1977-01-27 — Page 19

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2.由大會主席何襃費主持 技少應有三十個會,可 在假日酒店行研討會推出一個完全及格之

會三〇三案昨下午六時灤區會與檳會之品,

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180.

郭雅、少、謝

、各友會會長梁乃鹞、

【生、王翰,何榮高

、趾圍交、財寶章、岑關係發揚人類博愛精神

有正關總監黃翊盈、吳 潘約翰雷獅子會吳之一。 在男、瀬簽祺、劉天宏一務與查任,以透過國際S

·由陳錦司儀,到務應有之露紙。前總監

將若于闚陧檢倬得任大間,强公民來 會,一千二百會為,爲機會委任之。國際總會 使剧于會發科光大下珮事潘光迴露面臨幾個, 【强调共有十一個直行凱樂總監,否則由 -及對社會工作及社會服

怡,團結會員,追友

,促進社會福利互愛互

【良、黃世、雄政、男致詞。(守》 恩、宋北案、伯英,年學藝比賽均爲有意義 「機甲醚、憋輝立、李國之昂。末由副總監及商人 卞館淸、何文雜、花慶年康樂中心,張道,附 臣、宫永樂、陳狮街, 原則,如年來推動付 二陳乃拙、麥偉页、彭,强要有紀念品及永久性 力、伏金河、誠泰正,狮子發之服務計歡,主 馮彥、忡下、韋明、體。上座總監伍凌生點。 三〇三蓝猫子會研

「强、潘寶光、張孝、柯討會上篇盜監

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1977中學會考試題預習專欄

明德社主編

化潮(十二)・朱宏林

Chemistry (12)

Answer to Q.18

The method employed is called fractional dis- tillation because SiCl4

and PC13

are liquids,

with boiling Points close to each other.

g) Aluminum chloride. This

is because this compounds

A.

Air is a mixture.

B

The five main components

a)

dioxide,

of air are nitrogen, oxygen, carbon

noble gases and water. vapour.

(c) Water is run into the aspirator in order to force the air to Pass through the apparatus. d) i) caustic soda is used to absorb the carbon dioxide.

ii) Concentrated sulphuric acid is used to absorb the water vapour. iii) Copper Powder can re-

move the oxygen from air by combining with it.

iv) The lugs are to Pre- vent the copper powd- er from entering into the caustic soda. The gas is nitrogen. Copper (II) oxide is formed and it is black Dowder.

g) 1) 2NaOH+CONaco

H20

ii) 2Ca + 0,

+

The density is larger than its normal density because it consists of noble gases as the impurities.

i) It is because the first

few gas bubbles contain the air originally Present in the bottles and combustion tube.

j) No. It is because the air which have passed through the caustic soda will contain water vapour.

The formula is WX3. It consists of covalent bonds because both W and X are electronegative elements.

b) The formula iş ZX. it co consists of z` and X

ions because Z is elec- tropositive and x is electronegative. Ionic bonds are present.

c) The formula is x

X2. The

bonding is covalent.

a) There is no bonding

between atoms of Y because

Y is a noble element.

e) z is a metallic element

so the atoms of Z are held together by metallic bonds. Question 19

chlorine.

SiCl4 PC13 liquids

and scl2

c) Sodium chloride and

are

magnesium chloride. It is because both of them have high melting Points and boiling Points showing that the interparticulate forces are very great. d) Aluminium chloride

because its boiling Point is lower than its melting Point. That means, the substance can change into vapour state without passing through the li- quid state.

e) The molecular shape of

SiCl4 is,

Gi

l

consists of covalents bords but no ions so

there are no mobile ions

in molten state.

Add silver nitrate solu

tion acidified with dilute nitric acid to the two aqueous solution separately.

i) The one containing

iodide ions will give yellow Precipitates; i.e. Ag(ag)+I ̄(aq)→

ii) The one containing

chloride ions will

AgI(s)

give white Precipitates i.e. Ag+(aq)+cl ̄(aq)➜→→

AgCl (s).

1) Each of the solids is

heated separately with sodium hydroxide solution

i) If ammonia gas is

given off, which can "be tested by its

characteristic smell and its ability to turn the red wet litmus Paper blue, then the solid is ammonium . sulphate.

ẢNH

4

+2NaOH-

Na2SO + 2NH2 + 2H2O ii) There is no obvious

change in the case of sodium sulphate.

c) Pass each of the gases

into acidified Potassium

Permenganate solution separately.

i) The gas which can

decolourize the Po- tassium Dermanganate solution is ethene.- ii) The gas which cannot

turns the Potassium Permanganate solution colourless is ethane. d) Each of the liquids is. heated separately with a mixture of iodine and sodium hydroxide.

1) The liquid can Produce

a yellow Precipitate which is triiodomethane (or iodo form), CHI

CHI3' then the liquid is ethanol.

ii) If the liquid does

not give any Preci- pitate, then the liquid is methanol. This is known as iodoform test.

Answer to 20

A.

a) A is liquid dinitrogen

tetroxide, 2047

N.O

B is oxygen gas Solid Dis lead(Ir) oxide 2Pb(NO3)2

28b0 + 4NO2+ 0.

