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					         DEPARTMENT OF EDUCATION
         DEPARTEMENT VAN ONDERWYS


        PHYSICAL SCIENCE HG PAPER 2
     NATUUR- EN SKEIKUNDE HG VRAESTEL 2


               NOVEMBER 2001


               FINAL/FINAAL


      EXTENDED MARKING MEMORANDUM

UITGEBREIDE MEMORANDUM VAN PUNTETOEKENNING
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                             MEMORANDUM                                 2
                                            QUESTION 1 / VRAAG 1

1.1 C        1.2 D           1.3 B       1.4 B       1.5 D        1.6 A         1.7 C      1.8 B
1.9 D        1.10 C          1.11 B      1.12 D      1.13 C       1.14 A        1.15 D                           [60]

                                            QUESTION 2 / VRAAG 2

        p1V1 p 2 V2              p VT              440 x 150 x 453
2.1                    p2  1 1 2  p2                                       = 524,5 kPa
         T1     T2                V2 T1              200 x 285
          The tyre will not burst/Die band sal nie bars nie.                                                    (5)

Substituting both T values (in K) in equation - 
Other substitutions – 
Substitusie van beide T-waardes (in K) - 
Ander substitusies - 

                      p1V1  p 2 V2 440  150 p 2 x 200
                                        x
         OR/OF                    =                    p2              =   524,5 kPa
                       T1     T2      285      453
                                                          
                                    p1V1 440 x 103 x150 x 10-3
         OR/OF p1V1 = nRT1 n =                                   27,86 mol
                                      RT1         8,31 x 285
               nRT 2 27,86 x 8,31 x 453
         p2 =                            5,24 x 10 5 Pa  p2 = 524,5 kPa
                V2        200 x 10 3   

IF subject has been incorrectly changed – Give marks for correct initial formula and substitutions:
AS onderwerp verkeerd verander is – Gee punte vir aanvanklike korrekte formule en substitusies:
        p1V1 p 2 V2                     V2 T1                  200 x 285
                          p2                 p 2 
                                                                                = 1,9 x 10-3 kPa    
         T1   T2                       p1 V1T2              440 x 150 x 453

IF/AS p2 > 600 kPa the tyre will burst/sal die band bars (mark/merk pos) 
IF/AS p2 < 600 kPa the tyre will not burst/sal die band nie bars nie (mark/merk pos)  
IF pressure was not calculated (e.g. V or T) – only 1 mark can be given for the correct formula.
AS druk nie bereken is nie (bv V of T) – dan kan slegs 1 punt vir die korrekte formule gegee word.


2.2.1                                                     Positive gradient straight line
              pV
                                                          Positiewe gradient reguitlyn           (2 or/of 0)
                                  
                                                          Intersect / Afsnit
                         

                     0
                         273                      T (K)
                                                                  Unit/eenheid: J or/of Pa.m3 or/of               (3)
                                                              kPa.dm3 or/of N.m or/of kg.m2.s-2
2.2.2    pV = nRT = 2 x 8,31 x 273 = 4,537 x 103 J                                                                (4)
                                                                                           -3                    [12]
           pV = (101,3)(22,4 x 2) = 4,537 x 103 J                 BEWARE: NOT Pa.m
           will not be accepted/sal nie aanvaar word              PASOP: NIE Pa.m-3
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                    MEMORANDUM                               3
                                     QUESTION 3 / VRAAG 3

3.1    FeS  and/en HCℓ / H2SO4 /HNO3              (Formula incorrect/Formule verkeerd ( 1 2 )      (4)
      ALSO ZnS, PbS, CaS, Na2S, FeS2 or other sulphide             IF: salt is HCℓ      ( 34 )
      OOK ZnS, PbS, CaS, Na2S, FeS2 of ander sulfied               AS: sout is HCℓ

                               
        Iron sulphide and hydrochloric acid/sulphuric acid ( 2 4 )        Equation given ( 4 4 )
                                                                          Vergelyking gegee ( 4 4 )
        Ystersulfied en soutsuur/swawelsuur ( 2 4 )

3.2    H2S  S + 2H+ + 2e-                                                                           (2)

         H2S  S + 2H+ + 2e- ( 1 2 )           IF 3.2    Cr 2 O 7  + 14H+ + 6e-  2Cr3+ + 7H2O
                                                                2


         H2S  S + 2H+ + 2e- ( 0 2 )           and 3.3 H2S  S + 2H+ + 2e-
                                               give 2/4 marks, but then they both have to be correct
         S + 2H+ + 2e-  H2S         ( 22 )              Cr 2 O 7  + 14H+ + 6e-  2Cr3+ + 7H2O
                                                                2
                                               AS 3.2
         S + 2H+ + 2e-  H2S        ( 02 )     en 3.3    H2S  S + 2H+ + 2e-
                                               gee 2/4 punte, maar dan moet altwee hulle korrek wees

