G11 Chemistry Revision Course
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Transcript of G11 Chemistry Revision Course
Grade 11 Revision
K Warne
G12 Prior knowledge from 10 & 11
• the use of equations of motion in solving problems dealing with momentum, work, energy and power
• the use of Newton’s first, second and third laws of motion
• conservation of mechanical energy • sound waves and properties of sound• electromagnetism• Stoichiometric calculations • Concentration calculations• Balancing of chemical equations • Use of oxidation numbers • Identification and description of intermolecular forces
(Van der Waals forces and hydrogen bonds)
G12 Prior knowledge from 10 & 11
• the use of equations of motion in solving problems dealing with momentum, work, energy and power
• the use of Newton’s first, second and third laws of motion
• conservation of mechanical energy • sound waves and properties of sound• electromagnetism• Stoichiometric calculations • Concentration calculations• Balancing of chemical equations • Use of oxidation numbers • Identification and description of intermolecular forces
(Van der Waals forces and hydrogen bonds)
G12 Prior knowledge from 10 & 11
• the use of equations of motion in solving problems dealing with momentum, work, energy and power
• the use of Newton’s first, second and third laws of motion
• conservation of mechanical energy • sound waves and properties of sound• electromagnetism• Stoichiometric calculations • Concentration calculations• Balancing of chemical equations • Use of oxidation numbers • Identification and description of intermolecular forces
(Van der Waals forces and hydrogen bonds)
G11 Session 2
• Decimal conversions
• Concentration Calculations
• Calculations map
• No. of particles
• Redox reactions
Decimal Conversions
King Henry Died a miserable death calledmeasles
Kilo
Hecta Decca m(unit) deci centi milli
1000 100 10 1 1/101/100
1/1000
Volume Conversions
1 dm = ……. cm1 dm3 = ……… cm3
1 m3 = ……………… dm3 = ……………………………. cm3
….cm3
…………………………………
…… cm3
…… cm3
……. cm3 ... m3
Volume Conversions
1 dm = 10 cm1 dm3 = 1000 cm3
1 m3 = 1000 dm3 = 1 000 000 cm3
1cm3
1 dm3 (1 litre)
10 cm3
10 cm3
10 cm3 1 m3
Concentration - MolarityThe concentration of a solution is defined as the ………………. of
……………………… per ………………. (dm3) of ………………….
solute
solute
Final volume of ……………..
500cm3
=+
Concentration =Amount of ……… (……….)
Volume of ………………
30g of NaCl
C = n
v
Concentration - MolarityThe concentration of a solution is defined as the AMOUNT of SOLUTE
per LITRE (dm3) of SOLUTION.
c(NaCl) = (m/Mr)x1/v = ((30/(23+35.5))x1/0.5 =
solute
solute
Final volume of solution
500cm3
=+
Concentration =Amount of solute (moles)
Volume of solution
30g of NaCl
n
vC =
The Mole• The mole is defined as, “the amount of matter with
the same number of elementary particles as 12 grams of carbon 12”.
602 300 000 000 000 000 000 000Six hundred and two thousand, three hundred, billion billion !
6.023x1023 particles
12.00 g
CSymbol (L)
Number of particles = no of moles x no. particles in a mole
Particles = n x L
ASKEDGIVEN
Mole Calculations
MOLES MOLES
MASS MASS
VOLUME VOLUME
CONCENTRATIONCONCENTRATION
MOLARRATIO
Number Of
particles
Number Of
particles
Mixed example
Ammonia gas is made by reacting ammonium chloride with calcium hydroxide according to:
NH4Cl + Ca(OH)2 NH3 + CaCl2 + H2O
If 32.1 g of ammonium chloride reacts with 500 cm3 of a 0.75 M calcium hydroxide solution, Show by calculation; which is the limiting reagent, what volume of ammonia is produced at S.T.P in m3 and how many hydroxide ions are left after the reaction?
Redox Reactions• Involve electron transfer• Oxidation is ……….. Reduction is ………… O….R….
Na Na+ + e- Cl2 + 2e- 2Cl-
• Both processes ALWAYS occur ………………...• Oxidation is caused by …………… AGENTS – ……………!!
Na Na+ + e-, Na ……………… .: ………….. agent!! (Good one)
• Reduction is caused by …………… AGENTS – ……………!Cl2 + 2e- 2Cl-, Cl2 ………………. .: …………….. agent!
