SCH4U FINAL EXAM NOTES

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SHC4U1 Notes
Unit 1
The History of the Atomic theory
John Dalton
● Ballard ball model
● All matter consists of atoms
● Atoms of different elements have different properties
● Atoms cannot be destroyed, created or subdivided
J.J. Thomson
● Raisin bun model
●
​His discovery of the electron came from his work with the cathode ray tube
●
​If an electric current was passed through a vacuum tube, a stream of glowing material would be seen
● Thomson found that the glowing stream would bend towards a positive end of a magnet
● He theorized that the stream was made of pieces of atoms that carried a negative charge scattered inside a
smear of positive charges
Millikan’s Oil drop experiment
● He determined the charge (-1.6x10​-19​C) and mass (9.11x10​-31​kg) of an electron
Antoine Henri Becquerel
●
​The discovery of radiation
●
​Observed that a piece of mineral which contained uranium could produce its image on a photographic plate
in the absence of light
Ernest Rutherford
●
​He developed the language that is in use today to describe radioactivity and atomic theory
●
​Gold foil experiment
●
​Projected positive charged particles at a piece of gold foil and noticed that not all went straight through
● Theorized that the charged particles bounced off the nucleus of an atom.
James Chadwick
● Discovery of the Neutron
● Used a beam of positive particles (alpha particles) to hit a sheet of paraffin wax
● Discovered that neutral particles were released from the paraffin wax at the same rate that positive particles
were added
Key terms
Isotope – atoms with the same number of protons but different number of neutrons
Atomic number – the number of protons in a nucleus
Atomic mass – The total number of protons and neutrons in a nucleus
radioscope – An isotope that emits radioactive gamma rays and/or subatomic particles
Homework
page 142 #1-4,7
#1
79​
Br has 35 protons, 35 electrons and 44 neutrons
Br has 35 protons, 35 electrons and 46 neutrons
239​
Pu has 94 protons, 94 electrons and 145 neutrons
133​
Cs has 55 protons, 55 electrons and 78 neutrons
3​
H has 1 proton,1 electron and 2 neutrons
81​
56​
Fe has 26 protons, 26 electrons and 30 neutrons
#2
Oxygen ( A=17, Z=8)
Chlorine (A=37, Z=17)
Cobalt (A=60, Z=27)
Iron (A=31, Z=26)
Iodine (A=78, Z=53)
Lithium (A=1, Z=3)
#3
Ba has 56 protons and 56 electrons
Zn has 30 protons and 30 electrons
N has 7 protons and 7 electrons
Rb has 37 protons and 37 electrons
Co has 27 protons and 27 electrons
Te has 52 protons and 52 electrons
#4
S​ ​(A=34, Z=16)
S (A=32, Z=16)
#7
Symbol
Protons
Neutrons
Electrons
Net Charge
238 ​
U
92
146
92
0
40​
Ca
20
20
18
2+
51​
V
23
28
20
3+
89​
Y
39
50
39
0
79​
Br
35
44
36
1-
59​
Fe
26
33
23
3+
27​
Al
13
14
10
3+
92​
20 ​
23 ​
39 ​
39​
26 ​
13 ​
The Wave Nature of Light
Electromagnetic radiation carries energy through space and is referred to as radiant energy
Examples: Visible light, Radio waves, infrared radiation (heat), X rays
All forms of radiation move through a vacuum at the speed of light which is 3.00 x 10​8 ​M/S and have wavelike
characteristics
Quantized Energy and Photons
● When objects are heated they emit radiation
●
The wavelength distribution of the radiation depends on temperature, a red hot object being cooler than a
white hot object
● Max Planck discovered the relationship between temperature, intensity and wavelength radiation
● He assumed that energy can be released or absorbed by atoms only in “chunks” of some minimum size
● The term “Quantum” refers to the smallest quantity of energy that can be emitted or absorbed by
electromagnetic radiations
● The energy of a single quantum is equal to the frequency times a constant
● Light shining on a clean metal surface causes the surface to emit electrons
● For each metal, there is a minimum frequency of light below which no electrons are emitted
Example: light with a frequency of 4.6 x 10​14​S​-1​ or greater will cause Cesium metal to eject electrons but light of lower
frequency has no effect
How Einstein explained the photoelectric effect
● Radiant energy striking the metal surface is a stream of tiny energy packets called photons that behave like
tiny particles of light
● When a photon strikes the surface of metal, its energy is transferred to an electron in the metal
● If the energy is great enough, an electron can escape the surface of the metal
● If there is left over energy, the excess appears as the kinetic energy of the emitted electron
● Einstein’s theory has lead of the belief that light has both wavelike and particle like properties.
Homework
Read page 138-141
Bohr’s Model of the Hydrogen Atom
● If a current is passed through a tube containing hydrogen gas, the electrons in the hydrogen atom absorb
gas, the electrons in the hydrogen atom absorb energy and jump to a higher energy level.
● When the electrons fall down to a lower energy level they emit light producing an emission spectrum
Continuous Spectrum
● In a continuous spectrum, the colours merge into one another and there are no black spots
Line Spectrum
● The coloured lines are separated by black regions corresponding to wavelength that are absent in the light
Bohr's Model
Assumption one
● Only orbits of certain radii, corresponding to certain definite energies are permitted
● An electron in a permitted orbit has a specific energy and is in the “allowed” energy state thus, it will not
radiate energy and spiral into the nucleus
● Each orbit corresponds to a particular n and the radius of the orbit gets larger as the value of n increases
● R is proportional to n
● The lower the energy is, the more stable the atom. The lowest energy state is called the ground state. The
higher energy state is called the excited state.
● As n becomes infinitely larger, the orbit radius increases and it reaches a point in which the electron is
completely separated from the nucleus
Assumption two
●
●
●
Electrons can jump from one allowed energy state to another by absorbing or emitting photons of radiant
energy of certain specific frequencies
Energy must be absorbed for an electron to move to a higher energy state and emitted when it jumps down
to a lower energy state
The frequency of the radiant energy corresponds exactly to the difference in energy between two states.
Homework
Pages 143-147 #1-6
#1
The Rutherford model of the atom was unable to explain the stability of atoms. Bohr
discovered that atoms have discrete energy level that contribute to an atom's stability.
#2
When a molecule or atom absorbs light the electrons get excited and jump up energy levels.
#3
Scientists can use emission spectra to determine elements since each element has a unique spectra. This is
because a line spectrum arises when electrons get excited and scientists can compare the motion of the electrons
with those of known elements.
