The Basics
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Transcript of The Basics
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The Basics
• Elements, Molecules, Compounds, Ions• Parts of the Periodic Table• How to Name
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Classification of Elements• Metals – found on the left-side of the
Periodic Table•Metalloids (or semi-metals) – along the stair-step line
Properties are intermediate between metals and nonmetals
•Nonmetals – found on the right-hand side of the Periodic Table
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Elements , Molecules, Compounds, and Ions• Element – one single type of atoms
– Al Cu He – Naturally occurring elements that are Diatomic are still
elements – N2 O2 F2 Cl2 Br2 I2 H2
– How many elements are in Mn(SO4)2 ? 3– How many Atoms? 9
• Molecule - smallest electrically neutral unit of a substance that still has the properties of that substance, 2 or more different elements
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• Compounds – groups of atoms– Ionic and Molecular
• Molecular Compounds – share electrons typically 2 or more non-metals (hydrocarbons)
• Example H2S CO2 C5H10 • Ionic Compound (salts) – transfer electrons
typically metal and non metal (watch for poly atomic ions)
• FeS Mg(OH)2 (NH4)3PO4
Elements , Molecules, Compounds, and Ions
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Ions• Ions have either lost or gained electrons• Typically Metals lose electrons to become
positive• Example cations• Mono-valent Mg -> Mg2+ + 2e-
– Group 1A = Metal1+, 2A =Metal2+ and 3A = metal3+
– Multi-valent Fe --> Fe2+ + 2e- and Fe --> Fe3+ + 3e-
• Non-metals gain electrons - anion• S + 2e- -> S2-
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Poly Atomic Ions
• A molecular compound with a charge
• NH4+
• CO32-
• SO42-
• NO3-
• OH-
• H3O+
• Ammonium• Carbonate• Sulfate• Nitrate• Hydroxide• Hydronium
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Acids and Bases From Ions H+ or OH-
• Acids look for hydrogen up front (HA) or as COOH
• Example HF H3PO4 C4H6COOH• Strong Acids• HCl, H2SO4, HBr, HI, HClO3,HNO3
• Base look for hydroxide or NH group
• Example KOH C4H6NH2
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Naming Compounds1. Ionic or Covalent
2. Ionic – two ions or Poly atomic ions
Covalent 2 non-metals
Or a hydrocarbon
Type of Metal
Mono-valent metals groups 1A 2A 3A
Name the MetalName the Nonmetal + ide(if PAI use its name)
CaF2 calcium fluorideRbNO3 rubidium nitrateAl(OH)3 aluminum hydroxide
Multi-valent MetalTransitions metals and under the stairs
Find the Charge on the MetalTo make the compound neutralWrite the Charge with roman numeralName nonmetal + ide (if PAI use its name)
Ni+Cl- nickel(I) chloridePb2+SO4
2- lead(II) sulfatePb4+(SO4)2
2- lead(IV) sulfate
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Ionic naming
• Name to Formula– Final compound must be neutral based on subscripts and charges
• Magnesium Fluoride Mg2+ + F- MgF2
• Ammonium Sulfide NH4+ + S2- (NH4)2S
• Tin(II) Carbonate Sn2+ + CO32- SnCO3
• Iron(III) Oxide Fe3+ + O2- Fe2O3
• Iron(II) Oxide Fe2+ + O2- FeO
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Covalent 2 non-metals
Or a hydrocarbon
Ionic or Covalent
Ionic – two ions or Poly atomic ions
Use the prefix to tell how many of each atom there areMono is never used with the first element
ExamplePBr3 Phosphorous tribromideCCL4 Carbon tetrachlorideP2O5 diphosphorous pentoxideCO carbon monoxide
Hydrogen up frontMost likely an Acid you should have memorized
HCl - Hydrochloric acidHI - Hydroiodic acidHBr - Hydrobromic acidHNO3 Nitric AcidH2SO4 - Sulrufic acidHClO - Hypochlorous acid
HydrocarbonsLook for how many carbons
One – methane CH4
Two – ethane C2H6
Three – propane C3H8
Four – butane C4H10
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Molecular Compounds
• Name to formula – charge does not matter this time, just use the prefixes or memorize
• Tetraarsenic hexoxide As4O6
• Sulfur hexafluroide SF6
• Butane C4H10
• Nitric acid HNO3
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Lewis Dot Structures• Rules1. Fewest number of atoms goes in the middle or
C if present2. Connect remaining elements with single bonds3. Make sure all elements have 8 electrons (H only 2)
4. Count the number of electrons in structure5. Add up valence electrons from PT
– Too many e- in structure: remove 2 adjacent pairs fill in with one bond
– Too few e- in structure: add to central atom
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EXAMPLES:
• CH4
• 1. (1) C + (4) H (1)(4e-) + (4)(1e-) = 8e-
• 2. Spatial order
• 3. Draw bonds
• 4. Octet rule satisfied?• 5. # of e- match?
