2015 Diagnostic

YEE 6301 SMK MERADONG Kecergasan Untuk Kemajuan

NAME: CLASS: PRA U S

DIAGNOSTIC TEST [AUGUST 2015]

PAPER CHEMISTRY 1

CODE 962/1

COHORT STPM 2016

DURATION 1 hour 30 minutes

SUBJECT TEACHER MS UNG HIE HUONG

INSTRUCTION TO CANDIDATES: This paper consists of Section A, Section B and Section C. Answer ALL questions.Arrange and stapler your answers in numerical order.

SECTION A (15 marks)

Answer ALL the questions in Section A. Blacken the corresponding answer on the objective answer sheet provided.

1. The compound PCl3 is formed from 31P, 35Cl and 37Cl isotopes. The relative abundance of 35Cl to 37Cl is 3:1. Which statement about the mass spectrum of PCl3 is true?

A The base peak corresponds to P+ ion.

B The me

value for the last peak is 142.

C The number of peaks for PCl3+ ion is 5.

D The relative abundance of P35Cl3+ ion to P37Cl3+ ion is 3:1.

2. Which statement is true of the nuclides of silicon, Si1428 and Si14

30 ?

A The mass spectrum shows two peaks only.

B Isotope Si1428 is more reactive than isotope Si14

30 .

C The tetrachlorides, SiCl41428 and Si14

30 Cl4 have different shape.

D SiO21428 dissolves in hot concentrated solution of sodium hydroxide, SiO214

30 does not.

3. The radioisotope tritium, T13 , slowly turns into a helium isotope, He2

3 .

Which statement is true of the two isotopes?

A T13 and He2

3 have the same nuclear charge.

B T13 and He2

3 have the same number of neutral sub-atomic particles.

C T+¿13 ¿ and He+¿

23 ¿ have the same number of charged sub-atomic particles.

D T+¿13 ¿ and He+¿

23 ¿ are deflected to the same point in a magnetic field whose strength is

not varied.

4. 100 cm3 of liquid A and 100 cm3 of liquid B are allowed to evaporate. It is found that A evaporates first before B does. Which statement explains this observation?

A B is more volatile than A.B The density of A is higher than that of B.C The vapour pressure of A is lower than that of B.

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D The intermolecular forces of A is weaker than that of B.

5. Which hydride has the highest boiling point?

A HCl B HF C H2O D NH3

6. Gallium is a soft, silvery metal. The element gallium exists naturally as isotopes xGa and yGa

which have 38 and 40 neutrons respectively. What is the percentage abundance of xGa?

A 35.0 B 45.5 C 54.5 D 65.0

7. A phase diagram of water is shown in the margin.

What can be deduced from the phase diagram?

A Ice sublimes at a pressure higher than 611 Pa.

B An increase in pressure will decrease the boiling point of water.

C An increase in pressure will decrease the freezing point of water.

D Water exists as liquid at a pressure of 611 Pa and a temperature of 298K.

8. Electronic transition between energy levels in an atom will cause an absorption or emission of light. Which energy level diagram shows the transition of electrons that emits light with the shortest wavelength?

A B

C D

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9. Coordinate bond does not exist in

A BeCl2 B CO C PH4+ D [Fe(CN)6]3–

10. Polyatomic molecules and ions have varied geometries. Which species and geometry correspond correctly?

Species Geometry

A NH2– Linear

B H3O+ Trigonal planar

C SiCl4 Square planar

D ICl3 T-shaped

11. Based on the band theory, the conductivity of metal is due to delocalized electrons in the conduction band. How many electrons are found in the conduction band of a 10.0 g magnesium strip?

[Avogadro’s constant is 6.02 × 1023 mol–1 ]

A 2.48 × 1023 B 4.95 × 1023 C 2.48 × 1024 D 3.01 × 1024

12. An atom of element Z has nucleon number 55 and 30 fundamental uncharged particles in its nucleus. What is the electronic configuration of a Z2+ ion?