#4
As an element
returns to its ground state from its excited state it loses energy in the form of light
(photons), this light can be observed as spectra.
#5
The emission spectra of an atom or molecule depends on the electrons because
continuous spectra compares wavelengths and line spectra compare how electrons
increase and decrease energy levels.
#6
every element produces different colours when their electrons return to a ground state
results in them having a different electrospectrum.
behave when the
which
The Wave behaviour of Matter
● Louis de Broglie suggested that as an electron circles the nucleus, it has a particular wavelength
● The wavelength depends on the mass and velocity of the electron or any other particle
The Uncertainty Principle
● Physicist Werner Heisenberg stated that it is impossible to know both the exact momentum of the electron
and its exact location in space
The Quantum Mechanical Model
● The quantum-mechanical model predicts the approximate location of the electrons around the nucleus
Orbitals
● Wave functions are referred to as orbitals
● Each orbital describes a specific distribution of electron density in space
Quantum Numbers
Four quantum numbers used to describe quantum orbitals:
● N
● L
● Ml
● Ms
Principle Quantum number, n
● Can have values of 1, 2, 3, 4 etc…
● It describes the energy level
● An n increases the orbitals become larger and the electron spends more time farther from the nucleus
Azimuthal Quantum number, l
● Can have values of 0 to n-1
● It determines the angular momentum of an object with respect to a reference point
● It defines the shape of the orbital
● It is referred to as sublevels or subshells
Magnetic Quantum number, ml
● For each sub-orbital, l, the magnetic quantum number, ml, must lie in the range ml = +l->ml =-l including 0
● Within an orbital, the suborbital have the same energy level, but have different orientations in space
Spin Quantum Numbers, Ms
●
Electrons spin in opposite directions creating oppositely directed magnetic fields
● There are only two values for Ms (Ms=+½ or -½)
Pauli Exclusion Principle
● No two electrons in an atom can have the same set of 4 quantum numbers
● For an orbital, the values of n, I and ml are fixed
● If more than one electron is placed in an orbital, the electrons must have different values
Homework
Page 153-159 #1-9
#1
Quantum Number
Information that can be obtained
n
Describes energy level
l
Determines the momentum and shape of
the orbital
ml
Represents the orientations of the
subshells.
#2
The quantum numbers can relate to the motion of a beyblade or spinning top. ‘n’ would be
the
amount of power or energy it has.in the case ‘I’ would be is speed and how it orbits
the area. Ml
represents the movement compared to the others. Ms represent the amount
and direction the beyblade
or top is spinning.
#3
2p orbits can have different amount of electrons but have the same max capacity. 2p and
orbitals can hold a different amount of electrons.
#4
1S and 3S
2Px and 3Py
2S and 2Pz
#5
n has possible values of any real numbers, l has values from 0 to n-1, ml has possible
of -1 - +1.
#6
3p
values
4S - n is 2, l is 0 and ml is 0.
6S - n is 6, I is 0, ml is
5F - n is 5, l is 3, ml is +1
on the periodic table and specific n and l values cannot coexist
#7
1d,0p,4g and 2f do not exist since they don’t match the electron configuration of anything #9
quantum numbers for A aren't allowed because l must be lower than n
The quantum numbers for D aren’t allowed because ms can only be +½ or -½
Electron configuration
An “electronic configuration” of an atom is a particular distribution of electrons among available subshells
Example:
Lithium (atomic number: 3): 1S​2​, 2S​1
Shorthand Electron Configuration
● The core of the atom corresponds to the noble gas immediately preceding it
● The electron configuration is that of the remaining electrons
Example: Cobalt (atomic number 27)
Longhand: 1S​2​, 2S​2​, 2P​6​, 3S​2​, 4S​2​, 3D​7
Shorthand: [Ar], 4S​2​, 3D​7
The Electron Configuration of Transition metals
Consider Co​2+​ and Co​3+
Co: [Ar], 4S​2​, 3D​7
● The 4S orbitals have a lower energy than 3D orbitals
The
●
It is expected that cobalt would lose electrons from the higher energy level 3D orbitals but that is not what is
observed
Co​2+​: [Ar], 3D​7
Co​3+​: [Ar], 3D​6
Exceptions to Electron Configuration
Silver: [Kr], 4D​9​, 5S​2
Neon: 1S​2​, 2S​2​, 2P​6
Determining Elements from Quantum numbers
N - energy level
L - 0=S,1=P, 2=D, 3=F
Ml - alternates from negative, neutral, to positive
Ms - first half are negative and second half are positive
Aufbau Principle
● Electrons fill orbitals starting at the lowest available energy level before filling higher levels (e.g. 1S before
2S)
Hunds Rule
● When electrons are placed in a set of orbitals of equal energy, they are spread out as much as possible to
give few electron pairs possible
Homework
Hg: [Xe], 6S​2​,4F​14​, 5D​8
[Xe], 4F​14​, 5D​10​, 6S​2
Hg​1+​ [Xe], 6S​2​, 4F​14​ ,5D​7
[Xe], 4F​14​, 5D​10​, 6S​1
Hg​2+ ​ [Xe], 6S​2​,4F​14​, 5D​6
[Xe], 4F​14​, 5D​10
Magnetic Properties
● Although an electron behaves like a tiny magnet, two electrons that are opposite in a spin cancel each other
● A ​paramagnetic substance is one that is weakly attracted by a magnetic field, usually the result of at least
one unpaired electron
● A ​diamagnetic substance is not attracted by a magnetic field generally because it has only paired electrons
Ferromagnetism
● The property of being strongly attracted to either pole of a magnet
● Ferromagnetism materials contain unpaired electrons with a small magnetic field that align with each other
in response to an external field
● This alignment persists after the magnetic field is removed
Types of Chemical Bonds
Ionic compound
● Compound that contains positively charged ions and negatively charged ions
Covalent bond
● A chemical bond formed by sharing a pair of electrons
**Lewis diagrams and the Octet rule
Page 200 #1,2
Page 204 #1,2
Molecular Geometry and Bonding Theories
Lewis structures do not indicate the shapes of molecules, however they do show the numbers and types of bonds
between atoms
CCl4 has a tetrahedral structure
The overall shape of the molecule is determined
bond angles and bond lengths.
