華
WAH KIU YAT PO
報日僑
光陳愛意,季軍九龍協同中學李秋。在爲冠軍
4 1 3 6 1 1 1 7COK - THEZCIK
培街朱立文吹奏树形。
同鄉校易依年,遘瑪利精李其音、形、聖保衛 初級組現代曲樂犖溪哦,左至右季軍、東莞
·女围麗玲。一*報 BKEES BESCHKO RE - SHAR
- 期星
日四十月三年七七九一番公年六十六國民肇中
際音楽特
怀亮教,換上鄘信翦。季啉冉披萨件嘉志、喇沙
·翠右冠州保羅男女留學、呂淑杯。亞軍協恩 :「許會鼷潤女士盃」六級網驛二傘装,由左
1977中學會考試題預習專欄
明德社主編
Hydrogen sulphide can burn in air and turn lead acietate sol- ution black due to the formation of lead(II)sulphide. H2S+Fb(CH2C00)2
Answer to Q.30
A.
化學(十八) ·朱宏林·
Chemistry (18)
Answer to 0.28
a) Copper and silver, because
they are less reactive than hydrogen.
b) Potassium, because it is the
most reactive of the elements: listed.
b) Silver, because it is the least.
reactive of the elements shown. d) Silver, because its compounds. are less stable than those of: the other elements listed. Sodium and magnesium are usuáļ- ly extracted by electrolysis of their molten chlorides, be- cause the compounds of these strongly electropositive are so atable that chemical reduct- ion by smelting is not energe tie enough and electrons have to be supplied directly by electrolysis for reduction of the corresponding metallic ions. Zino will displace silver from silver nitrate and gray precip- itates of silver are observed.
Zn(s)+2Ag (aq)—>2n2+(aq)+2Ag(§) Potassium, being the most elec- tropositive of the metals in. the list, is the one most like- ly to reduce aluminium chloride on/heating, because the greater. the electropositive the metal, the greater is its reducing. power.
AlCl3 + 3K 3KC1 + A1
In this reaction the aluminium chloride is said to be reduced because the oxidation number of the aluminium has been decreased. from 3 in aluminium chloride to zero in. metallic aluminium. h) There is a reaction in heating
the mixture of aluminium and iron(II) oxide because aluminium being more reactive than iron, takes over the oxygen from the iron(II)oxide.
241 + 3Fe0 A123 + 32G 1) Potassium, because it is easy
to be oxidised by air as. tarn- ishing is the combination with the oxygen.
3) It is zinc oxide, no.
Question 29
A. Vol. of B used up ≈ 45 cm
The reaction equation is,
A25
Ba
45 cm2
30 cm3 2 vals
2 mole
cules 2 mole- cules
30 cm? 1.0.2 vols 3 vols i.e. 2 mole- 3 mole- culeg cules 1.8. 4 atoms 6 atoms
Hence, from the above it is found that 1 molecule of C consists of 2 atoms of A and 3 atoms of B; therefore, the molecular formula of C is A3z. Mixture A most probably consists of magnesium and sulphur,
B is hydrogen gas;
€ is sulphur powder;
Dis magnesium hydroxide;
E is magnesium sulphide;
F is hydrogen sulphide; G is lead(II) sulphide. c) Magnesium reacts with dilute
sulphuric agid to give a colour- less solution and hydrogen gás which can burn in air but does not react with lead acetate.
1504 + H2
Mg + H2504 Sulphur does not react with dilute sulphuric acid and re- mains as yellow residus.
The filtrate is magnesium sul- phate which forms magnesium hydroxide which appears as white precipitate, insoluble in NaOH··
too.
MgSO4 +2NaOH →→→→ Mɛ(OH)+Na2SO4
When the mixture A ia heated alone, magnesium reacts with sulphur to form magnesium sulp- hide.
kg + S
When magnesium sulphide is added to dilute sulphuric acid, a colourless gas which is hydrogen sulphide is given off.
MS+H250
Moso
4
+ H2S
➡2CH_COOH+PbS
a) Vol. of unused 02- 30 ou
(150 - 30)cm3
b) Vol. of used
=
← 120 cm2
a) vol. of CO2 = (110 30) cm
80 cm
d) According to the given equation, molecules. molecules, molecules
of 002
of C • of 02.
= 1 : (x + 2).
**vol of CH svol of Osvol of CO2
Cavol
y
- 20 cm3 : 120 cm3: 80 cm3
= 1 : 6 : 4
molecules, molecules, molecules
of O2
of CO2
The platinum acts as a catalyst. The reason is that the reaction is exothermic and the platinum is kept at a high temperature by the heat liberated during the reaction.
f). It turns from blue to red;
because the nitrogen dioxide formed can dissolve in water to form an acidic solution. 4NO.
+2H 0 4HNO3(aq)
The white precipitate is lead chloride which sivery soluble in hot water.
Fb2+(aq)+2Cl(aq)—→ PbCl (s). b) The blue precipitate is copper (II)hydroxide which decomposes
on heating to copper(II)oxide
which is black.
Ou2*(aq) + 2011 (eq)->Cu(011)2(8) ∙Cu(OH)
Cuo(s) + H20(6)
50.00
50r.
49.5
r + 50
49.5
* 4950 ohma. (Ans:
(11) From Charle's circuit, let the internal resistance
of the ammeter
Hence,
51 - 50
1 ohm, (Ans.)