CH
HHH
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EXAMPLES:
• CO2
• 1. (1) C + (2) O (1)(4e-) + (2)(6e-) = 16e-
• 2. Spatial order
• 3. Draw Bonds
• 4. Octet rule satisfied? • 5. # of e- match?
C OO
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EXAMPLES:• NH3
• 1. (1) N + (3)H(1)(5e-) + (3)(1e-) = 8e-
• 2. Spatial order
• 3. Draw bonds
• 4. Octet rule satisfied?• 5. # of e- match?
NH
HH
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EXAMPLES:• CCl4
• 1. (1) C + (4) Cl(1)(4e-) + (4)(7e-) = 32e-
• 2. Spatial Order
• 3. Draw bonds
• 4. Octet rule satisfied?• 5. # of e- match?
CCl
ClClCl
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EXAMPLES:
• NH4+
• 1. (1) N + (4) H - (1)(+) (1)(5e-)+(4)(1e-) - (1)(1e-) = 8e-
• 2. Spatial order
• 3. Draw bonds
• 4. Octet rule satisfied?• 5. # of e- match?
NH
HHH[ ]+
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EXAMPLES:• SO4
2-
• 1. (1) S + (4) O + (2)(-) (1)(6e-)+ (4)(6e-) + (2)(1e-) = 32e-
• 2. Spatial Order
• 3. Draw bonds
• 4. Octet rule satisfied?• 5. # of e- match?
SO
OOO[ ]2-
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EXAMPLES:
• CN-
• 1. (1) C +(1) N + (1)(-)(1)(4e-) + (1)(5e-)+ (1)(1e-) = 10e-
• 2. Spatial order
• 3. Draw Bonds
• 4. Octet rule satisfied? • 5. # of e- match?
C N[ ]-
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EXAMPLES:• CO3
2-
• 1. (1) C + (3) O + (2)(-) (1)(4e-)+ (3)(6e-) + (2)(1e-) = 24e-
• 2. Spatial Order
• 3. Draw bonds
• 4. Octet rule satisfied?• 5. # of e- match?
CO
OO[ ]2-
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VSEPR:• Regions of electron density (where pairs of electrons are
found) can be used to determine the shape of the molecule.
• CO2
• Here there are two regions of electron density.• The regions want to be as far apart as possible, so it is
linear.
C OO
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EXAMPLES:
• CH4
• There are four electron pairs.• You would expect that the bond angles would be 90°
but…• Because the molecule is three-dimensional, the
angles are 109.5°.• The molecule is of tetrahedral arrangement.
CH
HHH 1
23
4
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EXAMPLES:
• NH3
• Four regions of electron density• But one of the electron pairs is a lone pair• The shape is called trigonal pyramidal
NH
HH
1
23
4
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EXAMPLES:
• H2O• Four regions of electron density• But two are lone pairs • This structure is referred to as bent
O HH
1
23
4
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EXAMPLES:
• CO32-
• Three regions of electron density• This structure is referred to as trigonal planar
[ ]CO
OO
1
2
32-
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Practice determining molecular shape:
• H2S– 4 regions of e- density– 2 lone pairs– bent
SHHS HH
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Practice determining molecular shape:
• SO2
– 3 areas of e- density– 1 lone pair– bent
SOO
S OO
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Practice determining molecular shape:
• CCl4
– 4 areas of e- density– tetrahedral
CCl
ClClCl 3d
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Practice determining molecular shape:
• BF3
– 3 areas of e- density – trigonal planar
BFF
FB FF
F
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Practice determining molecular shape:
• NF3
– 4 areas of e- density – 1 lone pair– pyramidal
NF
FF 3d
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16.3 Polar Bonds and Molecules
• In covalent bonds, the sharing of electrons can be equal
• or it can be unequal.
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Nonpolar Covalent Bonds
• Nonpolar covalent bond - This is a covalent bond in which the electrons are shared equally.
• EXAMPLES:• H2
• Br2
• O2
• N2
• Cl2
• I2
• F2
H HBr BrO ON NCl ClI IF F
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Polar Bonds and Molecules
• If the sharing is unequal, the bond is referred to as a dipole.
• A dipole has 2 separated, equal but opposite charges.
• “∂” means partial
+_
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Polar Bonds and Molecules
• Polar covalent bond - a covalent bond that has a dipole
• It usually occurs when 2 different elements form a covalent bond.
• EXAMPLE:
• H + Cl H Cl
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• Electronegativity - This is the measure of the attraction an atom has for a shared pair of electrons in a bond.
• Electronegativity values increase across a period and up a group.