       IF ionic charges are omitted, 1 mark is forfeited per equation (not applicable to electrons)
       INDIEN ioonlading weggelaat is, word 1 punt per vergelyking verbeur. (nie van toepassing op e-)
        IF equation is unbalanced, 1 mark is forfeited per equation
        INDIEN vergelyking ongebalanseerd is, word 1 punt per vergelyking verbeur.
        IF equation is incomplete/INDIEN vergelyking onvolledig is - ( 0 2 )

3.3    Cr 2 O 7  + 14H+
              2
                            + 6e-            2Cr3+ + 7H2O                                           (2)

        Cr 2 O 7  + 14H+
               2
                            + 6e-  2Cr3+ + 7H2O         ( 12 )
                                                                            No positive marking from 3.2
        Cr 2 O 7  + 14H+
               2
                            + 6e-  2Cr3+ + 7H2O          ( 02 )            or 3.3 to 3.4
        2Cr3+ + 7H2O  Cr 2 O 7  + 14H+ + 6e- ( 2 2 )
                              2                                             Geen positiewe nasien van 3.2
                                                                            of 3.3 na 3.4
        2Cr3+ + 7H2O  Cr 2 O 7  + 14H+ + 6e- ( 0 2 )
                              2




3.4    3H2S + Cr 2 O 7  + 8H+
                     2
                                             3S + 2Cr3+ + 7H2O            reagents (1), bal (1)     (2)
        OR / OF 3H2S + K 2 Cr 2 O 7 + 8HCℓ              3S + 2CrCℓ3 + 7H2O + 2KCℓ 

        OR / OF 3H2S + K 2 Cr 2 O 7 + 4H2SO4             3S + Cr2(SO4)3 + 7H2O + K2SO4 

         OR/ OF : 3H2S + Cr 2 O 7  + 14H+ 
                                2
                                                      3S + 2Cr3+ + 7H2O + 6H+ 
                                     Accept  in 3.4/Aanvaar  in 3.4
3.5    X                                                                                              (1)
3.6    Sulphur / Swawel      S ( 12 )                                              (2)

3.7    Cr3+                 Chromium ion/Chromiun/Chromiun(III) ion) ( 1 2 )                         (2)
                                                                                                      [15]
       OR/OF 2Cr3+          Chroomioon/Chroom/Croom(III) ioon ( 1 2 )
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                   MEMORANDUM                             4
                                      QUESTION 4 / VRAAG 4

4.1     Nitrogen dioxide / Stikstofdioksied  OR nitrogen (IV) oxide/OF stikstof(IV)oksied         (2)
              NO2 ( 1 2 )
                                   Nitrogen oxide / stikstofoksied ( 0 2 )

4.2     4.2.1 Increases / Neem toe        Synonyms for Increases, etc. will be accepted    (2)
                                            Aanvaar sinonieme vir Neem toe ens. word aanvaar
        4.2.2 Increases / Neem toe                                                         (2)
        4.2.3 Stays the same / Bly dieselfde                                                      (2)
                                                  
4.3     Pt is not a strong enough RA to reduce NO 3 to NO2                                        (2)
                                           
        Pt is nie ‘n sterk genoeg RM om NO 3 tot NO2 te reduseer nie.
                       
        OR HNO3 (or NO 3 ) is not a strong enough OA to oxidise Pt to Pt2+
                       
        OF HNO3 (of NO 3 ) is nie ‘n sterk genoeg OM om Pt to Pt2+ te oksideer nie.

        OR Cu is a stronger RA (than Pt) OR Pt2+ is a stronger OA (than Cu2+)
        OF Cu is ‘n sterker RM (as Pt)     OF Pt2+ is ‘n sterker OM (as Cu2+)
                                 
        OR The cell Pt/Pt2+// NO 3 /NO2 has a negative emf
                                  
        OF Die sel Pt/Pt2+// NO 3 /NO2 het ‘n negatiewe emk
        Instead of OA electron acceptor can be used/ Ipv OM kan elektron akseptor gebruik word.

4.4.1   Exothermic / Eksotermies                                                                   (1)
4.4.2                                                             
        The T is decreased, therefore the system will try to increase T, (relatively) favouring the
        exothermic reaction.  / Die T word verlaag, dus sal die sisteem dit probeer
        verhoog, deur die eksotermiese reaksie (relatief) te bevoordeel.                            (3)
         The are marks awarded for the answers to the following questions:
         1. What is the disturbing factor?                        Decrease in T
         2. How does the system react to this disturbance?        Increases T
         3. Which reaction is (relatively)favoured?               Exothermic or forward or
                                                                  equilibrium shifts to the right
         Punte word toegeken vir die antwoord op die volgende vrae:
         1. Wat is die versteurende faktor?                      Verlaging in T
         2. Hoe reageer die sisteem op die versteuring?          Verhoog T
         3. Watter reaksie word (relatief) bevoordeel?           Eksotermies of voorwaarts of
                                                                 ewewig skuif na regs.


