Which one of the following has covalent as well as ionic bond A NaCl b b h2o C kcl d NaOH

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ANSWERS

Chapter 1

Exercises

 1 (a)

CuCO

3

→

 CuO + CO

2

 (b)

2Mg + O

2

→

 2MgO

 (c)

H

2

SO

4

 + 2NaOH

→

 Na

2

SO

4

 + 2H

2

O

 (d)

N

2

 + 3H

2

→

 2NH

3

 (e)

CH

4

 + 2O

2

→

 CO

2

 + 2H

2

O

 2 (a)

2K + 2H

2

O

→

 2KOH + H

2

 (b)

C

2

H

5

OH + 3O

2

→

 2CO

2

 + 3H

2

O

 (c)

Cl

2

 + 2KI

→

 2KCl + I

2

 (d)

4CrO

3

→

 2Cr

2

O

3

 + 3O

2

 (e)

Fe

2

O

3

 + 3C

→

 3CO + 2Fe

 3 (a)

2C

4

H

10

 + 13O

2

→

 8CO

2

 + 10H

2

O

 (b)

4NH

3

 + 5O

2

→

 4NO + 6H

2

O

 (c)

3Cu + 8HNO

3

→

 3Cu(NO

3

 )

2

 + 2NO + 4H

2

O

 (d)

6H

2

O

2

 + 2N

2

H

4

→

 2N

2

 + 10H

2

O + O

2

 (e)

4C

2

H

7

N + 15O

2

→

 8CO

2

 + 14H

2

O + 2N

2

 4 (a)

Sand and water: heterogeneous

 (b)

Smoke: heterogeneous

 (c)

Sugar and water: homogeneous

 (d)

Salt and iron filings: heterogeneous

 (e)

Ethanol and water: homogeneous

 (f)

Steel: homogeneous

 5 (a)

2KNO

3

(s)

→

 2KNO

2

(s) + O

2

(g)

 (b)

CaCO

3

(s) + H

2

SO

4

(aq)

→

 CaSO

4

(s) + CO

2

(g) + H

2

O(l)

 (c)

2Li(s) + 2H

2

O(l)

→

 2LiOH(aq) + H

2

(g)

 (d)

Pb(NO

3

 )

2

(aq) + 2NaCl(aq)

→

 PbCl

2

(s) + 2NaNO

3

(aq)

 (e)

2C

3

H

6

(g) + 9O

2

(g)

→

 6CO

2

(g) + 6H

2

O(l)

 6

 X has diffused more quickly, so it must be a lighter gas. Its particles have greater velocity than the particles of Y at the same temperature. (Note though that they will both have the same value for average kinetic energy.)

 7

 From the kinetic molecular theory we would expect a solid to be more dense than its liquid, and therefore that ice would sink in water.

 8

 Bubbles will be present through the volume of the liquid. A brown gas is visible above the brown liquid. As the two states are at the same temperature, the particles have the same average kinetic energy and are moving at the same speed. The inter-particle distances in the gas are significantly larger than those in the liquid.

 9

 At certain conditions of low temperature and low humidity, snow changes directly to water vapour by sublimation, without going through the liquid phase.

10

 Steam will condense on the skin, releasing energy as it forms liquid at the same temperature (e–d on Figure 1.4). This is additional to the energy released when both the boiling water and the condensed steam cool on the surface of the skin.

11

 B

12

liquidtime

   t  e  m  p  e  r  a   t  u  r  e   /   °   C

253580solid formingroom temperaturesolid cooling

13

 These calculations have used

L

 = 6.02 × 10

23

 (a)

7.2 × 10

22

 (b)

3.01 × 10

24

 (c)

1.2 × 10

23

14

 0.53 mol H

15

 0.250 mol

1

 Answers

16 (a)

262.87 g mol

−1

 (b)

176.14 g mol

−1

 (c)

164.10 g mol

−1

 (d)

248.22 g mol

−1

17

 189.1 g

18

 1.5 mol

19

 0.0074 mol Cl

−

20

 1.83 × 10

24

 C atoms

21

 171 g (integer value because no calculator)

22

 10.0 g H

2

O

23

 2.0 mol N

2

 > 3.0 mol NH

3

 > 25.0 mol H

2

 > 1.0 mol N

2

H

4

24 (a)

CH

 (b)

CH

2

O

 (c)

C

12

H

22

O

11

 (d)

C

4

H

9

 (e)

C

4

H

7

 (f)

CH

2

O

25

 Na

2

S

2

O

3

26

 CoSO

4

.7H

2

O

27

 C

17

H

25

N

28

 NH

3

29

 6.94 Li

30

 CdS

31

 empirical formula CH; molecular formula C

6

H

6

32

 empirical formula H

2

PO

3

; molecular formula H

4

P

2

O

6

33

 C

10

H

16

N

5

P

3

O

13

 for both empirical and molecular formulas

34

 C

3

H

8

O

35

 Let

 y 

 = mass of chalk in grams. moles of chalk used = mass used

M 

r

(CaCO

3

 ) 

=

 y 

 g100.09 g mol

–1

  This is the same as the number of moles of carbon atoms used. Therefore the number of carbon atoms used = moles of chalk × (6.02 × 10

23

 mol

–1

 ) = 6.02 × 10

23

 y 

100.09 

36 (a)

2.50 mol

 (b)

5.63 mol

 (c)

665.5 g

37 (a)

2C

4

H

10

 + 13O

2

→

 8CO

2

 + 10H

2

O

 (b)

1.59 g

38

 4.355 kg

39 (a)

CaCO

3

→

 CaO + CO

2

 (b)

92.8%

 (c)

CaCO

3

 is the only source of CO

2

; all the CaCO

3

 undergoes complete decomposition; all CO

2

 released is captured; heating does not cause any change in the mass of the other minerals present.

40 (a)

85.2 g

 (b)

1.3 g H

2

41

 5.23 g C

2

H

4

Cl

2

42

 254 g theoretical CaSO

3

; 77.9%

43

 3.16 g ester

44

 107 g of C

6

H

6

 needed

45 (a)

2.40 mol

 (b)

0.0110 mol

 (c)

44 mol

46 (a)

35.65 dm

3

 (b)

5.7 dm

3

47

 0.652 dm

3

48

 0.138 mol Br

2

 and 0.156 mol Cl

2

, so more molecules of Cl

2

49

 0.113 dm

3

50

 0.28 dm

3

51

 90 kPa

52

 16 °C

53

 3.0 dm

3

54

 2.8 dm

3

55

M 

 = 133 g mol

−1

 so gas is Xe

56

 90.4 g mol

−1

57

 Helium

58

 311 dm

3

59

 empirical formula and molecular formula = SO

3

2

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