Grade 11 Revision K Warne
G12 Prior knowledge from 10 & 11 <ul><li>the use of equations of motion in solving problems dealing with momentum, work, e...
G12 Prior knowledge from 10 & 11 <ul><li>the use of equations of motion in solving problems dealing with momentum, work, e...
G12 Prior knowledge from 10 & 11 <ul><li>the use of equations of motion in solving problems dealing with momentum, work, e...
G11 Session 2 <ul><li>Decimal conversions </li></ul><ul><li>Concentration Calculations  </li></ul><ul><li>Calculations map...
Decimal Conversions 1 / 100 c enti c alled 1 / 10 d eci d eath 1 / 1000 1 10 100 1000 m illi m (unit) D ecca H ecta K ilo ...
Volume Conversions <ul><li>1 dm = ……. cm </li></ul><ul><li>1 dm 3  = ……… cm 3 </li></ul><ul><li>1 m 3  = ……………… dm 3  =  …...
Volume Conversions <ul><li>1 dm = 10 cm </li></ul><ul><li>1 dm 3  = 1000 cm 3 </li></ul><ul><li>1 m 3  = 1000 dm 3  =  1 0...
Concentration - Molarity <ul><li>T h e concentration of a solution is defined as the  ……………….  of  ………………………  per  ………………....
Concentration - Molarity <ul><li>T h e concentration of a solution is defined as the  AMOUNT  of  SOLUTE  per  LITRE  (dm ...
The Mole <ul><li>The mole is defined as, “the  amount of matter with the same number of elementary particles as 12 grams o...
Mole Calculations ASKED GIVEN MOLES MOLES MASS MASS VOLUME VOLUME CONCENTRATION CONCENTRATION MOLAR RATIO Number  Of parti...
Mixed example <ul><li>Ammonia gas is made by reacting ammonium chloride with calcium hydroxide according to: </li></ul><ul...
Redox Reactions <ul><li>Involve electron transfer </li></ul><ul><li>Oxidation is ………..   Reduction is …………   O…. R…. </li>...
Redox Reactions <ul><li>Involve electron transfer </li></ul><ul><li>Oxidation is loss   Reduction is gain   OIL RIG </li><...
<ul><li>Reactions written as reductions. </li></ul><ul><li>Positive  potentials  accept electrons  are good OXIDISING AGEN...
<ul><li>Reactions written as reductions. </li></ul><ul><li>Positive  potentials  accept electrons  are good OXIDISING AGEN...
Oxidation numbers <ul><li>Oxidation numbers can be used to identify oxidation and reduction as well as balance equations –...
Oxidation numbers <ul><li>Oxidation numbers can be used to identify oxidation and reduction as well as balance equations –...
Oxidation number rules   <ul><li>The O.N. of a free element is 0 </li></ul><ul><li>Hydrogen is +1, (except hydrides -1) </...
BALANCING  HALF REACTIONS <ul><li>Balance oxygens by  adding H 2 O </li></ul><ul><li>Balance hydrogens by adding  H +  ion...
Redox Example <ul><li>An iron II solution can be standardized using a permanganate solution according to the reaction: </l...
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G11 Chemistry Revision Course

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A set of slides created to teach G11 Chemistry Revision Course to learners at Bishops Diocesan College in Cape Town.

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G11 Chemistry Revision Course

  1. 1. Grade 11 Revision K Warne
  2. 2. G12 Prior knowledge from 10 & 11 <ul><li>the use of equations of motion in solving problems dealing with momentum, work, energy and power </li></ul><ul><li>the use of Newton’s first, second and third laws of motion </li></ul><ul><li>conservation of mechanical energy </li></ul><ul><li>sound waves and properties of sound </li></ul><ul><li>electromagnetism </li></ul><ul><li>Stoichiometric calculations </li></ul><ul><li>Concentration calculations </li></ul><ul><li>Balancing of chemical equations </li></ul><ul><li>Use of oxidation numbers </li></ul><ul><li>Identification and description of intermolecular forces (Van der Waals forces and hydrogen bonds) </li></ul>