4.4.3   Decreases / Afneem                                                                       (2)
                                                                                                  [16]
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                           MEMORANDUM                         5
                                       QUESTION 5 / VRAAG 5

5.1     2Cℓ-           Cℓ2 + 2e-                       OR/OF 2HCℓ           Cℓ2 + 2H+      + 2e-    (2)

        2Cℓ-        Cℓ2 + 2e- ( 1 2 )
                                                          If Cℓ instead of Cℓ- -1 mark
        2Cℓ-        Cℓ2 + 2e- ( 0 2 )                    Indien Cℓ ipv Cℓ- -1 punt
        Cℓ2 + 2e-         2Cℓ- ( 2 2 )
        Cℓ2 + 2e-         2Cℓ-       ( 02 )

                          
5.2     2NaI + Cℓ2  2NaCℓ + I2                (Bal. )     OR/OF 2I- + Cℓ2  2Cℓ- + I2                 (3)
5.3     I2 (or iodine / of jodium)(in water)              Not iodide (I-)/ Nie jodied (I-)            (2)
5.4     I2 (or iodine) (dissolve in CHCℓ3, hence the purple colour)                                   (2)
        I2(of jodium) (los op in CHCℓ3, daarom die pers kleur)
5.5.1   C             [ionic / ionies]                                                                (2)
5.5.2   D             [Van der Waals]                                                                 (2)
5.5.3   B             [hydrogen bonding / waterstofbinding]                                           (2)
                                                                                                       [15]
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                             MEMORANDUM                           6
                                             QUESTION 6 / VRAAG 6

6.1                      2SO2(g)     +     O2(g)             2SO3(g)
        Begin (mol)      0,3               0,15
        Eq (mol)         0,1               0,05        0,2
        [Eq](mol.dm-3) 0,05               0,025       0,1             For calculating concentrations
                                                                        Vir berekening van konsentrasies

                                                           
                          [SO3 ]2                      (0,1)2
               Kc                                              = 160                                  (7)
                       [SO2 ]2 x [O2 ]            (0,05)2 x (0,025)
                                                        
 NB
 1. Marks at end of arrows are for substituting the calculated [O2] + [SO2] and [SO3], even if
    incorrectly calculated. No further positive marking.
    Punte by einde van pyle is vir substitusie van berekende [O2] + [SO2] en [SO3], selfs
    indien verkeerd bereken. Geen verdere positiewe nasien.
 2.      If concentration s were not calculated, but correct mole quantities used instead:
         Indien konsentrasies nie bereken is nie, maar korrekte mol-waardes in plaas daarvan
         gebruik is:

                   2SO2(g)     +      O2(g)            2SO3(g)
         Begin     0,3                0,15
         Eq        0,1                0,05              0,2
                   [SO3 ]2               
                                        (0,2)2
          Kc                                    = 80                 ( 57 )
                [SO2 ]2 x [O2 ]     (0,1)2 x (0,05)
                                            


                   
6.2.1     S + O2  SO2             Balancing incorrect/Balansering verkeerd ( 1 2 )                         (2)

                            
              OR/OF    4FeS2 + 11O2  2Fe2O3 + 8SO2


6.2.2     Sulphuric acid / Swawelsuur         H2SO4 ( 1 2 )                                               (2)

6.3.1     Decreases / Neem af                Synonyms for Decreases, etc. will be accepted                (2)
                                               Aanvaar sinonieme vir Neem af ens. word aanvaar
6.3.2     Decreases / Neem af                                                                             (2)
6.3.3     Stays the same / Bly dieselfde                                                                  (2)
6.3.4     Decreases / Neem af                                                                             (2)
6.3.5     Stays the same / Bly dieselfde                                                                  (2)
6.3.6     Stays the same / Bly dieselfde                                                                  (2)
                                                                                                           [23]
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                                     MEMORANDUM                           7


                                              QUESTION 7 / VRAAG 7
7.1     Hydrochloric acid / Soutsuur (HCℓ)                                                                       (2)