(III)(i) From Albert's cir
- 1 A. (Ans.) With reference; to fig.2
the equivalent resistance of
the circuit
6 + 6 - 12 ohms (Ans.)
the current through
1A. (Ann.).
(a)
resister
cuit
VAB = 11 (R + 40). 6511
+ 80)
13012
but VAB
16:4 Therefore, from the above dis- cussion, we have
物理(十八)
·魯榮家
651 13030
(1)
The current
and
+ = 6
since x=4, so y
*= (6
12
....(2)
— 4)04
From (1) and (2)
0.5A
8
Therefore the molecular formula of the hydrocarbon is
The possible structural formulae of hydrocarbon G are:
Ju
H H
but-1-ene
but-2-ene
2-methyl-m
cyclobutane
Air consists of water vapour and carbon dioxide so these two together can react with the bleaching powder to lib- erate chlorine gas, A. Ca0C1+H2O+CO-CaCO3+B2O+C] 2 So, bleaching powder smells strongly of chlorine and slow- ly deteriorates in air, Hanganese(IV)oxide acts as the catalyst in the decomposition of potassium chlorate because manganese(IV)oxide lowers the activation energy E of the decompostion reaction as shown in the following diagram:
Energy
KCIC
uncatalysed re:ction
catalysed rection
KC1+02
E-E
o) Nitric acid is easily decomp-
osed by light to give oxides of nitrogen such as nitrogen dioxide which, on dissolving in the nitric acid will impart
So a yellow colour to it. nitric acid is usually stored in brown bottles.
4150 420 + 21120 + 02
d) Freshly-prepared aluminium is. very reactive so it is very quickly is covered with a layer of aluminium oxide which is rather tough and insoluble in water, that is why it seems not very reactive.
4A1 + 302211203(0)
Answer to 0.31
The reaction conditions are: 1) high temperature, 400°C; 11) high pressuré, about
1100 to 1200 atmospheres; iii) a little oxygen as the
catalast.
b) Yes. The equation can be writ-
B.
ten as nCH=CHC1
-CH-CHC1-
The product can be named...aa polyvinylchloride or FVC.
a) It is blue because ammonia
solution is alkaline, Reddish-brown vapour is observ ved around the spiral, it is nitogen dioxide.
*) 4NE
50,- 4NO + 6H2O 2NO + 02
2N02
PHYSICS (18).
Answers to Exercise 9.
1(a)_(1) Since
IT
the charge deposited
on anode in one second
(-80 × 107 x 1) c
charge of an electron
1.6 x 10-19
number of electron reach-
ing the anode in one. Be conu
80 x 106
1,6 x 10
#5 x 10
(Ans.)
(ii) Kinetic energy of an ole -tron just before it reaches the anode
qv
(1,6x10~19)(18x103)J.
10-15J (Ans.)
- 2.88 x
(iii) Most of it will be con-
verted to beat and X- rays may be emitted,
(b)(1)
(i)
**
Ans: the current through X is
0.5A
(ii) Power consumed by X
=(1,)2x
(0.5) (50) W
(Ans.)
12,5W
2. Let the resistance of each
electric bulb be r.
(a) (i) The curreats flowing through La and are equal
and both equal to
both would give the same
brightness.
/ (ii) The currents flowing
through and Ly are the same and equal to
24
2r
both would give the same
brightness,
(b) (i) x¢ ¥1 * V2 and from the result in (a). the cur- rent flowing through each electric bulb in fig.l is 2
times that in fig. 2 Therefore the electric bulbs in fig, is brighter than that
in fig.2.
but the ugh L1 =
through
installed,
the brightness of I, is not affected but the bright- nass of L is reduced.
3. (a)
son
560
case II :
in series 10.0
502
does
case III (b) Case I:
: in parallel
R = 502
V
200V
Power
200
50
800 W (Ans.).
Case TT:
R = 5050 m 100
V 200 V
Pover
**
2002
100
Case III:
400 W (Ann).
Since both heating coils
are in parallel,
R
R R2 (50)(50)
50+
When the galvanometer shows zero reading,
- 250
i.R
180
-40.
(1) . (2)
fig. 1
2002
1600w (Ana
25
(c) Since
(ii)
X = 2R - 50 ohms (Ans.)
X = 50
H
V
Since R
X
1.02
(Ann.)
0.99 0.02 -49.5 ohms
(iii) X =
51 ohms (Ana 0.02 - (i) From Bernard's circuit, let'r be the internal resis-
tance of the voltmeter,
The equivalent resistance of
the mesh HKPQH' is
fig. 2
(ii) If V2
2V1, then
1. the current through each electric bulb in fig.2
2v.
1 2r
V
27 the current through each
electric balb in fig.1
the electric bulbs in fig.1 and that in fig.2 would give the same brightness. (c)(i) The resistance of each.
electric bulb
-
6 ohms. (Ans.)
(ii) with reference to fig.1
the current flowing through
in fig.1
Power
Required current
200
50
4A
The maximum number of
cookers allowed
22
*
- 5.5
Therefore 5 electric cookers
may be used at the same time,
(d) It is not advisable to replace a blown use wire with copper wire because
fuse made by copper wire will
not blow when excess current
passing through (as couper wire has higher welting poing.