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Transcript
CROSS COUNTRY EXAM 2014 CHEMISTRY 233/2 MARKING SCHEME MAY/JUNE 2014 1. (a) Halogen (b) L-2.8.8 P- 2.8.8 (c) Hydrogen A - weather balloon - Hydrogenatation - Rocket fuel - Welding /ox hydrogen - Manufacture of ammonia (any 1mk) M- Chlorine Main bleaching agent P.V.C manufacture /plastics HCl ” Water treatment (any 1mk) (d) G has a larger atomic radius than C hence √1mk the metallic bonds in C is stronger than in G √1mk T.T 2mks (e) Giant atomic structure / molecular structre√1mk (f) P has a smaller atomic radius than O due to a larger nuclear charge which pulls electrons strongly √1mk since added electrons enter the same energy level. (g) E (i) Smallest atomic radius hence readily gains electrons √1mk (ii) O√ ½mk - It loses electrons√ ½mk readily since it has the largest atomic radius√ ½ mk 2. (a) - cost availability effect on environment method of storage heating value (Any 2mks each 1mk)T.T 2mks (b) (i) Mass = Density x volume √ ½ mk -3 3 = 1gcm x 450cm = 450g = 450 = 0.45Kg 1000 ∆T = 46.5 – 25 = 21.5oC √ ½ mk Heat evolved = MC∆T √ ½ mk = (0.45Kg x 4.2KjKg-1K-1 x 21.5OK) = 40.635KJ √ ½ mk (ii) Mass of ethanol = (125.5 -124)g= 1.5g √ ½ mk R.M.M of ethanol = (24+16+6) = 46g Moles of ethanol = 1..5/46 = 0.0326moles √ ½ mk If 0.0326moles = 40.635KJ 1mole = 1mole x 40.635KJ√ ½ mk 0.0326moles Exothermic reaction = -1246.472KJmol-1 √ ½ mk -1 (iii) C2H5OH(l) + 7O2(g) → 2CO2(g) + 3H2O(l) ∆H=-1246.472KJmol √ 1 mk No heat change 0mks (iv) - Heat loss to surrounding /heat absorbed by apparatus - Error in reading temperature /mass - Incomplete ethanol combustion (any 2 = √ 2 mks) 1 (d) ∆H=-1246.472KJmol-1 Energy (KJ) √½mk √½mk √½mk √½mk Reactants Prod Enthalpy E C2H5OH(aq) + 7O2 (g) 2CO2 + 3H2 Reaction path √½mk 3. (a) (i) Solid A – Magnesium Oxide / MgO √ ½mk (ii) D – Hydrogen gas / H2 √ ½mk (iii) X-Ammonia / NH3 √ ½ mk (iv) B – Magnesium Nitride/ MO3N2 √ ½ mk (b) Oxygen is more reactive than nitrogen hence reacts faster and completely with Magnesium/Nitrogen √ 1mk (i) Less reactive, little nitrogen reacts with Magnesium √ ½mk (c) Mg3N2(s) + 6H2O(l) →3Mg(OH)2(aq) + 2NH3(g) √ 1mk (d) (i) Passing gas X/NH3 through heated √ 1mk Copper (III) oxide (CuO) where ammonia is √ 1mk oxidize to nitrogen/(Y) (ii) 2NH3(g) + 3CuO (s) → N2(g) + 3Cu(s) + 3H2O(l) √ 1mk (e) Carbon (IV) oxide/CO2 √ 1mk (f) Any noble gases (He, Ne, Ar) √ 1mk (g) (i) Haber process √ 1mk (ii) More of gas X formed √ 1mk lower temperature favours forward reaction √ 1mk in an exothermic reaction . 4. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2O Volume of gas produced (Cm3) Scale - √ ½mk Labeling √ ½mk Plotting 1mk 6 correct plots - √ ½mk at least three Shape – 1mk Shown above Total mks = 3mks 600 500 400 300 200 100 0 10 5. (a) 20 30 40 50 60 70 80 90 100 Volume of acid used (c) (i) 300cm3± 2 √ 1mk 3 (ii) 600cm (d) (i) Rate of reaction will be slow since the surface area could be reduced as compared to zinc powder. √ 1mk T.T 2mks (ii) Rate of reaction will increase√ ½mk since the number of hydrochloric acid molecules will increase√ ½mk thereby increasing the frequency of collision (i) B- Ammonia gas /formula√ ½mk (ii) C-Nitrogen (IV) oxide / NO2 √ ½mk (iii) E-Water/formula √ ½mk (iv) F-Air and Nitrogen (IV) oxide mixture √ ½mk 2 6. (b) Ammonia is oxidized to form Nitrogen (II) oxide√ 1mk in the presence of a catalyst √ 1mk to form nitrogen (II) oxide (c) Cooling of the gas (d) Concentrated acid is heated, Nitric acid being √ ½mk more volatile than water distills of fast√ ½mk and is condensed to form a more concentrated acid√ ½mk (e) Nitric (V) acid is a strong oxidizing agent√ 1mk and therefore reacts as a strong oxidizing agent and not as an acid √ 1mk (f) Prevent decomposing by light √ 1mk (i) Ƶ2/Z √ 1mk e- X X √ ½ mk Y√ ½ mk Salt bridge√½ mk Y2+(aq) √ ½ mk X2+(aq) √ ½ mk Correct half cell √ ½ mk Spoon Silver metal Silver nitrate (b) (i) Magnesium ions/formula√ ½mk sulphate ions/formula√ ½mk hydroxyl ions/formula√ ½mk hydrogen ions/ formula √ ½mk (ii) 4H+(aq) + 4e-(aq) →2H2(g) √1mk (iii) 4OH-(aq) →2H2O(l) + O2(g) + 4e-(aq) √1mk 4OH (aq) →2H2O(l) + O2(g) + 4e √1mk 4e = 4F = 4 x 96500C √1mk If 4 x 96500C = 24dm3 = 1.2dm3 1.2dm3 x 4 x 96500C √1mk = 19300 C √1mk 24dm3 7. (a) (i) 2-methyl pentane √1mk (ii) 3-chloro2-methylbut-1-ene √1mk (b) (i) Water/H2O (ii) Nickel catalyst √1mk (iii) H – C C – H √1mk (iv) Polymerization √1mk (v) Chloroethene/C2H3Cl √1mk (vi) H H C= C H Cl (viii) Non-biodegradable hence causes pollution 3 - releases poisonous gases when burnt(any 2 = 2mks) 4