7.2.1   Increases / Neem toe                   Synonyms for Decreases, etc. will be accepted                     (2)
7.2.2   Increases / Neem toe                   Aanvaar sinonieme vir Neem af ens. word aanvaar                   (2)
7.2.3   Decreases / Neem af                                                                                      (2)
                                                  [10 14 ]
7.3 [H+][OH-] = 10-14                 [H+] =                  = 10-8 
                                                     10 6

         pH = -log[H+] = -log(10-8) = 8                                                                         (5)
                                                                                  Only answer: ( 5 5 )
      OR/OF    pOH = -log[OH-] = -log(10-6) = 6 
               pOH + pH = 14                                                     Slegs antwoord: ( 5 5 )
                                       
                   pH         = 14 – 6 = 8 

                                                                                                                  [13]
                                              QUESTION 8 / VRAAG 8
                                                                                                                   (2)
8.1     H2O (water (   1
                             2 ))

                                                                            
        c a Va 1         c V                                 0,18 x 70
8.2              ca  b b =                                           = 0,21 mol.dm-3                   (4)
        c b Vb 2        2 x Va                                2 x 30
                                                                                       If answer is 0,42 mol.dm-3 ( 2 4 )
              -
  OR/OF: n(OH ) = cV = 0,18   x 0,07      = 0,0126 mol                                 As antwoord 0,42 mol.dm-3 is
                             
         n(H2SO4) = cV = 0,03 x c     = 0,03c mol
                    
         n(H+) = 2 x n(H2SO4) = 2 x 0,03.c = 0,06c mol
                        0,06c = 0,0126
                                                       0,0126
                                         c      =            = 0,21 mol.dm-3 
                                                        0,06



                                                                 
  R/OF        pcaVa = pcbVb                     2 x ca x 30 = 0,18 x 70
                                                          
                                                 ca = 0,21 mol.dm-3


                                    
8.3     m      = cMV= 0,18 x 40 x 0,07 = 0,504g                                                                  (4)

                                                                                                [10]
   
  40   for calculating and substituting M in correct formula/ Vir berekening en substitusie van M in
       korrekte formule
    
  0,07 For conversion to dm3 and substituting / Vir omskakling na dm3 en substitusie
                                     
        OR/OF      n = cV = 0,18 x 0,07 = 0,0126 mol                               0,0126 mol could have been
                                              
                   M  nM = 0,0126 x 40 = 0,504g
                      =                                                            calculated in 8.2
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                     MEMORANDUM                              8
                                         QUESTION 9 / VRAAG 9

9.1      c = 1 mol.dm-3       T = 298 K (25 °C)                                                    (4)
  Ignore p = 101,3 kPa, even it is one of the first two mentioned.
  Ignoreer p = 101,3 kPa, selfs al is dit een van die eerste twee wat genoem is.
  If only the conditions are mentioned (without values), no marks
  Indien slegs toestande genoem word (sonder waardes), geen punte

9.2 Ecell/sel        = EOA/OM - ERA/RM        IF unit of 0,8 V is omitted, the mark for 0,8 V is forfeited
                                              INDIEN die eenheid van 0,8 V uitgelaat, word die punt vir
         3,17  = EOA/OM - (-2,37)           0,8 V verbeur.
        EOA/OM = 0,8 V                      IF –0,8 V, no marks for Ag
                                              INDIEN –0,8 V, geen punte vir Ag
        X is Ag                      (4)    3 marks awarded for 0,8 V obtained by using any method
                                              3 punte toegeken vir 0,8 V op enige wyse bereken.
                                            IF only Ag (no working), only 1 mark
                 
9.3      MgMg2+Ag+Ag (3)                  INDIEN slegs Ag (geen bewerking), geen punte
                                              The mark for the final answer is only awarded for Ag
            
        Ag+AgMgMg2+ ( 1 3 )               Die punt vir die finale antwoord word slegs vir Ag toegeken


        Mg2+  +Ag
                    
            MgAg                ( 23 )       IF possible, mark positive from 9.2 to 9.3 and 9.4.
                                                This implies:
                
        MgMg2+AgAg+ ( 2 3 )                 The E° has to be higher than –2,37 V

               
                                             If a gas was used, an unreactive electrode (e.g. Pt)
        Mg MgAgAg+
           2+
                                   ( 13 )       has to be used in 9.3
        Coefficients not acceptable, e.g.:      INDIEN moontlik, merk positief van 9.2 na 9.3 en 9.4.
        Koëffisiënte nie aanvaarbaar, bv:
                                                Dit beteken:
                                             Die E°-waarde moet groter wees as –2,37 V
        MgMg 2Ag+2Ag
                2+
                                                Indien ‘n gas gebruik is, moet ‘n onreaktiewe electrode
      Examples for positive marking at 9.3:     (bv. Pt) in 9.3 gebruik word.
      Voorbeelde vir positiewe merkby 9.3:
                    