92

d) It is used to condense.

the dinitrogen. tetroxide

e)

solid D'is reddish- brown when hot but yellow when cold.

f) When solid D is added

to dilute hydrochloric,

sulphuric acid or dilute acid, an insoluble salts (i.e. lead chloride or lead sulphate) is form- ed over the solid D, thus stopping the reac tion as it prevents the acid from coming into contact with the solid. in the case. D_However, of dilute nitric acid, the salt formed is lead nitrate which is soluble in water, so the reaction can continuer.

g): On heating, liquid An

turns from Pale yellow to light brown and then dark-brown or nearly black. When the temperature is raised to 620°C, it- becomes colourless again.

a) The colour changes ob-

served are due to the following changes:-

it

No 24 (1)

42

pale yellow

2 NO2(s)

light brown.

150° 2 NO2 (8)

NO2 (e) 620

dark-brown

2NO + 02.

colourless

Answer to Question 21

A.

a) It is because in the molten state, the ions, Na" and

I, of sodium iodide are separated and free to move (i.e. mobile) and conduction of electric current through an elec- trolyte depends on such mobility of ions. b)i) Iodine is formed which

can be dissolved in water to give a yellow- ish-brown solution. ii) 21 -1 + 2e c) The metal is sodium. d) Sodium oxide.

e) It is hydrogen gas. fi) The gas, hydrogen,

comes from water which is slightly ionized to give H and OH

++

ii) 2 +2e

g) Time

2

= 60secX16=960sec. Quantity of electricity 0.1 X 960 coulombs.

= 96 coulombs.

96

96000

faradays.

≈ 0.001 faradays.

h) According to the equation,

+

24

+ 2e2

2 faradays can dis-

change 22400cm3 of H2

.'. Vol. of H2

dischanged

by 0.001 Faradays is, 22400cm3 0.001

X

= 11.2cm3

1) Yes. It is because gas B is insoluble in water and any dinitrogen tee troxide which has not. been condensed will be dissolved in the water.

j) If copper(II) nitrate

crystals are used, a mixture of nitric acid and nitrous acid is collected in the U-tube, because copper(II) nitrace crystals contain water of crystallization which is liberated at the same time on heating. N204

04+H20 HNO3+HNO, 2.

B.

and no. of moles of MO

1.0 obtained =

moles 16+ Y According to the equation in (a).

no. of moles of MCO3 no. of moles ofMO

60+y

1.0 16+y

2.1y + 33-6 60+y

1.1y 26.4.

y = 24

Hence,

the relative atomic. mass of M is 24.

e) The metallis carbonate

is most probably magnesium carbonate.

1977中學會考試題預習專機

明德社主編

物理

(十二)・御榮家

Physics (12)

Answers to exercise ó

the end

1. (i) By using two tuning forks instead of one, correction can be elimina teď easily by finding the first position of resonace. If only orie tuning fork is used, we have to find two different positions of resonance in order to eliminate the end correction, and it is rather difficult to obtain the arc- ond position of resonance accurately in the experiment.

the

(ii) let the length of re- sonating column be L cm the end correction be ̈€ wave-length be A, the freque- ncy be fand the velocity of sound he cms; Therefore

- L+€

A = H(LTE)

but, v†A

v

since

V =

also, f

So

a) Since M is divalent, the formulae for its carbonate and oxide are

and Mo respectively, MCO MCOMO + CO2

3

b) mass of metallic carbonate

= (44.4 - 42-3)g

2.1 g.

c) Mass of metallic oxide

(43.3-42-3)g.

=

-

1.0g.

d) Let the relative atomic mass of the metal be y. Then the formula mass

60 + y

and the formula mass of MO 16+ y

OF MCO 3

=

No. of moles of MCO3

2.1 60+y

moles.

used

44(6+6)

320 when I 26 4(320)(26 + € )·

480, when L - v = 4(480) (17 -+ € )

(1).

17

.(2)

Fliminating € and solve for v, from (1)

Pipe A, closed at one end, standing ware with lat over tone

.0.6m

Pipe B, open at hoth ends standing wave, first overtone

x

(ii) As the vel neity of sound in air does not change, there fore the wavelength should be the same for both pipes. Neglecting the end corɛection the length of pipe B = 0.6 x (4/3)

0.8m (Ans.

(iii)

x

For the fundamental tone in Pipe B, the wavelength of the standing wave

= 2 x 0.8.

1.6m

the frequency »

velocity wavelength

344 1.6

215 Hz (Ans.)

(c) Let the velocity of sound in air be v, the original distance of the man from the cliff be K..

Hence

(2)

-

20

2d

(1)

+(2)

(Ans.)

이

(ii) from (1)

1:

(2)(a)

(refractive

prism is

(ANS.)

As figure of the

the first

refraction takes place at the vertical face AB

By Snell's Law

sin i

sin a

. (1).

For the second refraction, since there is total internal

b must

reflection, therand

be the critical

€ = 90°.

Also, by Snell's law

sin b #in c

sin b

1

(2)

1.2

#in90"

1.

-0.8333

b

56

V

4 x 320 from (2)

26 + €

(3)

a = 90*

b

- 90

56"261

V

- 17 + É ... - (4)

4x 480

(3)

www

(4)

▼

V

4 x 320 4 x 480

13

1.2

Ans: the velocity of sound in air is $45,6 ma

(b) (1)

33-341

substitute înto (1)

ain i sin33340

min i = 1,2 x sin33

1.2 x 0.5529

= 0.6635

1°34 (Ans)

(To be cont. next week

* « 34560 cas

V

345.6 mm.