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PHYSICAL SCIENCES CHEMISTRY PAPER 2 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS NOVEMBER 2016

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PHYSICAL SCIENCES (CHEMISTRY)
PAPER TWO (P2) 
GRADE 12
NSC EXAM PAPERS AND MEMOS
NOVEMBER 2016

MEMORANDUM 

QUESTION 1
1.1 D ✓✓ (2)
1.2 C ✓✓ (2)
1.3 C ✓✓ (2)
1.4 D ✓✓ (2)
1.5 B ✓✓ (2)
1.6 D ✓✓ (2)
1.7 A ✓✓ (2)
1.8 A ✓✓ (2)
1.9 B ✓✓ (2) 
1.10 B ✓✓ (2)

[20] 

QUESTION 2
2.1 

2.1.1 A OR/OF D ✓ (1)
2.1.2 B ✓ (1)
2.1.3 E ✓ (1)
2.1.4 D ✓ (1)

2.2 

2.2.1  2.2.1 (3)

Marking criteria: 

  • Five C atoms in longest chain. ✓
  • Two Br and one methyl substituents. ✓ 
  • Whole structure correct.  ✓


2.2.2        2.2.2 (2)

Marking criteria: 

  • Whole structure correct: 2/2
  • Only functional group correct: Max 1/2
  • Accept -OH as condensed.
    IF:
    More than one functional group 0/2

2.2.3      2.2.3  (2)

 Marking criteria: 

  • Whole structure correct: 2/2
  • Only functional group correct 1/2

IF: More than one functional group 0/2

2.3

2.3.1 Hydrogen (gas) ✓ (1) 
2.3.2 Addition / Hydrogenation ✓  (1)

[13] 

QUESTION 3
3.1 Compounds with the same molecular formula ✓ but different structural  formulae.✓  (2) 
3.2 Chain✓ (1) 
3.3 From A to C:

  • Structure:
    Less branched / less compact / less spherical/longer chain length / larger  surface area (over which intermolecular forces act).✓ 
  • Intermolecular forces:
    Stronger / more intermolecular forces / Van der Waals forces / London  forces / dispersion forces.  ✓ 
  • Energy:
    More energy needed to overcome or break intermolecular forces / Van  der Waals forces. ✓ 
    OR
    From C to A
  •  Structure:
    More branched / more compact / more spherical / smaller surface area  (over which intermolecular forces act).✓ 
  • Intermolecular forces:
    Weaker / less intermolecular forces / Van der Waals forces / London  forces / dispersion forces. ✓
  • Energy:
    Less energy needed to overcome or break intermolecular forces / Van  der Waals forces. ✓ (3) 

3.4 A / 2,2-dimethylpropane ✓ →      Lowest boiling point. ✓ (2)
3.5 C5H12 + 8O2 ✓ ⭢ 5CO2 + 6H2O ✓ Bal ✓    (3) 

Notes: 

  • Reactants ✓ Products ✓ Balancing ✓ 
  • Ignore double arrows and phases.
  • Marking rule 6.3.10.
  • If condensed structural formulae used:Max.2/3 

[11] 

QUESTION 4
4.1 

4.1.1 High temperature / heat / high energy / high pressure ✓  (1)
4.1.2 C6H12 ✓    (1) 

Accept:  Condensed structural formula and structural formula.
E.g : CH3CH2CH2CH2CHCH2

4.1.3 Alkenes ✓ (1) 

4.2 X / C6H12 / Alkene / Hexene  ✓ 
OPTION 1

  • X is an alkene / has a double bond / unsaturated. ✓ 
  • X can undergo addition. ✓ 
  • X will react without light / heat / is more reactive. ✓ 

OPTION 2

  • Butane is an alkane OR butane is saturated. ✓ 
  • Butane can only undergo substitution. ✓ 
  • Butane will only react in the presence of light / heat OR butane is less  reactive. ✓  (4) 

4.3 

4.3.1 2-chloro✓butane ✓  (2) 
4.3.2 Substitution / Hydrolysis ✓  (1) 
4.3.3   4.3.3 (2)

Marking criteria: 

  • Whole structure correct 2/2
  • Only functional group correct 1/2 

IF: More than one functional group 0/2

4.3.4 Hydration ✓ (1)

[13] 

QUESTION 5
5.1 

5.1.1 The minimum energy needed for a reaction to take place. ✓✓ 
OR
Minimum energy needed to form the activated complex(2) 
5.1.2   5.1.2 (3) 

Marking criteria:

 

Shape of curve for exothermic reaction as shown. 

