# atomic number

```Physical Properties
A trait we can observe w/o changing the identity of the substance.
EXAMPLES:
 Color
 Shape
 Dimensions (size)
 Texture
 Taste
 Temperature
 Melting Point
 Density
 Smell
 Freezing Point
 Mass
Chemical Properties
A trait we can observe by changing the identity of the
substance.
EXAMPLES:
 Flammability
 Decomposable
 Digestible
 Toxicity
 Ability to Oxidization
 Reactivity/Inert
Physical Change
A change that affects only physical properties and does not
change the identity of the substance.
EXAMPLES:
 Rip
 Crumble
 Wet
 Throw
 Draw
 Fold
 Kick
 Mix
Chemical Change
A change that alters the identity of the substance.
EXAMPLES:
 Eating/Digesting
 Burning
 React w/ acid or anything
 Decompose
 Oxidization
Elements, Compounds, and Mixtures
Element: substances that cannot be separated by chemical
or physical means.
Atom: smallest unit of an element.
Compound: substance made of 2 or more elements. It can
only be separated by chemical means.
Molecule: smallest unit of a compound.
Mixture: A combination of substances that are not
chemically combined. Can be separated by physical means.
Homogeneous: looks the same throughout.
Heterogeneous: composed of different parts.
Density
A ratio of mass to volume.
D= m/v
D=Density
V=Volume
M=Mass
g/ml=g/cm^3
milliliters=cm^3
grams
Example Problem: Density
A rock has a mass of 8g. When Bob puts it in a
graduated cylinder it displaces 4L of water. What is the
density?
D=M/V
D=8g/4L
D=2L
Scientific Notation
 A way of writing numbers that are too big or too
small to be conveniently written in decimal form.
EXAMPLES:
300 = 3×102
6,720,000,000 = 6.72×109
Sig Figs
 Multiplication and Division Rules:
same as the least number of sig figs in
measurements.
EX: find the area from these measurements.
505cm long = 3 sig figs 10cm wide = 1 sig fig
A=LxW
A= 505 x 10 = 5050cm2
A= 5,000cm2
the
your
Extra Practice
Find the rate of measurement.
250.0mL
0.25s
Find the rate of measurements.
250.0 mL = 3 sig figs
0.25s = 2 sig figs
250.0/0.25 = 1000mL/s  1.0 x 103
Sig Figs Continued
located in the same place value column as the
estimated number in the
measurement’s that is
farthest to the left.
EX:
101.0g
+ 10.5g
111.5g  112 g
Extra Practice
212.25C and 12.3C
212.25C
+ 12.3C
199.95C  200.0C
Accuracy/Precision
Accuracy/Precision Continued
 the accuracy of a measurement system is the degree
of closeness of measurements of a quantity to that
quantity's actual value. The precision of a
measurement system, also called reproducibility or
repeatability, is the degree to which repeated
measurements under unchanged conditions show
the same results.
Quantitative/Qualitative
 Quantitative: Is expressed or related to in terms of
quantity as in numbers.
 Qualitative: Related to or involving quality or kind.
Ionic Compound
 Occurs
between One Metal and Non-Metal
 Crystilanial Solids (Made of Ions)
 High Melting and Boiling points
 Conduct with electricity when melted
 No Definite shape
Covalent Compound
 Occurs
Between Two Metals
 Gases, Liquids, Soft Solids (Made of Molecules)
 Low Melting and Boiling points
 Poor electrical conductors
 Define Shape
Periodic Table
 To
the left of the red line are non-metal
 To the right of the red line are metals
Example
…is which type of bond?
1. NaBr
2. HCL
3. AgCL
4. C2S
1. Ionic Bond- Na is a non-metal, Br is a
Metal
2. Ionic Bond- H is a non-metal, Cl is a metal
3. Ionic Bond- Ag is a non-metal, Cl is a
metal
4. Covalent Bond- C is a metal, S is a metal
To Tell The Difference
 Look
at periodic table
 Look for the stair case in the periodic table
 Under elements B,SI,AS,TE,AT
Isotopes
 An
isotope is an atoms with the same number of
protons, but differing numbers of neutrons
 Isotopes are different forms of a single element
 Different Atomic Mass
How To Calculate Average Atomic
Mass
 1.
