Final Exam Study Notes

advertisement
CP Chemistry
Period 5
Physical and Chemical Properties
Physical Property- a trait or characteristic that you can observe without changing the
identity of the substance
Chemical Property - a trait you can observe by changing the identity of the
substance
Physical Properties
Chemical Properties
•
•
•
•
•
•
•
•
•
•
•
•
•
Size
Color
Texture
Smell
Mass
Taste
Density
Volume
Area
Melting Point
Malleability
Elasticity
Solubility
•
•
•
•
•
Flammability
Digestibility
Decomposable
Ability to oxidize
Reactivity inert (not reactive)
Physical and Chemical Changes
Physical Change- a change that affects only physical properties and does not alter
the identity of the substance
Chemical Change- a change that alters the identity of our substance
Physical Changes
Chemical Changes
Examples in terms of a piece of paper
• Crumpled paper
• Ripped paper
• Drawn on paper
• Stomping on paper
Examples in terms of a piece of paper
• Eating it
• Digesting it
• Burning it
Elements, Compounds, and Mixtures
• Element- a substance that cannot be
separated by chemical or physical means
• Compound- a substance made up of two or
more elements only separated by chemical
means
• Atom- smallest unit of an element
• Molecule- smallest unit of a compound
Elements, Compounds and Mixtures
cont.
• Mixture- a combination of substances thhat are not
chemically combined. These can be separated physically
– Homogeneous: looks same throughout
– Heterogeneous: composed of different parts
Homogeneous
Heterogeneous
Accuracy and Precision
True Data is accurate.
Repeatable data is precise.
Accurate & Precise
3
2.5
2
1.5
Mass 1
1
Mass 2
0.5
Trial 5
Trial 4
Trial 3
Trial 2
Trial 1
0



The mass number of the element eqauls the
number of protons in an element
The number of protons is also the number of
electrons unless its an ion
To find the neutrons you subtract the atomic
number from the mass number

Are atoms that have lost or gained neutrons,
same element but different number of neutrons




Alpha- 42H- stopped by clothing or skin
Beta- 0 -1 e- stopped by a sheet of lead
Gamma- stopped by several inches of lead,
most dangerous
Nuclear reactions happened when there is an
unstable particle and eventually gives off a
particle of radiation

Atoms that have lost or gained electrons
atoms turn into ions when electrons move
 Ions have a charge



There are negative electrons and positive electrons
How Ions are formed:





Positive ions have lost electrons
Negative ions have gained electrons
Positive ions are called cations
Negative ions are called anions
When atoms are most stable they have an octet
 Octet- 8 electrons in the outer most energy level


Share electrons between two atoms
Properties






Low melting point
Molecule structure
Gases, liquids, soft solids
Poor conductors of heat
Poor conductors of electricity
Typically 2 non-metal atoms