      ✓      

Energy of activated complex shown as 75 kJ in line with the peak.

Energy of products shown as − 196 kJ below the zero. 

IF: Wrong shape, e.g. straight line.

0/3 

5.1.3 Marking criteria

  • Dotted line (---) on graph in QUESTION 5.1.2 showing lower energy for  activated complex. ✓ 
  • Dotted curve starts at/above energy of reactants and ends at/above energy  of products on the inside of the original curve. ✓  (2) 

Note: 
Allocate marks only if curve for either exothermic or endothermic reaction drawn in  QUESTION 5.1.2.  

5.1.4

  • A catalyst provides an alternative pathway of lower activation energy. ✓ 
  • More molecules have sufficient / enough (kinetic) energy. ✓
    OR
  • More molecules have kinetic energy equal to or greater than the  activation energy. 
  • More effective collisions per unit time / second. ✓
    OR
  • Rate / frequency of effective collisions increases.  (3) 

5.2

5.2 .1 Ave rate. tempo =   ΔV  
                                          Δt
= 52 - 16  ✓
   40 - 10
=1,2(dm3.s-1)✓  (3) 

Accept: 

  • Volume range:  16 to 17 cm3 
  • Answer range: 1,167 to 1,2 dm3∙s-1

5.2.2      (4) 

Marking criteria

  •  V(O2) = 60 dm3 AND divide volume by 24 ✓
  • Use ratio: n(H2O2) = 2n(O2) = 1:2 ✓
  • Use 34 g∙mol-1 in n =   m     or in ratio calculation. ✓ 
                                        M
  • Final answer: 170 g ✓

OPTION 1
n(O2) =  V  
            VM 
= 60   ✓
   24
= 2,5 mol  
n(H2O2) = 2n(O2)  
 = 2(2,5) ✓ 
 = 5 mol  
n(H2O2) =
                 M
 ∴5 = m   
        34 ✓
∴m = 170 g ✓ 

OPTION 2
24 dm3 : 1 mol 
60 dm3 : 2,5 mol ✓ 
n(H2O2) = 2n(O2)  
= 2(2,5) ✓ 
= 5 mol  
34 g ✓: 1 mol 
x : 5 mol  
x = 170 g ✓

OPTION 3 

n(O2) =  V  
            VM 
= 60   ✓
   24
= 2,5 mol 
 
n(O2) = 
               M
∴2,5 =  
           32
∴m = 80 g 
  ✓
2(34) g✓ H2O2 .......32 g O2 x g H2O2 ................ 80 g O2
m(H2O2) = 170 g ✓ 

5.2.3 Equal to  (1) 

5.3 

5.3.1 Q ✓ (1) 
5.3.2 P ✓ (1)

[20]

QUESTION 6
6.1 The stage in a chemical reaction when the rate of forward reaction equals the  rate of reverse reaction. ✓✓ (2 marks or no marks) 
OR
The state where the concentrations / quantities of reactants and products  remain constant. (2) 
6.2 

6.2.1 Remains the same✓ (1)
6.2.2 Decreases ✓ 

  • When the temperature is increased the reaction that will oppose this  increase / decrease the temperature will be favoured. ✓ 
    OR
  • The forward reaction is exothermic. 
  • An increase in temperature favours the endothermic reaction. ✓ 
  • The reverse reaction is favoured. ✓ (4) 

6.3

Marking criteria: 

  • Vertical parallel lines show a sudden increase in rate of both forward and reverse  reactions. ✓ 
  • Horisontal parallel lines showing a constant higher rate for both forward and  reverse catalysed reactions after time t1. ✓

    6.3    (2) 

6.4 CALCULATIONS USING NUMBER OF MOLES 

Marking criteria: 

  • Use M(PbS) = 239 g∙mol-1 in n =  or in ratio calculation ✓ 
                                                          M
  • Use ratio: n(H2S)equil = n(PbS) ✓
  • n(H2S)formed = n(H2S)equilibrium
  • USING ratio: H2 : H2S = 1 : 1 ✓
  • n(H2)equilibrium = n(H2)initia – n(H2)formed
  • Divide equilibrium n(H2S) & n(H2) by 2 dm3. ✓
  • Correct Kc expression ✓
  • Substitution of concentrations into Kc expression. ✓
  • Final answer: 0,07 ✓
    NB.: If not rounded: 0,067

 

