basic chemistry knowledge

1. SELARAS 2 FORM 5 2011

2. Question 1
b) Write the formula of the following
substances:
• Hydrochloric acid – HCl
• Sulphuric acid – H2SO4
• Barium hydroxide – Ba(OH)2
• Barium sulphate – BaSO
4

3. b) Hydrochloric acid is a strong acid.
Why?
HCl ionised completely in water to
produce high concentration of
hydrogen ion

4. c)Silver chloride and copper (II)sulphate
are two salts that are required to be
prepared in the lab
i) Which one is the soluble salt?
Copper (II) sulphate
ii) State the name of acid needed to
prepare copper (II) sulphate
Sulphuric acid

5. iii) Silver chloride is prepared by the
precipitation method.Write the chemical
Equation to prepare silver chloride salt
AgNO3 + NaCl → AgCl + NaNO3

6. d) Calculate the following:
i) What is the concentration of 50 cm3
NaOH solution needed to neutralise
100 cm3 0.5 moldm-3 H2SO4?
2 NaOH + H 2 SO4 → Na2 SO4 + 2 H 2O
Where
M 1V1 M 2V2
= M1 = molarity of acid
V1 = volume of acid
a b M2 = Molarity of alkali
V2 = volume of alkali
a = no. of mol of acid in the equatn
b = no. of mol of alkali in the equatn

7. 2 NaOH + H 2 SO4 → Na2 SO4 + 2 H 2O
M 1V1 M 2V2 Where
= M1 = molarity of acid
a b V1 = volume of acid
0.5 x100 M 2 x50 M2 = Molarity of alkali
= V2 = volume of alkali
1 2 a = no. of mol of acid in
the equatn
M 2 = 0.5 x100 x 2 / 50 b = no. of mol of alkali
−3 in the equatn
M NaOH = 2moldm

8. i) How much water is needed to form a
solution of 1.0moldm-3 NaOH solution
from a solution of 50cm3 sodium
hydroxide 2.0moldm-3
Where
M 1V1 = M 2V2 M1 = molarity of original
2(50) = 1(V2 ) solution
V1 = volume of original
V2 = 100 solution
M2 = Molarity of diluted
∴VolumeH 2O = solution
V2 = total volume of
100 − 50 = 50cm 3 diluted solution

9. 2.Diagram 2 shows the apparatus set
up for two types of cells, P and Q

10. a) State one difference in energy
change between cell P and cell Q:
In cell P energy change is from
electrical energy to chemical
energy
In cell Q energy change is from
chemical energy to electrical
energy

11. b) What is the product at the anode in
cell P?
Oxygen gas
c) Write the half equation for the
discharge of cation in cell P
2+ −
Cu + 2e → Cu

12. d) State one observation at electrode
zinc in cell Q
Zinc electrode becomes thinner

13. e) Diagram 2.2 shows the apparatus
set-up for the electrolysis of sodium
chloride solution using carbon
electrodes

14. i) State all ions attracted towards
electrode P and Q
Electrode P Electrode Q
Na , H
+ +
Cl , OH
- -
ii) State which ion discharge at
electrode Q. Explain why
Chloride ion because the
concentration of Cl- is higher than
concentration of OH-

15. iii) Describe one chemical test to identify
gas X
Place burning wooden splinter into
mouth of test tube, ‘pop’ sound will
be produced

16. Diagram 3 shows a series of reaction of
compound P. Compound P and
compound Q are shown in condensed
structural formulae.

17. Compound Q
CO2 + H2O
CH3CH2 CH3 Combustion
I
II
Compound P Compound R
Steam and
CH2=CHCH3
Phosphoric acid
+ Ethanoic acid III
+ concentrated
sulphuric acid
Compound S
Diagram 3

18. (a) Compound P is propene . State
the homologous series of compound P
ALKENE

19. (b)Compound Q is used as a cooking
gas.
• i) State the molecular formula of
compound Q
C3H8
ii) Write a chemical equation for the
combustion of compound Q to produce
carbon dioxide and water
C3H8 + 5O2 3CO2 + 4H2O

20. (C) Compound P can be changed to
compound R which is an alcohol through
process II.
• Name process II
HYDRATION
• State the general formula for alcohol
CnH2n+1OH

21. (d) Explain briefly how compound S
can be prepared from compound R in
laboratory.
In your answer include the
appropriate reagents and condition for
the preparation.

22. • Add absolute propanol and glacial
ethanoic acid into a beaker.
• Add slowly concentrated H2SO4.
• Heat under reflux for 6 hours
• Distill the mixture to get the
products.

23. e) An organic compound W has a
molecular formula of C4 H8 . Compound
W has 3 isomers. Draw two structural
formulae of two isomers of W.

24. e) isomers of W.

25. 4) Natural rubber is a polymer, where the
monomer is isoprene or 2- methylbuta-1,3-
diene. Natural rubber is obtained from latex,
which is a colloid made up of rubber particles
and water.
Draw Structural formula of isoprene (1m)

26. Draw labelled diagram of latex
particle in water (2m)

27. Explain why latex in the original form
does not coagulate

28. However when latex is exposed to air or when acid is
added, coagulation of latex occur. Explain(3m)
• Air contains bacteria that acts on latex to produce
lactic acid

29. The properties of natural rubber can be
improved by treating it with sulphur.
Name this process (1m)
VULCANISATION

30. State ONE use of the treated rubber
above (1m)
 TYRE
 GLOVE
 RUBBER HOSE

31. QUESTION 5
Initial temperature of
both solutions = 29.5 o C
Highest temp of the
50 cm3 of 1.0 mol mixture = 34.5 o C
dm-3 Potassium
sulphate solution
50 cm3 of 1.0
mol dm-3
Barium
Chloride
solution

32. a. What is meant by the heat of
precipitation of barium sulphate?
• Heat change when 1 mole
of barium sulphate is
formed from barium ion
and sulphate ion in
aqueous solution 1m

33. b. Write an ionic equation for the
reaction
Method:
First, write the formula of the
precipitate on the right side of the
equation
Second, write the ions of the
precipitate on the left side of equation
• Ba 2+
+ SO4 -2
 BaSO4 1m

34. c. State one observation for the
experiment
• White precipitate forms
or
the containers become
warm 1m

35. d. Calculate
i. The heat released in the
experiment. [Specific heat capacity
of water = 4.2 Jg-1oC-1, Density of water
= 1.0 g cm-3]
Heat release
= mc
=100x4.2x(34.5- 29.5)J
= 2100J 2m

36. ii. The number of moles of barium
sulphate formed
• Mole of BaSO4
= MV/1000
= 50 x 1.0 /1000
= 0.05 1m

37. iii. The heat of precipitation of barium
sulphate
0.05 mole BaSO4 produce heat 2100 J
1 mole BaSO4 2100 / 0.05 J
= 42000 J
∆ H = - 42 kJ mol-1

38. e.Draw the energy level diagram for
the precipitation of barium sulphate
Energy Ba + SO 4
2+
-2
∆ H = -42 kJ
mol-1
BaSO
4
2m