|
No.
|
Marking
criteria
|
Marks
|
||||||||||
|
8(a)
|
(i)
|
1. A structural material that is formed by combining two
or more different materials
2. with properties
superior to those of the original components
|
1
1
|
2
|
||||||||
|
|
(ii)
|
|
1
1
1
|
3
|
||||||||
|
|
(iii)
|
|
1
1
1
|
3
|
||||||||
|
(b)
|
(i)
|
Saturated hydrocarbons – hydrocarbons that contain only
carbon-carbon single bonds or single covalent bond.
Example : hexane
Unsaturated hydrocarbons – hydrocarbons that contain at
least one carbon-carbon double or triple bond.
Example : Propene
|
1+1
1+1
|
4
|
||||||||
|
|
(ii)
|
By Hydrogenation process.
Ethene reacts with hydrogen
at 1800 C in the
presence of nickel or platinum catalyst
to form ethane.
 + H2 → 
accepted: chemical equation
|
1
1+1+1
(max 2)
1
|
4
|
||||||||
|
(c)
|
|
Latex coagulate in acid because:
1
Acid contain hydrogen ion and ammonia
solution contain hydroxide ion
2
Hydrogen ion from acid will
neutralises the negative charge on the
protein membrane of the rubber
particle and
3
cause the membrane to break when
collided thus releasing the rubber molecules/polymers
4
The rubber molecules/polymers entangled
with one another and the latex coagulates
5. Hydroxide ion can neutralise any acid
that are
produced
by bacteria in latex, thus the negative
charges on protein membrane of rubber
particle is
maintained and prevented from coagulating
|
1
1
1
1
1
|
Max 4
|
||||||||
|
|
|
Total
|
20
|
|||||||||
|
10
(a)
|
(i)
|
Compound formed
when hydrogen ion from an acid is
replaced by metal
ions or ammonium ions
|
1
|
|
|
(ii)
|
Lead(II) chloride
|
1
|
|
|
|
Double
decomposition
|
1
|
|
|
(iii)
|
1. Pour [50-100 cm3]
[0.5-2.0 mol dm-3] lead(II) nitrate solution in a beaker
|
1
|
|
|
|
2. Add [50-100 cm3]
[0.5-2.0 mol dm-3] sodium chloride / any soluble salts solution
contain chloride ions into the beaker
|
1
|
|
|
|
3. Stir the mixture
|
1
|
|
|
|
4. filter the
mixture
|
1
|
|
|
|
5. Rinse the
residue with distilled water
|
1
|
|
|
|
6. dry the salt
with filter paper
|
1
|
|
|
|
7. Equation:
Pb2+ + Cl
- → PbCl2
|
1
|
|
(b)
|
[Calculate the mass
of KCl required]
1. Molar mass of KCl = 39 + 35.5 = 74.5 g mol-1
|
1
|
|
|
[Calculate number
of mole of KCl require]
2. No. of mole = (MV)/1000 = (0.5 x
250)/1000 = 0.125 mol
|
1
|
|
|
[Calculate
mass of KCl required]
3. Mass = 0.125 x 74.5 = 9.3125 g
|
1
|
|
|
4. Weigh out 9.3125 g of KCl
|
1
|
|
|
5. Dissolve the solid KCl with a little
distilled water in a beaker
|
1
|
|
|
6. Transfer the solution into 250 cm3
volumetric flask using filter funnel.
|
1
|
|
|
7. Rince the beaker and filter funnel with distilled water and add
the washing to the flask
|
1
|
|
|
8. Add distilled water into the flask slowly
until the graduation mark.
|
1
|
|
|
9. Closed the flask with stopper
|
1
|
|
|
10. Shake well//inverte several time until the
solution mixed well
|
1
|
|
|
|
10
|
|
|
|
|
|
|
Total
|
20
|
No comments:
Post a Comment