      MgMg2+Hg2+Hg
                
      MgMg2+NO3-NO2 .Pt
                                    ( 23 )                     Only NO3- can be used as anion
                 2+ 
9.4 Mg + 2Ag  Mg + 2Ag (bal)
            +
                                                                Slegs NO3- kan as anioon       (3)
        
    Mg + 2AgNO3  Mg(NO3)2 + 2Ag (bal)                        gebruik word

      Examples for positive marking at 9.4:/Voorbeelde vir positiewe merk by 9.4:
                              
      Mg  Hg2+  Mg2+ + Hg (bal)
         +
      Mg + 2NO3- + 4H+ Mg2+  + 2NO2 + 2 H2O (bal)

9.5      Internal resistance / Interne weerstand                                                  (2)
                                                                                                   [16]
 OR an explanation of the causes of internal resistance. (the term resistance has to be used.
 NOT only more resistance or lost volts)
 OF ‘n verduideliking van die oorsake van interne weerstand (die term weerstand moet
 gebruik word. NIE slegs meer weerstand of verlore volt nie.)
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                  MEMORANDUM                             9
                                   QUESTION 10 / VRAAG 10
10.1   D or/of E                                                                                (2)
                                                              for  deduct 1 mark
10.2   2C2H6 + 7O2   4CO2 + 6H2O            (Bal.  )                                          (3)
                                                              vir  trek 1 punt af
10.3   D                                                                                        (2)

10.4   A      .                                                                                 (2)

10.5   B will discolour  a Br2 solution  / B sal in ‘n Br2-oplossing ontkleur.               (4)
                     OR/OF I2         B will undergo addition ( 2 4 )/B sal addissie ondergaan ( 2 4 )

            O       H H                                        H O         H
                                                                      
10.6    H  C  O  C  C - H  (3)      IF/INDIEN: 10.6 H  C  C  O  C  H
                                                                        
                    H H                                        H           H
            H H                                            H       O H
                        O
                                                                
  OR/OF H  C  C  O - C  H              OR/OF: 10.6 H  C  O  C  C  H
                                                                    
            H H                                            H           H

                                           AND/EN: 10.7 Methyl ethanoate / Metieletanoaat
  Award ( 1 3 ) for:                          Then award 2 marks for 10.7
                                              Ken twee punte toe vir 10.7
  Gee ( 1 3 ) vir:

  CH3 CH2 COOH
  HCOO CH2 CH3
  CH3-CH2-COOH
  HCOO-CH2-CH3


10.7   Ethyl methanoate / Etielmetanoaat                                                          (2)
10.8   Sulphuric acid / swawelsuur(H2SO4 ( 1 2 )                                                (2)
                                                                                                 [20]
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                    MEMORANDUM                               10
16. GUIDELINES FOR MARKING
         This section provides guidelines for the way in which marks will be allocated and marking
         will take place in the national Senior Certificate Examinations in Physical Science. They
         are broad principles and the examiners may in specific questions decide to deviate from
         them.

16.1     No marks will be awarded for an answer that is based on an incorrect or inappropriate
         formula, even though there may be relevant symbols and applicable substitutions.

16.2     FORMULAE AND SUBSTITUTIONS:
16.2.1 Mathematical manipulations and change of subject of appropriate formulae carry no marks,
       but if a candidate starts off with the correct formula and then changes the subject of the
       formula incorrectly, marks will be awarded for the formula and the correct substitutions.
       The mark for the incorrect numerical answer is forfeited.

16.2.2 When an error is made during substitution into a correct formula, a mark will be awarded
       for the correct formula and of the correct substitutions, but no further marks will be given.

16.2.3 Marks are only awarded for a formula if a calculation had been attempted.                    i.e.
       Substitutions have been made or a numerical answer given.

16.2.4 Marks can only be allocated for substitutions when values are substituted into formulae and
       not when listed before a calculation starts.

16.3. UNITS:
16.3.1. Candidates will only be penalised once for the repeated use of an incorrect unit within a
        question.

16.3.2     Units are only required in the final answer to a calculation.

16.3.3 Marks are only awarded for an answer, and not for a unit per se. Candidates will therefore
       forfeit the mark allocated for the answer in each of the following situations:

             Correct answer + wrong unit
             Wrong answer + correct unit
             Correct answer + no unit.

16.3.4 S.I. units must be used except for certain exceptions, e.g. V.m-1 instead of N.C-1 , or cm.s-1
       or km.h-1 where the question warrants this. (Physics Paper 1 only.)
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                       MEMORANDUM                         11



16.4. POSITIVE MARKING regarding calculations will be followed in the following cases:

16.4.1 Subquestion to subquestion: When a certain variable is calculated in one subquestion (e.g.
       3.1) and needs to be substituted in another (3.2 or 3.3). e.g If the answer for 3.1 is incorrect
       and is substituted correctly in 3.2 or 3.3, full marks are to be awarded for the subsequent
       sub-questions.