They trade electrons between the two atoms


Ions must form from the atom
Properties







High melting point
Crystal lattice structure
Hard solids
Brittle
Good conductors of heat
Good conductors of electricity
Typically 1 metal and 1 non metal
Wavelength
• Waves of Light: electromagnetic radiation
(light) moves as a wave
Crest
Wave
Trough
Wavelength/Frequency
• λ= Wavelength: Distance from crest to crest on
a wave
• v= Frequency: How often a wave passes by in
a second (s-1)
• Wavelength and Frequency are inversely
related
– Wavelength increases, Frequency decreases
– Wavelength decreases, Frequency increases
Calculations
• E=h𝜐
• E= Energy (Joules)
• h= Plank’s Constant (6.626 x10-34)
• 𝜈= Frequency
• Example: A yellow light has a wavelength of
600nm
– a) What is the frequency of the light?
– b) What is the Energy of the light?
• Answers on next slide
Answer – Frequency
• a) 𝜆= 600nmn
v= x
c= 3.0 x 108 m/s
𝜆= c/v
600nm x 1n/10 x 9nm= 6.0 x 10-7m
6.0 x 10-7m= 3.0 x 108 m/s /v
V= 3.0 x 108 m/s / 6.0 x 10-7m = 5.0 x 1014 s-1
Answer- Energy
•
•
•
•
•
E=hv
E= ?
h= 6.626 x10-34 J(s)
v= 5.0 x 1014 s-1
E= (6.626 x10-34 ) 5.0 x 1014
– 3.31 x 10-19 J
Ionization Energy
• The amount of energy
needed to remove one
electron from an atom
• Size of atom determines
how easily electrons are
removed
• Big atoms lose e- with
minimal effort
• Little atoms lose e- with a
huge amount of energy
needed
- Increases up and to the right on the
Periodic Table
• Noble gases all have elect
negativities equal to zero
• If an atom needs a lot of
energy to remove an
electron its because it
really wants the e-
Periodic Trends
• Properties of elements can be predicted using
the location on the periodic table
– Electron Configuration
– Family and Periods
– Densities
– Reactivity
– Atomic radius
– Ionization Energy
– Electronegativity
Atomic Radius
• The distance from the nucleus to the outermost elections (in the highest energy orbital
filled)
• As electrons fill into higher energy orbitals,
Rb
5s1
the radius of the atoms gets bigger!
K
4s1
Na
3s1
Rb
Na
K
Atomic Radius
• Within an energy level adding more protons
makes the radius of atoms smaller because
the protons can hold the electrons in closer
P
3P3
P
S
3 P4
S
Cl
3P5
Cl
Question
• Put in order smallest to largest:
– Rb, P, Na
• Answer on next slide
Answer
• Na, Rb, P
Periodic Table Families and Periods
• Groups (families) = The columns on the Periodic Table
• Periods = Rows on the Periodic Table
• Elements arranged by atomic number
– Column 1: #1
• Silvery White
– Column 2:
• React with water to form a base
– Column 3: 5,8,9,10,11,12
•
•
•
•
•
•
All metals
Form colorful solutions
Hard Brittle
Metallic
Versatile in bonding ability
Charges: +2, +3
Periodic Table Families and Periods
• Column 14
– Charge +4 -4
– Non-metallic
– Can bond with positive or negative
ions
– Solid at room temperature
– Relatively low reactivity
• Column 16: Oxygen
Family
–
–
–
–
Charge: -2
Non-metallic
Bonds with itself
Shares electrons with other
elements
• Column 17: Halogens
-
Non-metallic
Charge -1
Very reactive with positive ions
Not solid at room temperature
Colored Gas
• Column 18: Noble Gases
-
Non-Metallic
Non Reactive
Charge: 0



VSEPR is a model of molecular structures
based on the idea that ideal structures
minimize electron pair repulsions
Used to draw and evaluate Lewis Structures
Bare electrons are the most repulsive!




Looking at the molecular geometry of a
single atom, not of an entire molecule
3D Figures to represent Lewis Structures
Constituent groups are the things bonded to
the atom under scrutiny
Dashed lines represent a bond behind the
plane of the paper; wedged lines represent a
bond coming toward you
•
•
•
•
Linear
1-2 Constituents
0 Lone Pairs
Bond Angle: 180
•
•
•
•
Trigonal Planar
3 Constituents
0 Lone Pair
Bond Angle: 120
•
•
•
•
Bent
2 Constituents
1 Lone Pair
Bond Angle: <120
•
•
•
•
Tetrahedral
4 Constituents
0 Lone Pair
Bond Angle: 109.5
•
•
•
•
Trigonal Pyramidal
3 Constituents
1 Lone Pair
Bond Angle: 107.3
•
•
•
•
Bent
2 Constituents
2 Lone Pairs
Bond Angle: 104.5