Convert the percentages into decimals. (This
percentage is known as its relative abundance or
percent abundance)
 2. Multiply the percentage of each isotope by its
respective mass
 3. Add the numbers from step two together
Example
mass number
exact weight
percent abundance
12
12.000000
98.90
13
13.003355
1.10
 This
is the solution for carbon:
(12.000000) (0.9890) + (13.003355) (0.0110) =
12.011 amu
Basics
 Find
your element on the Period Table
 element's atomic number and atomic weight.


atomic number is the number in the upper left corner
atomic weight is the number on the bottom
Protons and Electrons
 The
atomic number is the number of protons in
an atom of an element.

krypton's atomic number is 36, so an atom of
krypton has 36 protons
 Electrons
have the same quantity as protons
Neutrons
 Mass
Number = (Number of Protons) + (Number
of Neutrons)
 For krypton, this equation becomes:
 84 = (Number of Protons) + (Number of Neutrons)
 84 = 36 + (Number of Neutrons)
 NUETRONS = 48
Summary
 For
any element:
 Number of Protons = Atomic Number
 Number of Electrons = Number of Protons =
Atomic Number
 Number of Neutrons = Mass Number - Atomic
Number
Ions
 Ions
are formed when an atom gains or loses
electrons and gets charged
 If an atom gains electrons, number of electrons are
higher than protons

an atom gets negatively charged
 Similarly
if an atom loses electrons the total
number of protons become more than the number
of electrons

atom becomes positively charged
What is an ion?
 Ion


electrically charged atom or atom group:
an atom or group of atoms that has acquired an
electric charge by losing or gaining one or more
electrons
Knowing the charge
So…


Determine the charge of each element when present in an ionic
compound.
Use the table above to determine these charges. For example, O =
-2, Rb = +1.
Use the appropriate number of each ions such that:






For example, if magnesium (Mg) and bromine (Br) are mixed:
Metal (Mg) + Non-metal (Br) IS an ionic compound.
Mg ⇒ Mg+2 and Br ⇒ Br-1
For the final steps:



The sum of all charges adds up to zero.
The simplest ratio of ions is used.
One +2 ion is exactly balanced by two -1 ions.
2:1 is the simplest possible ratio.
Thus, the formula of the ionic compound formed is MgBr2
Alpha
 There
 Alpha
are three primary types of radiation:
- these are fast moving helium atoms. They
have high energy, but due to their large mass, they
are stopped by just a few inches of air, or a piece
of paper.
Beta
 These
are fast moving electrons. Since electrons
are lighter than helium atoms, they are able to
penetrate further, through several feet of air, or
several millimeters of plastic or less of very light
metals.
Gamma
 Gamma
- these are photons, just like light, except
of much higher energy, X-Rays and gamma rays
are really the same thing, the difference is how
they were produced. Depending on their energy,
they can be stopped by a thin piece of aluminum
foil, or they can penetrate several inches of lead
Review




A nuclear reaction can be described by an equation,
which must be balanced.
The symbol for an atom or atomic particle includes the
symbol of the element, the mass number, and the
atomic number.
The mass number, which describes the number of
protons and neutrons, is attached at the upper right of
the symbol.
The atomic number, which describes the number of
protons in the nucleus, is attached at the lower left of
the symbol
Electron Configurations
Diagram way/ Orbital notation
1.List all of the orbitals with boxes
2.Find out how many electrons are in the element
3. Electrons fill the lowest energy levels first (follow diagonal rule)
4.Reminder* Only 2 electrons fit into a degenerate orbital
Long way/List way
1.Write out the orbitals
2.Depending on the orbital write the amount of electrons in the exponent
3.Do this until your configuration is complete
4.Reminder * S orbital- up to 2 electrons, P orbital- up to 6,
D orbital- up to 10, F orbital- up to 14
Noble gas notation
1.Refer to noble gas before element
2.Write noble gas in brackets
3.Figure out next orbital and fill like usual
Sample Problem:
Be- Beryllium
Write the:
Orbital Notation
List way
Noble gas notation
Orbital Notation:
2s
1s
Electron Configuration:
1s^2 2s^2
Noble Gas: [He]2s^2
Wavelength, Energy & Frequency Calculations
Wavelength- distance from crest to crest of a wave, one full cycle of a wave-measured in meters
Frequency- how often a wave passes per second-- measured in 1/s per second
As wavelength gets smaller, wave frequency gets larger
Equation:
c= speed of light= 3.0x 10^8 m/s
Energy of a wave is related to its frequency:
E= energy in joules
h= plank's constant= 6.626 x 10^-34 Joules/second
Sample Problem:
Ex. A light at a stoplight has a wavelength of 5m
1. What is the frequency of light?
2. What is the energy of light?
1. wavelength= speed/frequency
5 m= 3.0 x 10^-8 m/s
___________
frequency
frequency= 3.08 x 10^-8/ 5
frequency= 6.16 x 10^-9 1/s
2. E= hv
h= 6.6 x 10^-34 J x S
E= (6.6 x 10^-34)(6.16 x 10^-9)
E= 4.07 x 10^-42 J
Trends
Electronegativity-chemical property describing
tendency of an atom to attract electrons to itself, which is
affected by both its atomic number and the distance that
its valence electrons reside from the charged nucleus.
Ionization Energy- describes the amount of energy
required to remove an electron from the atom or
molecule.
Atomic Radius- measure of the size of an element's
atoms.
Electronegativity, Ionization Energy & Atomic
Periodic Table Families and Periods
Periods- go across the table; there are 7
Groups- go vertically down the table; 8 are classified
Families- represent elements of the same nature ex. group 8 is
called the Noble Gas family because all of these elements have
something in common, they all have complete valence shells
Valence Electron trends on periodic
table
The Valence Electron trends show the amount of
valence electrons in each element. Each column
has a different number of valence electrons.
Transition
Metals
Ion Trends on the Periodic
Table
The Ion trends on the periodic table show the
charge of each element on the periodic table.
Each column has a different charge.
Ion Trends
1)
2)
Name the 1st element written in the formula
Use a prefix to tell how many we have
1) Mono
2) Di
3) Tri
4) Tetra
5) Penta
6) Hexa
7) Hecta
8) Octa
9) Nona
10) Deca
Naming Covalent Compounds
3) Name the 2nd element, change the
ending to -ide
4) Add a prefix to tell how many
◦
◦
◦
◦
◦
CO- Carbon Monoxide
CO₂- Carbon Dioxide
SO₃- Sulfur Trioxide
H₂O- Dihydrogen Oxide
S₆F₂- Hexasulfur Diflouride
Naming Covalent Compounds
Cont.
1)
2)
Name the cation (First element in
compound)
Is the cation a transition metal
1) If yes, use a roman numeral to tell the change
2) If no, move on
Name the anion
Change the ending to –ide unless it’s a
polyatomic ion
Ex. 1) Cu₂O₃Copper III Oxide
Ex. 2) Na₃PSodium Phosphide
3)
4)
Naming Ionic Compounds
Hydrogen ChloridePut anion root in
formula Hydro_ic acidHydrochloric Acid
 Hydrogen ChloratePut anion root into
formula ___ic acid Chloric Acid
 Hydrogen Chlorite Put anion root into
formula ___ous acid  Chlorous acid

Naming Acids
CuCl₂•4H₂O
 CuCl₂  Name normal
 4H₂O  Name the Prefix
 Copper II Chlorine • Tetra Hydrate

Naming Hydrated Crystals
5 types of chemical reactions
•
Synthesis: 2Cu+O2 2CuO
•
•
Decomposition: MgCl2Mg+Cl2
•
•
The single element is switched with the “ “ from the compound
Double replacement: MgCl2+CuOMgO+CuCl2
•
•
The Compound decomposes into two elements
Single replacement: MgCl2+CuCuCl2+Mg
•
•
The two elements combine
One element from each compound is switched
Combustion: CH3OH+O2CO2+H2O
•
Hydrocarbon + oxygen = Carbon dioxide+ water
Predicting Products
•
If there are two individual elements as the reactants, The reaction
will be synthesis.
– The product will be a compound.
•
If the reactant is a compound, the reaction will be decomposition.
– The product will be two separate elements.
•
If the reactants are a compound and a single element, the element
will switch with one of the elements in the compound. The reaction
will be single replacement.
–
•
The product will be a new element and a new compound
If the reactants are two compounds (respectively) an element from
each compound will switch. The reaction will be double
replacement.
– The product will be two new compounds.
•
If the reactants are a hydrocarbon and an oxygen, the reaction will
be combustion.
– The product will be carbon dioxide and water.
Law of Conservation of Matter
• The law of conservation of matter states
that matter cannot be created or
destroyed, only transferred.
– This is why we must balance equations.
Balancing Equations
• When balancing equations, you must
have the same amount of each element
on either side of the reaction.