Polar Bonds and Molecules
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Examples:
• Identify the type of bond for each of the following compounds:
• HBrBr = 2.8H = 2.1
0.7 Polar Covalent.1 < < 1.9
H Br
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Examples:
• NaFF = 4.0Na = 0.9
3.1• N2
N = 3.0N = 3.0
0.0
Ionic
Non-PolarCovalent
> 2
Na F
N N
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Molecular Polarity• If there is only one bond in the molecule, the bond type
and polarity will be the same.• If the molecule consists of more than 2 atoms, you
must consider the shape. To determine its polarity, consider the following:– Lone pairs on central atom
• If so… it is polar– Spatial arrangement of atoms
• Do bonds cancel each other out (symmetrical)? – If so… nonpolar
• Do all bonds around the central element have the same difference of electronegativity?
– If so… nonpolar
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Polar Molecules
• If the molecule is symmetrical it will be nonpolar.– Exception hydrocarbons are nonpolar
• If the molecule is not symmetrical itwill be polar, with a different atom or with lone pair(s)
CCl
ClClBr
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Attractions Between Molecules
• Van der Waals forces – the weakest of the intermolecular forces. These include London dispersion and dipole-dipole forces.– London dispersion forces – between nonpolar
molecules and is caused by movement of electrons
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Attractions Between Molecules
• van der Waals forces(cont.)-– Dipole interactions – between polar molecules
and is caused by a difference in electronegativity.
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Attractions Between Molecules• Hydrogen bonds – attractive forces in which hydrogen,
covalently bonded to a very electronegative atom (N, O, or F) is also weakly bonded to an unshared (lone) pair of electrons on another electronegative atom.
OHH
OHH
OHH
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• Ionic Bonding-occurs between metals and nonmetals when electron are transferred from one atom to another.
• These bonds are very strong.
Attractions Between Molecules
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Summary of the Strengths of Attractive Forces
Ionic bonds hydrogen bonds dipole-dipole attractions LDF
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Writing and Balancing Chemical Equations
Example:
Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water.
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Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water.
1. Write the formulas of all reactants to the left of the arrow and all products to the right of the arrow.
Sodium + water
Translate the equation and be sure the formulas are correct.
Na + H2O NaOH + H2
sodium hydroxide + hydrogen
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Write the equation for the formation of sodium hydroxide and hydrogen, from the reaction of sodium with water.
2. Once the formulas are correctly written, DO NOT change them. Use coefficients (numbers in front of the formulas), to balance the equation. DO NOT CHANGE THE SUBSCRIPTS!
_____Na + _____H2O ____NaOH + _____H2
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3. Begin balancing with an element that occurs only once on each side of the arrow.
Ex: Na
_____Na + _____H2O ____NaOH + _____H2
Na
H
O
Na
H
O
222
4
2
2
2
4
2
When you are finished, you should have equal numbers of each element on either side of the equation
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4. To determine the number of atoms of a given element in one term of the equation, multiply the coefficient by the subscript of the element.
Ex: In the previous equation (below), how many hydrogen atoms are there?
4
____Na + _____H2O ____NaOH + _____H22 2 2
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• Balance elements one at a time.• Balance polyatomic ions that appear on both
sides of the equation as single units. (Ex: Count sulfate ions, not sulfur and oxygen separately)
• Balance H and O last. Save the one that is in the most places for last…
• Use Pencil!
(NH4)2SO4 (aq) + BaCl2 (aq) BaSO4 (s) + 2NH4Cl (aq)
Helpful Hints:
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Practice:• Balance the equation for the formation of
magnesium nitride from its elements.
____Mg + ____N2
Mg2+ N3-
Mg3N2
3 ____Mg3N2
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Ex: NH3 + O2 NO2 + H2O
• H can be balanced by placing a 2 in front of NH3 and a 3 in front of H2O. Then put a 2 in front of NO2 for nitrogen to balance.
_____NH3 + _____O2 ____NO2 + ____H2O2 32
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• Now all that is left to balance is the oxygen. There are 2 O on the reactant side and 7 on the product side. Our only source of oxygen is the O2. Any whole number we place in front of the O2 will result in an even number of atoms. The only way to balance the equation is to use a coefficient of 7/2.
_____NH3 + _____O2 ____NO2 + ____H2O2 327/2
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Stoichiometry – study of calculations of quantities in
chemical reactions using balanced chemical equations.
2Mg + O2 2MgO
2 moles Mg + 1mole O2 2 moles MgO
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• The mole ratios can be obtained from the coefficients in the balanced chemical equation.
• What are the mole ratios in this problem?• Mole ratios can be used as conversion factors to
predict the amount of any reactant or product involved in a reaction if the amount of another reactant and/or product is known.
2Mg + O2 2MgO
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What’s that mean?Well, a stoichiometry problem gives you an amount of one chemical and asks you to solve for a different chemical.
To get from one type of chemical to another, a MOLE RATIO must be
found between the two chemicals. You get the MOLE RATIO from the BALANCED CHEMICAL EQUATION!