All 6 Cl-C-Cl bond angles are 109.5°
by
The VSEPR Model
The best rearrangement of electron pairs is one that minimizes electron repulsion
Electron pair geometry
● The arrangement of electron pairs around the centre atom (a) in the AB​n ​ molecule
Molecular geometry
● The arrangement of the atoms within a molecule
● The electron pair geometry is used to predict the molecular geometry is used to describe the shape of
molecules
Example: NH​3
Lewis Structure
Electron pair Geometry
Molecular Geometry
Practice problems page 214
1)
CH​3​BH​2
CH​3​OCH​3
CH​3​CH​2​OH
Multiple Bonds
● Multiple bonds contain a higher electronic charge density than do single bonds
● The double bond acts like a nonbonding pair
Page 215 #1,2
Page 216 #2-10
Electronegativity and Polarity of Bonds
Consider CO2● Since both CO bonds are polar, is the CO2 molecule polar?
● Look at the linear shape of the CO2 molecule
● Bonds dipoles are vector quantities, they have both magnitude and direction
● When added together the total is zero (non-polar)
For AB​n​ molecules, where all of the B atoms are the same, certain symmetrical geometries including linear (AB​2​),
trigonal planar (AB​3​), tetrahedral (AB​4​), square planar (AB​4​), trigonal bipyramidal (AB​5​) and octahedral (AB​6​) will result
in non polar molecules.
Examples:
NF​3
BCl​3
Page #227 1,2
Page #229 1-8
Hybridization
BeF​2
F: [He], 2S​2​, 2P
F: [He], 2S​2​, 2P
Be: [He], 2S​2
Split the electron pair
Be to make two orbitals
in
BF​3
F: [He], 2S​2​, 2P
F: [He], 2S​2​, 2P
F: [He], 2S​2​, 2P
B: [He], 2S​2​, 2P​1
Orbital Diagrams***
SF​6
Sigma bonds
● The overlap of orbitals occurs close to the nucleus (ieintermolecular axis)
● Sigma bonds are the first bonds to be made between
the
atoms
●
They are strong
Pi bonds
● It always occurs above or below the nucleus (intermolecular axis)
● Only occurs above with pure P orbitals
● It is a weaker bond
1.
2.
3.
Draw the lewis structure
Use Vsepr to predict the hybridization b/w all center atoms
Label all Sigma and Pi bonds
CH​3​CN
In this example, the single bonds are Sigma bonds and the triple bonds are made up
of two Pi bonds and another Sigma bond.
Homework:
Page 238 #1-11
Central atoms often have the highest electronegativity
Intramolecular bond
● The chemical bond within a molecule
Intermolecular Forces
● The force that causes one molecule to interact with another molecule; occurs between molecules
Ionic-Dipole Forces:
● Forces that exist between an ion and a partial charge of the end of a polar molecule
Dipole-Dipole Forces
● Forces that exist between neutral polar molecules
● They attract each other when the positive end of one molecule is to the negative end of another molecule
● They are effective only when polar molecules are close together, generally in solution
London Dispersion Forces:
● At some instance, the 2e’ of Helium will be on the same side creating a partial dipole
● When it comes in contact with another Helium atom to move to the opposite end of the molecule
● The negatively charged original helium atom will be attracted to the positively charged: nucleus of the
second helium atom
● The strength of London Dispersion forces increases with increasing molecular size, mass and number of
electrons
Hydrogen Bonding:
● Attractions that exists between the hydrogen atoms in a polar bond (particularly an H-F, H-O or H-N bond)
and an unshared electron pair or a nearby electronegative ion or atom (usually O,F or N atom)
● They are stronger than both dipole and dispersion forces
● They are significant to biological activities such as the structure of proteins
Homework:
Read 239-246 and Answer 1-7 on 247
Properties and Structure of Solids
● Crystalline solids are solids whose atoms, ions or molecules are ordered in well defined arrangements
● The physical properties of crystalline solids, such as melting point, hardness, depends on both the
arrangement of particles and on the attractive forces between them
Ionic crystals
● Ionic solids are held together by ionic bonds
● The strength of the bond depends on the charge of the ions
Typical properties
● Hard, high melting point, non-conductors of electricity as solids but good conductors when melted
Metallic crystals
● These crystals consist entirely of metal atoms
● The bonding is due to valence electrons ​delocalized ​throughout the entire solid
● The strength of the bonding increases as the number of electrons available for bonding increases
● Table 1 page 250
Typical properties
● Range from hard to soft, melting points range from high to low, conduct electricity well, have characteristic
lustre
Molecular crystals
● Consists of atoms or molecules held together by intermolecular forces
●
The properties depend on both the strength of the forces that operate between molecules and also on the
ability of the molecule to pack efficiently into three dimensions
Typical properties
● Soft, low melting points, non-conductors of electricity.
