Rxns

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Rxns
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This representation containsreactants and
products. Distinguishbetween them

• HCl NaOH ? NaCl H2O
• reactants
• products

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What are the different Classifications we need to
know

• Synthesis
• Decomposition
• Single Replacement
• Double Replacement
• A special one called Combustion

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How do you predict the Products of a reaction

• If you look at the reactants, you can determine
  the type of reaction. From the type you can
  determine the potential products

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Synthesis

• You can recognize a synthesis rxn by
• Elements are the reactants
• You predict the products by applying nomenclature
  rules

2

• Na Cl2 ?

2

• NaCl

C O2 ?
CO2
Metals go first, nonmetals last. Cross valences
for subscripts. If you get 2 nonmetals, use
experience (put O last usually)(most
electronegative element last)
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Synthesis

• K F2 ?

KF
2K F2 ? 2KF
Na S ?
Na2S
2Na S ? Na2S
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Single Replacement
You can recognize by One reactant is a compound
and one is an element You predict the products by
trading the elements( w/ or - w/ -)

• Ag2O Cu

• Ag CuO ?

2Ag CuO ? Ag2O Cu
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Double Replacement

• You can recognize by
• Two compounds react
• You predict the products by
• trading the metals( w/ or - w/ -)

• NaCl BaO ?

• BaCl2 Na2O

2NaCl BaO ? BaCl2 Na2O
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Decomposition
You can recognize a decomp rxn by One compound
is the reactant You predict the products by
separating the cmpd

• Ag O2

• Ag2O ?

2Ag2O ? 4Ag O2
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Generalized Equation Form

• A B ?(AxB-y ) AyBx synthesis
• AyBx ? yA xB simple decomp
• AyBx C ?( CzB-y ) CyBz A S.R. If C not
  above A on activity seriesN.R.
• AyBx CwDz ? AwDx CyBz D.R. IF AD or CB
  not insoluble N.R.
• C (CyBx) O2 ? CO2 H2O combustion
• MUST ALWAYS BALANCE AFTER COMPLETION

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Stoichiometry
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Steps to solving Stoichiometry

• Complete the equation( synthesis, single
  replacement, double replacement, decomposition).
• Balance the equation
• Know what the equation tells you (The number of
  moles of each substance reacting and produced).
• Solve by proportion or factor-label

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Here are your rules

• Factor-Label
• Mass-Mass
• Given(g) x 1 mole x Coeff want x PT g
  PT(g) Coeff given 1 mole
• Mass-VolGiven(g) x 1 mole x Coeff want x
  22.4 L PT(g) Coeff
  given 1 mole
• Vol-Vol Given(L) x 1 mole x Coeff want x
  22.4 L 22.4L Coeff given
  1 mole

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Here are your rules

• Math (Proportion)
• Mass-Mass
• Given(g) want(g) Coeff x PT
  Coeff x PT
• Mass-Vol Given(g) want(L) Coeff x
  PT Coeff x 22.4L
• Vol-Vol Given(L) want(L) Coeff
  x22.4L Coeff x 22.4L

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Here are your rules

• Limiting Rgt
• You will have 2 givens.
• Work two problems, use the smallest answer
• Yield
• Work problem-Use your given amount of reactant to
  calculate the amt of product formed. Take this
  answer divide it into the amt given of the
  product in the equation

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Here are your rules-Variations

• Mole-Mole
• Given mol x Coeff want
  Coeff given
• Mole-Mass
• Given(mol) Coeff want x PT g
  Coeff given 1 mole

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Limiting Reactants(Reagents)

• These are special stoichiometry problems. You
  can recognize these stoichiometry problems by the
  givens of amounts of all the reactants and then
  they will ask for an amount of any or all of the
  products.

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• The most straightforward way to work this kind is
  to work two problems and the smaller answer is
  the amount of product you can make. The reactant
  that produces that smaller amount is the limiting
  reactant.(Method 1)

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• Another way is to determine which substance is
  the smaller number of moles. If g use P.T. and
  vol use 22.4. then use the ratio of the
  reactants to find which substance is the smaller.
  Use that one to solve the problem. (Method 2)

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Method 1 Solution.

• 1. N2 H2 ? NH3
• 2. N2 3 H2 ? 2 NH3
• 3. 3 moles H2 make 2 mole NH314g 6g
  x g
• N2 3 H2 ? 2 NH328g 3(2g) 2(17g)14g
  x 17 g ammonia produced by nitrogen28 g
  34g

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Now do the other part

• 6g x 6 g 34g
• 34 g ammonia produced by hydrogen
• Since 17g is the smaller number, it is the amount
  produced and nitrogen is the limiting reactant

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Factor-Label

• 14g N2 x 1 mole N2 x 2 mole NH3 x 17g NH3
  28g N2 1 mole N2 1 mole NH3
  17g
• 6g H2 x 1 mole H2 x 2 mole NH3 x 17g NH3
  2g H2 3 mole H3 1 mole
  NH3
• 34g

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Method 2 Determine Limiting Rgt

• 14 g nitrogen 0.5 moles and
• 6g hydrogen 3 moles
• the coefficients tell us that we need 3 moles of
  hydrogen to react with 1 mole of nitrogen.
  Therefore
• 0.5 moles need 1.5 moles of hydrogen.
• We have 3 moles, which is an excess of what is
  needed. Therefore it is the limiting reactant.
• Now work the problem like any mass-mass
  problem.

