1

COLLEGE OF FOUNDATION AND GENERAL STUDIES

PUTRAJAYA CAMPUS

TEST 1

TRIMESTER 3, 2011/2012

PROGRAMME : Foundation in Engineering

SUBJECT CODE : CHEF 114

COURSE CODE : Chemistry I

DATE : 9 March 2012

DURATION : 3.00pm – 4.45pm

NAME : _______________________________ STUDENT ID: ___________________

SECTION: ________________ LECTURER : ______________

Instructions For Candidates :

1. Answer ALL the questions.

2. Write the answers in the space provided.

3. All calculations must be shown clearly in order to get full marks.

4. No reference books, papers and mobile phones are allowed into the examination hall.

5. Periodic Table of the elements and Table of Constants are provided.

DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE INSTRUCTED TO DO SO

This question paper has 13 pages including this page.

Attendance List CHEF114 Test 1 (9 March 2012) Seat No:

NAME : _______________________________ STUDENT ID: ___________________

SECTION: ________________ LECTURER : _________________

SIGNATURE: ________________

2

QUESTION 1 [15 marks]

(a) (i) Name the three important subatomic particles. Which of these subatomic particles

have negligible mass?

[2 marks]

Electron ; proton and neutron. 3 x ½=1½

Electron ½

(ii) The modern periodic table arranges element in order of increasing ________

[1 mark]

Number of protons.

(b) (i) Define isotopes [1 mark]

Ans: Atoms with the same number of protons ½ but different number of neutrons. ½

(ii) Given the following data, calculate the average atomic mass of magnesium:

Isotope Mass of Isotope Abundance

Magnesium-24 23.985 78.70%

Magnesium-25 24.986 10.13%

Magnesium-26 25.983 11.17%

[2 marks]

Ans:

Avg atomic mass = (23.985 x 0.787) + (24.986 x 0.1013) + (25.983 x 0.1117)

= 24.3096

3

(c) (i) What is the correct formula for potassium sulfite and iron (III) nitrate?.

[1 mark]

Ans: K2SO3 ½ and Fe(NO3)3½

(ii) What is the correct name for HNO2 and Ba3(PO4)2 ?

[1 mark]

Ans: nitrous acid ½ and barium phosphate ½

(iii) Give the formulas of the following nonmetal compounds

[1 mark]

Xenon difluoride; diphosphorus tetrafluoride

Ans: XeF2 ½ ; P2F4 ½

(d) (i) A cation of element X forms an ionic compound with bromine, having the

formula XBr2. If the ion of element X has a mass number of 230 and 86

electrons, how many neutrons does the element X have?

[3 marks]

Ans:

The formula XBr2 ; Br is Br- so X is X2+

Given that the ion X contains 86 e; So X2+ must have 2 positive protons more

than it has negative electrons: p = 86 + 2 = 88

Since mass number 230 = n + p, ½

Then the number of neutrons is n = 230 - 88 = 142 ½

4

(ii) What are metalloids?

[1 mark]

Elements with both behavior of metal and nonmetals.

(iii) All elements which are metals are solids except ________.

[1 mark]

mercury

(iv) Which fourth period transition element has the highest number of proton?

[1 mark]

Zinc

QUESTION 2 [15 marks]

(a) (i) From the following four solutions given, select the best and the worst in terms of

electrical conductivity.

Solution: 0.10 M CH3COOH 0.10 M NH3

0.10 M KNO3 0.10 M C6H12O6 (glucose)

[2 marks]

Ans = Best (source of most ions in solution): - KNO3.

Worst (source of zero ions in solution): - glucose - C6H12O6.

(ii) What is the term for a type of reaction in which two substances react to produce a

single compound?

[1 mark]

Combination reaction

(b) (i) What substance is reduced in the following redox reaction?

H+(aq) + Fe(s) + NO3

–(aq) → Fe3+

(aq) + NO(aq) + H2O(l)

[1 mark]

NO3-

5

(ii) What is the classification of the types of chemical reaction illustrated by the

following reaction: Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g)

[1 mark]

single displacement and redox

(iii) What is the oxidation number of carbon in C2O42– ?

[1 mark]

2y + -2(4) = -2; y = +3

(c) Transition metals can combine with carbon monoxide (CO) to form compounds such as

Fe(CO)5. Assume that you combine 0.125 g of nickel with CO and isolate 0.364 g of

Ni(CO)x. What is the value of x?