OPTION 1
n(PbS) = m = 2,39 = 0,01 mol
               M     239 ✓
n(H2S)equilibrium = n(PbS) ✓ = 0,01 mol  

  H2   H2S  
Initial quantity (mol)   0,16  0  
 Chnage (mol)  0,01  0,01✔ ratio✔
 Quantity at equilibrium (mol)  0,15 ✔  0,01  
 Equlibrium concentration (mol.dm-3)  0,075  0,005 divide by 2✔

Kc  = [H2S ] ✔
          [H2  ]
= 0,005  ✔
   0,075
=0,067 = 0,07  ✔

 No Kc expression, correct substitution: Max : 8/9
 Wrong Kc expression: Max : 6/9
 IF: [S] = 1 in Kc  = [H2S ] 
                                [H2][S ]
No mark for Kc expression, but continue marking substitution and answer

 

OPTION 2
n(PbS) = m = 2,39 = 0,01 mol
               M     239 ✓
n(H2S)reacted = n(PbS) ✓ = 0,01 mol  
 = n(H2S)equilibrium
n(H2S)formed= n(H2S)equilibrium – n(H2S)initial
 = 0,01 – 0 ✓ 
 = 0,01 mol  
n(H2)reacted = n(H2S)formed✓ = 0,01 mol  
n(H2)equilibrium = n(H2)initial - n(H2)reacted
= 0,16 - 0,01 ✓ 
 = 0,15 mol  
c(H2) =          c(H2S)  = n
            V                          V
= 0,15               = 0,01
     2                       2
= 0, 075 mol.dm-3    = 0,005 mol.dm-3   

Kc  = [H2S ] ✔
          [H2  ]
= 0,005  ✔
   0,075
=0,067 = 0,07  ✔

 No Kc expression, correct substitution: Max : 8/9  Wrong Kc expression: Max : 6/9  IF: [S] = 1 in Kc  = [H2S ] 
                                [H2][S]
No mark for Kc expression, but continue marking substitution and answer

 

OPTION 3

  H2   H2S  
Initial quantity (mol)   0,16  0  
 Change (mol)  x  x✔ ratio✔
 Quantity at equilibrium (mol)  0,16 - x ✔  x  
 Equlibrium concentration (mol.dm-3) 0,16 - x
      2
 x  
 2
divide by 2✔

n(PbS) = m = 2,39 = 0,01 mol
               M     239 ✓
n(H2S)equilibrium = n(PbS)  ✔ ∴ x  = 0,01 mol
[H2]equilibrium 0,16 - 0,01 = 0,075  mol.dm-3   
                                 2
[H2S]equilibrium 0,01 = 0,005  mol.dm-3   
                                 2

Kc  = [H2S ] ✔
          [H2  ]
= 0,005  ✔
   0,075
=0,067 = 0,07  ✔

 No Kc expression, correct substitution: Max : 8/9  Wrong Kc expression: Max : 6/9  IF: [S] = 1 in Kc  = [H2S ] 
                                [H2] [S]
No mark for Kc expression, but continue marking substitution and answer

CALCULATIONS USING CONCENTRATION 

Marking criteria: 

  • Use M(PbS) = 239 g∙mol-1 in n =  m or in ratio calculation ✓
  • Use ratio: n(H2S)equil = n(PbS) ✓
  • Divide equilibrium n(H2S)equil & n(H2)initial by 2 dm3. ✓ 
  • (H2S) formed = (H2S)equal  ✓ 
  • USING ratio: H2:H2S =1 :1
  • [H2]equilibrium = [H2]initial – [H2]formed ✓ 
  • Correct Kc expression ✓
  • Substitution of concentrations into Kc expression. ✓
  • Final answer: 0,07 ✓ 

Note: If not rounded: 0,067

 

OPTION 4

n(PbS) = m = 2,39 = 0,01 mol
               M     239 ✓
 n(H2S)equilibrium = n(PbS) ✓ = 0,01 mol   

  H2   H2S  
Initia concentration ( mol.dm-3   )   0,16 / 2 = 0,08   0  
 Change in concentration ( mol.dm-3   )  0,05  0,05✔ ratio✔
 Equlibrium concentration (mol.dm-3)  0,075  0,005 divide by 2✔

Kc  = [H2S ] ✔
          [H2  ]
= 0,005  ✔
   0,075
=0,067 = 0,07  ✔

 No Kc expression, correct substitution: Max : 8/9
 Wrong Kc expression: Max : 6/9
 IF: [S] = 1 in Kc  = [H2S ] 
                                [H2][S ]
No mark for Kc expression, but continue marking substitution and answer