A 1 s22 s2 2 p6 3 s23 p6 3 d5 B 1 s22 s2 2 p6 3 s23 p6 3 d3 4 s2

C 1 s22 s2 2 p6 3 s23 p6 3 d4 4 s1 D 1 s22 s2 2 p6 3 s23 p6 3 d5 4 s2

13. The boiling points of CH3OH and CH3SH are 64.5°C and 5.8°C respectively. What is the cause of the difference in the boiling points?

A The O―H bond is stronger than the S―H bond.

B CH3SH molecule is bigger than CH3OH molecule.

C Hydrogen bonds exist between CH3OH molecules.

D The electronegativity of oxygen is higher than that of Sulphur.

14. The unit cell of an oxide of X is shown below.

The formula of the oxide of X is

A X2 O4 B XO3 C X3 O4 D X2 O5

15. Compounds have either ionic or covalent bonds or both. Which statement is not true of ionic bonds?

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A They involve the transfer of one or more electrons from s, p or d orbitals.

B The strength of ionic bonds is proportional to the size of the ions.

C They involve ions with stable electronic configurations.

D They result in the formation of solid compounds.

SECTION B (15 marks)

Answer ALL the questions in Section B. Write your answers in the spaces provided.

16. (a) Hydrogen cyanide, HCN is a colourless and poisonous gas.

(i) Draw the Lewis diagram for the hydrogen cyanide molecule and predict its shape.

Lewis diagram: [2 marks]

Shape:

(ii) State the type of hybridization the carbon atom undergo in the molecule. [1 mark]

(iii) Draw a labelled diagram to show the overlapping of the bonding orbitals in hydrogen cyanide. [2 marks]

(b) A sample of carbon with relative atomic mass 12.01112 consists of two isotopes 12C and 13C. Calculate the percentage abundance of 12C and 13C isotopes to two decimal places. [3 marks]

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17. A real gas X behaves almost like an ideal gas. For n mol of gas X at pressure P, the graph of volume V versus temperature T is shown below.

(a) What is the most probable identity of gas X? Explain your answer. [3 marks]

Gas X:

Explanation:

(b) On the graph above, sketch and label a graph of variation in volume with temperature at a lower pressure P’ while other conditions remain constant. [1 mark]

(c) At pressure P and temperature 0°C, a gas Y shows a negative deviation from an ideal gas. Mark the expected volume of gas Y on the above graph.

Explain your answer. [3 marks]

Explanation:

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SECTION C (30 marks)

Answer ALL the questions in Section C.Write your answers in the additional answer sheets on page 7-9.

18. (a) The table below lists the temperature and pressure for the critical point and the triple point of carbon dioxide.

Temperature/ °C Pressure/ atm

Critical point 31 73

Triple point –57 5

Carbon dioxide sublimes at –78°C under atmosphere pressure. The freezing point of carbon dioxide increases by 2°C for every increase of 10 atm in pressure.

(i) Based on the information given above, sketch the phase diagram of carbon dioxide.[5 marks]

(ii) Calculate the freezing point of CO2, in °C, under a pressure of 75 atm. [2 marks]

(iii) Explain why the freezing point of CO2 increases with pressure. [1 mark]

(iv) Solid carbon dioxide is known as dry ice. How can liquid CO2 be obtained from dry ice? [1 mark]

(v) Dry ice has the advantage of being relatively cheap and non-toxic. It is commonly used in concerts to form fog. Explain the formation of the fog. [3 marks]

(b) The boiling point of hydrogen halides are given in the following table:

Hydrogen halide HF HCl HBr HI

Boiling point/ K 293 188 206 238

Explain the variation in boiling points of the hydrogen halides. [3 marks]

19. (a) (i) Ammonium nitrate is an explosive compound and it decomposes at a high temperature according to the following equation:

2NH4NO3(s) → 4H2O(g) + 2N2(g) + O2(g)

Calculate the total volume of gases collected from the decomposition of 100 g of ammonium nitrate at 1.01 × 105 Pa and 25 °C.[Gas constant, R = 8.31 J g–1 °C–1] [3 marks]

(ii) Sketch a graph of PVRT

against p for 1.0 mol of ammonia gas at 0°C.