– __Mg+__O2__MgO
• There is too much O and too little Mg and MgO,
so this equation must be balanced to follow the
law of conservation of matter
– 2 Mg+O22MgO
Balancing Equations
• There are 7 diatonic elements.
• When listed alone, these elements
always occur with the subscript of 2
• The elements are Br2, I2, N2, Cl2, H2, O2
and F2.
– These can be remembered using the word
BrINClHOF.
Net Ionic Equations
• First start by writing an ionic equation
– This includes aqueous substances that
dissociate into ions
– Zn(s)+CuSO4(aq)ZnSO4(aq)+Cu(s)
• The aqueous substances dissociate
– Zn(s)+Cu+2 (aq) +SO4-2(aq)Zn=2 (aq) +SO42(aq)+Cu(s)
• (aq)=aqueous, (s)=solid, (l )=liquid
Net Ionic Equations
• Eliminate the substances that are the
same on either side of the reaction (the
spectator ions)
– Zn(s)+Cu+2(aq) +SO4-2(aq)Zn+2(aq) +SO42(aq)+Cu(s)
– Zn(s)+C+2(aq)Zn(s)+Cu(s)
• Above is the net ionic equation since the
spectator ions have been removed
Factor Label Method
• The factor-label method was made to keep
track of units in conversion problems. In the
method, conversion factors are set up in
fraction form. To get the unit you want, you
put it on the top of the fraction next to it. You
put the unit on the bottom that you are trying
to get rid of.
Factor Label Method Formulas
Factor Label Method Examples
Molar Mass
Molar Mass Example
Empirical Formula
• An empirical formula is the lowest term ratio
of elements
• Its not always the true formula
• When given the % composition, you can find
the empirical formula
•
•
•
•
Step 1: % to mass
Step 2: Mass to Moles
Step 3: Divide by small
Step 4: Multiply till whole
Empirical Formula Examples
Molecular Formulas
• The ‘true’ formulas
• Not in lowest terms
1.
2.
3.
4.
Get atomic mass
Multiply by number of subscript
Divide by molecular mass
Molecular Formula Example
Percent Composition
Mole Ratios
• The ratio between the amounts in moles of any two
compounds involved in a chemical reaction.
• Used as conversion factors between products and reactants in
many chemistry problems.
• Example:
2 H2(g) + O2(g) → 2 H2O(g)
The mole ratio between O2 and H2O is 1:2. For every 1 mole
of O2 used, 2 moles of H2O are formed.
The mole ratio between H2 and H2O is 1:1. For every two
moles of H2 used, 2 moles of H2O is formed.
Stoichiometry
•
•
•
•
The part of chemistry that studies amounts of substances that are
involved in reactions
Helps you figure out how much of a compound you will need, or how
much you started with.
To solve:
1.Balance the equation.
2.Convert units of a given substance to moles.
3.Using the mole ratio, calculate the moles of substance yielded by
the reaction.
4.Convert moles of wanted substance to desired units.
Ex:
2Mg + O2
2MgO
If you have 1.2 moles of Mg, how many moles of O2 will you get?
1.2 mol Mg x O2/2Mg = .6 mol O2
Limiting and Excess
Reactants
• If you have more than needed, you have excess.
• If you need more than you have, you have limiting.
• Example:
2Al2O3+3C
3CO2+4Al
If 2.3 g AlO3 and 3.2 g of C mix, which is the limiting and
which is the excess?
2.3 g AlO3 x 1 mol/101.96 g = .02 mol Al2O3 (have)
3.2 g C x 1 mol/12.01 g = .27 mol C
.02 x 3C/2Al2O3 = .03 mol C (need)
• Excess : C
• Limiting: Al2O3
Theoretical Yield
•
•
•
The quantity of a product obtained from the complete
conversion of the limiting reactant in a chemical reaction.
Commonly expressed in grams or moles
To calculate:
1. Use the molar mass of your reactant to convert grams of
2. Use the mole ratio between reactant and product to
convert moles reactant to moles product.
3. Use the molar mass of the product to convert moles
product to grams of product.
Theoretical yield (cont.)
• Na2S(aq) + 2 AgNO3(aq) → Ag2S(s) + 2 NaNO3(aq)
• How many grams of Ag2S will form when 3.94 g of AgNO3
and an excess of Na2S are reacted together?