 

 

OPTION 5

n(PbS) = m = 2,39 = 0,01 mol
               M     239 ✓
 n(H2S)equilibrium = n(PbS) ✓ = 0,01 mol   
[H2S]equilibrium = n   
                            V
= 0.01
     2
= 0,005  mol.dm-3
[H2]initial =   n   
                   V
= 0.16
     2
= 0,008  mol.dm-3
[H2S]
formed= [H2S]equilibrium - [H2S]initial
= 0,005 - 0  ✓
= 0,005
 mol.dm-3
[H2]reacted = [H2S]formed = 0,005 mol
[H2]equilibrium= [H2]initial- [H2]reacted
= 0,008 - 0,005
= 0,075 mol

Kc  = [H2S ] ✔
          [H2  ]
= 0,005  ✔
   0,075
=0,067 = 0,07  ✔

 No Kc expression, correct substitution: Max : 8/9
 Wrong Kc expression: Max : 6/9
 IF: [S] = 1 in Kc  = [H2S ] 
                                [H2][S ]
No mark for Kc expression, but continue marking substitution and answer

(9) 
[18]

QUESTION 7
7.1 

7.1.1 Hydrolysis is the reaction (of a salt) with water. ✓✓ (2 or 0) 
Accept: 
A chemical reaction in which water is a reactant.  (2) 

7.1.2 Smaller than (7) ✓
NH4+ + H2O ✓ → NH3 + H3O+ ✓  
Accept: 
NH4Cℓ + H2O → NH3 + H3O+ + Cℓ- 
NH4+ → NH3 + H+     ((3)

Note: 

  • Mark equation independently of first  answer.
  • If incorrect balancing: Max 2/3 

7.2 

Marking criteria for equation: 

  • Reactants ✓ Products ✓
  • Ignore double arrows and phases.
  • Marking rule 6.3.10

7.2.1    (2)

Marking guidelines: 

  • Substitution of 98 g∙mol-1. ✓
  • Final answer: 0,08 mol ✓
  • Note:
    If not rounded: (0,075 mol) 

OPTION 1 

n = m = 7,35 = 0,08 mol
      M     98 ✓
(0,075 mol) 

OPTION 2
98 g ✓: 1 mol 
7,35 :0,08 mol ✓

OPTION 3 

n = m   = 7,35 
     MV   98 × 0,5✓
 = 0,15 mol∙dm-3 
n = cV 
 = 0,15 x 0,5 
 = 0,08 mol ✓

7.2.2 POSITIVE MARKING FROM QUESTION 7.2.1. 

OPTION 1

pH = −log[H3O+] ✓ 
1,3 ✓ = −log[H3O+]  
[H3O+] = 0,05 mol∙dm-3 
[H2SO4] = ½[H3O+
 = ½ x 0,05 ✓ 
 = 0,025 mol∙dm-3 (0,03)  

n(H2SO4)ex = cV ✓ 
 = 0,025 x 0,5 ✓ 
 = 0,0125 mol (0,02) 
n(H2SO4)react = 0,075 – 0,0125 ✓  
[the highlighted part is from Q7.2.1]
= 0,0625 mol (0,06) 
n(NaOH) = 2n(H2SO4
 = 2 x 0,0625 ✓ 
 = 0,125 mol (0,12)

OR EITHER⇒⇒⇒


n(NaOH) = m = 0,125 = m   
                  M                  40 ✓

 m = 5 g ✓ (4,8 g) 

Marking guidelines: 

  • Formula: pH = −log[H3O+] ✓
  • Substitution of 1,3 ✓
  • Use [H2SO4] : [H3O+] = 1 : 2 ✓
  • Formula: c =  ✓ 
                          V
  • Multiply by 0,5 dm ✓
  • Subtract ninitial – nexcess ✓ 
  • Use n(NaOH) : n(H2SO4) = 2:1 ✓ 
  • Substitution of 40 g∙mol-1
  • Final answer: m = 5 g ✓
    Range: 4,8 – 5,6 g

 

 

 

1 mol : 40 g ✓ 
0,125 mol : 5 g ✓

 

 

 

 