Lewis dot structure is a drawing of how the
atoms are bonded together covalently using
valence electrons.
You need to know
 Shared pair= 2 electrons shared by 2 atoms
(bond)
 Lone pair= 2 electrons not shared by atoms
(unshared pair)
1. Count the total valence electrons for the
molecule
Ex: SCl2=20 valence electrons
2. Select a central atom.
look for= *the only one of its kind.
*less electronegative
Ex: SCl2= S is central atom because it’s alone
3. Set up the elements as symmetrical as
possible.
Ex: SCl2= Cl S Cl
4. Draw in shared pairs by drawing a line.
Ex: SCl2= Cl-S-Cl
5. Account for electrons used from total you
started with.(Shared pairs=2 electrons)
Ex: 20 valence electrons
-4 shared electrons
___
16 unshared electrons
6. Fill in unshared pairs around the outside of
elements
Ex:
7. When there isn’t enough electrons for every
element to have an octet, we share more pairs.
Ex:

When finding the polarity in molecules you
need to find out if the bonds are polar or nonpolar first.
 Polar bond- when 2 atoms share electrons
unequally
 Non-Polar bond- share electrons completly even.

Polar Molecules- Molecules where one side of
the molecule has more electrons than the
other.
 1. if there is a lone pair on the center atom, it is
polar
 2. If bonds are unequal polarity, than the molecule
is polar

Non-Polar Molecules
 If there is no lone pairs on the center atom
 If the bonds are equal polarity
• Synthesis:
• Ex: Cu+3 + O2  Cu2O3
• Decomposition
• Ex: MgCl2  Mg+2 + Cl
• Single Replacement
• Ex: MgCl2 + Cu+2  Mg + CuCl2
• Double Replacement
• Ex: 3MgCl2 + Cu2O3  3MgO + 2CuCl3
• Combustion
• Hydrocarbon + oxygen  CO2 +H2O
• CH3OH + O2  CO2 + H2O
• Synthesis all you have to do is combine the two reactants to get
your products
• Decomposition you break up your reactants and get two
products
• Single replacement take either the positive or negative ion (by
itself) and replace it with the positive or negative ion from a
formula in the equation
• Double replacement you take the positive ion from one formula
and put the negative ion from the other formula to create a new
formula, do this again with the left over positive and negative
ions. (positive come first)
• Combustion always ends up with carbon dioxide and water
• Matter can neither be created nor destroyed
• In chemical equation it’s crucial to make sure its balanced
because, if its not balances it goes against the law of
conservation of matter because it creates (or destroys) matter
• According to the law of conservation of matter you have to
balance all of your equations so that you don’t create or
destroy matter.
• NaOH + Cl2  NaCl + OH
• This is not balance because you have two chlorines in the
reactants and only one on the product side
• So, all you have to do is add a coefficient in order to balance
it:
• 2NaOH + Cl2  2NaCl + 2OH
• Aqueous substance dissociate
• A complete ionic equation shows all the ions and molecules in a
reaction
• Zn (s) + CuSO4 (aq)  ZnSO4 (aq)
• Complete ionic equation:
• Zn (s) + Cu+2 (aq) + SO4-2 (aq)  Zn+2 (aq) + SO4-2 (aq) +Cu (s)
• NET IONIC:
• Anything that’s AQUEOUS that’s the same on both sides you can cancel out
• So you can get rid of SO4-2 (reactant side) AND SO4-2 (product side)
• Final net ionic equation:
• Zn (s) + Cu (aq)  Zn (aq) + Cu (s)
