1978-03-04 — Page 30

頁二第張八第二日六廿月正年午戊屬夏

育教僑華

NIB & Date Presi

1978

【中學會考試題預習專欄

WAH KIU YAT PO

(iv) It turns milky or cloudy.

CO2 + Ca(011), →→ Caco2+ H2 (v) No; it is because when dilute

sulphuric acid is add to lime- stone, the insoluble calcium sulphate formed on the chips stops the r action as the acid and the chips are no longer in contact.

reagent: phosphorus(V)chloride condition: in cold

OH

化學

(廿二 )

朱宏林

1 C1+FOCI 2+ICT

(ii) reagents hydrogen gas

condition: high temperature,

high pressure, platiram as catalyst

CHCH

Chemistry (23)

Solution to

21126

(ii) t. of ammonia liberated

increase in weight of acid 3.7.6

Formula mash of (31)220=132 Formula rass of 17 According to the equation

4

132 g of (11) 30, give

(17 x 2) of

(iii) rearent acidified Totassium

dichromate or potas- sium permangan: te condition: prolonged heating

90+2(0) - CCH CHCOCK

(iv) reagert: conc sulphuric acid

condition: heated to 180°C,

excess cono 1,504

報日僑華

(ii) If

then the

current through each bulb in

2V

V1 tik. 2

六期星

日四月三年八七九一层公年七十六圈民中青教傿螯

4.375% (ABB.) (iii) Let the resistivity

Johm a

Since

the current through each bulb in Figure 1.

Therefore, the electric bulbs

in figure 1 and figure 2 would give the same brightne (c)(i) The resistance of each

electric bulb.

R

(Ans;)

(11) From figure. 1, the cur

rent flowing through Lygin

figure 1.

* 1 A (Ana.)

From Figure 2, the equivalent.

resistance "of, the circuit

6 = 12 A

The current through.

GIF, OH, OH

IA (Ans.)

(v) reacent: methanol

condition: heat, H250, or FC1

(i)

is the catalyst

Solution to 3.33

vt. of (

to give

(a)

of I

1.7

132 g x

resistor

Force

As shown in the figure above, furrent is sent through the

A

coil,

Among the four sides (or ti.

the coil, AB and C are per- pendicular to the field, there. is no fores acting on AD- And

BCA and CD are acted: by two equal and opposite forces

Those two forces. form a con- ple and the coil will then rotate about the axis XY. (ii) The magnitude of deflect- ion of thecoil is proportiona)

to

the magnitude of the current flowing through the coil.

2. the strength of the magnetic

Field

the number of turns of the coila

4.37 x

100⋅

4.37 x

10-9

ohm.m

(Ans.)

(1) The graph shows that the potential difference against

current is linear therefore

it obeys Ohm's Law.

(is) Resistance of the wire

slope of the linear gra ph

17.5 – 2.2

40,5

4.370 (Ana.)

(iii) Let the resistivity be

ohmm

-6.6 c

ence, by weight of pure

80, in the sample is. 4/24

6.6

x100%

82.3.

(111)

6.6.

132

of moles of (11,),30,

-.05 moles

According to the equation

+220

amount of NU needed to re- act with (1)

2 x no. of moles of annon

ium sulphate.

2 x 0.01 moles

0.02 moles

But, the amount of a

Cizally present is-

50

12.1 x molas

2:265 males

amount of Mat in excess

(0.105 -0.02)moles 0,085,

Row, according to the equation

2014, SC a 50+220 amount of 250 required

4

1x no. of moles of excess

sodium hydroxide

4 x 0.085 moles

= 0.0245 Loles.

Let the volume of 0.1M 1,50

required be en

Then, 0.1 x

200+2Na

cm

1000

-.0.0425

2021 ONa+

sodium hydro-

ethoxide gen

CHCOC1+

(1) Nitrogen monoxide is formed.

440 + 610

4 + 50

(ii) Redish-brown gas which is

nitrogen dioxide is seen..

2N0 + -202

(iii) Platinum coil acts as

catalyst.

(iv) The platinum coil is heated initially in order to start the reaction, Yowever, once when there ction is started, the heat liberated by the rểm: action can maintain the coil at a high temperature, so no no ore heting is required.

(v) The litmus is blue at the start,

but at the end of the experiment, It turns red, This is because all the ammonia is oxidised and the nitrogen dioxide dissolves in water to give an acidic sol- ution.

2102 + H20 - TEIG + HNO

It continues to purn toive white powder and yellow parti cles of suphur. This is because the heat liberated by the burn- ing magnesium can deco pose sul- phur dioxide into slpur and oxygen which can support the. combustion of magnesium,"

2 g + 502 21.60 + $

(ii) A vigorous effervescence occurs.

Reddish-brown gas of nitrogen. dioxide is given off and a green ish-blue solution is obtained. It is copper(I)nitrate

This is because concent Solution

nitric acid oxidises copper to copper(II)oxide which is, then at once neutralised by the excess nitric acid to form copper(II) nitrate and water.