16.4.2       In one subquestion itself: e.g. The candidate first has to calculate acceleration
         (using v = u + at) and then has to substitute it into F = ma to find the force (which is the
         final answer). The substitution of an incorrect answer of acceleration in F = ma will be
         awarded full marks. This only applies where two separate calculations are required in
         the same sub-question.

16.5     If a final answer to a calculation is correct according to the memorandum, full marks will not
         automatically be awarded. Markers will always ensure that the formula used and workings,
         including substitutions, are correct.

16.6. Questions where a series of calculations has to be made (e.g. a circuit diagram question) do
      not necessary always have to follow the same order. FULL MARKS will be awarded
      provided it is a valid solution to the problem. However, any calculation that will not bring
      the candidate closer to the answer than the original data, will not count any marks.

16.7. If one answer or calculation is required, but two given by the candidate, only the first one
      will be marked, irrespective of which one is correct. If two answers are required, only the
      first two will be marked, etc.

16.8. Normally, if based on a conceptual mistake, an incorrect answer cannot be correctly
      motivated. If the candidate is therefore required to motivate in question 3.2 the answer given
      to question 3.1, and 3.1 is incorrect, no marks can be awarded. However, if the answer for
      e.g. 3.1. is based on a calculation, the motivation for the incorrect answer in 3.2 could be
      considered.

16.9. If instructions regarding method of answering are not followed, e.g. the candidate do a
      calculation when the instruction was to solve by construction and measurement, a candidate
      may forfeit all the marks.

16.10.       For an error of principle no marks are awarded (Rule 1) e.g.:
       EXAMPLE (Rule 16.10)
         Calculate the force between two point charges of respectively 2 x 10-9 C and 3 x 10-9 C,
         if they are 0,5 m apart.
         Correct answer (1)                         Possible working (2)
               
              kq q                                       Gm 1m2
         F  1 2                                 F 
                r2                                         r2
                           
           9 x 10 9 x 2 x 10 -9 x 3 x 10 -9           9 x 10 9 x 2 x 10 -9 x 3 x 10 -9
                                                   
                        0,5 2                                      0,5 2
          2 x 10 7 C                        (5)    2 x 10 7 C                        (0)
   PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                             MEMORANDUM                      12


   EXAMPLE 2 (Rule 16.2)
   If the potential difference is 200 V and resistance is 25 , calculate the current.

   CORRECT    POSSIBLE                            POSSIBLE                 POSSIBLE           POSSIBLE
   ANSWER (1) ANSWER (2)                          ANSWER (3)               ANSWER (4)         ANSWER (5)
         
          
        
        
        
      V 
                                                            I
                 V                                 R                           V                    V
   I         R                                                           R                 I      
      R   
                I
                                                            V
                                                                                I                    R
                                                            200
          200                  200                                             R                =8A      (2)
                                                          25             I
          25                   25                                               V
                                                       I=8A          (0)
        =8A        (4)      =8A         (2)                                     25
                                                                            
                                                                                200
                                                                            = 0,125 A   (3)

   17.     GENERAL PRINCIPLES OF MARKING IN CHEMISTRY (PAPER 2)
             The following are a number of guidelines that specifically applies to Paper 2.

   17.1.     When using the formula pV = nRT, all substitutions have to be made in SI units. Should
             kPa and dm3 be used, the correct answer would be obtained, but the marks allocated for the
             substitutions of p and V would be forfeited.

   17.2.     When a chemical FORMULA is asked, and the NAME is given as answer, only one of the
             two marks will be given. The same rule applies when the NAME is asked and the
             FORMULA is given.

17.3.      When redox half-reactions are to be written, the correct arrow should be used. If the equation
                    H2S  S + 2H+ + 2e-                     ( 22 )
             is the correct answer, the following marks will be given:
             H2S  S + 2H+ + 2e-              ( 12 )
             H2S  S + 2H+ + 2e-              ( 02 )
             S + 2H+ + 2e-           H2S              (2 )
                                                         2
             S + 2H + 2e 
                     +       -
                                      H2S               0 )
                                                       ( 2

   17.4.     When candidates are required to give an explanation involving the relative strength of
             oxidising and reducing agents, the following is unacceptable:

           Stating the position of a substance on table 4 only (e.g. Cu is above Mg).
           Using relative reactivity only (e.g. Mg is more reactive than Cu).