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A balanced chemical equation tells the quantity of reactants and products as well as what they are.
2 mol 1 mol 2 mol
*the coefficients are*
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The MOLE RATIO is your mechanism of transition between the chemical that is your starting given
and the chemical you are solving for.
The MOLE RATIO is the bridge between the two different
chemicals
given
moles moles
? want(given) (want)
MoleRatio
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EXAMPLE• How many grams of KClO3 must decompose to
produce KCl and 1.45 moles O2?
2KClO3 → 2KCl + 3O2
1.45 moles O2
3 mol O2
2 mol KClO3
1 mol KClO3
122.6 g KClO3 =
119 g KClO3
GFM
mol-mol ratio
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• CaCO3, limestone, is heated to produce calcium oxide, CaO, and CO2. What mass of limestone is required to produce 156.0 g of CaO?
156.0 g CaO56.1 g CaO
GFM
1 mol CaO1 mol CaO
CaCO3 (s) CaO (s) + CO2 (g)
mol-mol ratio
1 mol CaCO3
1 mol CaCO3
GFM
100.1 g CaCO3
278.4 g CaCo3=
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EXAMPLE
• Calculate the number of joules released by the oxidation of 5.00 moles of Na completely react with oxygen gas. ΔH = -416 kJ/mol
4Na + O2 2Na2O
5.00 mol Na4 mol Na2 mol Na2O
mol-mol ratio
1 mol Na2O-416 kJ
1 kJ
1000 J
enthalpy 1.04x106 J
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Atomic orbital – the region in space where the electron is likely to be
found
A quantum mechanical model of a hydrogen atom, which has one electron, in its state of lowest energy. The varying density of the spots indicates the relative likelihood of finding the electron in any particular region.
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Electrons can be described by a series of 4 quantum numbers.
You must be familiar with all of these!
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1. Principle quantum number (n)
-describes the principal energy level an electron occupies
-values of 1,2,3,4,etc
1234567
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2. Azimuthal quantum number (l)-describes the shape of atomic orbitals-s orbitals are spherical, p orbitals are peanut shaped, d orbitals are daisies and f orbitals are fancy-designates a sublevel-values of 0 up to and including n-1
0=s, 1=p, 2=d, 3=f
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Spheres
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67Peanuts
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Daisies or Doughnuts
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Fancy
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Where do I find the orbital shapes?
SP
D
F
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3. magnetic quantum number (ml)-Designates the spatial orientation of an atomic orbital in space-values of -l to +l so s has 1 orbital
p has 3 orbitals (x, y, and z)d had 5 orbitals (xy, xz, yz, x2-y2 and z2) f has 7 orbitals
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4. spin quantum number -describes the orientation of the individual electrons; values of +1/2 and -1/2-each orbital can hold 2 electrons with opposite spins
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Symbol Shape Orbitals Electrons
s sphere 1 2
p peanut 3 6
d Daisy/ donut 5 10
f fancy 7 14
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Quick and Easy Electron Configuration
SP
D
F
1234567
3456
45
Now, lets try to do an electron configuration for carbon.Begin with the first quantum number and use the periodic table to write the configuration.
Now, try the three in your notes.He 1s2
Na 1s2 2s2 2p6 3s1
Ti 1s2 2s2 2p6 3s2 3p6 4s2 3d2
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For a shorter way to write electron configuration, write the nearest noble gas and then continue. AKA “Shorthand Notation”.
Ex: Ti can be written as
OR1s2 2s2 2p6 3s2 3p6 4s2 3d2 [Ar] 4s2 3d2
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Electron Configuration of Ions
• For the loss of an electron remove electrons from the last orbital
Ge 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2
Ge2+ 1s2 2s2 2p6 3s2 3p6 4s2 3d10
For the gain of electrons add them in the last orbital filledAt [Xe] 6s2 4f14 5d10 6p5
At- [Xe] 6s2 4f14 5d10 6p6
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Atomic Radius
Increasing Atomic Radius
Incr
easi
ng A
tom
ic R
adiu
s
BIGGEST
SMALLEST
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• Ion Size• Anions are larger than the atoms from which they
were formed.• The negative charge means more electrons are
present causing the size of the ion to be larger.• Cations are smaller than the atoms from which they
were formed.• The positive charge means fewer electrons are
surrounding the nucleus, thus pulling the existing electrons closer and causing the ion to be smaller.