• 1. What type of reaction is this?
• Mg + KCl  MgCl + K
• 2. Balance the following equations:
• ___ Al + ___ O2  _____ Al2O3
• ___CuS + ____ O2  ____CuO + _____ SO2
• ____ Ca3P2 + ____ H2O  ____ Ca(OH)2 + ___ PH3
• 3. Write the net ionic equation for:
• AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)
• 1. single replacement
• 2. 4 Al + 3 O2  2 Al2O3
• 3. 2 CuS + 3 O2  2 CuO + 2 SO2
• 4. (1) Ca3P2 + 6 H2O  3 Ca(OH)2 + 2 PH3
• 5. AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)
• Complete ionic =
• Ag+1 (aq) + NO3-1 (aq) + Na+1(aq) + Cl-1(aq)  Ag+1 (aq) + Cl-1 (s)
+ Na+1 (aq) + NO3-1 (aq)
• Net Ionic =
• Ag+1 (aq) +Cl-1 (aq)  AgCl (s)
The Factor Label Method
• A ratio used to convert the unit you have into
the desired unit
• Example: If you are given one day, how do you
convert it into the amount of seconds in a
day?
Answer:
1 day* 24 hours*60 minutes*60 sec
1 day
1 hour
1 min
The Factor Label Method Cont.
• The factor label method is useful in converting
metric prefixes
• The Metric Prefixes are:
– TGMKHDBDCMMNP
– The Great Mister King Henry Died By Drinking Chocolate Milk Monday Night Partying
– Tetra Giga Mega Kilo Hecto Deca BASE Deci Centi Mili Micro Nano Pico
You can use Metric conversions to change
from prefix to prefix!
Converting Moles, Liters, Grams & Particles
• 1 Mole = 6.02 X 1023 “things”
– 6.02 X 1023 = Avogadro’s Number
• Moles to Particles/Liters/Grams:
mole of element X 6.02*1023 = # with desired unit
1 Mole
Molar Mass
• To find the molar mass you must refer to the
periodic table
• Look up each atomic mass of the element and
add them all together to find the molar mass
• If there is a subscript then you must multiply
the atomic mass by the subscript
Empirical Formula
• Empirical formula= the lowest terms ratio of
elements in a formula (Not the true formula)
• To calculate the empirical formula you must
find
1.
2.
3.
4.
Percent to Mass
Mass to Mole
Divide by small
Multiply until whole
Molecular Formula
• Molecular Formula= The true formulas for a
compound (Not lowest terms)
• To find the molecular formula you must divide
the actual molecule mass by the empirical
formula mass
– Example: OH is the empirical formula, the actual formula has a mass
of 34 g/mol, what is the molecular formula?
– 34 g/mol = 2
17 g/mol
so OH becomes
OH
2
2
Percent Composition
• To calculate the % composition you must use
the following equation:
Mass of particle
Mass of whole
* 100
States of Matter
 Gas
 Forms to shape of container
 Spread apart atoms
 Solid
 Atoms tightly compacted
 Definite shape and volume
 Liquid
 Definite volume
 Moderate spacing of atoms
Parts of Solutions
 Solute
 Dissolved by the solvent in the solution
 Ex: Salt in salt water
 Solvent
 Substance that does the dissolving
 Ex: Water in salt water
Phase Diagrams/ Heating Curves
 Phase Diagrams
 Shows what temperature and pressure combinations can
create each state of matter for a particular chemical.
 Heating Curves
 Shows the temperatures at which changes in states of matter
occur and describe how a substance uses heat.
Molarity Calculations
 Molarity= moles of substance/ volume(L)
 Ex: The chemical Carbon Dioxide has a volume of 2L. Find
the concentration of Carbon Dioxide if it has a mass of
24.02g
 Grams to Moles using Molar Mass conversions.