Covalent Network crystal
● Consists of atoms held together in large interlocking networks or chains by covalent bonds
● Can be one non-metal (graphite or silicon) or two non-metals (SiC, SiO2)
● 2D or 3D
Typical properties
● Very hard, very high melting point, non-conductors of electricity
Semiconductors
● Increase temperature causes the vibration of silicon atoms
● Electricity can be controlled through doping
● N-type doping involves adding electrons to the network by replacing Si with atoms in group 5
● N-type doping involves removing electrons from the network by replacing Si with atoms is group 3
Homework
Page 254 #1-8
Properties ​v ​Solids​ >
Ionic crystal
Metallic crystal
Molecular crystal
Covalent Network
crystal
Particles involved
Metal and
Nonmetal
Metals
Non Metals
Non Metals
Primary force of
attraction
Oppositely charged
ions
Sea of electrons
Dipole-dipole bonds
Covalent bonds
Boiling point
high
varies
low
Very high
Conductivity
Only when
dissolved
yes
no
no
Examples
Sodium chloride
Aluminum
Crystal iodine
diamond
Test Outline
❏ 15 multiple choice definitions
❏ Matching vsepr shapes with molecules
❏ Lewis diagram showing hybridization, bond angles
❏ Matching properties to compounds (crystal structures)
❏ Intermolecular forces and properties***
❏
❏
❏
❏
❏
❏
Melting points, freezing point and boiling points from textbook
Electronegativity and polarity***
Symmetrical and asymmetrical molecules and polarity**
Electron configuration
Quantum numbers
Drawing molecules with hybrid orbitals and bond types*****
Unit 2
Alkanes
Prefixes of hydrocarbons (ane)
● Meth (1)
● Eth (2)
● Prop (3)
● But (4)
● Pent (5)
● Hex (6)
● Hept (7)
● Oct (8)
● Non (9)
● Dec (10)
Structural Isomers
● Structures with the same
molecular formula but different bonding/spatial arrangements
Steps for naming Alkanes
1. Select the longest continuous chain of carbon for the parent chain
2. Number the carbons in the chain using the point of first difference rule:
a. Number the chain from either end so that the substituents are given the lowest numbers possible
b. If the substituents from either end have the same number, continue numbering so that the second
substituent has the smaller number
3. The substituents are assigned the number of the carbon to which they are attached
4. The name of the compound is
composed of the name of the parent chain preceded by the name and
the number of the substituents are arranged in alphabetical order
● If a substituent occurs more than once in the molecule the prefixes, “di”, “tri” and “tetra” are used to describe
how many times it occurs
● When determining the alpha order, numerical prefixes are ignored and structural isomers/orientation prefixes
are honoured
● Numbers are always separated by a comma and numbers are separated by a dash
●
●
Alkyl Halides
Alkyl halides are named in the same manner as alkanes except that the halogen is treated as a substituent
The halogen name is shortened and produces the parent name (floro, chloro, bromo, rodo)
Cyclic Hydrocarbons
● When naming cyclic hydrocarbons by adding the prefix “cyclo” to the parent name which consists of the
corresponding hydrocarbon having the same number of carbon atoms as the ring
● Substituents on the ring are named and their positions are indicated by numbers using the lowest possible
combination of numbers
Reaction of Alkanes
● High temperature and pressure conditions are necessary for alkanes to react
● Oxidation of alkanes (reaction with O2 in a combustion reaction)
Substitution (halogenation)
● CH4 + Cl2 + energy = CH3CL + HCl
● C2 is excess
Alkenes and Alkynes
Alkenes
Alkynes
Unsaturated hydrocarbons with one or more
carbon-carbon double bonds
General Formula: C​n​H​2n
Unsaturated hydrocarbons containing triple bonds
General Formula: C​n​H​2n-2
Steps for naming Alkenes and Alkynes
● Follow the same steps as alkanes except:
○ Select the longest continuous chain of carbons containing a triple or double bond. This is the parent
structure and is assigned the name of the corresponding alkane with the suffix changed from “ane
to “ene” for a double bond and “yne” for a triple bond
Markovnikov's rule
● In addition of unsymmetrical reagents to alkenes or alkynes, the negative atom of the reagent ends up on
the carbon atom that has the fewest hydrogen. That is the M atom bonds to the carbon atom with the most H
atoms
Reactions of Alkenes and Alkynes
● Hydrogenation - the addition of hydrogen (H2)
● Halogenation - the addition of a halogen (F2, Br2, Cl2)
● Hydrohalogenation - addition of hydrogen and a halogen (HCl, HBr, HI)
● Hydration - the addition of H2O with H2SO4
Page 17 and 27
Aromatic Hydrocarbons
● Most are derivatives of benzene
● They have a high degree of unsaturation
Properties
● Regular hexagon
● Planar molecule
● All bond angles are 120°
● Is a resonance structure with delocalized pi- electrons
Rules for naming Aromatic Hydrocarbons
● Use benzene as the parent chain
● Identify the groups attached and number accordingly
● For compounds with two groups attached , the following prefixes can be used instead of the following
combinations of positions
○ 1, 2 = Ortho
○ 1, 3 = Meta
○ 1, 4 = Paramount
Phenyl Group
● When the benzene ring is considered a substituent, it is called a phenol group
● Example: 2 Phenylbutane or s-butylbenzene
Common Names
● Methyl Benzene or Toluene
Reactions of Aromatic Compounds
● Halogenation
○ Adding two halogen atoms (Br2, Cl2, F2)
○ causes a hydrogen to leave the shape and be replaced by a halogen atom
● Nitration
○ Adding Nitric Acid (HNO3)
○ Causes a hydrogen to leave the shape and NO2 and H2O to be added
● Alkylation
○ Adding a Halogen attached to a carbon chain of any length
○ Results in a hydrogen leaving the shape and a Hydrogen attached to a Halogen added to the
shape
Alcohols
● Hydrocarbon derivatives where OH replaces a H atom
● General Formula: R-OH, where R may be either alkyl or aryl
Properties of Alcohols
● Liquid at room temperature
● Experience hydrogen bonding with other alcohols and water
● Are polar and have a much higher boiling point than their parent alkanes
● The are good solvents due to their ability to dissolve polar and nonpolar compounds
Rules for naming Alcohols
● Alcohols are named by replacing the “e” ending of the corresponding alkane with the suffix “ol”
● Number the chain so that the hydroxyl group has the lowest possible position number
● If there is more than one OH group, leave the “e” in the parent name of the Alkane, and put the appropriate
prefix before the suffix “ol”
Reactions
● Oxidation of Alcohols (a combustion reaction)
○ Similar to hydrocarbons, Alcohols undergo complete combustions in the presence of excess
oxygen
○ CH3CH2CH2OH + 6O2 = H2O + CO2 + energy
● Dehydration of Alcohols
○ Elimination reaction producing water catalyzed by conc. H2SO4
○ CH3CHOHCH3 = H2O + CH3CH CH2
Ethers
General Formula
● ROR’, Where R and R’ could be from the same alkyl group
Properties of Ethers
● Higher polarity and boiling points than hydrocarbons (but less than alcohols) due to C-O bonding and their V
shape
● Good solvents due to their ability to dissolve polar and nonpolar compounds and their low reactivity
Naming Ethers
● Ethers are named as aloxyalkanes, the largest alkyl group is chosen as the parent chain and it follows the
following format:
○ Substituent - Oxy - Parent
●
Drop the “ane” from the corresponding alkane when naming the Substituent of an ether
● If the parent chain is not attached by the terminal carbon, the carbon # where this attachment occurs
precedes the Substituent name
● List the alkyl groups that are attached to the oxygen in alphabetical order
● Place the suffix “ether” at the end of the name
Preparation
●
Ethers are prepared via condensation reactions in which two alcohols come together forming water as a
by-product
General Reaction
● R-OH (alcohol) + HO-R’ (alcohol) = R-O-R’ + H2O (Ether)
Thiols
●
A thiol is an organosulfur compound that contains-bonded sulfhydryl (-C-SH or R-SH) group (where R
represents an Alkane, Alkene or other carbon-containing group of atoms)
● Thiols are used in natural gas and gives it its scent