• Trend: ionic radius increases from right to left and top to bottom
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Ionic Radius
Increasing Ionic Radius
Incr
easi
ng Io
nic
Rad
ius
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• Ionization energy - energy required to overcome the attraction of the nuclear charge and remove an electron from a gaseous atom
• 1st ionization energy: the energy required to remove the first electron
• 2nd ionization energy: the energy required to remove the second electron
• 3rd ionization energy: the energy required removing the third electron
• Trend: ionization energy increases from left to right and bottom to top
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81Ionization Energy
Increasing Ionization Energy
Incr
easi
ng Io
niza
tion
Ener
gy
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• Electronegativity – the tendency for the atoms of the element to attract electrons when they are chemically combined with atoms of another element
• Note: Noble gases don’t have values for electronegativity because their outer orbitals are full and they do not need to gain or lose electrons to be stable.
• Trend: electronegativity increases from left to right and bottom to top
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83Electronegativity
Increasing Electronegativity
Incr
easi
ng E
lect
rone
gativ
ityHIGHEST
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We can measure not only the length of each wave of light, but also its frequency of occurrence.
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Parts of a wave:
Amplitude
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amplitude- height of the wave from the origin to the crestwavelength - - distance between the crestsfrequency - - the number of wave cycles to pass a given point per unit of time.The units of frequency are 1/s, s-1, or Hertz (Hz)
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= c/ where c = speed of light
c= 3.00 x 108 m/s
As increases, decreases.
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Ex. A certain wavelength of yellow light has a frequency of 2.73 x 1016s-1. Calculate its wavelength.
= c/= c/ = 3.00 x 108m/s 2.73 x 1016s-1
= 1.10 x 10-8 m
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Spectrum- series of colors produced when sunlight is separated by a diffraction gradient.
ROY G. BIVRed: has the longest wavelength, lowest frequency Lowest energyViolet: has the shortest wavelength, highest frequency highest energy
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Kinetic theory- The tiny particles in all forms of matter are in constant motion.
1. A gas is composed of particles, usually molecules or atoms. We treat them as, Hard spheres, Insignificant volume, and Far from each other
2. The particles in a gas move rapidly in constant random motion.
3. All collisions are perfectly elastic.
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Gas Laws
• One single set of conditions• PV = nRT• Two Sets of conditions• P1 V1 = P2 V2 T1 T2
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V1 = 3.40L V2 = ?P1 = 120.0 kPa P2 = 1 atm = 101.3 kPaT1 = 25.0oC + 273 = 298KT2 = 0oC + 273 = 273KP1V1 = P2V2 (120.0 kPa)(3.40L) = (101.3 kPa)V2
T1 T2 298K 273KV2 = 3.69L
Ex. If a helium-filled balloon has a volume of 3.40 L at 25.0oC and 120.0 kPa, what is its
volume at STP = 0°C, 1 ATM?
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P=? V= 5.0L R = 0.0821 (L.atm/mol.K)T = 22°C +273= 295Kn = 0.60g O2 1 mol O2 = 0.01875 mol 32.0g O2
PV = nRT P(5.0L) = (0.01875mol)(0.0821Latm/molK)(295K) P = 0.091 atm
Ex. A 5.0 L flask contains 0.60 g O2 at a temperature of 22oC. What is the pressure (in atm) inside the flask?
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Factors affecting solubility:
• The nature of the solute and the solvent– “like dissolves like”
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• Miscible - liquids that are soluble in each other– Ex. ethanol and water
• Immiscible- liquids that are not soluble in each other– Ex. oil and water
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Molarity(M) = moles of solute liters of solution
• A 1 M solution contains 1 mole of solute per 1 L of solution. A 0.5 M NaCl solution has 0.5 mol NaCl in 1 L total solution.
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EXAMPLES:• What is the concentration in molarity of a
solution made with 1.25 mol NaOH in 4.0 L of solution? # mol
# L
#mol = 1.25 molV = 4.0 L
M = ?
M = 1.25 mol 4.0 L
= 0.3125 M = 0.31 M
M =
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Exothermic reactions have - H
Endothermic reactions have + H
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Enthalpy (H)- the amount of heat in a system at a given temperature
Enthalpy change:
H = q = m C T
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Ex. A 25 g sample of a metal at 75.0oC is placed in a calorimeter containing 25 g of H2O at 20.0oC. The temperature stopped changing at 29.4oC. What is the specific heat of the metal?
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Standard Heat of Formation of a compound( Hf
o)
* Hfo of a free element in its standard state is
zero. H = Hf
o products - Hf
oreactants
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Ex. 1. Calculate H for the following reaction:(endo or exo thermic?)CaCO3(s) CaO(s) + CO2(g)
∆Hfo values:
CaCO3 = -1207.0 kJ/molCaO = -635.1 kJ/molCO2 = -393.5 kJ/mol
∆H = [-635.1 + (-393.5)] – [-1207.0] ∆H = 178.4 kJ endothermic, heat going in
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2. Calculate the heat of reaction for the following reaction: (endo or exo thermic?) 2H2(g) + O2(g) 2H2O(g)
∆Hfo values:
H2O(g) = -241.8 kJ/mol
∆H = [2(-241.8)] – [2(0) + 0] ∆H = -483.6 kJExothermicHeat leaving
Remember to multiply heat values by coefficients!!!
