2008 YTSS 4E Chemistry Ans

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(c) NaH (s) + H 2O (l) → NaOH (aq) + H 2 (g) 1m for eqn1m for statesymbols

(d) Universal Indicator changes from green to blue/purple 1

A4(a)

Group 2/4/6 not accepted

½ m each

(b) YO is likely to be basic. Y is a silvery solid so it is likely to be a metal henceYO is a metal oxide and is basicOR

Y is in Group II and all Group II elements are metalshence YO is likely to be a metal oxide

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A5 (a) C 2H5OH + 3O 2 → 2CO 2 + 3H 2O 1(b) Mass of ethanol burnt = 85.3 – 84.8

= 0.5g1Deduct ½ m if nostatements

(c) Amount of heat produced = 200 × 4.2 × 12= 10080 J

1

(d) No of moles of ethanol burnt = 0.5 ÷ 46= 0.0109 mols

1Deduct ½ m if no

X Y ZAppearanceof element

Shinyblacksolid

Silverysolid

Yellowsolid

Formula of oxide

XO2 YO ZO 3

Oxidationnumber of

the elementin the oxide

+4 +2 +6

Group IV II VI



3 sig fig or units(e) Energy given out to burn 1 mole of ethanol

= 10.08 ÷ 0.0109= 925 kJ/mol

1

(f) Some of the heat energy is lost to the surroundings 1A6 (a) A weak tribasic acid when dissolved in water

dissociates/ionises partially to form 3 moles of hydrogen ions per mole of acid.

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(b) The pH of the acid is about 3 1(c) Cresol red 1(di) No of moles of sodium hydroxide = 20.0/1000 × 0.300

= 0.00600 molNo of moles of citric acid = 0.00600 ÷ 3

= 0.00200 molConcentration of citric acid = 0.00200/20.0 × 1000

= 0.100 mol/dm 3

½

½

1

(dii) Concentration of citric acid in g/dm 3 = 9.6/500 × 1000= 19.2g/dm 3

1

(diii) Relative molecular mass of citric acid = 19.2/0.100= 192

1

A7 (a) A: Copper (II) carbonate CuCO 3

B: Copper (II) oxide CuOC: carbon dioxide CO 2

D: Copper (II) chloride CuCl 2

E: copper CuF: chlorine gas Cl 2

1 each

(b) CuCO 3 → CuO + CO 2 1

(c) Cathode: Cu 2+ (aq) +2e → Cu (s)Anode: 2Cl - (aq) → Cl 2 (g) + 2e

½ , ½½ , ½

A8 (a)Polymer:

H Cl H Cl| | | |

– C – C – C – C –| | | |

H H H H

Monomer F F| |

C = C| |

F F

1

1



(b) Addition polymerisationNOTE:Additional polymerisation not accepted

1

(c) When burnt, incomplete combustion of the polymersproduces carbon monoxide and can cause air pollutionWhen buried in landfills, the non-biodegradablepolymers do not decompose and can cause landpollution

1

1

SECTION BB9 (a) Oxygen 1(b) Gold and silver are unreactive metals 1(c) Although aluminium is more abundant than iron, it costs

more to produce aluminium.This is because production of aluminium requireselectrolysis which is a very expensive process however,the production of iron involves reduction with coke whichis a cheaper process.

NOTE : Reason must show comparison between iron andaluminium

1

1

(d) Both copper and tin are scarce so the cost of production is higher.

1

(e) Oil, gas and coal are finite and will be used up in time.Burning oil, gas and coal produces carbon dioxide whichcontributes to global warming

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(f) HydrogenAdvantage: No pollutionDisadvantage: In gaseous state which makes it difficult tostore.

Ethanol (biomass):Advantage: RenewableDisadvantage: produces carbon dioxide which leads toglobal warming

Accept any other alternative sources

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B10(ai)

Negative electrode: Zn (s) → Zn 2+ (aq) + 2ePositive electrode: Cu 2+ (aq) + 2e → Cu (s)Overall eqn: Zn (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu (s)

1m eqn, 1m ss1m eqn, 1m ss½ , ½

(aii) Solution turned from blue to colourless/ blue solutionfades/Blue solution gets lighter

1

(aiii) Change the Zn electrode to more a more reactive metallike magnesium OR

1



Change Cu electrode to a less reactive metal like silver.(Do not accept metals like Na, K and Ca as they are tooreactive)

(b) 1m for labelledelectrodes

1m for correctelectrolyte

1m for labelingobject

B11(a) EITHER

(b) CaCO 3 + 2HCl → CaCl 2 + CO 2 + H 2O(c) Carbon dioxide.

Bubble the gas into limewater. If white precipitate formsthen carbon dioxide is present

11,1

(d) See part (a) answer 1,1,1(e) When sulphuric acid is used, calcium sulphate is formed

which is insoluble. It will form a layer around the calciumcarbonate and prevent further reaction

1

(ai)OR

C H% composition 86% 14%No of moles 86/12

= 7.17 mol14/1= 14

÷ by smallest 7.17/7.17= 1

14/7.17= 1.95

Ratio 1 2Empirical formula = CH 2

1m for working

1(aii) M: propene

N: propanolO: propanoic acid

111

(aiii) Warm with concentrated sulphuric acidC 3H7OH + C 2H5COOH → C 2H5COOC 3H7 + H 2O

½ , ½1

(bi) Amide linkage 1(bii) 1,1

2008 YTSS 4E Chemistry Ans

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