Answering scheme paper 2 Trg 2010

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					Answering scheme paper 2 Trg

Section A

Q1          a. Proton number is the number that represent the number of protons
               in nucleus of atom                                                     1 mrk
            b. electron                                                               1 mrk
            c. i. 6                                                                   1 mrk
               ii. Has same proton number//has same number of valence electron        1 mrk
            d. i. 14                                                                  1 mrk

                                                                                      1 mrk

            e. i. 2.8                                                                 1 mrk
               ii. 1+                                                                 1 mrk
Q2          a. i. Any monoprotic acid eg. HCl// HNO3                                  1 mrk
                  Any Suitable metal that have 2+ charge + anion of chosen acid)      1 mrk
                  eg. ZnCl2//MgCl2//Zn(NO3)2//Mg(NO3)2
               ii. 1 .... 2 ....1 .....1                                              1 mrk
            b. i. Copper(II)oxide and carbon dioxide                                  1 mrk
               ii. CuCO3  CuO + CO2                                                  1 mrk
            c. i. Na2SO4 + BaCl2  BaSO4 + 2NaCl                                      1 mrk
               ii. mol of Na2SO4
                    n = MV/1000
                      = ( 1 x 500)/1000
                     = 0.5 mol                                                        1 mrk

                   Mol ratio                       mass of BaSO4
                   Na2SO4 : BaSO4                 = 0.5 x (137 + 32 + 4(16))          1 mrk
                      1     : 1                    = 136.5 g                          1 mrk
                     0.5    : 0.5
Q3          a.   i. porous pot                                                        1 mrk
                 ii. to allow ion exchange take place                                 1 mrk
            b.   i. zinc become thinner//corrode//mass decrease                       1 mrk
                 ii. Zn  Zn2+ + 2e                                                   1 mrk
            c.   reduction                                                            1 mrk
            d.   electron from Zn to Cu through the circuit                           1 mrk
            e.   i. Inverse                                                           1mrk
                 ii. electron travel from Mg to Zn because electron was released by
                 Mg because Mg is more electropositive than Zn//higher position       1 mrk
                 compare to Zn in electrochemical series
            f.   i. Cu, Q , P                                                         1 mrk
                 ii. 1.5 V                                                            1 mrk
Q4    a. Strong alkali is a alkali that fully ionised in water to produced high     1 mrk
         concentration of hydroxide ion
      b. i. Sulphuric acid                                                          1 mrk
         ii. The pH value of sulphuric acid is lower than ethanoic acid             1 mrk
      c. i. Phenolthalein//methyl orange                                            1 mrk
         ii. the solution turn from pink to colourless//                            1 mrk
         iii. H2SO4 + 2NaOH  Na2SO4 + 2H2O                                         2 mrk
         iv. from equation
              mol ratio H2SO4 : NaOH
                          1 : 2                                                     1 mrk

                 M 1V 1 1
                                                                                   1 mrk
                 M 2V 2 2

               2(0.5 x 20) = M2 x 25
               M2 = 0.8 mol dm-3                                                    1 mrk
Q5.   a.   i. To detect//indicate the present of ion Fe2+                           1 mrk
           ii. Fe  Fe2+ + 2e                                                       1 mrk
           iii. B                                                                   1 mrk
                the present of dark blue precipitate is more than test tube A and
                C                                                                   1 mrk
           iv. Q, Fe, P                                                             1 mrk
      b.   i. 5 Fe2+ .......... 4 H2O                                               1 mrk
           ii. purple to colourless                                                 1 mrk
           iii. Fe2+                                                                1mrk
                Fe2+ was released and electron to form Fe3+ // oxidation
           number of Fe change from +2  +3                                         1 mrk
           iv. Put in the solution from electrode Q into sodium hydroxide
           solution, brown precipitate form, Fe3+ was presence//any correct         2 mrk
           chemical test of ion Fe3+ .
Q6.   a.   The combustion of 1 mol propanol will produced 50.425 kJ of              1 mrk
      b.   Exothermic reaction                                                      1 mrk
      c.   i. C3H7OH + 5O2  3CO2 +4H2O                                             2 mrk
           ii. carbon dioxide and water                                             2 mrk
      d.   Product contain less energy compare to reactant                          1 mrk
      e.   i. Number of mol propanol
           3g/((12)3+1(8) +16) = 0.05 mol                                           1 mrk

           1 mol propanol produced 50.425 kJ
           0.05 mol produced
           0.05 x 50.425 = 2.512 kJ                                                 1 mrk

                                                                                    3 mrk

           ii.                        + 3 labels

Q7      a. The position of U in Periodic table
           Group 1 Period 2                                                            2 mrk
           In group 1 because the number of valence electron is 1                      1 mrk
           In period 2 because U has two shell that filled with electron               1 mrk
        b. i. V2 + Fe  FeV2                                                           2 mrk
           ii. V is more reactive compare to w                                         1 mrk
               because the ability to attract an electron the its valence shell is
               higher compare to W
              distance of valence shell and nucleus V is nearer compare to W
              because V has 2 shell but W has 3 shell so the ability of V to           1 mrk
              attract electron to its valence shell is higher compare to W, thus its
              more reactive.
        c. i. Compound X                                                               1 mrk
           ii. formed by sharing electron between element A and element B              1 mrk
           iii. Physical properties of compound X and Y

             Compound X              Physical properties     Compound Y                Any two
             Lower                   Melting point           higher                    correct
             Lower                   Boiling point           higher                    comparison of
             Insoluble               Solubility in water     soluble                   physical
             soluble                 Solubility in organic   insoluble                 properties for
                                     solvent                                           compound X
             Cant conduct            Electrical              Able to conduct           and Y
             electricity in both     conductivity            electricity in liquid
             state                                           state but not in          4 mrk
                                                             solid state

            Explanation 2 of given physical properties                                 (4mrk)
            The different on the melting point and boiling point of compound X
            and Y is affected by the force that bind the particles in the both
            In compound X is Van Der Walls Force which is weaker compare to            2 mrk
            Electrostatic Force in compound Y so the melting point of
            compound Y is higher compare to compound X

            The different on their ability to conduct of electricity is depend on
            the presence on moving ion and type of particle.
            In solid state both compound X and Y do not conduct an electricity
            because the absent of free moving particle. All the particle was held      1 mrk
            in their fix position by the intermolecular force.

