked to example to practice the upcoming problems!
35Cl
has atomic mass 34.97 amu (75.76%) and 37C has atomic
mass 36.97 amu (24.24%).
• Use atomic mass and relative abundance of each isotope to
calculate the contribution of each isotope to the weighted
average.
34.97 x
36.97 x
75.76
100
24.24
100
=
=
• Sum is atomic mass of Cl
26.49 amu
+
8.961 amu
35.45 amu
Average Atomic Mass of Magnesium
Isotopes Mass of Isotope Abundance
24Mg
=
24.0 amu
78.70%
= ______
25Mg
26Mg
=
25.0 amu
10.13%
= ______
=
26.0 amu
11.17%
= ______
Find the Average Atomic mass of Magnesium
28
Average Atomic Mass of Boron
Isotopes
Mass of Isotope Abundance
10B
=
10.013 amu
19.8%
= ______
11B
=
11.009 amu
80.02%
= ______
Find the Average Atomic mass of Boron
29
Contrast & Compare Fission vs. Fusion
Fission
Fusion
ALPHA & BETA DECAY (12B, Level 3)
What is the alpha particle?__________________________________________
What is the beta particle?___________________________________________
Define gamma rays________________________________________________