24.02g*1mol/12.01g/mol= 2mol
Then use the formula M(Molarity)=mol(Moles)/Volume(L)
M=2mol/2L
 M=1M

Intermolecular Forces
 Intermolecular Forces
 The forces of attraction between molecules.
 Vander Waal's(London Dispersion)
 Hydrogen Bonding
 Dipole-Dipole
SPECIFIC HEAT
 Specific Heat Capacity- The amount of heat needed to raise the
temperature of one gram of substance one degree Celsius or one
degree Kelvin.
 Molar Heat Capacity- The amount of heat needed to raise the
temperature of one mole of substance one degree Celsius or one
degree Kelvin.
SPECIFIC HEAT
CALCULATIONS
q = m C
Heat (Joules)
Mass (g)
or moles
(mols)
T
Change in
temperature
(kelvin or celsius)
Tf – Ti
Specific or
molar heat
capacity
EXAMPLE PROBLEM
 12g of water is heated from 15 degrees Celsius to 35 degrees
Celsius. How much heat was absorbed by the water? (The specific
heat capacity for water is 4.184 J/g degrees Celsius).
Q=?
 m = 12 g
 C = 4.184 J/g

T = 20 degrees Celsius
q = 12 * 4.184 * 20
q = 1004.16 J
PROPERTIES OF ACIDS
 Sour
 Burn/ sting
 React with metal
 Electrolyte (conducts electricity)
 pH less than 7
 Releases hydrogen ions in water
 Accept an electron pair
PROPERTIES OF BASES
 Bitter
 Slippery
 Non-reactive with metals
 Electrolyte
 Releases hydroxide ions in water
 Donate an electron pair
PH
CALCULATIONS
 The pH scale is a numerical system that expresses the acidity of a
solution
What is the pH of a solution that has [H+]
of 1.38 * 10-11 ?
(plug into your calculator)
Ans: pH = 10.86
P OH
CALCULATIONS
 A numerical scale that measures solutions by basicity.
pOH = -log [OH-]
What is the pOH of the solution you used in the last slide?
(plug into your calculator)
Ans: 3.14
Hint: pH + pOH = 14
H + CALCULATIONS
 [H+] = 10-pH
 Example:
Determine the concentration of [H+] in the solution. pH = 3.0
 Plug into your into your calculator by clicking “2nd” and log
 Ans: 1 x 10-3 M
[OH-] CALCULATIONS
 [OH-] = 10-pOH
 Example:
Determine the [OH-] in the solution given the pOH is 4.0.
Ans: 1 x 10-4 M

Definition: A theory concerning the
thermodynamic behavior of matter, especially
the relationships among pressure, volume, and
temperature in gases.


STP is used for measuring gas temperature and
volume. STP means standard temperature
pressure.
Absolute zero is used for Absolute zero is the
point where no more heat can be removed from
a system, according to the absolute or
thermodynamic temperature scale. This
corresponds to 0 K or -273.15°C.

Pressure is measured
using a barometer.
The glass tube on the
barometer contains a
vacuum that allows
mercury flow up it
when pressure is
excreted on the
surface of the mercury






Pressure is usually
measured in:
Atmospheres (atm)
Bar (bar)
Pascals (pa)
Millimeter of Mercury
(mmHg)
Torr (torr)
PRESSURE CONVERSION
PROBLEM





1 atm= 101.325 Pa
1 bar= 100.025 Pa
1 Torr= 133.32 Pa
1 MMHg= 133.32 Pa
1 MMhg= 1 Torr

A radio station
announcer reports the
atmospheric pressure to
be 99.6 kPa. What is the
pressure in
atmospheres? In
millimeters of mercury?
99.6 kPa x 1 atm/101.3 kPa = 0.983 atm
0.983 atm x 760 mm Hg/1 atm = 747 mm Hg
Answer
0.983 atm; 747 mm Hg


Gas laws describes observed behaviors of
gasses.
𝐵𝑜𝑦𝑙𝑒𝑠 𝑙𝑎𝑤 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑃𝑉 = 𝑃2𝑉2
𝑉
𝑇
𝑉2
 Charles law equation:
=
𝑇2
𝑃
𝑃2
 Gay-lussac law equation:
=
𝑇
𝑇2
𝑃𝑉
𝑃2𝑉2
 𝐶𝑜𝑚𝑏𝑖𝑛𝑒𝑑:
=
𝑇
𝑇2



Grahams Law:
Ideal Law:
Dalton Law:
𝑀𝑀1
𝑀𝑀2
=
𝑉2
𝑉1
1) In a thermonuclear
device, the pressure of 0.050
liters of gas within the
bomb casing reaches 4.0 x
106 atm. When the bomb
casing is destroyed by the
explosion, the gas is
released into the
atmosphere where it
reaches a pressure of 1.00
atm. What is the volume of
the gas after the explosion?
2) On hot days, you may
have noticed that potato
chip bags seem to “inflate”,
even though they have not
been opened. If I have a 250
mL bag at a temperature of
19 0C, and I leave it in my
car which has a temperature
of 600 C, what will the new
volume of the bag be?
1) 2.0 x 105 L
2) 285 mL
Download