Cu + 4003 Cu(NO3)2 +2002+22

8

(ii) CH COOH+FC

POC13

物理

acetyl

phos-

chloride

HC1 hydrogen chloride

phorus

oxych-

loride

CHCHHSO4

ethyl hydrogen sulphate

(iii) CH2=CH2

(iv) C2O+CH COOH CH, COOCH5

(v) 20H

2CH_COCNa+Ca(OH)2

ethyl ethanoate.

water

me thane

+ Ca303

calcium carbonate

s-dium cartonate:

Solution to Q.32

is darbon aloxide. CaCO2+2PC1→→→ CaCl2 + 10+ 0 0.

(ii) Y is carbon monoxide.

+ C - 200

012

(iii) Greys beads of lead metal are

obtained in hout And, reddish brown copper powder in hoat B.

2004 CO

CuO CO

C

+

2

+ Cu

(十二) 榮家

PHYSICS (22)

Answers to Exercise 11.

The current throngh

but the current through

โร

the brightness of 1 not affected but the bright- ness of ln is reduced.

2.(1) Let the current through

the battery be 1, the current through the voltameter be I and the current through the

heater be l

Since

3.3

x 3600

The current I taken from the hattery is given by

10)

10 + 20

3 x 9.75 A 2.25A

(ing.)

(b)(i)

As shown in the above. Figure;

in order to use it to measure

"corrent up to 1,0 ampere,

low resistance shunt R

must be connected in

parallel

witli the ammoter ameter

given.

Let the resistance of the

he Ra

currents.

through the meter and the

resistance Rhe

respectively..

and

If a current 1 - 1 A is

sent through the whole system

then

By Ohm's Law

(ii) The current through the

each

(11)

heater

is given by.

I la

म

- 2.25 0.75

1.5A

12 Let the resistance of electric bulb bera (a)(i) The currents flawing through and Ly are equa and hoth equal to"

r.

bath would give

same brightness, (ii) The currents flowing

through Ly and

same and equal to

samo

JAR

the

hoth would give the

(brightness,

•

then-

3 V V the current flowing through the electric baths in figure 1

V

2 times the current flowing through the electric bulbs in figure 2..

Therefore, the electric bulba in fig. 1 are brighter than "that in figure 2.

to

·Energy given out by the heat ing coil

(1.5)

10 x 3600 (Ans.)

81000-J

(iii) Let the rise in temp- erature of copper vessel and its content bec Total heat gained -0.5x4200x# + 0,5x420x#

- 23108 J

2310081000

9-15

the rise in temperature

of the "copper vessel and its content is 35°c.

3.(a)

Since

+37

100

4.37 x 10 ohm. m.

(Ano.)

育教僑華

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學能推理練習專欄

文字推理

文字推理辣營四

·英文編號富在答案讓內。

(四)

(二)

下列师有五個項目,其中一個和其餘四個 如不同類;就還出這個不同類的項目,僅把它的

1. AM.

C魍

D憋

E魑

2. A

B*

DZ

E終:

3. Att

B棒

D劍

E

4. AÆ

B.

D

5. A M

D性

10 x 107

0.01 A

6. P佈量 S騆飾

·TEM

7. PG

R

- 2.50

珍珠

8. P

Q將會

R**

0.01 x 2.5

S假使

TER

9. P

Q公佈

R***

2.525 x 10 'n (ins.)

SWE

T宣传

10. P##

Q ##

R 奇怪.

/ S

T懼伯

10.

1,0 = 0,01

As shown in the above figures in order to measure voltage" up to 1.0 volt, a high

sistance R must be connected

in series with the meter.

The potential difference: across the system is:

11., A MOB]

D刷 E

12. A 4-

C#

鈺漂亮

CAL

D

E鮑變

12.

13. A合

B厪 C大厦

洋房

E木屋

13.

14. A*4

̇D 尬

15. A肥胖

D 健碩 16. P黃花魚Q讲寐 SAR T** 17.P時間 SA

BBQ E備课: B.寬版 E瘦削

CALA

14.

C矮小

R虹衫熊

16.

R韶光

18. P

ESME

T⭑# Q搶劫 R 藏慈

17, (

T盜竊

18.

19. P

SAM

Q問漂

R俐落

T明快

19.

20. P飛彈

21. A

Q核彈

燃燒彈

S导弹

T麖明垾

20.

BÝV

C效法

V

D參照

E批評

21.

22. A

B证是

C.月亮

llance,

V - 1 V

DAI

E太陽

22.

10-31 -0.01A

23. A · B***

***

D 行書

T楷書

23. ()

24 A¶X

B俏父

C叔父

2:50

D##

E姑大

24. (

25. A**

B##

C奸詐

R

JD 繪 T

25.

1-2.5 x 0,01

0.01

QA * ®Q

OB OA OD

OS

B

D JA DA

B

- 97.5.2

(Ans.)

* OT

DE

B OC

VD BE

-

.OR OF FOT

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