             The correct answer would for instance be: Mg is a stronger reducing agent than Cu, and
             therefore Mg will be able to reduce Cu2+ ions to Cu (in terms of loss and gain of electrons).
   17.5         If ionic charges are omitted -1 mark.
   17.6         If a halfreaction is incomplete, no marks are awarded
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                 MEMORANDUM                          13
16. RIGLYNE VIR NASIEN

         Hierdie afdeling voorsien riglyne vir die wyse waarop punte toegeken en nasien geskied in
         die Nasionale Senior Sertifikaat Eksamens in Natuur- en Skeikunde. Dit is breë beginsels
         en die eksaminatore mag by spesifieke vrae besluit om daarvan af te wyk.

16.1.    Geen punte word toegeken vir ‘n antwoord wat op ‘n verkeerde of ontoepaslike formule
         gebaseer is nie, selfs al mag daar relevante simbole en toepaslike substitusies wees.

16.2.    FORMULES EN SUBSTITUSIES:
16.2.1 Geen punte word toegeken vir wiskundige manipulasies en veradering van formules nie,
       maar indien ‘n kandidaat met ‘n korrekte formule begin en dan die onderwerp verkeerd
       verander, word punte vir die korrekte formule en substitusies toegeken. Die punt vir die
       verkeerde numeriese antwoord word verloor.

16.2.2 Wanneer ‘n fout tydens substitusie in ‘n korrekte formule gemaak word, word ‘n punt
       toegeken vir die korrekte formule en die korrekte substitusies, maar geen verdere punte sal
       toegeken word nie.

16.2.3 Punte word slegs vir ‘n formule toegeken indien daar probeer is om ‘n bewerking te doen
       d.w.s waardes is vervang of 'n numeriese antwoord is gegee.

16.2.4 Punte kan slegs toegeken word vir substitusies wanneer waardes gesubstitueer word, en nie
       wanneer dit voor die berekening begin, gelys word nie.

16.3. EENHEDE:
16.3.1 Kandidate sal slegs een maal in ‘n vraag gepenaliseer word vir die herhaaldelike verkeerde
       gebruik van ‘n eenheid.

16.3.2     Eenhede word slegs in die finale antwoord van ‘n berekening vereis.

16.3.3 Punte word slegs vir ‘n antwoord toegeken, en nie vir ‘n eenheid as sulks nie. Kandidate sal
       daarom die punt wat vir die antwoord toegeken word, soos volg in verskillende situasies
       verloor:

             Korrekte antwoord + verkeerde eenheid
             verkeerde antwoord + Korrekte eenheid
             Korrekte antwoord + geen eenheid.

16.3.4 S.I. eenhede moet gebruik word, behalwe in sekere gevalle bv. V.m-1 in pleks van N.C-1 of
       cm.s-1 of km.h-1 waar die vraag homself daartoe leen (slegs Fisika Vraestel 1).
PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                       MEMORANDUM                      14
                                                         -11-

16.4. POSITIEWE NASIEN van bewerkings sal in die volgende gevalle toegepas word:
16.4.1 Tussen onderafdelings van vrae: Wanneer ‘n sekere verandering in een onderafdeling (bv.
       3.1) in ‘n ander (3.2 of 3.3) gesubstitueer moet word.
       Indien die antwoord in 3.1 verkeerd is maar korrek in 3.2 of 3.3 vervang word, sal volpunte
       vir die daaropvolgende onderafdelings van vrae toegeken word.

16.4.2       Binne een onderafdeling van ‘n vraag: (bv. Die kandidaat moet eers die versnelling
          (m.b.v. v = u + at) bereken en dit dan in F = ma substitueer om die krag te vind (wat die
         finale antwoord is). Vir die substitusie van die verkeerde waarde van die versnelling in F =
         ma sal volpunte toegeken word. Dit is slegs van toepassing waar twee afsonderlike
         berekenings in dieselfde onderafdeling van ‘n vraag vereis word.

16.5. Volpunte sal nie outomaties toegeken word indien die finale antwoord van ‘n berekening
      volgens die memorandum korrek is nie. Nasieners sal altyd verseker dat die formule gebruik
      en die bewerkings, insluitend die substitusies, reg is.

16.6. Vrae waar ‘n reeks van berekenings gedoen moet word (bv. ‘n stroombaanvraag) hoef nie
      altyd in dieselfde volgorde beantwoord te word nie. VOLPUNTE sal toegeken word vir ‘n
      geldige oplossing van die probleem. Geen punte sal egter toegeken word vir enige
      bewerking wat die kandidaat nie nader aan die antwoord bring as die oorspronklike
      gegewens nie.

16.7. Indien slegs een antwoord of berekening vereis word, maar die kandidaat twee gee, sal slegs
      die eerste een gemerk word, ongeag van watter een korrek is. Indien twee antwoorde vereis
      word, word net die eerste twee gemerk, ensovoorts.