OPTION 2

pH = −log[H3O+] ✓ 
1,3 ✓ = −log[H3O+]  
[H3O+] = 0,05 mol∙dm-3 
n(H3O+)ex= cV ✓ 
 = (0,05)(0,5) ✓ 
 = 0,025 mol (0,03) 
n(H3O+)in = 2n(H2SO4
[the highlighted part is from Q7.2.1]

 = 0,075 x 2 ✓ 
 = 0,15 mol (0,16)  
n(H3O+)react = 0,15 – 0,025 ✓ 
= 0,125 mol (0,13) 
n(NaOH) = n(H3O+) ✓
 = 0,125 mol (0,13) 

OR EITHER⇒⇒⇒


n(NaOH) = m = 0,125 = m   
                  M                  40 ✓

 m = 5 g ✓ (5,2 g)

Marking guidelines: 

  • Formula: pH = −log[H3O+] ✓
  • Substitution of 1,3 ✓ 
  • Formula/Formule: c =  ✓ 
                                        V
  • Multiply by 0,5 dm3 ✓ 
  • Use n(H2SO4) : n(H3O+) = 1 : 2 ✓ 
  • Subtract ninitial – nexcess
  • Use n(H3O+) : n(NaOH) = 1 : 1 ✓
  • Substitution of 40 g∙mol-1
  • Final answer: m = 5 g ✓ 
    Range: 4,8 – 5,6 g

 

  

1 mol : 40 g ✓ 
0,125 mol : 5 g ✓

 

 

OPTION 3
 [H2S]ein =   =   0,075
                  V          0,5
[the highlighted part is from Q7.2.1]
= 0,15 mol∙dm-3  (0,16)
[H3O+]in = 2[H2SO4
 = 2 x 0,15 ✓ 

 = 0,3 mol∙dm-3 (0,32) 
pH = −log[H3O+] ✓ 
1,3 ✓ = -log[H3O+]  
[H3O+] = 0,05 mol∙dm-3 
[H3O+]react = 0,3 – 0,05✓ 
 = 0,25 mol∙dm-3 (0,27) 
[H2SO4]react = ½[H3O+]  
 = ½ x 0,25  
 = 0,125 mol∙dm-3 (0,14) 

Marking guidelines:

  • Formula:  c =  ✓ 
                          V
  • Divide by 0,5 dm3  ✓ 
  • Use [H3O+] : [H2SO4] = 2:1 ✓
  • Formula: pH = −log[H3O+] ✓ 
  • Substitution of 1,3 ✓ 
  • Subtract [H3O+]initial – [H3O+]excess
  • Use n(NaOH) : n(H2SO4) = 2:1 ✓
    OR
  • Use [H2SO4] : [NaOH] = 1 : 2 ✓
  • Substitution of 40 g∙mol-1 ✓ 
  • Final answer: m = 5 g ✓
    Range: 4,8 – 5,6 g
 

n(H2SO4)react = cV  
= (0,125)(0,5)  
= 0,0625 mol (0,07) 
n(NaOH) = 2n(H2SO4)  
 = 2 x 0,0625 ✓ 
 = 0,125 mol (0,14) 

n(NaOH) =
                  M 
0,125 =  m  
             40✓
 m = 5 g ✓ (5,6 g) 

 [H2SO4] : [NaOH]  

            1 : 2 
     0,125 : 0,25 ✓ (0,28) 

m = cMV  
 = 0,25 x 40 ✓x 0,5  
 = 5 g ✓ (5,6 g)

(9) 
[16] 

QUESTION 8

8.1 

8.1.1 AgNO3 / Silver nitrate ✓ (1)
8.1.2 Ni → Ni2+ + 2e- ✓✓   (2) 

Marking guidelines: 

  • Ni ⇌ Ni2+ + 2e-   ½                                      Ni2+ + 2e- ⇌ Ni  0/2 
    2 Ni2+ + 2e- → Ni ½                                    Ni2+ + 2e- ← Ni 0/2
  • Ignore if charge omitted on electron.
  • If charge (+) omitted on Ni2+: Max: 21 Example: Ni → Ni2 + 2e- ✓ 

8.1.3 Ni + 2Ag+ ✓ → Ni2+ + 2Ag ✓     Bal ✓ 
OR
Ni + 2 AgNO3 → Ni(NO3)2 + 2Ag    (3) 

Notes: 

  • Reactants ✓ Products ✓ Balancing: ✓
  • Ignore double arrows.
  • Marking rule 6.3.10/ 

8.2 

8.2.1 Ni ✓ -  Ni is a stronger reducing agent. / Ni has a higher reducing ability. / Ni is the  anode. / Ni loses electrons. / Ni is oxidised. ✓  (2) ✓ ✓ ✓ 