[3 marks]

Mass of CO = 0.364-0.125 = 0.239g; ½ hence 0.239/28 = 8.5357 x 10-3mol ½

0.125 g/58.7 = 2.129 x 10-3mol of Ni ½

Ni : CO

2.129 x 10-3 8.5357 x 10-3

2.129 x 10-3 2.129 x 10-3

1 : 4;

So x= 4 ½

6

(d) Sodium hydroxide (NaOH) reacts with phosphoric acid (H3PO4) to form sodium

phosphate (Na3PO4) and water (H2O) by the reaction:

3 NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3 H2O(l)

If 35.60 grams of NaOH is reacted with 30.80 grams of H3PO4,

(i) What is the limiting reactant?

(ii) Amount of Na3PO4 formed in grams?

(iii) How many grams of the excess reactant remains when the reaction is

completed?

[6 marks]

To determine the limiting reactant, calculate the amount of product formed by each reactant.

The reactant the produces the least amount of product is the limiting reactant.

Amount of Na3PO4 formed from 35.60 grams of NaOH

grams Na3PO4 = (35.60 g NaOH) x (1 mol NaOH/40.00 g NaOH) x (1 mol Na3PO4/3 mol

NaOH) x (163.94 g Na3PO4/1 mol Na3PO4)

= 48.64 grams

Amount of Na3PO4 formed from 30.80 grams of H3PO4

grams Na3PO4 = (30.80 g H3PO4) x (1 mol H3PO4/98. 00 grams H3PO4) x (1 mol Na3PO4/1

mol H3PO4) x (163.94 g Na3PO4/1 mol Na3PO4)

= 51.52 grams

The sodium hydroxide formed less product than the phosphoric acid. This means the sodium

hydroxide is the limiting reactant and 48.64 grams of sodium phosphate is formed.

+

grams of H3PO4 used = (48.64 grams Na3PO4) x (1 mol Na3PO4/163.94 g Na3PO4) x (1 mol

H3PO4/1 mol Na3PO4) x (98 g H3PO4/1 mol) = 29.08 grams

grams H3PO4 remaining = 30.80 grams - 29.08 grams

= 1.72 grams

7

QUESTION 3 [15 marks]

(a) (i) State the Charles’s Gas law

[1 mark]

Charles law state that, at constant pressure, the volume of a fixed mass of an ideal

gas is directly proportional to its temperature on the Kelvin scale.

(ii) A sample of neon gas occupies 4.8L at temperature 25°C. If the temperature is

raised to 298°C, what is the change in the volume of neon gas at constant

pressure?

[2 marks]

V1 / T1 = V2 / T2 ½

4.8 / (25+273) = V2 / (298+273) ½

V2 = 9.20mL

Change in volume = 9.2mL - 4.8 mL= 4.4 mL

(b) (i) Use the kinetic molecular theory of gases to explain why two gases mix more

rapidly at high temperature than at low temperature.

[1 mark]

The average kinetic energy of the gas particle is proportional to absolute temperature.

Therefore, dissimilar gases mix at high temperature.

8

(ii) When 0.17 gram of a liquid is heated, it vaporized and occupied a volume of

59 mL at a pressure of 1.2 atm and 40°C. Calculate the molar mass of the liquid.

[2 marks]

PV=nRT ½

(1.2 atm)(0.059L)=0.17g (0.08021 L.atm/mol.K) (40+273K)

MM

MM=61.7 g/mol ½

(c) When solid carbon dioxide (also known as dry ice), is heated to 400 K it becomes

gaseous. An 88.0 g sample of solid CO2 is placed into a sealed 100 L container that is

initially at a pressure of 1.00 atm and a temperature of 298 K. The container is heated to

400 K. What will be the final pressure inside the container?

[4 marks]

The molar mass of CO2 is = 44.01. The sample of 88.0 g ;

moles of CO2 = 88/44.01 = 2mol 1

Pco2 = nRT/V = 2 x 0.08206 x 400 / 100 = 0.656 atm. 1

The container initially at 1 atm and 298K. Then heated to 400K, so new pressure

P2/P1 = T2/T1 ½

P2T1 = P1T2

P2 x 298 K = 400 K x 1 atm ½

P2= 1.34 atm

Total pressure at 400K is 0.656 atm + 1.34 atm= 2.0 atm. 1

9

(c) (i) State Dalton’s law of gas

[1 mark]

The total pressure of a mixture of ideal gases that do not react is the sum

of each constituent gas in the mixture.

(ii) A gas mixture containing 2.45 gram of N2 and 3.10 gram of Ne are introduce in a

container. If the total pressure of the gaseous mixture is 420 mmHg, what are the

partial pressures of N2 and Ne?

