Chemistry Nov/Dec 2010

ANSWER

In part (a)(i) and (ii), candidates correctly defined atomic number and isotopy thus:
– atomic number is the number of protons (in the nucleus) of an atom of an element/the number of electrons in a neutral atom.
– isotopy is a phenomenon whereby atoms of an element exhibit different mass numbers but have the same atomic number/the same number of protons but different number of neutrons.

In part bli], candidates correctly stated the sub atomic particles and the corresponding number present in the two isotopes as follows:

                Proton         Electron          Neutron
³⁵₁₇R          17                  17                   18
³⁵₁₇R          17                   17                   20

In (b)(iiL I and II, candidates correctly calculated the relative atomic mass of the element but most of them could not infer that the element is diatomic and hence lost mark in II. The expected calculations from candidates were as follows:
(I)          ReI. atomic mass of R = (75.5 x 35 + (24.5 x 37)
                                                            100
                                   =35.49 or = 35.5
(II)         Molar mass of R₂ = 2 x 35.5
                                      = 71 gmol-1
                   71g → 602 x 10²³
                   No. of molecules of R₂
                   = 6.02 X 1023 x 17.75
                                 71
                   1.51 X 10²³ molecules

In part (c)(i),(ii), (Hi) and (Iv), candidates correctly defined natural radioactivity, stated the differences between natural and artificial radioactivity, copied, and completed the given nuclear reaction and stated the type of reaction exhibited by the reaction thus:

   (i)           It is spontaneous disintegration/decay of a nucleus to produce radiations/particles (and energy)
(ii)

²³⁵₉₂U + ¹₀n →²³⁶₉₂U →⁹²₃₆Kr + ¹⁴¹₅₆Ba + ³¹₀n +Energy

(iv)               Fission

In part(d), only few candidates knew that ethanol is completely miscible with water because the two compounds contain hydroxyl group (OH)/both are polar which allows for (intermolecular) hydrogen bonding.

Question 2

(a) (i) Mention the various states of matter (ii) Arrange the states of matter in decreasing order of:                I.           kinetic energy of particles;                II.          disorderliness between particles;                III.         forces of cohesion between particles. (iii) A pure sample of a substance melts at 35°C and boils at 100°C.                  I.            Sketch a graph to show the heating curve                   II.          What is the state of the substance at normal room temperature? [9 marks] (b) Consider the following equation: 2KHC03(s)     What volume of CO2(g) measured at s.t.p. would be produced when 25.0 g of KHC03(s) was completely decomposed? [K = 39.0 = 16.0; C = 12.0; H = 1.00; Molar volume of a gas ats.t.p. = 22.4 drn” mol”] [ 5 marks] (c) When ammonium chloride was heated in a dry test tube, it was observed that the damp red litmus paper placed at the tip of the test tube immediately turned blue and after some seconds turned red. (i) State the gas that was responsible for the: I.           red litmus paper turning blue; II.           blue colour of the litmus paper changing back to red. (ii) Give reason for your answer in 2 (c)(i) (iii) Write an equation for the decomposition [6 marks]          (d)   (i) Name two salts that could react with dilute acids to produce sulphur (IV) oxide (ii) Mention                  I. two uses of sulphur (IV) oxide;                  II. one chemical property of sulphur (IV) oxide. [5 marks]

ANSWER

In(a)(i) and (ii), Candidates correctly mentioned solid, liquid and gas as the states of
matter and gave the arrangement of the states of matter as follows:

(ii) I Gas > liquid > solid
     II       Gas > liquid > solid
     III       Solid > liquid > Gas

In (a)(iii) I and II, most candidates could not sketch the required heating curve graph and hence, did not know the substance. The expected answer from candidates were as follow:



II. Solid.

In part (b), candidates correctly determined the volume of CO₂(g) at S.T.P thus:

Molar mass of KHCO₃ = 39 + 1 + 12 + ( 3 x 16) = 100gmorl^-1
No. of mole of KHCO₃ = 25.0 = 0.25 mol
                                         100
From the equation
     2 moles of KHC0₃ = Imole CO₂
     No. of mole of CO₂ = ½ x 0.25 mol
                                       = 0.125 mol.
Volume of CO₂ at s.t.p
     = 0.125 mol x 22.4 dm³ mol^-1
     = 2.8 dm3

