1977-11-29 — Page 25

育教僑華頁一第張七第日九十月十年巳丁磨复 WAH KIU YAT PO

智華

報日僑

·模主版

日擬編辦

擬試卷猁枪各一坐,薄刻行各時間爲四十五分鐘,內容及份量均携仿 NVERSAZNANDING CAKE ,增加對參加食盐孕对能力测劃的例心 - 特根某本月廿二日教署公佈, 日版 協助小學六年极考生在武林柏有-

份準

*HEERKREE+R+BATI 並送孔,日於本月(廿二日)分發一杯水力商村镇髻录演本一份,使 行致育局爲使應小六學生對行持行之爭視谋,有進一步 第一年只能力滚, - 童年在十二月六日 【星期三 } 下午二陵.

定本月三十日刊出

學能推理模擬試卷

正式发行之船民,上述兩科

期三季刊出 2

JAIZERENKE) VZELIN

本月三十日刊出之đ接就迷對動,仍由來向爲本版「升中民事

水枪,定期在本月三十日(E

二期星

a 三第三名:蕙理上午校

日九廿月一十年七七九一腊公年六十六國民中

29

初小月英诗集拼第一名錦田公會小學。第二名:公立

,

£ • DIN--JOERE WE IN 373–1

香港學校朗誦節特刊

育敦僑華

1978

中學會考試題預習專機|

明德社主講

化

學 (八) 朱宏林

Chemistry (8).

Solution to' 0.10

G is aniondej

1 is water;

X is nitrogen,

(b) (NH)2SO + Ca(OH)2

4

CaSO4 + 2NH ̧ + 2H 0

(c) Quickline(calcium oxide) can

be used to dry gas G.

(a) Reddish-brown copper powder

is obtained.

(e) (1)

Formula mass of (NH4)2SO4

is 132.

1.32 g of ammonium sulphate is equivalent to 0.01 moles. According to the equation in (b), 1 mole of ammonium gulp hate gives (2 x 22.4) am of ammonia (gas G) at s.t.p.

is

volume of gas G produced by 0.01 moles of (NH),50

2 x 22.4 dm x 0.01

dm3

0.448 dm

448 cm3

(ii).

According to the

low:-

quation be-

2NH2+300-N2+3H0+3Cu

2 vols.

1 vol.

.vol. of nitrogen (gas X)

2x val. of ammonia

x448 cm3

224 cm

(f) The solid is magnesium nitride.

3Mg +

14832

(g) The gas is ammonia.

MgzN2 + 6H20-3Mg(OH)2+2NH,

Solution to §.12

The yield will decrease; It.

is because a rise in temper- ature favours the backward reaction which is endothermic thus the equilibrium is shifted to the left.

(ii) The yield will decrease; it is because when the pressure is lowered, the backward re- action is favoured as the

reactants occupy a large vol- ume; so the equilibrium

18 shifted to the left.

(iii) The yield remains unchanged;

it is because a catalyst only speeds up the attainment of

the equilibrium but does not shift the equilibrium in any direction.

(iv) The yield will increase; it

is because the equilibrium is shifted to the right when the concentration of oxygen is increased.

8

A dark-blue solution is seen. Chlorine oxidises the iodide ions to iodine which then at once combines with the starch to form a dark-blue complex,

012 + 21 →→ I1⁄2 ★ 201 (ii) White precipitate is seen

when the first few drops of sodium hydroxide are added, The precipitate is zinc hy- droxide.

• Zn(OH)2(s)

2++ 201 Zn

However, when excess sodium hydroxide is added, the pre- cipitate is dissolved due to

the formation of a complex

salt, sodium zincate.

Zn(OH)2+2NaOH →→→ Ne, Zn(OH)

(c) P is ethanoic acid (or acetic

acid) and its structural for mula is

CHY LOH

Mg +20E COOH (CH_COO), Mg+H2

CH COOH+PC1CH COC1+POC13

HCL

The gas observed in (1) is... hydrogen while the white fumes

in (ii) is moisted hydrogen chloride gas.

Solution to Q.' 11

(a) Pb(NO3)2 + K2CQz

Pb00 (8)+2KNO

(b) Formula mass of lead(II):

nitrate=331.

..3.31 g of lead(II)nitrate

is equivalent to 251 = 0.01

moles.

According to the following equation

Pb(NO3)2+K2 CO->Phco,

no. of moles of PbC0z formed = no, of moles of Pb(NO3)2

= 0.01 moles

•*formula mass of PbC0z=307

wt. of the precipitate = 0,01 moles x 307 g/mole.

2.07 €

(c) Furthermore, according to the

equation in (a),

no. of moles of K,CO, used

up in forming the ppt. no. of moles of Pb(NO3)2 0.01 moles

But, the original amount of

KC

50

1000

present

x 0.25 moles

0.0125 moles

amount of K2CO

in excess

(0.0125 - 0.01) moles 0.0025 moles

Let y cn be the volume of

0.5M HNO required to react with excess KCO3.