            In liquid state ions that held in their fix position in solid Y can move
            freely so compound Y can conduct an electricity, but in compound X
            even the particle is move freely but there are non ions presence           1 mrk
            because it made up from molecule(covalent compound) so its cant
            conduct electricity.
Q8.       a. i.
       Hydrocarbon A              Properties            Hydrocarbon B
       Single bond                Type of bond          Double bond
       Alkane                     Homologous Series     Alkene                     6 mrk
       CnH2n+2                    General Formula       CnH2n

           ii. C4H10 + 2 O2  4CO2 + 5H2O// 2C4H10 +13O2  8CO2 + 10H2O            2 mrk
              Gas C : Carbon dioxide
                                                                                   1 mrk
          iii. Hydrocarbon B
                                                                                   1 mrk
              Hydrocarbon B has double bond so the addition reaction will occur
             not in hydrocarbon A which has single bond, the substitution
                                                                                   2 mrk
             reaction only occur by the presence of UV light. So hydrocarbon B
             will change the colour of the bromine water from brown to

      iv. The hydrocarbon B produced more soot compare to hydrocarbon A
                                                                                   1 mrk
          The percentage of carbon in B is higher compare to A
          Percentage of Carbon in hydrocarbon A
        (5 12 +10(1)
                       x 100 = 85.71%                                              1mrk

         Percentage of Carbon in hydrocarbon B
                      x 100 = 88.25%                                               1 mrk
        (5 12 +8(1)

         b. Name of carboxylic acid: Propanoic acid                                1 mrk
            Name of alcohol : Ethanol                                              1 mrk
            Correct structural formula for both named carbon compound
            above.                                                                 2 mrk

9         a. Gas Y: Carbon dioxide                                                 1 mrk
             Chemical test: Bubble/channelled the gas through lime water, lime
             water turn cloudy, carbon dioxide was presence.                       2 mrk
             Anion in solution X: nitrate ion                                      1 mark
             Add in dilute sulphuric acid solution into test tube that contain
             anion of X than add in ferum(ii)sulphate solution and lastly slowly   2 mrk
             drip the concentrated sulphuric acid into the solution. Brown ring
             will form.
             Nitrate ion was presence.
             Solid X : Lead(II)nitrate
             The heating of lead(II)nitrate                                        2 mrk
             Observation: Brown gas was release, the solid was brown in colour
             during the hot but when cooled its turn yellow
             The heating of lead(II)carbonate
             Observation: Colourless gas produces that turn the lime water to      1 mrk
             cloudy, the solid was brown in colour during the hot but when
             cooled its turn yellow
             The gas that produce during heating of lead(II)nitrate is nitrogen
             dioxide(brown gas) and oxygen
             The gas produces during heating of lead(II)carbonate is carbon
          b. Suggestion of substance: any soluble carbonate compound eg
             Sodium carbonate//potassium carbonate//ammonium carbonate                   1 mrk

              Procedure of lead(II)carbonate preparation
              Name of reaction: Precipitation reaction                                   1 mrk

              1. Put in 50cm3 of sodium carbonate 0.5 moldm-3 solution into the          1 mrk
                 beaker of 100 ml.
              2. Put in 50cm3 of lead(II)nitrate 0.5 moldm-3 solution into the           1 mrk
                 beaker .
              3. White precipitate form in the beaker.                                   1 mrk
              4. Filtered the solution to separated the precipitate from the
                 solution.                                                               1mrk
              5. Rinse the precipitate using distilled water.                            1mrk
              6. Dry the precipitate with the filter paper.                              1mrk

                Chemical equation:
                Na2SO4 + Pb(NO3)2 2NaNO3 + PbCO3                                        2mrk
          c. Using baking powder that contain carbonate compound                         1 mrk
                Carbonate compound will produce carbon dioxide gas during the
                heating and the cake will rise nicely// using yeast during the           2mrk
                preparation of doh, leave it for 30 minute, the activity of yeast will
                produce carbon dioxide that trapped in the doh so the cake will rise
                nicely when it was baked.
Q10.   a. i. Any acid and metal eg: HCl//HNO3//H2SO4 and Mg//Zn                          2 mrk
               Eg. 2HCl + Mg  MgCl2 + H2                                                2 mrk
          ii. Experiment 1 30/10 = 0.3 cm3s-1
               Experiment 2 30/20 = 0.15 cm3s-1                                          2 mrk
           iii. the concentration of acid that use in experiment 1 is higher
               compare to experiment 2.
               So in exp. 1 number of particle in the volume of the solution is
               higher compare to exp. 2 .                                                4 mrk
              Thus, Frequency of the collision increase in exp 1,
                frequencies of effective collision also increase.
              Rate of reaction is higher compare to exp2.

       b. Check exp. Sodium thiosulphate                                                 10 mrk

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