[5 marks]

Number of mole of N2 = 2.45 g / 28.02 gmol-1 ½

= 0.0874mol

Number of mole of Ne = 3.10 g / 20.18 gmol-1 ½

= 0.153mol

Mole fraction of N2 = 0.0874 mol = 0.363

(0.087+0.153) mol

Mole fraction of Ne = 0.153 mol = 0.636

(0.087+0.153) mol

Partial pressure of N2 = 420mmHg × 0.363 = 152.46 mmHg

Partial pressure of Ne = 420mmHg - 152.46mmHg = 267.54 mmHg

= - (1.0 atm) (0.242 L)

N2H4 (l) + O2 (g) → N2 (g) + 2H2O (g) ΔH = -534 kJ

2 [H2O (l) → H2O (g)] ΔH = (+40.7 kJ) 2 = 81.4 kJ ½

2 H2O (g) → 2 H2O (l) ΔH = - 81.4 kJ ½

N2H4 (l) + O2 (g) → N2 (g) + 2H2O (l)

ΔHrxn = -534 kJ + (- 81.4 kJ ) = -615.4 kJ

10

========END OF QUESTION PAPER=========

PERIODIC TABLE

1A

1

H 1.008

IIA

(2)

IIIA

(13)

IVA

(14)

VA

(15)

VIA

(16) VIIA (17)

2

He 4.003

3

Li 6.941

4

Be 9.012

5

B

10.81

6

C 12.01

7

N 14.01

8

O 16.00

9

F 19.00

10

Ne 20.18

11

Na 22.99

12

Mg 24.31

IIIB

(3)

IVB

(4)

VB

(5)

VIB

(6)

VIIB

(7)

VIIIB

(8) (9) (10)

IB

(11)

1IB

(12)

13

Al 26.98

14

Si 28.09

15

P 30.97

16

S 32.06

17

Cl 35.45

18

Ar 39.95

19

K 39.10

20

Ca 40.08

21

Sc 44.96

22

Ti 47.90

23

V 50.94

24

Cr 52.00

25

Mn 54.94

26

Fe 55.85

27

Co 58.93

28

Ni 58.70

29

Cu 63.55

30

Zn 65.39

31

Ga 69.72

32

Ge 72.61

33

As 74.92

34

Se 78.96

35

Br 79.90

36

Kr 83.80

37

Rb 85.47

38

Sr 87.62

39

Y 88.91

40

Zr 91.22

41

Nb 92.91

42

Mo 95.94

43

Tc 98

44

Ru 101.1

45

Rh 102.9

46

Pd 106.4

47

Ag 107.9

48

Cd 112.4

49

In 114.8

50

Sn 118.7

51

Sb 121.8

52

Te 127.6

53

I 126.9

54

Xe 131.3

55

Cs 132.9

56

Ba 137.3.

57

La 138.9

72.

Hf 178.5

73

Ta 180.9

74

W 183.9

75

Re 186.2

76

Os 190.2

77

Ir 192.2

78

Pt 195.1

79

Au 197.0

80

Hg 200.6

81

Tl 204.4

82

Pb 207.2

83

Bi 209.0

84

Po (209)

85

At (210)

86

Rn (222)

87

Fr (223)

88

Ra (226)

89

Ac (227)

104

Rf (261)

105

Db (262)

106

Sg (266)

107

Bh (264)

108

Hs (269)

109

Mt (268)

110

Ds (269)

58

Ce 140.1

59

Pr 140.9

60

Nd 144.2

61

Pm (145)

62

Sm 150.4

63

Eu 152.0

64

Gd 157.3

65

Tb 158.9

66

Dy 162.5

67

Ho 164.9

68

Er 167.3

69

Tm 168.9

70

Yb 173.0

71

Lu 175.0

90

Th 232.0

91

Pa (231)

92

U 238.0

93

Np (237)

94

Pu (244)

95

Am (243)

96

Cm (247)

97

Bk (247)

98

Cf (251)

99

Es (252)

100

Fm (257)

101

Md (258)

102

No (259)

103

Lr (260)

TABLE OF CONSTANTS

Avogadro's number = 6.022 x 10 23 /mol

1 atm = 760 mmHg = 760 torr = 101,325 Pa

1 J = 1 kg. m2 / s2

Specific heat of water = 4.184 J/g.C

1 amu = 1.49 x 10-10 J

Planck constant, h = 6.63 x 10-34 J.s

Rydberg constant, RH = 2.18 x 10-18 J

Gas constant, R = 0.0821 L.atm/ K.mol

Gas constant, R = 8.314 J/ K.mol

1 L.atm = 101.3 J

Speed of light, c = 3.00 x 108 m/s

1 kg = 6.022 x 1026 amu.

For an equation of the form:

ax2 + bx + c = 0

VIIIA

(18)