Nomenclature 3
● Adds the ending - Thiol to the name of the Alkane, without dropping the final -e
Aldehydes
Properties
● Lower boiling point and less soluble in water than alcohols but more than hydrocarbons
● Can dissolve polar and nonpolar substances
Rules for naming Aldehydes
● Select the longest carbon chain which includes the aldehyde as a derivative of the hydrocarbon by changing
the “e” ending of the parent hydrocarbon to “al”
● Number the chain to give the carbonyl group carbon atom #1
● The aldehyde group has priority over hydroxyl groups, halogens, double bonds, triple bonds and alkyl
substituents
Ketones
Rules for naming Ketones
● Select the longest carbon chain which includes the carbonyl carbon atom
● Name the ketone as the derivative of that hydrocarbon by changing the “e” ending to “one”
● Number the chain to give the carbonyl group the lowest possible number
● The carbonyl group has priority over others
Carboxylic Acids
Properties
● Polar
● Strong Hydrogen bonding between molecules
● Small carboxylic acids are soluble in water but solubility decreases as the carbon number increases
Naming Carboxylic Acids
● The largest chain carrying the carboxyl group is considered the parent chain
●
Replace the “e” at the end of the name of the parent chain to “oic acid”
● Example: Pentanoic Acid
Esters
Properties
● Less polar than carboxylic due to the absence of the hydroxyl group
● Do not have hydrogen bonds
● Small esters are soluble in water due to polarity of carbon-oxygen bonds
● They are less soluble than carboxylic acids
Naming Esters
●
●
The alcohol group is named first, followed by the name of the of the acid with the “ic acid” ending replaced
by “ate”
Example: Propyl Ethanoate
Reactions
● Controlled oxidation of an aldehyde in
the presence of KMnO4 or Cr2O7(2-) in
H2SO4 will yield a carboxylic acid
● Carboxylic acids can be used to create organic salts (esters) in a process called esterification
(condensation)
● Carboxylic acid + Alcohol = Ester
● The reverse of esterification occurs when Esters are treated with a base (NaOH) in a hydrolysis reaction
Amines
Properties
● Polar
● Presence of N-H bonds allows for hydrogen bonds
● Smaller amines are soluble in water
● Higher boiling point and melting point than same sized hydrocarbons
● Fishy odor
Rules for naming Amines
● Name the amines as the derivative of that hydrocarbon by changing the ending of the parent hydrocarbon to
“amine”
● If it's a secondary or tertiary amine than the “N-” prefix is given
● If the attachment point of the amino group is on the terminal carbon than the amino group is numbered
● If groups are present that have a higher priority, the -NH2 is the amino substituent
Amides
Properties
● Amides are weak bases and are insoluble in water (low molecular weight makes amides slightly soluble)
● Acetaminophen - in pain killers
● Urea - in the urine of mammals
Rules for naming Amides
● Primary amides are named by changing the name of the acid by dropping the “oic acid” or “ic acid” endings
and adding “amide”
● Secondary and tertiary amides are named by using an uppercase N to designate that the alkyl group is on
the nitrogen atom
●
Reactions
Properties
● Amines are Prepared when an Alkyl group reacts with ammonia generating a primary amine
● Carboxylic acids can be used to create amides in a condensation reaction
Polymers
● A large molecule (usually a hydrocarbon or a derivative thereof) made up of smaller repeating hydrocarbon
molecules.
● Natural and synthetic polymers
● Monomer: the small repeating unit of a polymer
●
●
●
●
●
●
The process of forming polymers is called polymerization
If an alkene or alkyne monomer has more than one saturated (i.e. double or triple bond) then
C-C bonds form polymers of plastic
C-O bonds form polymers of esters
C-N bonds form polymers of amine
Table 1 on page 85******
Functional group
Prefix
Suffix
carboxylic acids
none
-oic acid
aldehydes
none
-al
ketones
none
-one
alcohols
hydroxy-
-ol
amines
amino-
-amine
ethers
alkoxy-
-ether
fluorine
fluoro-
none
chlorine
chloro-
none
bromine
bromo-
none
iodine
iodo-
none
Unit 2 Test
● Synthesis reactions
● Polymers
Unit 3
● System - particular part of the universe we wish to study
● Surroundings - whatever is entirely outside the defined system
● Boundary - real or imaginary, separates the system from its surroundings
Types of systems
● Open system
● Closed system
● Isolated system
Energy
● The ability to do work
● Measured in J in KJ
● There are two forms of energy, kinetic Ek and potential Ep
● The total Et = Ek + Ep
● Ek is derived from the movement of atoms
● Ep is derived from substances condition, position, composition or electrical charge
Thermal energy
● The form of kinetic energy that comes from the movement of atoms at the molecular level
Heat
● The flow of thermal energy between substances or to the surroundings
Temperature
● The yardstick that measures the flow of thermal energy
Endothermic vs Exothermic
● Endothermic is the heat absorbed from the surroundings
● Exothermic is the heat released into the surroundings
Calorimetry
● A method of determining the energy released from a reaction
● A reaction is carried out in an isolated system containing pure water
● Any transfers in energy can be measured by an increase/decrease in the temperature of water
● Heat is denoted by the symbol q
●
●
●
●
●
Since energy is a constant, and energy released/absorbed from a reaction must be released/absorbed
elsewhere
q System + q Surroundings = 0
q System = -q Surroundings
q rxn = -q Soln
q rxn = ΔH rxn
Enthalpy
● The enthalpy of reaction is the net energy difference between products and reactants based on the quantity
of reactants consumed
● Often the enthalpy of reaction is standardized based on one mole of reactants consumed
● This is called molar enthalpy
Types of Calorimeters
● Coffee cup calorimeter
● Bomb calorimeter
Coffee Cup Calorimeter
● q=MCΔT
● M= the mass of the liquid in the calorimeter
● C= the specific heat capacity (constant for each liquid) measured in the amount of energy to increase the
temperature of a liquid by 1 degree
● ΔT = the difference of observed temperature
Assumptions - Coffee cup calorimeter
● Ho heat transfer between the calorimeter and the outside environment
● Any heat absorbed or released by the calorimeter materials is negligible
Bomb calorimeters
● When determining the heat of reaction using a bomb calorimeter a different formula is used
●
●
No mass of liquid is needed because the heat capacity of the entire system is is determined by the
manufacturer
q=CΔT
Enthalpy Change
● The energy released to or absorbed from the surroundings during a chemical or physical change
● Measured as (ΔH)
● When it's greater than 0 its endothermic
● When it's less than 0 its exothermic
Molar enthalpy change (ΔHr)
● The change that occurs when 1 mol of that substance undergoes a physical, chemical or nuclear change
(kJ/mol)
Standard enthalpy change of formation (ΔHf)
● The enthalpy change when one mole of a compound is formed from its elements
To calculate enthalpy change for some amount of substance other than 1 mol, the molar enthalpy value is obtained
from a reference source, and then the following formula is used:
(ΔH = nΔHr)
Bond Energies
● The bond association energy is a measure of the strength of a particular bond
● Table on page 307 and 308
● The enthalpy of the reaction is (approximately) equal to the sum of the bond energies for the bonds broken
minus the sum of bond energies for bonds formed
● ΔH = Σnumber x D (bond broken) - Σnumber xD (bonds formed)
Example:
Use bond energies to estimate the enthalpy of ethene in the following combustion reaction
C​2​H​4​ + 3O​2​ →2CO​2​ + 2H​2​O
H-C=C-H + 3 O=3 → 2 O=C=O + 2 H-O-H
H H
ΔH = [Σ(C=C)+4(C-H)+3(O=O)] - [Σ4(C=O)+4(O-H)]
ΔH = [(614 KJ)+4(413 KJ)+3(495 KJ)] - [4(799 KJ)+4(467 KJ)]
ΔH = -1313 KJ (exothermic)
Standard enthalpy of formation
● The standard enthalpy of formation is a measure of the energy released or consumed when one mole of a
substance is created under standard conditions (25°C and 1000 Kpa) from its pure elements.