Atomic
Atomic
Atomic
Atomic
Atomic
Atomic
Atomic
Atomic
Atomic
Atomic
weight
weight
weight
weight
weight
weight
weight
weight
weight
weight
of
of
of
of
of
of
of
of
of
of
Ag = 107.87 g
N = 14 g
O = 16 g
S = 32.01 g
AgNO3 = (107.87 g) + (14.01 g) + 3(16.00 g)
AgNO3 = 107.87 g + 14.01 g + 48.00 g
AgNO3 = 169.88 g
Ag2S = 2(107.87 g) + 32.01 g
Ag2S = 215.74 g + 32.01 g
Ag2S = 247.75 g
• Two moles of AgNO3 is needed to produce one mole of Ag2S.
Theoretical yield (cont.)
• The excess of Na2S means all of the 3.94 g of AgNO3 will be
used to complete the reaction.
grams Ag2S = 3.94 g AgNO3 x 1 mol AgNO3/169.88 g AgNO3
x 1 mol Ag2S/2 mol AgNO3 x 247.75 g Ag2S/1 mol Ag2S
• 2.87 g of Ag2S will be produced from 3.94 g of AgNO3.
Percent yield
• To find percent yield:
1. Balance the chemical equation
2. Find the limiting reagent
3. Find the theoretical yield
4. Find the actual yield
5. Find the percentage yield
• Percent yield = mass of actual yield/mass of theoretical yield x
100
KINETIC MOLECULAR THEORY
(COLLISION THEORY)
Gas particles exhibit constant random motion
 Collisions happen
 All collisions are perfectly elastic (can bounce off each
other)
What Collisions Mean in the Real World Collisions between molecules of gas are what creates air
pressure
 The more collisions that occur, the more pressure is excited
 All gas particles move in straight lines, but they bounce off
anything they collide with
 As the temperature increases the particles move faster, and
more collisions occur, that increases the pressure
 As you increases the number of particles, the number of
collisions increases, which increases the pressure

ABSOLUTE ZERO AND STP

Absolute ZeroAbsolute zero is the point where no more heat can be
removed from a system, according to the absolute or
thermodynamic temperature scale.
 0 K or -273.15°C.


STP0˚C
 1 atm
 273.15 K
 101,325 Pa

PRESSURE CONVERSIONS

Pressure is Usually Measured in:






Conversions:






Atmosphere (atm)
Bar (Bar)
Pascals (Pa)
Millimeters of Mercury (mmHg)
Torr (torr)
1 atm=101325 Pa
1 bar= 100025 Pa
1 torr= 133.32 Pa
1mmHg= 133.32 Pa
1mmHg= 1 torr
Practice:

14.7KPa= _______ atm

14.7KPa*1000Pa/1KPa=14700Pa
BOYLE’S GAS LAW
A relationship between pressure and volume
 Inverse relationship:

When pressure goes up, volume goes down.
 When volume goes up, pressure goes down
 PV=P2V2


Practice problem:
A balloon with a volume of 2.0 L is filled with a gas at 3
atmospheres. If the pressure is reduced to 0.5 atmospheres
without a change in temperature, what would be the volume of
the balloon?
 V1 = 2.0 L
P1 = 3 atm
P2 = 0.5 atm
V2 = (2.0 L)(3 atm)/(0.5 atm)

CHARLES’S GAS LAW
A relationship between volume and temperature
 When temperature goes up, volume goes up
 When temperature goes down, volume goes down
 Direct relationship
 V/T=V2/T2
 Practice problem:

Given 300.0 mL of a gas at 17.0 °C. What is its
volume at 10.0 °C?
 (300.0) / (290.0 K) = (x) / (283.0 K)
 292.76mL

GAY-LUSSAC GAS LAW





Relates pressure to temperature
When temperature goes up, pressure goes up
When temperature goes down, pressure goes down
P/T=P2/T2
Practice problem:






Find the final pressure of gas at 150 K, if the pressure of
gas is 210 kPa at 120 K.
P1= 210 kPa
T1= 120 K
P2= ?
T2= 150 K
P2= (210 kPa) (150 K) / 120 K
P2= 262.5 kPa
COMBINED GAS LAW
Combines Boyles, Charles and Gay-Lussac’s gas
laws into one equation.
 PV/T=P2V2/T2
 Practice Problem:

Oxygen occupies a fixed container of 5.5L at STP.
What will happen to the pressure if
the temperature rises to 300K?