8.2.2 Ni (s) | Ni2+ (aq) || Ag+(aq) | Ag(s) 
OR 
Ni (s) | Ni2+ (1 mol∙dm-3) || Ag+(1 mol∙dm-3) | Ag(s) 
Accept: 
Ni | Ni2+ || Ag+ | Ag (3) 

8.2.3     (4) 

OPTION 1:
Eθcell = Eθreduction −Eθoxidation✓ 
= 0,80 ✓ – (-0,27) ✓  = 1,07 V ✓

Notes

  • Accept any other correct formula from the data  sheet
  • Any other formula using unconventional  abbreviations, e.g. E°cell = E°OA - E°RA followed by  correct substitutions: ¾ 

OPTION 2
Ag+ + e- ⭢ Ag✓                 Eθ = 0,80 V ✓ 
Ni ⭢ Ni2+ + 2e-                  Eθ = +0,27 V ✓ 
Ag+ + Ni ⭢ Ag + Ni2+         Eθ = +1,07 V ✓ 

8.2.4 Increases  ✓ (1)

[16] 

QUESTION 9
9.1 Endothermic ✓ (1) 
9.2 Anode ✓ - Connected to the positive terminal of the battery. ✓ (2) 

9.3 

9.3.1 Chlorine (gas) / Cℓ2✓ (1)
9.3.2 Hydrogen (gas) /H2  ✓ (1)
9.3.3 2H2O(ℓ) + 2e- ⭢ H2(g) + 2OH-(aq) ✓✓   (2) 
Ignore phases 

Notes
H2(g) + 2OH-(aq) ← 2H2O(ℓ) + 2e-  2/2        2H2O(ℓ) + 2e- ⇌ H2(g) + 2OH-(aq) 1/2 
H2(g) + 2OH-(aq) ⇌ 2H2O(ℓ) + 2e 0/2         2H2O(ℓ) + 2e- ← H2(g) + 2OH-(aq)  0/2 

9.4 Basic  ✓ -  OR Alkaline 
OH (ions) / NaOH / Strong base forms.✓  (2) 

[9]

QUESTION 10
10.1 

10.1.1 Haber (process)  ✓ (1) 
10.1.2 Contact process / Catalytic oxidation of SO2 ✓  (1)
10.1.3 Sulphur trioxide / SO3  ✓ (1)
10.1.4 SO3 + H2SO4 ✓ ⭢ H2S2O7 ✓ Bal. ✓   (3) 

Notes

  • Reactants ✓ Products ✓ Balancing ✓
  • Ignore ⇌ and phases
  • Marking rule 6.3.10

10.1.5 H2SO4 ✓ + 2NH3 ✓ ⭢ (NH4)2SO4 ✓ Bal. ✓   (4)

Notes

  • Reactants ✓✓ Products ✓ Balancing ✓
  • Ignorer ⇌ and phases
  • Marking rule 6.3.10

10.2 (4) 

Marking guidelines: 

  • Calculate the mass of fertiliser.
  • Add %N and %P OR mass N and mass P.
  • Subtraction: 100 – (%N + %P) 
    OR m(fertiliser) – [m(N) + m(P)]
    OR %fertiliser – [%N + %P] 
  • Final answer: 8:1:5

 

OPTION 1:

m(fertiliser) = 36/100  x 20 
 = 7,2 kg  
%N = 4,11  x 100 
           7,2 
 = 57,08% 
 %P = 0,51  x 100 
           7,2 
 = 7,08% 
%K = 100 – ✓ (57,08 + 7,08) ✓ 
= 35,84%  
57,08 : 7,08 : 35,84 
       8 :    1   : 5 ✓ 

OPTION 2:

m(fertiliser) = 36/100 x 20 ✓
 = 7,2 kg  
m(K) = 7,2 – ✓ (4,11 + 0,51) ✓ 
= 2,58 kg 
4,11 : 0,51 : 2,58  
     8 :    1   : 5 ✓ 

OPTION 3

%N =4,11 x 100 = 20,55%  ✓
          20 
%P =0,51 x 100  = 2,55% 
          20 
%K = 36 –✓ (20,55 + 2,55) ✓ = 12,9%  

20,55 : 2,55 : 12,9 
       8 :     1  : 5 ✓ 

[14] 
TOTAL: 150

Last modified on Tuesday, 15 June 2021 12:40