● The symbol of the standard enthalpy of formation is ΔHf°.
● ΔH = A change in enthalpy
● f = The f indicates that the substance is formed from its elements
● o = A degree signifies that it's a standard enthalpy change.
● Reaction Equation: ΔHf° = Σ ΔHf°(products) - Σ ΔHf° (reactants)
● If an element can exist in multiple phases at STP the more stable configuration is said to have an enthalpy
of zero
Page 320 in Textbook!
Hess’s Law
● Enthalpy is a state function
● The enthalpy change for a chemical reaction is independent of the path by which the products are obtained
● If a set of reactions occurs in different steps and the initial reactants and final products are the same, the
overall enthalpy change is the same
Hess’s Law of heat summation
● ΔH ​Overall​ = ΔH1 + ΔH2 + ΔH3 + …
● ΔH ​Overall =
​ ΣΔH​Known
● It allows for the determination of enthalpy changes without the direct use of calorimetry
Two rules must be followed
● If a chemical equation is reversed, then the sign of ΔH changes
● If the coefficients of a chemical equation are altered by multiplying or dividing by a constant factor, then the
ΔH is altered in the same way
Example:
Determine the ΔH for the reaction
C(s) + ½ O​2 (g)
→ CO(g)
​
Given the following information
C(s) + O2(g) → CO2 (g)
CO (g) + ½ O2 (g) → CO2 (g)
ΔH=-393.5 KJ
ΔH=-283.0 KJ
Example 2:
Determine the ΔH for the reaction
NO(g) + O(g) → NO2(g)
Given the following information
NO(g) + O3(g) → NO2(g) + O2 (g) ΔH=-198.9 KJ
O3(g) → 3/2O2(g)
O2(g) → 2O(g)
ΔH=-142.3 KJ
ΔH=495.0 KJ
Rates of Reactions
Chemical kinetics
● Area of chemistry that is concerned with the speeds or rates at which reactions occur
Reaction rate
● The rate at which a reaction occurs when reactants become products
Calculating average reaction rates
● Rate = Δ[A]/Δt
● A refers to any reactant or product
● [ ] = concentration
● Units: mol/L.S
C4H9Cl + H20 → C4H9OH (aq) + HCl (aq)
For every 1 mole of C4H9OH formed, 1 mole of C4H9Cl is consumed
Therefore the rate of appearance of the product is equal to the rate of disappearance for the reactant
What happens when the ratio is not 1:1?
2HI → H2 + I2
2 moles of HI disappears for every one mole of H2 or I2 that forms
Therefore the rate of disappearance of HI is twice the rate of appearance of H2 and I2
Examining Reaction rates
Collision theory
● Particles are always in motion
● For a successful collision, particles must: have a minimum energy called activation and have the correct
orientation
● A chemical system consists of particles that are constantly moving
● The average kinetic energy of particles is proportional to the temperature of the sample
Activation energy
●
The minimum energy that reactant molecules must posses for a reaction to be successful
Transition state
● Bonds are partly formed and/or partly broken
Temperature
● Increased temperature increases the rate of reaction by increasing collision frequency and fraction of
effective collisions
Catalysts
● Catalysts increase the rate of reaction by providing an alternate way for the reaction to take place with less
energy
Rate Law
● The rate constant is not affected by concentration changes but does very with temperature changes
● The values of “m” and ‘n” are not the stoichiometric numbers obtained from the balanced equation unless
the equation is a one step reaction
The sum of the exponents is called the reaction order
H2(g) + I2(g) → 2HI(g)
r=k[H2][I2]
This is a second order reaction. The sum of the component is 2
In this case the values of “m” and “n” just happen to be stoichiometric numbers meaning it's a one step reaction
Reaction Mechanisms
● Multistep mechanisms - consists of a sequence of elementary steps
● Some multistep mechanisms will include intermediates
● Intermediates are formed in one elementary step and consumed in another
A valid mechanism must meet the following criteria:
● It must agree with the overall stoichiometry
● It must agree with the experimentally determined rate law
Rate determining steps
● Most reactions occur by mechanism with more than one elementary step
● The slow steps limit the overall reaction rate
● This is called the rate determining step
● This step governs the overall rate law for the overall reaction
Decomposition reactions are often slower than synthesis reactions
The more molecules in a reaction the slower the reaction will be
Rate laws of elementary reactions
● The rate laws of elementary steps determine the overall rate law of the reaction
Elementary reaction
Rate law
A → products
Rate = k[A]
A + B → products
Rate = k[A][B]
2A → products
Rate = k[A]​2
2A + B → products
Rate = k[A]​2​[B]
The rate law for a chemical reaction was found to be:
Rate = k[NO2(g)][F2(g)]
The following elementary steps have been proposed:
NO2(g) + F(g) → NO2F(g)
NO2(g) + F2(g) → NO2F(g) + F(g)
The Final reaction is:
2NO2(g) + F2(g) → 2NO2F(g)
Unit 3 Test
● Factors that speed up reactions
● Isolated, open and closed systems
● Bomb/Coffee Cup
●
Diagram for endothermic and exothermic reactions
● Bond energies
● Hess’s Law
● Rate Law
● Mechanisms (communication)
Unit 4
(test on december 22nd)
Equilibrium
Phase Equilibrium
● the state of balance between two phases of a molecule
● Ex: H2O(l) ⇌ H2O(g)
Dynamic Equilibrium
● Does not mean that the number of reactant molecules equals the number of product molecules
Chemical Equilibrium
● Condition in which the concentration of all reactants and products eventually stop changing
● For equilibrium to occur neither products or reactants can leave the system
● Can occur as a synthesis reaction or a decomposition reaction
Requirements for Equilibrium
● System must be closed
● System must be at a constant temperature
● No change in the macroscopic properties of the system, such as color change, emission of gas bubbles,
release of heat
● Opposite reactions occur at the same rate
● Equilibrium is attained by starting with either reactants or products
Three types of reactions
1.