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Entropy Calculation Example: Mn(s) + 2O2(g) MnO4(s)
S° J mol-1 K-1 Mn (s) = 32.8 O2 = 205.0 MnO4 = 120.5
∆S = [S° Products] – [S° Reactants]∆S = [120.5] – [2(205.0) + 32.8] ∆S = -322.3 J/mol K
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Entropy, S - a measure of randomness or disorder
• associated with probability (There are more ways for something to be disorganized than organized.)
• Entropy increases going from a solid to a liquid to a gas.• Entropy increases when solutions are formed.• Entropy increases in a reaction when more atoms or molecules
are formed.• The entropy of a substance increases with temperature.
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Gibbs free energy, G• Energy available to do work • Go = standard free energy change
– change in free energy that occurs if the reactants in their standard states are converted to products in their standard states
Go =Ho -T So When Go is negative the reaction is spontaneous is the forward directionWhen Go is positive the reaction is nonspontaneous is the forward direction
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• What temperature would a reaction be spontaneous if ΔH = 9500 J/mol and ΔS = 6.5J/mol K?
Go =Ho -T So 9500 = T(6.5)T = 1461 KAbove 1461 K this reaction will be spontaneous and below it will not
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• Will this reaction be spontaneous at 100°C?ΔH = -18 KJ/mol and ΔS = 94.3 J/mol K
Go =Ho -T So Yes at all temperatures Go =(Ho) -T (So) Go =(-) - (+)(+)Go = can only be negativeDon’t believe me try putting in the values and craze high/low number for temperature
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Go =Ho -T So
• A spontaneous reaction has a negative G. For example, when ice melts H is positive (endothermic), S is positive and G = 0 at 0oC.
• If...
Entropy, ΔS Enthalpy, ΔH Spontaneity
Positive Positive Yes at high temp
Negative Positive Never spontaneous
Positive Negative Always spontaneous
Negative Negative Yes at low temp
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Collision Theory:
• In order to react, two or more particles must collide with sufficient energy (called the activation energy) and with the proper molecular orientation. If the colliding particles do not have either of these two prerequisites, no product is formed.
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Factors that affect reaction rate:
• Temperature- Reactions go faster at higher temperatures. Particles have more kinetic energy. More colliding particles have enough energy to overcome the activation energy barrier.
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Factors that affect reaction rate:
• Concentration- Increasing the concentration of reactants usually increases the reaction rate. If there are more particles to collide, there should be a greater number of collisions that produce products.
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Factors that affect reaction rate:
• Catalysts- A catalyst is a substance that speeds up a reaction by lowering the activation energy barrier. It is not a product or reactant and it is not used up or changed itself.
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Rate Law
• equation that is written that expresses how the reaction rate of a particular reaction is dependent upon the concentrations of its reactants.
• For the reaction aA + bB cC + dD, the general form of the rate law would be:
Rate = k [A]a[B]b
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• Rate is usually expressed as mol/L time.∙• k is the specific rate constant. It is constant for a given
reaction at a given temperature. The faster a reaction, the larger the k value.
• [A] and [B] represent the concentrations of reactants A and B in moles per liter (M).
• x and y are the order of the reactant. They can only be determined by analyzing experimental data. These exponents are usually positive integers.
Rate = k [A]x[B]y
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( )0.20.4
0.10.8( ) 23 =8
0.10.2
EXAMPLES
2A + B 2CEx. [A] [B] Rate1 0.1 0.2 0.102 0.1 0.4 0.203 0.2 0.4 0.80Determine the rate law:
( ) =x
rate = k
[A]
[A] [B]3 1
=x[B]
0.10.2
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Half-life
Based on how much time does it take ½ of the substance to change into productsx = time / half-life (number of half-lives)mf = (mi)(1/2)x
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If you start with 2.00 g of nitrogen-13 how many grams will remain after 4 half lives?
mf = 2.00 (1/2)4
mf =0.125 gPhosphorous-32 has a half-life of 14.3 yr. How many grams remain after 57.2 yr from a 4.0 g samplemf = 4.00 (1/2)(57.2/14.3)
mf = 0.25 g
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• Acid: hydrogen-ion donor (proton donor)• Base: hydrogen-ion acceptor (proton acceptor) NH3 + H2O NH4
+ + OH-
base acid
BrØnsted-Lowry
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• Acid: hydrogen-ion donor (proton donor)• Base: hydrogen-ion acceptor (proton acceptor) NH3 + H2O NH4
+ + OH-
base acid
BrØnsted-Lowry
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Working Ka and Kb problems
• 1st: Write the equation• 2nd: Set up a reaction
diagram (RICE diagram) • 3rd: Set up Ka or Kb expression
• 4th: Substitute values into Ka expression
• 5th: Solve Ka expression for X.• 6th: Calculate pH from H+ or OH-
concentration.