(Good)
OVERALL: 2Na + Cl2 Na+ + 2Cl-
Redox Reactions• Involve electron transfer• Oxidation is loss Reduction is gain OILRIG
• Na Na+ + e- Cl2 + 2e- 2Cl-
• Both processes ALWAYS occur together.• Oxidation is caused by OXIDIZNG AGENTS – REDUCED!!
Na Na+ + e-, Na oxidised .: reducing agent!! (Good one)
• Reduction is caused by REDUCING AGENTS – OXIDISED!Cl2 + 2e- 2Cl-, Cl2 reduced .: Oxidizing agent!
(Good)
0 0 +1 -1OVERALL: 2Na + Cl2 Na+ + 2Cl-
Reactions written as reductions.
Positive potentials accept electrons are good OXIDISING AGENTS.
Negative potentials donate electrons are good REDUCING AGENTS.
ELECTROCHEMICAL SSERIESequilibrium E° (volts)
1.5
0.8
0.34
0
-0.13
-0.44
-0.76
-1.66
-2.37
-2.71
-2.87
-2.92
-3.03
Electrochemical half-cell Electrochemical half-cell potentials are listed from +ve to potentials are listed from +ve to –ve E–ve Eθθ values. values.
Reactions take place
Reactions written as reductions.
Positive potentials accept electrons are good OXIDISING AGENTS.
Negative potentials donate electrons are good REDUCING AGENTS.
ELECTROCHEMICAL SERIESequilibrium E° (volts)
1.5
0.8
0.34
0
-0.13
-0.44
-0.76
-1.66
-2.37
-2.71
-2.87
-2.92
-3.03Ele
ctro
ns
flow
in e
xter
nal
cir
cuit
.
Electrochemical half-cell Electrochemical half-cell potentials are listed from +ve to potentials are listed from +ve to –ve E–ve Eθθ values. values.
Oxidizing Agents
REDUCINGREDUCING AGENTSAGENTS
Reactions take placeReactions take placeTop LEFT to bottom RIGHTTop LEFT to bottom RIGHT
Oxidation numbersOxidation numbersOxidation numbers can be used to identify
oxidation and reduction as well as balance equations – particularly when it is difficult to identify e- transfer.
Oxidation numbers are ………………………… charges charges that an atom of an element that an atom of an element …………….……………. in a compound, if all bonds were …………..bonds were …………...
OOHH
HH++ --
H = …., O = …..
NH3
H = ….., N = …..
Oxidation numbersOxidation numbersOxidation numbers can be used to identify
oxidation and reduction as well as balance equations – particularly when it is difficult to identify e- transfer.
Oxidation numbers are imaginaryimaginary charges charges that an atom of an element that an atom of an element would havewould have in a compound, if all bonds were ionicbonds were ionic.
OOHH
HH++ --
H = +1, O = -2
NH3
H = +1, N = -3
Oxidation number rulesOxidation number rules • The O.N. of a free element is 0• Hydrogen is +1, (except hydrides -1)• Metals: GI +1, GII +2, GIII +3, Zn +2• Halides -1, Oxides -2, sulphides -2, nitrides
-3• The O.N. of a simple ion is equal to its ionic
charge i.e. Mg2+ O.N. is +2
• In allocation of O.N., charge is conserved i.e. – the sum of the total O.N. of the atoms in a
compound is zero, Na+Cl- +1 = (-1) = 0– while that for a polyatomic ion is equal to ionic
charge on the ion. MnO4-
Mn +7 + (4xO -2) = -1
BALANCING HALF REACTIONS
1. Balance oxygens by adding H2O2. Balance hydrogens by adding H+ ions3. Balance charges by adding electrons
(to most + side)(Balanced M’s & NM’s first)
To combine 1/2 reactions1. Multiply by suitable factors to equal
out the electrons.2. Add the reactions and cancel things
appearing on both sides.3. Add spectator ions.
Redox ExampleAn iron II solution can be standardized using a
permanganate solution according to the reaction:
FeCl2 + KMnO4 FeCl3 + Mn2+
Determine; • the oxidation states of all reacting species• The forumula of the oxidising and reducing
agents • the balanced overall (ionic) reaction• The identity of spectator ions in the given
reaction