2.
3.
Reactants favoured when % reaction is <1%
Products favoured when % reaction >99%
Reaction appears to be at equilibrium when the % reaction is between 1% and 99%
Equilibrium Constant Expression
● Calculating the % reaction is only useful in determining whether the products or reactants are favoured for
specific reaction conditions
● Instead the equilibrium constant expression is used to relate to the concentration of reactants and products
in a mathematical and predictable way
Finding the equilibrium constant (K)
● K = [Products]/[Reactants]
● aA + bB ⇌ cC + dD
● K = [C]​c ​[D]​d​ / [A]​a​[B]​b
● K depends upon the way the reaction was written
● Liquids and solids are not included in the equilibrium expression because their concentration is constant and
has already been incorporated into the equilibrium product constant
Effect of Temperature on K​eq
● Changing the temperature favours either the forward or reverse reaction
● This causes a shift the the value of K​eq​ as well as a change in the [participants]
● A new equilibrium will be reached with its own K​eq​ value for the reaction at that temperature
● At constant temperature, changing the equilibrium constant does not affect K​eq​ because the rate constants
are not affected by concentration changes
● When the concentration of one of the participants is changed the concentration of the others vary in such a
way as to maintain a constant value for the K​eq
Magnitude of the equilibrium constants
● The magnitude of K​eq​ is a measure of the extent to which a given reaction has or will take place
K > 1 products are favoured, equilibrium lies to the right
K < 1 reactants are favoured, equilibrium lies to the left
K = 1 half the reactants become products
Le Chatelier's Principle
● If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that
tends to reduce that change
Altering the equilibrium composition of a gaseous mixture
● The addition of a reactant or product increases the concentration of the reactant or product
● The equilibrium will shift in the direction that will consume the added component
● The addition of a pure solid or liquid does not cause a shift of equilibrium
Effect of a pressure change
There are two ways to change the pressure in a system:
1. Change of Volume - as volume increases pressure decreases, equilibrium shifts to side with more moles of
gas
2. Adding Gas - adding an inert gas causes no shift in the equilibrium position as long as it does not affect the
concentration or partial pressure
Effect of a temperature change
● As temperature decreases and the reaction is exothermic, equilibrium shifts towards the products
● As temperature increases and the reaction is endothermic, equilibrium shifts towards the reactants
Effect of a Catalyst
● A catalyst increases the rate of reaction, but for a reversible reaction it must increase both the forward and
the reverse reactions equally since it lowers their E​a​ values by the same amount
● Adding a catalyst will establish equilibrium more quickly since reaction rates are increased, but it will not shift
the position of the equilibrium
Predicting the direction of reaction
● The reaction quotient Q, is used to predict whether a system is at equilibrium
● Reaction quotient(Q) = [Products]/[Reactants]
Comparing Q and K
● Q<K - in order to attain equilibrium, the reaction will move from left to right
● Q>K - in order to attain equilibrium, the reaction will move from right to left
● Q=K the system is at equilibrium
Example:
The following reaction occurs in a closed container at 445 degrees celsius. The equilibrium constant (K) is 0.02
2HI(g) ⇌ H​2​(g) + I​2​(g)
Is the system at equilibrium if
HI(g) = 0.14M H​2​(g) = 0.04M I​2​ = 0.01M
Q= ​[H​2​][I​2​]
[HI]​2
=​ (0.04)(0.01)
(0.14)​2
= 0.020
∴ the system is at equilibrium
Example 2:
Phosphorus pentachloride, PCl​5​, decomposes when heated.
PCl​5​(g) ⇌ PCl​3​(g) + Cl​2​(g)
If the initial concentration of PCl​5​ is 1.00mol/L what is the equilibrium concentration of the gaseous mixture at 160℃?
The equilibrium constant K is 0.0211.