           OR

Alternate Method

Molar mass of KHCO₃ = 39 + 1 + 12 + (3 x 16) = 100gmol^-1
2 x 100g KHCO₃ = 22.4 dm³
., 25g = 22.4 x 25
                 200
           = 2.8dm³

In part (c), most candidates correctly inferred ammonia and HCL in (i) I and II. However, most of them could not explain their answer in (ii). In (c)(iii), most candidates correctly wrote an eqation for the decomposition

The expected responses from candidates in each of (ii) and (iii) were as follows:

(ii) NH₃ being lighter than HCI/NH₃ travels faster than HCL
(iii) NH₄CI(s) ~ NH₃(g) + HCI(g)

In part (d) (i), candidates named potassium trioxosulphate(IV) and sodium trioxosulphate(IV) as the two salts that would react with dilute acids to produce sulphur(IV)oxide. In(d)(ii), candidates mentioned two uses and one chemical property of sulphur(IV)oxide from among the following:

1. – as germicide

– manufacture of tetraoxosulphate(VI) acid
– as preservative
– as refrigerant
– as bleaching agent.
II.               Sulphur (IV) oxide combines with water to form solution of trioxosulphate (IV) acid
reacts with oxygen in the presence of catalyst to form sulphur(VI)oxide
reacts with alkali to form salt and water
reacts with H2S to form water
bleaching action.

.

ANSWER

In(a)(i), candidates correctly responded to I and II as follows:
I – pressure/volume of vessel
temperature

II – Catalyst
Catalyst affects both forward and backward reactions equally

In (a) (ii], most candidates gave the effect on the equilibrium position instead of the effect on the equilibrium concentration of Cl2• They therefore lost the marks. The required answers for I – III were as follows:
(a) (ii)      I – Equilibrium concentration of Cl2 decreases.
Increase in pressure shifts equilibrium position to the right where there is
lower number of moles of species.

II – Equilibrium concentration of Clz increases
Increase in temperature shifts equilibrium to the left where the reaction is
endothermic.

III – Equilibrium concentration of Cl2 decreases
Increase in concentration of PCl3 (thereby decreasing the
concentration of Cl2)

In (b)(i), most candidates could not explain that when a standard hydrogen electrode is connected to a standard copper half cell, the voltmeter reads an emf of O.34V and that the positive sign indicates the flow of electrons from the hydrogen electrode to the copper electrode in the external circuit.

In (b) (ii) I – IV, candidates showed shallow knowledge of electrochemistry and hence, lost
most marks for the question. The expected answers from candidates were:
                                (iii)                At Anode
Zn(s) —-+ Zn2+(aq) + 2e-

At Cathode

                                           II      Anode – oxidation
                                                   Cathode        reduction
                                          III Zn(s) + Cu2+ (aq) → Zn2+ (aq) + CU(s)

                     (iv)         E^O cell = E^Oreduction – E^O oxidation
                                    = 0.34V – (- 0.76V)
                                    = 1.10V
Alternative method

 + 0.76V
Cu2+ 2e- → Cu(s) + 0.34V

e.m.f. = +0.76 + 0.34
           =1.10 V

In part (c), candidates correctly named one chemical industry, gave three factors that should be considered when sitting a chemical industry and stated two effects of a chemical industry on the community in which it is sited in (i) – (iii) as follows:
    (i)  Named chemical industry
       e.g Textile
       Tannery
       Brewery
       Bottling Company
       Soaps and detergents
       Infant formula/Baby food
       Food seasoning.

(ii)     – nearness to raw material source/feedstock
          – nearness to market
          – labour supply
          – transportation
          – nearness to power supply
          – government policy
          – away from residential areas.

(ii)      – improvement in the standard of living of people
          – employment opportunities
          – development in that community
          – pollution (air, land water etc)

ANSWER

In(a)(i) I-II, candidates could not give the chemical formula and IUPAC names of the
compounds represented as A,B,C,D and E in the scheme. The expected answers from
candidates were as follows:
(i)      A – CH3CH2COOH
B – CH3 CH = CH2
C – CH3 CH2 COOCH2 CH3
o – CH3 CH2 COOH/CH3 CH2 OH
E – CH3 CH2 OH/CH3 CH2 COOH

  (ii)     A – propanoic acid
B – propene
C – ethyl propanoate
0- propanoic acid/ethanol
E – ethanol/propanoic acid

In(a)(iii).most candidates did not know that conversion to A is oxidation, B is dehydration and C is esterification in I -III.