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Yield

• These are normal stoichiometry problems with a
  equation added. In these it might be best to
  remember (actual/theo) times 100.
• The answer to the stoich problem is the amount
  you are supposed to get (theoretical) and the
  amount actually produced is the actual.
• If you like to think part/whole. The calculated
  number is the whole (the total amt you are
  supposed to get) and the amt given as produced is
  the part

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Calculate the yield if when 14 g of nitrogen
reacts with hydrogen and 15 g of ammonia is
produced

• Solution. 1. N2 H2 ? NH3
• 2. N2 3 H2 ? 2 NH3
• 3. 3 moles H2 make 2 mole NH3
• 4. 14g x g
• N2 3 H2 ? 2 NH3
• 28g 3(2g) 2(17g)

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Calculate the yield if when 14 g of nitrogen
reacts with hydrogen and 15 g of ammonia is
produced

• Solution.     15g is the actual yield
•      actual x 100
• theoretical
• 15g x 100 88
• 17g

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Calculate the yield if when 14 g of nitrogen
reacts with hydrogen and 15 g of ammonia is
produced

• Factor-Label
• 14g N2 x 1 mole N2 x 2 mole NH3 x 17g NH3
  28g N2 1 mole N2 1 mole NH3
• 17g
• 15g is the actual yield
•       actual x 100 15 X 100
  88
• theoretical 17

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Atomic Theory
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Bohrs Planetary Model
Electrons in specific circular orbits. If enery
is absorbed they move out (excited) when they
fall back the energy is released
1st
1
2
e-
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Bohr was a success only for Hydrogen

• He was able to explain the line spectrum for
  hydrogen (p. 137).

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Who were de Broglie and Schrodinger and what
contribution did they bring to atomic theory?

• De Broglie reasoned that if light had a dual
  nature possibly matter could also(it could have
  wave properties). Schrodinger devised a theory to
  find the wave properties in atoms and
  molecules(Quantum or Wave Mechanics). This
  changes the way we can think about an electron
  and its movement about the nucleus.

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De Broglie and Schrodinger cont.

• To describe the path we need to know both the
  speed and location of an electron. Heisenberg
  used quantum mechanics to show it is impossible
  to know both the location and the speed of an
  e-(Uncertainty principle). This makes it
  impossible to describe or know how the electron
  moves in an atom. It does allow us to calculate
  the probability of finding an electron at any
  given point.

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Which model of the atom do we associate with them?

• Wave mechanical

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Whats the deal

• We now look at the atom as a system and the
  electron movements related to the others in the
  system.
• We can calculate the probability of finding an
  electron

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How is this different than Bohr?

• p s
• space filled p

Now instead of definite paths electrons are
positioned by probable location. We guess they
are in certain areas. levels, sublevels,
orbitals
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What is the probability map foran electron
called?

• Atomic orbital

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What determines the size of this orbital?

• The 90 chance of finding an electron here

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How could you create a chart to show the places
electrons could be?
Remember there are 7 levels, 4 sublevels, and 4
types of orbitals. The sublevels add as we add a
level. There is 1 s orbital in a s
sublevel. There are3 p orbitals in a p
sublevel. There are 5d orbitals in a d sublevel
and 7 fs in an f sublevel

• 1s2
• 2s2 2p6
• 3s2 3p6 3d10
• 4s2 4p6 4d10 4f14
• 5s2 5p6 5d10 5f14
• 6s2 6p6 6d10 6f 14
• 7s2 7p6 7d10 7f14

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Electron configurations

• Remember to use your diagonal rule to determine
  the configurations
• Representations are done by boxes and arrows or
• Circles and diagonal lines
• See following

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What is Hunds Rule?

• When electrons fill degenerate(identical)
  orbitals(ps ds or fs) each orbital must have
  one electron before any orbital has two
• (electrons stay unpaired as long as possible)

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How does this work?

• Lets look at a series of atoms
• B 1s2 2s2 2p1 C 1s2
  2s2 2p2
• ?? ?? ? ?? ??
  ??
• N 1s2 2s2 2p3 O
  1s2 2s2 2p4
• ?? ?? ??? ?? ??
  ????
• F 1s2 2s2 2p5 Ne
  1s2 2s2 2p6
• ?? ?? ????? ?? ??
  ??????

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How does this work with the other method?

• B
• 1s2 2s2 2p1 C 1s2 2s2 2p2
• ? ? ??? ? ? ???
• N 1s2 2s2 2p3 O 1s2 2s2
  2p4
• ? ? ??? ? ?
  ???
• F 1s2 2s2 2p5 Ne 1s2 2s2 2p6
• ? ? ??? ? ? ???

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How does this work contd?

• Lets look at a series of atoms
• Ti 1s2 2s2 2p6 3s2 3p6 4s2
  3d2
• ? ? ??? ? ??? ?
  ?????
• V 1s2 2s2 2p6 3s2 3p6 4s2
  3d3
• ? ? ??? ? ???
  ? ?????
• Ni 1s2 2s2 2p6 3s2 3p6 4s2
  3d8
• ? ? ??? ? ???
  ? ?????

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Ground state vs Excited State

• Ground state is when all the electrons are in the
  lowest energy levels available.
• (Most stable state)
• Excited electrons have moved to outer energy
  level
• (Unstable)

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Periodic Trends
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