I
1 mol/L
0
0
C
-X
X
X
E
1-X
X
X
(X)(X) ​= 0.0211
(X-1)
X​2​ = 0.0211
X-1
X​2​ = 0.021(X-1)
X​2​ = 0.021 - 0.0211X
0 = X​2​ + 0.0211X - 0.0211
Quadratic Formula
X= 0.135 ​X=-0.156
The Hundred Rule
● If the resultant number is greater than 100, you can approximate the x value as zero
Validation (the 5% rule)
When making assumptions it must be within 5%
X/[HA]​i x​ 100%
Solubility
● Solubility is the concentration of a saturated solution of a solute in a particular solvent at a specific
temperature
The Solubility Product Constant
●
●
The solubility product constant (K​sp​) is the equilibrium constant for the solubility equilibrium of a slightly
soluble ionic compound
Multiply mol/l by mol/g to solve for solubility
Trial Ion Product (Q)
Q > Ksp
Supersaturated solution
Precipitate will form
Q=Ksp
Saturated solution
No precipitate
Q < Ksp
Undersaturated Solution
No precipitate
Common ion effect
● The presence of a common ion will reduce solubility
Calculate the molar solubility of CaF​2​ in a solution containing:
a) Pure water
b) 0.01M Ca(NO​3​)​2
c) Ksp at 25 degrees is 3.9 x 10​-11
CaF​2​(s)
Ca
2F
I
NA
0
0
C
NA
E
NA
+
X
X
2X
CaF​2​(s)
Ca
2F
I
NA
0.01
0
C
NA
E
NA
+
2X
+
2X
Ksp = [Ca] [F​2​]​2
3.9 x 10​-11​ = (x) (2X)​2
3.9 x 10​-11​ = 4X​3
X = 2.136 x 10​-4​ mol/l
+
0.01+X
X
2X
Ksp = (0.01)(2X)​2
3.9 x 10​-11​ = (0.01)(2X)​2
X = 3.122 x 10​-5 ​mol/l
Acids and Bases
Arrhenius Theory
● An acid is a substance that when dissolved in water increases the [H​+​(aq)]
● A base is a substance that when dissolved in water increases the [OH​-​(aq)]
Bronsted-Lowry Theory
● In a proton transfer reaction; an acid species donates protons and a base species receive protons
● A bronsted-lowry acid must have a removeable proton
● A bronsted-lowry acid must have a pain of unbonded electrons
Conjugate Acid-Base Pair
● A conjugate acid base pair consists of two species in an acid-base reaction , one acid and one base that
differ by the loss or gain of a proton
Ionization
● When acids or bases are added to water, they are ionized
Strong Acids/Bases
● These are strong electrolytes and exist as ions in aqueous solutions
● Their equilibrium lies far to the ​right
● They will yield very weak conjugates
Types of Acids
● Monoprotics - have 1 proton to donate
● Diprotic - have 2 protons to donate
● Triprotic - have 3 protons to donate
● Oxyacids - acidic proton attached to an oxygen
● Organic Acids - have a carbon atom backbone
Weak Acids/Bases
● These only partially ionize in water
● They exist in a solution as a mixture of molecules and component ions
● They are weak electrolytes
● Their equilibrium lies far to the left
Self Ionization of water
● Self ionization (autoionization) is a reaction
Kw relates to Ka and Kb
● The relationship between Ka and Kb is Kw=(Ka)(Kb)
● If Ka or Kb is given the dissociation constant for the conjugate acid/base can be calculated
The pH scale
● A numerical scale is a numerical scale which extends from 0 through to 14
● A ph of 4 is 10 times more than a ph of 3
● pH= -log​10 ​[H+]
● pH + pOH = 14
Measuring pH
● Most accurate is a pH meter which measures voltage, more electrolytes means higher voltage
Percent ionization
p=[H​+​] / [HA] x 100%
Acid Base properties of Salt Solutions
● Salt solutions can be acidic, basic or neutral
● It depends on the ions that are produced in hydrolysis
General Rules
● Strong acids and strong bases have weak conjugates
○ Salts that consist of cations of strong bases and the anions of strong acids have no effect on pH
○ Group 1 and 2 metals ion exhibit no acidic or base properties
● Water will have relatively little acid-base effect
●
●
Anions act as Bases
Cations act as Acids
Calculate the pH of a solution in which 15 ml of 0.1m NaOH has been added to 25 ml of 0.01 HCl
N=CxV
N = 0.0015 mol of NaOH
N = 0.0025 mol of HCl
0.0025 - 0.0015 = 0.001
0.001mol / 0.4 L
= 0.025 mol/L
Calculate the pH when 30 ml of 0.15m NaOH is added to 50ml of 0.2m HCHO2
K= 1.8 x 10​-4
V= 0.03 l
C = 0.15 mol/l
n= 0.0043 mol
V= 0.05
C = 0.2 mol/l
n= 0.01 mol
Calculate the pH at the equivalence point when 0.05m KOH is added to 50 ml of 0.025m benzoic acid, HC​7​H​5​O​2
when Ka = 6.5 x 10​-5
KOH + HC​7​H​5​O​2​ → HOH + C​7​H​5​O​2
N of HC​7​H​5​O​2​ = 0.025m x 0.05L
= 0.00125
I
0.00125 mol
0.00125 mol
NA
0
C
-0.00125
-0.00125
NA
+0.00125
E
0
0
NA
0.00125 mol
V=n/C
= 0.00125 mol / 0.05m
= 0.025 L
Kw/Ka = Kb
1.0 x 10​-1​/ 6.5 x 10​-5​ = Kb
= 1.5 x 10​-10
Calculate the pH of the titration of ammonia by hydrochloric acid at the equivalence point when 0.1M of HCl is added
to 25 ml of 0.1m NH3 when Kb = 1.8 x 10​-5
H​+​ + NH​3​ → NH4​+
I
0.0025
0.0025
0
C
-0.0025
-0.0025
+0.0025
E
0
0
0.05
C=n/v
V=n/c
= 0.0025mol /0.1m
=
Unit 4 Test
● Multiple choice - buffers, bronsted-lowry, arrhenius
● Percent ionization
● Percent dissociation
● Acids/Bases
● Le chatelier's principle
● Solubility (common ion effect)
● Titration
● Define equilibrium
● Difference in k value for forward/reverse
● Haber process
Unit 5: Electrochemistry
Oxidation
The process in which one or more electrons is lost by a chemical entity
Reduction
Half reaction method
1. Divide the skeleton equation into half reactions
2. Balance atoms other than hydrogen and oxygen
3. Balance oxygen atoms by adding water to the side that needs oxygen
4. Balance hydrogen by adding it to the other side
5. Balance the charges by adding electrons
6. Make electrons gained equal to the electrons lost and then add the two half reactions
7. Cancel out what is common to both sides
Galvanic cells
● A galvanic cell consists of two separate half
cells
● The two electrodes are connected by a wire
● The two solutions are connected by a salt
bridge
● The stronger oxidizing agent is reduced at the
cathode
● The stronger reducing agent is oxidized at the
anode
● As the cell operates electrons travel from the
anode to the cathode
Line Notation and Galvanic Cells
Anode (-) I Electrolyte II Electrolyte I Cathode (+)
Calculating a Standard cell potential given a net ionic equation
E cell = E cathode - E anode
The difference between the two half reactions
A table of half cell potentials is based on the following conventions:
a) Each half cell is written as a reduction
b) The numerical values of cell potentials and half cell potentials depends on various
conditions, thus, the standard reduction potentials are used when ions and molecules
are in their standard state
c) Because potential differences not individual reduction potentials are measured, all of the
values in the table are relative and each half cell reduction potential is given relative to
the reduction potential of the standard hydrogen electrode, which has been assigned a
value of zero
Unit 5 test
● Corrosion***
● Banacling Methods (acidic and basic)
● Oxidation numbers
● Line Notation
● Galvanic cell
maxwell boltzmann distribution - communication
Intermolecular forces - communication
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