CHEMISTRY PAPER 1 Marking Scheme - 2019 KCSE Prediction Answers Set 2

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  1.  
    1. T3N2   √1
    2. Ionic bond √1
  2. Ethanol forms hydrogen bond√1 while butane has no√1 hydrogen bond
  3.  
    1. X Calcium Carbide / CaC2 √½
      Y Copper (II) sulphate √½
    2. b) CaC2(s) + 2H2 O(l) → C2 H2(g) + Ca(OH)2(aq)
  4.  
  5. Iron (II) sulphate change from green to yellow / brown √1 iron (iii) sulphate 
    Chlorine oxidizes Iron (ii) ions to Iron (iii) ions √1
    Fe2+ + Cl
  6.  
    1. Cations   Zn2+ √½
      Anion     NO3-   √½
    2. 2HCl(aq)   +   ZnO(s) → ZnCl2(aq)   + H2O(l) √1
  7. SiO2 form giant atomic structure √½
    SO2 simple molecular structure √½
    Giant atomic structure requires strong heating to break √1
  8. Heat the mixture and collect NH4Cl√½ in a cold surface. Add water √½ to the remaining mixture stir then filter. Obtain filtrate√½ Na2CO3 and residue√½ as CaCO3. Heat filtrate √½in a water bath to saturation and allow it to crystallize. √½
  9.  
    1. MnO4- = -1
      x + -2 (4) = -1 √1
      x-8 = -1
      x = 7   √1
    2. MnO-4 √1
      Oxidation number of manganese changes from +7 to +2 √1
  10.  
    1. Iron (iii) chloride / FeCl3   √1
    2. 2Fe(s) + 3Cl(g) → 2FeCl3(s)   √1
  11.  
    1. Non metallic √1
      Ironic radius is higher than the atomic radius √1
    2. Y<X<Z<W √1
  12.  
    1. Mg2+ + 2e- → Mg                  -2.36  √1
      Ag+ +  e- → Ag                     +0.80 √1
      Or
      Mg(s) / Mg2+(aq) // Ag+(aq) / Ag(s)   √1
    2. Ered - Eox
      0.80 v - -2.36   √1
      3.16v √1
  13. It reacts with ammonia to form Ammonium Sulphate √1

  14.  
    1. N ≡ N = 944      
      3(H-H)           3x436 = 1308
      Total energy required to break bonds
      944 + 1308
      = 2252 (endotherimic)   √1

      Bond formation
      2x(3x388) = -2328   √1
      ∆Hr = -2328 + 2252
      = -76kJmcl-  √1
    2. Exothermic√½. The heat of reaction is negative √½
  15.  
    1. Propyl ethanoate
    2. Alcohol - Prupanol
      Alkanic acid – Ethanoic acid
  16. Q = It
    = 2 x 4 x 60
    = 480 Columbs   √1
    Cu2+ + 2e → Cu
    Moles of copper deposited
    =480 x 1
    96500 x 2
    =0.002487gm √1

    Mass deposited = 0.002487 Mol x 63.5
    = 0.1579grams
  17. React excess copper √½ metal with concentrated nitric (v) acid to form Copper (II) nitrate√½ solution.
    React the solution obtained above with a soluble carbonate√½
    Filter to obtain copper carbonate as the residue. √½
    Wash the residue with distilled water and dry it between filter papers. √½
  18.  
    1. MgO √½ and Mg3N2 √½
    2.  
      1. 2Mg(s) + O2(g) → 2MgO(s) √1
      2. 3Mg(s) + N2(g) → Mg3N2(s)√1
  19.  
    1. Different crystalline forms of the same element in the same physical state √1
    2. In graphite, the fourth electron is delocalized √1 (unbounded) while all the four electrons in diamond are used for bonding√ The delocalized electron is responsible for conductivity
  20. Moles of MgSO4 X H2O in 500cm3.
    0.011M x 500
    1000
    = 0.0055mol √1
    Mole   =  Mass in gramm/R.F.M
    RFM =  1.23/0.0055
    = 224 √1
    MgSO4. XH2O = (24+32+64+18X) = 224 √½
    = 120 +18X = 224
    = 18X = 104
    X = 6 √½
  21.  
    1.  
      a - Alpha particle
      b - Gamma rays
      c - Beta particles
    2. beta particles are lighter√½ than alpha particles hence more deflected√½
  22.  
    1. Hydrolysis of ethane
    2. Fuel√½
      - Solvent√½
      - Beverage
  23.  
    1. Curve II √1
    2. Increases the kinetic energy√½ of the reacting particles hence increasing the rate of successful collisions. √½
    3. Surface area
      Catalyst
  24.  
    1. ∆H2 Enthalpy of formation of Carbon (II) Oxide (∆Hf CO)
      ∆H3 Enthalpy of combustion of CO
      (∆Hc   CO)
    2.  ∆H1 = ∆H2 + ∆H3
      = -110 + -283 √1
      = -393kJmol-1   √1
  25. Colourless droplets were formed at the cooler part of the test tube√½
    A gas that turns a moist red litmus paper blue was produced√½
    NH4SO4, FeSO4. 6H2O is a hydrated√1 salt and ammonium sulphate decomposes to form Ammonia gas √1
  26. Concentration of KOH          =          4/56
    = 0.0714M√½
    Moles of KOH used=
    0.0714 x 25
    1000
    = 0.001786Mol√½
    Mole ratio
    2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)
    2:1√½
    Moles of H2SO4 Used =
    0.001786
    2
     = 0.00089 Moles√½
    Concentration of H2SO4 =
    0.00089 x 1000√½
    22
    = 0.04√½
  27.  
    1. It does not form scum with hard water √1
    2.  it causes environmental pollution
      Since it is non-biodegradable
  28.  
    1. It is denser than air√½
      It does not burn neither does it support combustion √½
    2. Refrigeration as dry ice
      manufacture of sodium carbonate in Solvay process
      Used in preservation of soft drinks
  29. Ethanoic acid forms more√1 hydrogen bonds between molecules√1 than ethanol
  30.  
    1. Alcohol
      Tobbaco
      Bhang
      Miraa
      Cocaine
      Madrax
    2. Hallucinations
      Addiction
      Health problems
      Family break ups
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