Based on the graph, explain the negative deviation of ammonia gas compared to an ideal gas. [5 marks]

(b) (i) With the aid of a diagram, describe the structures of diamond and graphite. [5 marks]

(ii) What is the relationship between diamond and graphite in term of structure? Explain your answer. [2 marks]

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ANSWER SHEETSECTION C Examiner’s

use onlyQuestion Answers18

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SECTION C Examiner’s use onlyQuestion Answers

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SECTION C Examiner’s use onlyQuestion Answers

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Periodic Table (Jadual Berkala)

Group (Kumpulan)1(I)

2(II)

3 4 5 6 7 8 9 10 11 12 13(III)

14(IV)

15(V)

16(VI)

17(VII)

18(VIII)

1.0H

1

4.0He

26.9Li

3

9.0Be

4

aX

b

a = relative atomic mass (jisim atom relatif)X = atomic symbol (symbol atom)b = atomic number (nombor atom)

10.8B

5

12.0C

6

14.0N

7

16.0O

8

19.0F

9

20.2Ne

1023.0Na

11

24.3Mg

12

27.0Al

13

28.1Si

14

31.0P

15

32.1S

16

35.5Cl

17

40.0Ar

1839.1

K19

40.1Ca

20

45.0Sc

21

47.9Ti

22

50.9V

23

52.0Cr

24

54.9Mn

25

55.8Fe

26

58.9Co

27

58.7Ni

28

63.5Cu

29

65.4Zn

30

69.7Ga

31

72.6Ge

32

74.9As

33

79.0Se

34

79.9Br

35

83.8Kr

3685.5Rb

37

87.6Sr

38

88.9Y

39

91.2Zr

40

92.9Nb

41

95.9Mo

42

[98]Tc

43

101Ru

44

103Rh

45

106Pd

46

108Ag

47

112Cd

48

115In

49

119Sn

50

122Sb

51

128Te

52

127I

53

131Xe

54133Cs

55

137Ba

56

139La

57

178Hf

72

181Ta

73

184W

74

186Re

75

190Os

76

192Ir

77

195Pt

78

197Au

79

201Hg

80

204Tl

81

207Pb

82

209Bi

83

[209]Po

84

[210]At

85

[222]Rn

86[223]

Fr87

[226]Ra

88

227Ac

89

[261]Rf

104

[262]Db

105

[266]Sg

106

[264]Bh

107

[269]Hs

108

[268]Mt

109

[281]Ds

110

[272]Rg

111

[285]Cn

112

140Ce

58

141Pr

59

144Nd

60

[145]Pm

61

150Sm

62

152Eu

63

157Gd

64

159Tb

65

163Dy

66

165Ho

67

167Er

68

169Tm

69

173Yb

70

175Lu

71232Th

90

231Pa

91

238U

92

237Np

93

[244]Pu

94

[243]Am

95

[247]Cm

96

[247]Bk

97

[251]Cf

98

[252]Es

99

[257]Fm

100

[258]Md

101

[259]No

102

[262]Lr

103

The proton numbers and approximate relative atomic masses shown in the table are for use in the examination unless stated otherwise in an individual question. (Nombor proton dan anggaran jisim atom relatif yang ditunjukkan dalam jadual adalah untuk digunakan dalam peperiksaan kecuali yang sebaliknya dinyatakan dalam soalan

yang tertentu.)