R = reactionI = initialC = changeE = equilibrium
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Example: Calculate the pH of a 0.10 M solution of
acetic acid. The Ka for acetic acid is 1.8 x 10-5.
R HC2H3O2 H+ + C2H3O2-
I C
E
.1-x
.1- x
0+x
x
0+x
x
Ka =[H+] [C2H3O2
-] [HC2H3O2]
=[x] [x]
[0.1-x]= 1.8 x 10-5
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[x] [x][.1-x] = 1.8 x 10-5
x2
.1-x = 1.8 x 10-5
x2
.1 = 1.8 x 10-5
x2 = 1.8 x 10-6
x = 1.3 x 10-3 = [H+]
pH = -log(1.3 x 10-3)
pH = -log [H+]
pH= 2.87
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Example:Calculate the pH of a 0.25 M solution of HCN. The Ka for HCN is 6.2 x 10-10.
R HCN H+ + CN-
I C
E
.25-x
.25 - x
0+xx
0+xx
Ka = 6.2 x 10-10 = [H+][CN-] [HCN]
[x][x].25 – x
=
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Ka = 6.2 x 10-10 = [H+][CN-] = [x][x]
[HCN] .25 – x X2
.25= 6.2 x 10-10
= [H+]1.55 x 10-10X2 =
x = 1.24 x 10-5
pH = -log(1.24 x 10-5)pH= 4.9
pH = -log [H+]
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• Calculate the pH of a 0.15 M solution of ammonia. The Ka of ammonia is 1.8 x 10-9.
R NH3 + H2O (l) NH4+ + OH-
ICE
.15-x
.15 –x
0+x
x
0+x
x
Ka = [NH4+] [OH-]
[NH3] [x][x].15 – x=
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Ka = [NH4+] [OH-] = [x][x]
[NH3] .15 – x
X2 = 1.8 x 10-9
.15
X2 = 2.7 x 10-10
X = 0.000016
pH = -log (H+)
pH = 4.78
= [H+]
pH = -log (0.000016)
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LeChatelier's Principle
• When a stress is applied to a system, the equilibrium will shift in the direction that will relieve the stress.
Henry Louis Le Chatelier
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Changes in concentration
• An increase in concentration of:– a reactant will cause equilibrium to shift to the right
to form more products. – a product will cause equilibrium to shift to the left to
form more reactants.• A decrease in concentration of:
– a product will cause equilibrium to shift to the right to form more products.
– a reactant will cause equilibrium to shift to the left to make more reactants.
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A + B C + D• Remove A or B ← • Add C or D ←• Remove C or D → • Add A or B →
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Changes in temperature
• Treat energy as a product or reactant and temperature changes work just like changes in concentration!
• An increase in temperature of an exothermic reaction (H is negative) will cause equilibrium to shift to the left. A decrease in temperature of an exothermic reaction will cause equilibrium to shift to the right.
• An increase in temperature of an endothermic reaction (H is positive) will cause equilibrium to shift to the right. A decrease in temperature of an endothermic reaction will cause equilibrium to shift to the left.
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Example:
N2 + O2 2NO
H = 181 kJ (endothermic)
• addition of heat• lower temperature
181 kJ +
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Example:
2SO2 + O2 2SO3
H= -198 kJ (exothermic)
• increase temperature• remove heat
+ 181 kJ
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Example:
CaCO3 + 556 kJ CaO + CO2
• decrease temperature
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Changes in pressure
• Changes in pressure only affect equilibrium systems having gaseous products and/or reactants.
• Increasing the pressure of a gaseous system will cause equilibrium to shift to the side with fewer gas particles.
• Decreasing the pressure of a gaseous system will cause equilibrium to shift to the side with more gas particles.
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Addition of a catalyst
• Adding a catalyst does not affect equilibrium. Catalysts speed up the forward and reverse reactions equally.
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Example:P4(s) + 6Cl2(g) 4PCl3(l)
• increase container volume– Shifts to side with more gas
• decrease container volume– Shifts to side with less gass
• add a catalyst– Inert gases have no effect on equilibrium
N/C
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Example:Consider the reaction:
2NO2(g) N2(g) + 2O2(g)which is exothermic• NO2 is added• N2 is removed• The volume is halved• He (g) is added• The temperature is increased• A catalyst is added
+ HEAT
N/C
N/C
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Mechanisms
• Uni molecularA → BAB → A + BA2 → 2A
• Bi molecular A + B → AB AB + C → AC + B Only one thing changing at a time
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143
Overall reactionsNO2 + F2 NO2F + F (slow step)NO2 + F NO2F (fast step)
Simplify F on both sides2NO2 + F2 NO2FIntermediate FRate = k[NO2]1[F2]1
Rate depends on both NO2 and F2
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144
Overall reactionsPO2 + Cl PClO2 (fast step)PClO2 + PO2 P2O4 + Cl (slow step)
Simplify Cl and PClO2 on both sides2PO2 P2O4 Intermediate PClO2 Catalyst ClRate = k[PO2]2
Rate depends on only PO2
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Oxidation Reduction Reactions
• Oxidation-reduction reactions- chemical changes that occur when electrons are transferred between reactants.