  3. 3. G12 Prior knowledge from 10 & 11 <ul><li>the use of equations of motion in solving problems dealing with momentum, work, energy and power </li></ul><ul><li>the use of Newton’s first, second and third laws of motion </li></ul><ul><li>conservation of mechanical energy </li></ul><ul><li>sound waves and properties of sound </li></ul><ul><li>electromagnetism </li></ul><ul><li>Stoichiometric calculations </li></ul><ul><li>Concentration calculations </li></ul><ul><li>Balancing of chemical equations </li></ul><ul><li>Use of oxidation numbers </li></ul><ul><li>Identification and description of intermolecular forces (Van der Waals forces and hydrogen bonds) </li></ul>
  4. 4. G12 Prior knowledge from 10 & 11 <ul><li>the use of equations of motion in solving problems dealing with momentum, work, energy and power </li></ul><ul><li>the use of Newton’s first, second and third laws of motion </li></ul><ul><li>conservation of mechanical energy </li></ul><ul><li>sound waves and properties of sound </li></ul><ul><li>electromagnetism </li></ul><ul><li>Stoichiometric calculations </li></ul><ul><li>Concentration calculations </li></ul><ul><li>Balancing of chemical equations </li></ul><ul><li>Use of oxidation numbers </li></ul><ul><li>Identification and description of intermolecular forces (Van der Waals forces and hydrogen bonds) </li></ul>
  5. 5. G11 Session 2 <ul><li>Decimal conversions </li></ul><ul><li>Concentration Calculations </li></ul><ul><li>Calculations map </li></ul><ul><li>No. of particles </li></ul><ul><li>Redox reactions </li></ul>
  6. 6. Decimal Conversions 1 / 100 c enti c alled 1 / 10 d eci d eath 1 / 1000 1 10 100 1000 m illi m (unit) D ecca H ecta K ilo m easles a m iserable D ied H enry K ing
  7. 7. Volume Conversions <ul><li>1 dm = ……. cm </li></ul><ul><li>1 dm 3 = ……… cm 3 </li></ul><ul><li>1 m 3 = ……………… dm 3 = ……………………………. cm 3 </li></ul>… .cm 3 ………………………………… …… cm 3 …… cm 3 …… . cm 3 ... m 3
  8. 8. Volume Conversions <ul><li>1 dm = 10 cm </li></ul><ul><li>1 dm 3 = 1000 cm 3 </li></ul><ul><li>1 m 3 = 1000 dm 3 = 1 000 000 cm 3 </li></ul>1cm 3 1 dm 3 (1 litre) 10 cm 3 10 cm 3 10 cm 3 1 m 3
  9. 9. Concentration - Molarity <ul><li>T h e concentration of a solution is defined as the ………………. of ……………………… per ………………. (dm 3 ) of ………………… . </li></ul>Final volume of …………….. 500cm 3 = + Concentration = Amount of ……… (……….) Volume of ……………… 30g of NaCl C = n v solute solute
  10. 10. Concentration - Molarity <ul><li>T h e concentration of a solution is defined as the AMOUNT of SOLUTE per LITRE (dm 3 ) of SOLUTION . </li></ul>c(NaCl) = ( m / Mr ) x 1 / v = ( (30 / (23+35.5) )x 1 / 0.5 = Final volume of solution 500cm 3 = + Concentration = Amount of solute (moles) Volume of solution 30g of NaCl n v C = solute solute
  11. 11. The Mole <ul><li>The mole is defined as, “the amount of matter with the same number of elementary particles as 12 grams of carbon 12”. </li></ul><ul><li>602 300 000 000 000 000 000 000 </li></ul><ul><li>Six hundred and two thousand, three hundred, billion billion ! </li></ul><ul><li>6.023x10 23 particles </li></ul>12.00 g C Symbol (L) Number of particles = no of moles x no. particles in a mole Particles = n x L
  12. 12. Mole Calculations ASKED GIVEN MOLES MOLES MASS MASS VOLUME VOLUME CONCENTRATION CONCENTRATION MOLAR RATIO Number Of particles Number Of particles
  13. 13. Mixed example <ul><li>Ammonia gas is made by reacting ammonium chloride with calcium hydroxide according to: </li></ul><ul><li>NH 4 Cl + Ca(OH) 2  NH 3 + CaCl 2 + H 2 O </li></ul><ul><li>If 32.1 g of ammonium chloride reacts with 500 cm 3 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 m 3 and how many hydroxide ions are left after the reaction? </li></ul>
  14. 14. Redox Reactions <ul><li>Involve electron transfer </li></ul><ul><li>Oxidation is ……….. Reduction is ………… O…. R…. </li></ul><ul><li>Na  Na + + e - Cl 2 + 2e -  2Cl - </li></ul><ul><li>Both processes ALWAYS occur ………………... </li></ul><ul><li>Oxidation is caused by …………… AGENTS – …………… !! </li></ul><ul><li>Na  Na + + e - , Na ……………… .: ………….. agent!! (Good one) </li></ul><ul><li>Reduction is caused by …………… AGENTS – …………… ! </li></ul><ul><li>Cl 2 + 2e -  2Cl - , Cl 2 ………………. .: …………….. agent! (Good) </li></ul><ul><li>OVERALL: 2Na + Cl 2  Na + + 2Cl - </li></ul>
