233/1 – CHEMISTRY PAPER 1 – MARKING SCHEME MANGU HIGH SCHOOL TRIAL 2 MOCK 2021

1. (a)  Liebig’s condenser.

(b)  Determine the point at which one of the liquids in a mixture has evaporated completely.  Temperature tends to remain constant when one liquid in a mixture is evaporating.

(c)  Liquid ‘C’ since it is more volatile.

• (a)  Pass the mixture of gas “D” and “E” through sulphuric acid.  Gas “D” will react to form salt – leaving behind gas “E”. Collect gas E by downward delivery / upward displacement of air since it is heavier than air.

OR

Pass a mixture of gas “D” and “E” over sodium hydroxide.  Gas “D” will dissolve but gas “E” will not be affected.  Collect gas “E” by downward delivery.

• Ammonia gas (HN₃)

Ammonia is lighter than air.  It reacts with acids to form salt since itself is basic.  It does react with sodium hydroxide since both are basic but will dissolve in it without any reaction.

• (a)  A strong base is a substance that dissociate, completely to produce many hydroxide ions.

(b) 

                                               HCl

Volume of

CO² in cm³                                     CH₃COOH

                              Time (min)

• 2C_(s)  +  O_2(g)                     2CO_(g)

2CO_(s)  +  Fe₂O_3(s)              2Fe_(s)  +  CO_2(g) 

• Magnesium phosphide.

• IV, II, I, III

• (a)  (i)  B                (ii)  C
• D

• LUMINOUS                                   NON LUMINOUS

– Sooty Flame                               Non-sooty

–  Produce more light                    Less light

–  Less heat                                   Very hot

–  Weavy flame                              Steady flame               (any 2 mentioned @ 1mk)

• (a)  Colour of solid changed from black to reddish brown.

(b)  CuO_(s)  +  CO_(g)               Cu_(s)  +  CO_2(g)

(c)  CO is poisonous.

1. (a)  Under the same condition of temperature and pressure, the rate of diffusion of gas is inversely proportional to the square root of its density.

(b)  RCO₂              =          MNO₂              But RCO₂        =      100cm³ = 3.33cm³/per sec.

      RNO₂                           MCO₂                                                  30

                  3.33     =          46

                  RNO₂                      44

                  RNO₂               =            3.33               =          3.26cm³ per second

                                                      1.0225

      Time taken      =          150cm³            =          46 seconds

                                          3.26cm³sec^-1

1. (a)  I. Propan – I – ol

II.  2-Methylbutan-I-oic

(b) 

H                     C₆H₅

            C = C

H                     H

1. (a)  A bright yellow precipitate formed.

• 8cm³ – the graph flattens //

         –  PbNO₃ used up.

• Volume lower than 8cm³of NaI, Pb(NO₃)₂ was in excess.

–  At 8cm³ the ratio of moles of Pb²⁺ ions to Cl^– ions was 1:2

–  After 8cm³ the I^– ions are in excess, all Pb²⁺ ions are used up.

1. (a)  Blue solution changes to colourless // pale blue.P½

–  Brown solid is deposited. P½

• ΔH      =          MCΔT

                        =          0.03kg x 4.2kJkg^-1k^-1 x 15k

                                                =          1.89kJ

                                                =          -1.89kJ

1. IncreasedP1

HCl is reduced, thus the rate of forward reaction is favoured. P1

// NaOH neutralizes HCl hence the concentration of HCl is reduced and the reaction proceeds faster in the forward direction to produce more HCl and hence the precipitate.

1. (a)  Lead sulphide

Air / Oxygen                                                    ( ½ mk each)

• SO₂ leads to formation of acid rain

SO₂ causes respiratory infection.                  (any mentioned 1mk)

•  Reduces Lead (II) oxide into lead metal.

1. (a)  216      =          208 + 4M                     84        =          82 + 4 + n

216 – 208 = 4M

4M = 8                                                 84 – 86 =         n

M = 2                                                   n = -2 = 2

•            t½                    t½                    t½                    t½                   t½

      48                    24                    12                    6                      3                  1.5

      5t ½     =          270

      t ½       =          270      =          54 days

                                5

      Or

      N1       =          No(½) ^T/_t½                    5t ½ = 270

      1.5       =          48 (½) ²⁷⁰/_t½                 t½        =          270

      48                                                                                  5

       1         =          (½)²⁷⁰/_t½                       t ½       =          54 days

      32

      ( ½)⁵    =          ( ½)²⁷⁰/_t½ 

1.  

1. No. of moles          =          m

R.f

=          15

            58.5

                              =          0.25641 moles

Molarity                 =          No. of moles

                                          Volume in litres

                              =          0.25641

                                             0.12

                              =          2.13675M

1. (a)  Boyles law states that the volume V₁ of a fixed mass of a gas is inversely proportional to its pressure P₁ when temperature is kept constant.

(b)  P₁V₁    =          P₂V₂

         T₁                    T₂

780 x 400        =          600 x V₂

    303                              323

• V₂  =          323 x 780 x 400

    303 x 600

                        =          554.3234cm₃

• (a)  Allotropy – existence of an element in more than one structural form in the same physical state.

(b)  Rhombic /         P½

      Monoclini // βP½

•  (i)  Blue flameP(1mk)  // pungent smell

(ii)  AcidicP(1mk) // low pH

• (a)  Produced when unburnt carbon glows. P1

(b)  –  It is hot.

      –  Does not produce soot which makes apparatus dirty.

• (a)  Electrolytes are melts or aqueous solutions which allow electric current to pass through them and are decomposed by it while non-electrolyte are melts or aqueous solutions which do not conduct electric current.

–  Electrolytes contain mobile ions while non-electrolytes contain molecules.

(b)  Graphite has delocalized electrons in its structure which carry electric current.

• Ratio                                  M                                 O

                  Moles              0.254                           0.64P½

                                          63.5                             16

                                          0.04                             0.04P½

                  Ratio                            1          :           1P½

                  E.F      =          MOP1

• (i)  Prevent rusting P½

– Improve appearance P½

(ii) Q = 0.5 x 60 x 60P½

                  =          1800C

      Ag⁺  +  e          AgP½

      96500C           108g

      1800                1800 x 108P½

                                   95600

                              =          2.0145gP½

• –  Add water to Lead (II) nitrate to obtain Lead (II) nitrate solution. P½
  • Add water to sodium carbonate to obtain sodium carbonate solution. P½
    • Mix the two solutions to ppt Lead (II) carbonate. P1
    • Filter to obtain Lead (II) carbonate as a residue.P½
    • Wash the residue and dry between filter papers. P½

• (a)  Propane.
• CH₃CH₂CH₂COONa + NaOH                     Na₂CO₃ + C₃H₈        ignore state symbols.
• (i)  Green – yellow chlorine turns colourless.
  •  Chloropropane

• (a)  (i)

(ii)

• Aluminium has more protons than Mg, hence a greater nuclear charge than Mg, resulting in stronger metallic bases in Al than in Mg.

–  During the manufacture of nitrogen and oxygen gases.                             (2mks)

• (i)  To displace the air in the aspirator.

(ii) A black solid (1mk); copper (II) oxide is formed // copper is oxidized to copper (II) oxide.

(iii) Magnesium nitride. (1mk).