16.8. Waar ‘n verkeerde antwoord op 'n konseptuele fout gebaseer is kan dit nie reg gemotiveer
      word nie. Indien dit dus vereis word dat die kandidaat in vraag 3.2 die antwoord op vraag
      3.1 moet motiveer, maar 3.1 is verkeerd, kan geen punte by 3.2 toegeken word nie. Indien
      die antwoord op bv. 3.1 egter op 'n berekening berus, kan die motivering in 3.2 in ag geneem
      word.

16.9. Indien instruksies omtrent die metode van beantwoording van vrae nie gevolg word nie, bv.
      Die kandidaat doen ‘n berekening terwyl die opdrag was om op te los deur konstruksie en
      meting, mag ‘n kandidaat al die punte verbeur.

16.10.       Vir ‘n beginselfout word GEEN PUNTE toegeken nie. (Reël 1) bv.:

     VOORBEELD (Reël 16.10)
         Bereken die krag tussen twee puntladings van onderskeidelik 2 x 10-9 C en 3 x 10-9 C,
         Indien hulle 0,5 m van mekaar af is.
         Korrekte antwoord (1)                      Moontlike bewerking (2)
               
              kq q                                       Gm 1m2
         F  1 2                                 F 
                r2                                         r2
                           
           9 x 10 9 x 2 x 10 -9 x 3 x 10 -9           9 x 10 9 x 2 x 10 -9 x 3 x 10 -9
                                                   
                        0,5 2                                      0,5 2
          2 x 10 7 N                        (5)    2 x 10 7 N                        (0)
   PHYSICAL SCIENCE HG PAPER 2 – NOVEMBER 2001                               MEMORANDUM                      15
                                                                   -12-
   VOORBEELD 2 (Reël 16.2)
   Indien die potensiaalverskil 200 V en die weerstand 25  is, bereken die stroom.

   KORREKTE                    MOONTLIKE                     MOONTLIKE             MOONTLIKE          MOONTLIKE
   ANTWOOD (1)                 ANTWOORD (2)                  ANTWOORD (3)          ANTWOORD (4)       ANWOORD (5)
         
          
        
        
         
                                                                   I
      V                             V                        R                         V                  V
   I                          R                                                  R                 I     
      R  
                                                                      V
                                    I                                                   I                  R
                                                                       200
          200                        200                                               R              =8A       (2)
                                                                     25          I
          25                         25                                                 V
                                                                 I=8A        (0)
        =8A              (4)       =8A             (2)                                  25
                                                                                    
                                                                                        200
                                                                                    = 0,125 A   (3)

   17.     ALGEMENE BEGINSELS VIR DIE NASIEN VAN CHEMIE (VRAESTEL 2)
             Die volgende is ‘n aantal riglyne wat spesifiek van toepassing is op Vraestel 2.

   17.1.     Wanneer die formule pV = nRT gebruik word, moet alle substitusies in SI-eenhede gemaak
             word. Indien kPa en dm3 gebruik word, mag die korrekte antwoord verkry word, maar
             punte vir die substitusies van p en V sal verbeur word.

   17.2.     Wanneer ‘n chemiese FORMULE gevra word en die NAAM gegee word as antwoord, sal
             slegs een van die twee punte toegeken word. Dieselfde reël is van toepassing wanneer die
             NAAM gevra word en die FORMULE gegee word.

17.3.      Wanneer ‘n redoks-halfreaksie neergeskryf moet word, moet die korrekte pyle gebruik word.
         Indien die vergelyking
                    H2S  S + 2H+ + 2e-                    ( 22 )
             die korrekte antwoord is, sal punte soos volg toegeken word:
             H2S  S + 2H+ + 2e-            ( 12 )
             H2S  S + 2H+ + 2e-                 ( 02 )
             S + 2H+ + 2e-                H2S            (2 )
                                                             2
             S + 2H + 2e 
                     +         -
                                           H2S            ( 02 )
   17.4.     Indien kandidate ‘n verduideliking in verband met die relatiewe sterkte van oksideer- en
             reduseermiddels moet gee, is die volgende onaanvaarbaar:
           Die noem van die posisie van stowwe slegs op tabel 4 (bv. Cu bokant Mg).
           Gebruik van slegs relatiewe reaktiwiteit (bv. Mg is meer reaktief as Cu).
             Die korrekte antwoord sal byvoorbeeld wees: Mg is ‘n sterker reduseermiddel as Cu, en Mg
             sal daarom in staat wees om Cu2+-ione tot Cu te reduseer (in terme van verlies en wins van
             elektrone).

   17.5         As ioniese ladings uitgelaat is –1 punt.

				
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