 P2 = (5.5 L) (101.3 Kpa) (300 K) = 111.3 KPa
(5.5 L) (273 K)

IDEAL GAS LAW




Ideal Gases DO NOT EXIST in the real world
But we pretend real gases act like ideal gases so we can predict what
will happen
Properties
 Infintetly small (no volume)
 No attraction to other particles (No intermolecular force)
 Don’t condense to liquids
PV=nRT


P is the pressure of the gas (in atmospheres, ATM) V is the volume of
the container (in liters, L)
n is the number of moles of gas in the container (in moles, mol)
R is Universal Gas Constant (which is 0.0820574587 L · ATM · K-1 · mol1)
T is the temperature of the gas (in Kelvin)
Practice:



6.2 liters of an ideal gas are contained at 3.0 atm and 37 °C. How many
moles of this gas are present?
n = PV / RT
n = ( 3.0 atm x 6.2 L ) / ( 0.08 L atm /mol K x 310 K)
n = 0.75 mol
DALTON’S GAS LAW
At a particular temperature, the total pressure of
a mixture of two or more non-interacting (ideal)
gases is equal to the sum of partial pressures of
the individual gases
 P1+P2+P3…
 Practice:






The pressure of a mixture of nitrogen, carbon dioxide,
and oxygen is 150 kPa. What is the partial pressure
of oxygen if the partial pressures of the nitrogen and
carbon dioxide are 100 kPA and 24 kPa, respectively?
P = Pnitrogen + Pcarbon dioxide + Poxygen
150 kPa = 100 kPa + 24 kPa + Poxygen
Poxygen = 150 kPa - 100 kPa - 24 kPa
Poxygen = 26 kPa
GRAHAM’S GAS LAW

States of Matter
Parts of Solutions
 Solution- a homogeneous mixture (same
throughout)
 Solute- part of solution that gets dissolved
 Solvent- part that surrounds the solute to make it
dissolve
Phase
Diagram
Shows what
temperature and
pressure combinations
result in each state of
matter for a particular
chemical
Critical point- the last
pressure and
temperature
combination where
liquids and gases exist
separately
Heating Curve
Shows the temperature
at which the state
changes occur and
describes how the
substances uses heat
 Opposite phases of the heating
Cooling Curve
curve diagram, work backwards
and they decrease
 Gas-condensation-liquidfreezing-solid
 Photos were not working, see
Solubility
Curve
The relationship
between temperature
and solubility
notes(class handout), know how
to interpret
 M= m/V
Molarity
Calculations
 Molarity= moles/liters
 Use unit (M)
 Example: You make a solution




using 47.9 g of HCl. The volume
of the solution is 753 mL. What is
the Molarity of the solution?
47.9g*1 mole/36.46g=1.31mol
753mL*1L/1000mL=.75L
M= 1.31mol/.75L
M=1.75M
 Dipole-dipole forces: the
Intermolecular
Forces
attraction found between two
polar molecules
 Vanderwaals forces- attraction
between two nonpolar molecules
 Hydrogen bonding- a strong force
of attraction, not considered a
bond. Requires a very
electronegative element (Oxygen,
Nitrogen, Fluorine) bonded to a
Hydrogen
 Cohesive forces- attracted to
Intermolecular
Forces
other molecules of the same type


Surface tension
Pepper flakes float in a glass of water
until a spoon is inserted into the cup
and they sink
attracted to other molecules of a
different type


Capillary action
Pulls the liquid up the side of a tube
creating a meniscus
Specific heat calculations
The specific heat is the amount of heat per unit mass
required to raise the temperature by one degree
Celsius
• Delta, Δ or D= change of a variable.
• M=Mass of the sample
• Q= amount of heat that must be added.
• T= temperature of the substance.
• C= specific heat
Formulas:
c = Q/mΔt
ΔQ = mcΔt
Properties of Acids and Bases
Acids
•
•
•
•
Sour
Burns/Stings
Reacts with metal
Electrolyte (conducts
electricity)
• Corrosive
Examples: citric acid, vinegar
Bases
•
•
•
•
Bitter
Slippery
Non-reactive with metals
Electrolyte (conducts
electricity)
• Corrosive
Examples: ammonia, baking
soda
pH/OH Calculations
•
•
•
•
•
pH=-log[H+]
[H+]=10-pH
pOH=-log[OH-]
[OH-]=10-pOH
pH+pOH=14
```

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