  15. 15. Redox Reactions <ul><li>Involve electron transfer </li></ul><ul><li>Oxidation is loss Reduction is gain OIL RIG </li></ul><ul><li>Na  Na + + e - Cl 2 + 2e -  2Cl - </li></ul><ul><li>Both processes ALWAYS occur together. </li></ul><ul><li>Oxidation is caused by OXIDIZNG AGENTS – REDUCED !! </li></ul><ul><li>Na  Na + + e - , Na oxidised .: reducing agent!! (Good one) </li></ul><ul><li>Reduction is caused by REDUCING AGENTS – OXIDISED ! </li></ul><ul><li>Cl 2 + 2e -  2Cl - , Cl 2 reduced .: Oxidizing agent! (Good) </li></ul><ul><li> 0 0 +1 -1 </li></ul><ul><li>OVERALL: 2Na + Cl 2  Na + + 2Cl - </li></ul>
  16. 16. <ul><li>Reactions written as reductions. </li></ul><ul><li>Positive potentials accept electrons are good OXIDISING AGENTS . </li></ul><ul><li>Negativ e potentials donate electrons are good REDUCING AGENTS . </li></ul>ELECTROCHEMICAL SSERIES Electrochemical half-cell potentials are listed from +ve to –ve E θ values. Reactions take place
  17. 17. <ul><li>Reactions written as reductions. </li></ul><ul><li>Positive potentials accept electrons are good OXIDISING AGENTS . </li></ul><ul><li>Negativ e potentials donate electrons are good REDUCING AGENTS . </li></ul>ELECTROCHEMICAL SERIES Electrons flow in external circuit. Electrochemical half-cell potentials are listed from +ve to –ve E θ values. Oxidizing Agents REDUCING AGENTS Reactions take place Top LEFT to bottom RIGHT
  18. 18. Oxidation numbers <ul><li>Oxidation numbers can be used to identify oxidation and reduction as well as balance equations – particularly when it is difficult to identify e - transfer. </li></ul><ul><li>Oxidation numbers are …………… charges that an atom of an element ……………. in a compound, if all bonds were ………….. . </li></ul>H = …., O = ….. NH 3 H = ….., N = ….. O H H  +  -
  19. 19. Oxidation numbers <ul><li>Oxidation numbers can be used to identify oxidation and reduction as well as balance equations – particularly when it is difficult to identify e - transfer. </li></ul><ul><li>Oxidation numbers are imaginary charges that an atom of an element would have in a compound, if all bonds were ionic . </li></ul>H = +1, O = -2 NH 3 H = +1, N = -3 O H H  +  -
  20. 20. Oxidation number rules <ul><li>The O.N. of a free element is 0 </li></ul><ul><li>Hydrogen is +1, (except hydrides -1) </li></ul><ul><li>Metals: GI +1, GII +2, GIII +3, Zn +2 </li></ul><ul><li>Halides -1, Oxides -2, sulphides -2, nitrides -3 </li></ul><ul><li>The O.N. of a simple ion is equal to its ionic charge </li></ul><ul><li>i.e. Mg 2+ O.N. is +2 </li></ul><ul><li>In allocation of O.N., charge is conserved i.e. </li></ul><ul><ul><li>the sum of the total O.N. of the atoms in a compound is zero, Na + Cl - +1 = (-1) = 0 </li></ul></ul><ul><ul><li>while that for a polyatomic ion is equal to ionic charge on the ion. MnO 4 - </li></ul></ul><ul><ul><li>Mn +7 + (4xO -2) = -1 </li></ul></ul>
  21. 21. BALANCING HALF REACTIONS <ul><li>Balance oxygens by adding H 2 O </li></ul><ul><li>Balance hydrogens by adding H + ions </li></ul><ul><li>Balance charges by adding electrons ( to most + side ) </li></ul><ul><li>(Balanced M’s & NM’s first) </li></ul><ul><li>To combine 1 / 2 reactions </li></ul><ul><li>Multiply by suitable factors to equal out the electrons. </li></ul><ul><li>Add the reactions and cancel things appearing on both sides. </li></ul><ul><li>Add spectator ions. </li></ul>
  22. 22. Redox Example <ul><li>An iron II solution can be standardized using a permanganate solution according to the reaction: </li></ul><ul><li>FeCl 2 + KMnO 4  FeCl 3 + Mn 2+ </li></ul><ul><li>Determine; </li></ul><ul><li>the oxidation states of all reacting species </li></ul><ul><li>The forumula of the oxidising and reducing agents </li></ul><ul><li>the balanced overall (ionic) reaction </li></ul><ul><li>The identity of spectator ions in the given reaction </li></ul>
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