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H C Nspsp spsp

1s

py

py

py

py

pz

pz

pz

pz

bondbond

bond

bond


MARKING SCHEME

DIAGNOSTIC TEST [AUGUST 2015]

Q RUBRIC M

SECTION A [15 marks]

1 B 1

2 A 1

3 D 1

4 D 1

5 C 1

6 D 1

7 C 1

8 B 1

9 A 1

10 D 1

11 B 1

12 A 1

13 C 1

14 A 1

15 B 1

16 a(i)

Linear

1

1

a(ii) sp 1

a(iii) Orbitals Labels (Orbitals and bonds)

NOTE:

11

b Let the % abundance of 12C = x% ; % abundance of 13C = (100 – x)%

12 x+13 (100−x )100

¿12.01112

x=98.89 %

1

1

12


Q RUBRIC M

y=1.11 % 1

17 a Hydrogen gas// H2 // Helium// He

Non-polar, small H2 molecule/ He atom/ gas particle has … negligible volume/ size of particle relative to volume of container. Weak, negligible intermolecular van der Waals forces.

1

11

bc

B1

C1

c Negative deviation at lower pressure is due to: Stronger intermolecular forces between gas particles. Gas particles are drawn closer to each other, volume occupied by gas is smaller

than expected/ smaller than ideal gas.

11

18 a(i) Drawn and labelled axes + boiling curve + sublimation curve Correct melting line (straight line with positive gradient) Mark and state coordinates for Triple point = (-57°C, 5 atm)

Critical point = (31°C, 73 atm) Normal sublimation point = (-78°C, 1 atm)

NOTE:1. If axes are not labelled 0 mark2. Curves must have positive gradient (upwards from left to right)

11111

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Q RUBRIC M

3. The three phase transition lines/ curves must meet at triple point

18 a(ii) Under a pressure of 75 atm,

Freezing point ¿−57+¿ (75−5)

10 ×2

¿−57+14¿−43℃

1

1

a(iii) As external pressure increases, the melting point/ freezing point of CO2 increases.

REASON (Either one): The density of solid CO2 (dry ice) is higher than the density of liquid CO2. When solid CO2 (dry ice) melts, its volume expands.

NOTE: Accept reverse argument

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a(iv) By increasing the temperature of dry ice to above –57°C and increasing the pressure to above 5 atm.

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a(v) Dry ice sublimes to form CO2 gas.The sublimation process absorbs heat from surrounding/ is endothermic, thereby lowering surrounding temperature.Moisture/ water vapour in air condense into fine water droplets or “fog”.

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1

(b) Boiling point of HF is exceptionally high due to strong hydrogen bonding exist between HF molecules

Weak intermolecular van der Waals forces exist in HCl, HBr and HI. Molecular size/ mass increases from HCl to HI, therefore the strength of

intermolecular van der Waals forces increases from HCl to HI.

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19 a(i)Number of moles of gases formed =

32

× 10080.0

= 1.875 mole

pV=nRT @ V=¿ nRT

p

Volume of gases formed, V = (1.875 ) (8.31 )(273+25)

1.01×105 m3

= 0.046 m3

1

1

1

a(ii)

Ideal gas has no particle/ molecular volume and no intermolecular (attractive or repulsive) forces.

Ammonia is a non-ideal gas @ real gas because it is a polar molecule. It has a large molecular size/ volume and strong intermolecular attractive forces (hydrogen bonds) between the molecules.

The attractive force causes molecules to be nearer to each other, causing the measured volume to be smaller than expected.

The attractive forces also lowers the speed of the molecules colliding with the walls of the vessel. Therefore, the measured pressure is lower than expected.

1

1

1

1

1

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Q RUBRIC M

Both factors cause pVRT

< 1, giving negative deviation.

19 b(i) Structures:

Both diamond and graphite have giant 3-dimensional covalent structures.

In diamond, each cabon atom undergoes sp3 hybridisation and is covalently bonded to four

other carbon atoms in a tetrahedral manner.

In graphite, each cabon atom undergoes sp2 hybridisation and is covalently bonded to three

other carbon atoms in a trigonal planar shape. This forms hexagonal rings in a layered structure. The layers are held together by

weak van der Waals forces.

NOTE: Maximum 5 marks

1

1

1

1

1

1

b(ii) Both are allotropes of carbon.The physical properties are different due to different arrangement of carbon atoms.

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2015 Diagnostic

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