• Also called REDOX reactions
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• Oxidation• Modern definition - loss of electrons • Examples• 4Fe + 3O2 2Fe2O3 (rusting of iron)
• C + O2 CO2 (burning of carbon)
• C2H5OH + 3O2 2CO2 + 3H2O (burning of ethanol)
Oxidation Reduction Reactions
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OxidationIsLoss
ReductionIsGain
LEO (Lose Electrons-Oxidation)
the lion goes
GER (Gain Electrons-Reduction)
To help remember these definitions, use one of these mnemonic devices:
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Formation of Ions
• Ex. 2Na + S Na2S• Sodium goes from the neutral atom to the 1+
ion. Therefore, it has lost an electron (It was oxidized). Sulfur goes from the neutral atom to the 2- ion. Therefore, it has gained two electrons. (It was reduced)
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Question
• Lead loses four electrons. It take on a charge of ___. Does this mean that it is oxidized or reduced?
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REDOX Reaction Examples• Identify the element oxidized, the element
reduced, the oxidizing agent and the reducing agent for each of the following:
MnO2 + 4HCl MnCl2 + Cl2 + 2H2O+4 +2+1 -1-2 0-1 +1 -2
Mn
Cl
+4 +2 Gained 2 e-
Oxidized
If it was Reduced, then the reactant that contains Mn acts as the “Oxidizing Agent”. (MnO2)
-1 0 Lost 1 e-
Reduced
If it was Oxidized, then the reactant that contains Cl acts as the “Reducing Agent”. (HCl)
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Batteries• Electrochemical cells that convert
chemical energy into electrical energy are called voltaic cells.
• The energy is produced by spontaneous redox reactions.
• Voltaic cells can be separated into two half cells.
• A half cell consists of a metal rod or strip immersed in a solution of its ions.
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Batteries
• We write half reactions to show what happens in each part of the cell.– Example Write the half reactions that occur in the
Fe2+/Ni2+ cell.– Oxidation Fe Fe2+ + 2e-
– Reduction Ni2+ + 2e- Ni
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Diagram of voltaic cell for the reaction of zinc
and copper.• Diagram of voltaic cell for the reaction of
zinc and copper.• Oxidized: Zn Zn2+ + 2e-
• Reduced: Cu2+ + 2e- CuDirection of electron flow
Solution of Solution of
From anode to cathode
anode cathode
Zinc ions Copper ions
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Half-Cells
• The half cells are connected by a salt bridge. A salt bridge is a tube containing a solution of ions.
• Ions pass through the salt bridge to keep the charges balanced.
• Electrons pass through an external wire.• The metal rods in voltaic cells are called electrodes.• Oxidation occurs at the anode and reduction occurs
at the cathode. (An Ox and Red Cat)• The direction of electron flow is from the anode to
the cathode. (FAT CAT )
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Calculating the Charge of a Battery
• The potential charge of a battery can be calculated with a set of values from a table of reduction potentials. – To do this, write the oxidation and reduction half
reactions.– Look up the cell potentials from the data table.– Flip the sign of the cell potential for oxidation.– Add the potentials together.
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Example A common battery is made with nickel and cadmium. What is the cell potential
of this battery? (E0Cd = -0.40V, E0
Ni = -0.25V)
Oxidation Cd Cd2+ + 2e- E0 = 0.40VReduction Ni2+ + 2e- Ni + E0 = -0.25V
Total = E0 = 0.15V
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Spontaneous Electrochemical Reactions
ΔG = -nFERecall the NiCd batteryTotal = E0 = 0.15V
Positive Voltage, -ΔG, spontaneous reaction Negative Voltage, +ΔG, nonspontaneous reaction
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The word, polymer, implies that polymers are constructed from pieces (monomers) that can be easily connected into long chains (polymer). When you look at the above shapes, your mind should see that they could easily fit together.
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There are two types of polyethylene polymers (plastics). One is when
the polyethylene exists as long straight chains. The
picture here shows the chains of one carbon with
two hydrogen atoms repeating. The chain can
be as long as 20,000 carbons to 35,000
carbons. This is called high density polyethylene
(HDPE).
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Low density polyethylene (LDPE) is made by causing the long chains of ethylene to branch. That way they cannot lie next each other, which reduces the density and strength of the polyethylene. This makes the plastic lighter and more flexible.