1977-03-14 — Page 26

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.