Then, according to the equat-

ion below

CO+2HNO

2KNO+H

no, of moles of

NO+H2O+CO2 needed

HINO

2 x no. of moles of excess

potassium carbonate

2 x 0.0025 moles

=0.005 moles

1000 x 0.5 moles

0.05 x 1000

100

0.5

Therefore, the volume of 0.5M nitric acid required to reagt

with excess KCO, is 100 em

物

理(八) 魯榮家

PHYSICS (8)

Answers to Exercise 4

1. (1) Let the tension. in

X be T and the tension in T be T2

When a load of mass 1 kg is added to X alone, the stretching force is LON, the extension of X is. 60 50 10cm. When the metal FQ is added, the extension of X is 55-50

5cm

=

= SN (Ans.)

Then a load of mass 1 kg

is added to Y alone the stretching force is 1ON, the extension in Y is 75 - 50 = 25cm. When PQ is added, the extension is 55 50 5cm

=

(Ans.)

22N

(11) Let W be the weight of the metal

W = 1 + 2

= 7N (Ans.)

(111) Let the distance of the centre of gravity of

from P be (cm).

By taking moment about F, we have

7x= 2(35)

x = 10(cm)

Ans: The distance of the

centre of gravity of

P from P is 10 cm.

2. (1) Take the direction

toward the right as +ve.

Total momentum, before impact

= 2(8) + 10(-1)

= 6 kgms-l

Total momentum aftes Impact

= (2+10)v = 12v

where v is the common

velocity of the box of

sand and the sphere after

impact.

By conservation of moment-

um,

21

The

required raulo,

- 1 : 2 (Ans.)

(11) Extension of the

spring in figure 1

0.23m-0.2m = 0.03m tension of the spring in figure 2

= 0.360 - 0.2m 0.13m

(Tension in figure 1)

(Tension in figure 2)

= 0.03: 0.13

313 (Ans.) (lil) Let T2 and T2 be the tensions in springs in (figure 1 and figure 2

ruspectively. U,, and un be the unthrusts on Can figurel and figure 2 respectively.

There 7 is the weight of

the cube C.

Fron part (1) uy From part (11) 1 =

Putting

these into (1)

(3)

solving for

2 from (2), ()

142 = 201

The relative density of

30

0.35 (Ans.)

4a)Total weight of oil

(Ans.) A

12v 6

+0.5 ms (ii) Total kinetic ene gy

Before impact

=

*(2)(8)2 + 1(10)(−1)2;

69J

Total kinetic energy after impact

= 3(2 + 10)(0.5)2

1.5J

Since there is no change

in potential energy before.

and after the impact,

mass of oil x 6

- density of oil

x volume of of:

= (0.75 x 103) x 6x104

4.5 (Ane.):

(b) Reading of the balance

weight of oil

weight of cylinder

≈ 4.5 + 0.1N 4.6N (Ans.).

(c) Total thrust

(pressure at bottom)

xbaue area)

= (density of oil)x (6) (depth of oil)x(base arca)

(0.75 x 103)x(10)x(0.1) x(0.01)

-

therefore the energy loss

in the collision is

697 -

1.50

7.5 (ans.)

(a)

= 67.5J (Ans.)

(iii) The energy lost in this im act will be diss-

ipated mainly in the form of heat and little in the

form of sound.

3(1) Let V volume of the

cube.C.

d = density of water Upthrust

weight of water displaced Upthrust in figure 1 =(v)ag

Upthrust in figure 2

ΣΕ

As shom in the figure, since the pressure acts normally to the surface in contact. The horizon- tal components will can- cel each other and the rosultant thrust vil act vortically upwards.

(e) The resultant thrust

(Total thrust)-(reaction

on flask)

= result of (c)

- result of (b) =7.5N

-

4.6N

= 2.9N (Ans,)

(f) Let a be the relative

density of the wooden cylinder and A be the cross-sectional area of the cylinder.

By law of floatation, weight of cylinder

- weight of oil displaced

(d x10) x0.06A x 6 =0.04A x(0.75x 103):

a = 0.5 (Ans.) (s) No, the same volume oil will displaced in

both cases, so that the liquid level will be the sane. Hence same pressure at the bottom and there- fore same thrust exert on the botton.

5(a) It shows that

(hair is made up or small

particles.

•

2. These senll particles are in constant ire- gular motion and never stop.

(11) When temperature rises,

the smoke particles nove nore vigorously. It is. because the air moleculos

more vigorously with rise. in temperature. They cause the smoke particles to

nove more vigorously when making collisions with them.

(b)(1) Volume of oleic acid in 0.03 cm3 of the solution

= 0.03% 1000

= 3 x 10-5cm3

Area of the patch of

oleic acid formed

(3)(10) 2

= 300cm2

The aproximate length of

a molecule

= thickness of the patch

=

10-5

300

107cm

10m (Ans.).

(11) The assumptions are (1) the oleic acid forms

mono-molecular layer on the water surface. (11) The oleic molecules

occupy the volume of the whole patch.

(3) Oleic acid is complete-

ly dissolved in methanol and the solution so form- ed is uniform

(4) Oleic acid does not. dis olved in water and does not evaporate.

Page 25Page 26