In part(b)(i)1 and II, candidates correctly wrote balanced equation for the reaction of chlorine with ethane and ethene respectively thus:

  I.            C2H6 + Cl2 -‘CH3 CH2 CI + HCL
  II.           CH2 = CH2 —-+ CH2 ClCH2 CL

In(b)(ii), candidates know that the type of reaction in (i) I and II were substitution and addition, respectively.

In part(c), candidates correctly gave one product obtained from refining petroleum that is gaseous, liquid and solid as butane/propane/ethane/methane/ethane,kerosene/dieselOil/lubricating oil/gasoline and asphalt/bitumen/paraffin wax respectively.

In part(d) only few candidates could write a balanced chemical equation for the oxidation of glucose in living organisms thus:

.

ANSWER

In(a)(i), candidates correctly stated law of multiple proportions thus:

If two elements, A and B combine to form more than one compound, then the various masses of one element A which combine separately with a fixed mass of the other element B are in simple ratio.

In(a)(ii), most candidates failed to correctly calculate and verify the law of multiple proportion and hence, write the correct formula of P and Q. The expected answers from candidates were as follows:

              = 1.5 x 2                             1.0 x 2
           =        3           :                           2                                                                       

Since the masses of chlorine which combine with LOg of M in P and Q are in simple ratio 3:2 then the law of multiple proportion is obeyed

            III.      P – MCl3
                    Q – MCl2

In part (b), candidates has shallow knowledge of the IUPAC nomenclature hence could not name each of the substances in (i) and (ii). The expected answers from candidates were:
                                  (i)         Potassium hexacyanoferrate(II);
                                 (ii)       Sodium trioxocarbonate (IV) monohydrate.

In(c)(i), candidates correctly stated that salts are compounds formed when all or part of the ionizable hydrogen of an acid are replaced by metallic or ammonium ions/when all part of the OW
ions of a base are replaced by non metallic ions.

In (cHii), candidates knew that either solubility in water (solubility) or stability to heat as a factor that influences the method used for the preparation of a salt.

In (cHiii), only few candidates could classify the given salts accordingly. The required answers from candidates were as follows:

                                                    I.       Na2Zn(OH) – complex salt,
                                                    II.      Zn(OH)CI – basic salt;
                                                    III. (NH4)2Fe(S04)2.6H20.- double salt.

In(c)(iv), most candidates could not write a two-step equation to show the preparation of sodium hydrogen trioxocarbonate(lV) from sodium hydroxide. The expected answers from candidates
were:

ANSWER

In (a)(i), candidates knew that the law illustrated was Charles’ law.

n (a)(ii), they correctly stated that the volume of a given mass of gas is directly proportional to its temperature in (Kelvin) provided that the pressure remains constant was the statement of the law.

In(a)(iii), only few candidates correctly sketched a graph to illustrate the law as follows:
         



In(a)(iv), most candidates applied the law to calculate the new volume of oxygen thus:

V1 = 28.0cm³
T1 = 27⁰C = 27 x 273 = 300k
T2 = 0⁰C + 273 = 273k
V2 = ?
V1/T1 = V2/T2
V2 = V1T2/T2 = 28 x 273/300
= 25.48cm³

In part (b), candidates could not give the required explanation in (i)-iv). The expected answer for each of (i) – Iv) was as follows:
      (i)        Chlorine bleaches item, (example garment) as dilute acid
                  During rinsing the water removes acid that can damage the material
      (ii)        Dilute HN03 oxidizes the hydrogen formed immediately to water
      (iii)        CO2 is heavier than air and does not support combustion
      (iv)       To remove the traces of acid fumes/HCl.

In (c)(i), candidates named oxygen and hydrochloric acid as the two products formed when chlorine water is exposed to sunlight. However, some of them lost the marks because they wrote O2 and
HCl which were not names but formula.

In (c)(ii), only very few candidates knew that NaOCI/NaCI was the main product of chlorine with cold dilute NaOH.
In(c)(iii), candidates could not state the variation of boiling point and oxidizing ability among the
halogen nor give reason for the variation.

The expected answer from candidates were as follow:
           I.          Boiling point increases down the halogen group due to increase in molecular mass/van der Waals forces

           II.          Oxidizing ability of the halogen decreases down the group due to decrease in effective nuclear charge as a result of increase in atomic size/decrease in electronegativity.