1967-04-05 — Page 17

第五集

CITY HAL

中文中學會考試題預習專欄

敦學科

(廿三)

第廿二次預習試題解答:

·喬仲强。

(1)(a)將一個4吋x2李时×号时的長方体,分割成相同之最大 正方体,間说)正方体每边長?(i)每個正方体立体?(

正方体若干個?

(解)此長方体之三度為4吋,H.CE(最大公約數)意 此即為好求正方体每边長吋數,故体積((立方吋) *** (÷) × (≈£÷†)× (/÷÷)=24×15×10=36001 答:正方体边去时,体積為25立方吋,共有正方体3600個、 忍)某人的收入增加20%,但其入息稅則由每鎊4先令增35 先令問此人之纯收入增加之百分率若干?

(解)每鎊4光令之稅率為20%,每鎊5先令之税率骂 嘉=25%設此人原有收入為100%,則繳稅後凈得鸢 100%×(1-20%)=80%;增加收入後,其收入120%镦税 後凈得為 120%x(1-25%)=90%。

該收入增加後,其纯收入為原来纯收入之號112套90 即增加.12元% (答)

(註)本題不可以 90%-80%—10%為答案,因此百分率係原来 未缴税前之收入(非纯收入)為“母數(100%)故也。

(2){a} 一空心鉄管,其內面直径2号呎,其重量典等長而直徑為 湯之鉄柱相等,問此铁管之厚為若干時?(準確至小數两位) (解)因→32吋,13呎=16吋,

談鉄管厚吋,則其妹面半篤1648吋,長度為见吋,則 ( == π { ( 1b + x)2 = 1 6 *] × & ે‡‡ 4‡ =π × 8 × £ £t.

π [ ( 16+ x)2=162] × l = + x 82 x 7.

(16+ x)*—16* ---- §

16+%=184 -₤320-

=17.89(查平方根

x=1.89

答:铁管 1.89吋

湯)有密度為每立方厘米0.6克及1.1克的兩種液体,混合後密度 為每立方厘米0.75克,求其混合比

(解)設此两種液体各用x,y立方厘米,則天重0.6x+1.14元

依題意

0.6x+1.19=0.75 (x+Y)

0.35y=0.15x

0.35

ONS

答:其混合比為7:3.

(註)本題亦可依混合比舊損蓋成反比之方法求出

(3)64x+ax+82-40x416為完全方式,試求af之值

(解)其平方為x之二次三項式,且其首項為x(但未項可能為 ±4)現假設平方根為200+px+g/其中48文字當數

4x2+ax38x~40x+16=(2x2+px+8),

— 4x2 + 4px + 48 px + 2 p q x + q z

“比較两边係數?a=p

*(2)

AH KIU

眺榮榮權

·術。該院因接紐約來,「業 」客劇版權,已由百老湘舞合床 有,

【衆者並無顯路云。[]

財院對帽盛,握持,來入場

「船塢工人铯女和共二個堂兄無愛故事, 演出。社會人士對該院社號線-

節曲折道癌,係一動人聖劇。在該院戲劇

·大会堂音樂姬阿出。整翡述一個綁約 【「橋頭滾跳」-康五日起一三晚,仍在

公演中國名「蠢配」, 六一晚,假香港大會堂音樂

該院

AT PO

明起一連三晚在大會堂演出

浸會學院劇社

演 橋頭遠眺

▶欄專習預題試

化學科 (

J

九

時自由題會,七情六』七。各唆友知識

一饭店。席金-母位十健雅定席苫。〔珧〕

CHEMISTRY (23)

六。時間:下午四 年四月十五日(B肼◎大*];歌闳深 | 加。茲將裡則列後- 高-英舘-道院大E 一,希望各换发踴躍參四四七五八四;洪接 日期 王一九六七 九栩K率,铖話 - -七

| 五日假六國店舉行 榮生辦辦館,電話: 「八大會,定期木月十弈,小閣患嘅街二號 中學校友會本年度 獎。售票處-蹴亦 〔舂訊〕金文樂,二元。餘興:幸運抽 十五日般六國酒店舉行 堅-大道店三九號

上 定期同人大會

英

•

王錦釗 •

1.1.An intimate mixture of amimandum chloride and slaked

lime is put into a round-bottomed flask, and the. apparatus is set up as shown in the diagram.

JAMNUM CHLORIDE

To DAY the

When the-flask is heated gently, ammohia gas is given off. The gas is dried by passing it through a drying tower containing calcium oxide as a drying agent. The ls then dollected in gas jars by upward delivery; gas.

Ca(OH)2 + 2NHLCY CaCl2 + 2H2O + 2NH3↑

slaked Time

ammonium chloride

(1) when a gas jar

jar of chlorine chloride are: -action of aEM

8NH= +-3C+

Ammonia

=

ammonia

da is inverted over a gas

cite fumes of ammonium

result of the reducing.

期 金文泰校友會

Q.2(87

日五月四年七六九一层公年六十五國民中

你先·柏怨,符合

元面板老力,本稻越。州

A

,仍年後宣行 師資 加

使費二期大校。院試圖

人數仍擠

(特選} 本年度]餘人,其中三千

·今年共達七千八百餘人

|投考師範|

•今年三開篮學

[本年度報考三批|性,要有超卓的这

四季校青年,才可望有取錄機 是引考試有高飛的胡

• (E)

Test\ (i)Substanceyx

s heated' gently

(ii)Solution or:

x + hydrom chloric acid.

*barium

chloride

solution

Observation

A gas is evolved which turns damp litmus blue/

A white pPLE 19 faman

Hence X is ammonium sulphate]

擠範

"Inference

The gas evolved Is Ammonia since it is

the only common alkline gas.

x is therefore an ammonium salt

The precipitate 19 harium sulphate, which is insoluble)

in hydrochloric acid

Ba

穿

** 504 Ba304

* is therefore at

sulphate

B.2(b)The common drying agents used. in the laboratory are:

CONR

(1) Concentrated sulphuric acid,

(ii) fused calcium chloride ~

(iii) quicklime,

Yiv) phosphorus pentoxide;

(v) ailica gel.

Two ways in which gases" are brought into contact

with the drying material are:

(a) by bubbling the gas through the drying material:

if it is a liquid such as sulphuric acid; . (b) by passing the gas through the drying material,

: contained in a suitable piece of apparatus,

such as a U-tube, if it is a solid such as fused calcium chloride.

NET

GAS

DRY

ÚAS

FUSED

由(4) 得

2pg=-40

2016

代入(3)得本三

-20

5208209.

=4,代入门(2)得

-20 b=41 £ a=20, b=9

(注):其方根萬+(20=50+(4)

(4) 解方程式

+A+&O

(26) £54 ali+ bx)+ f (i+ax) + (a+b) (1+ ax) (1+ bx)

£3 a+abx+b+ab x+ (a+b)+ (a+b) (bx+ax)+(a+b) ab x'= 0

ablars) x2+ { (a+b)+ 2a8]x+2(a+b);

分解 (abx+(a+b)][(a+b)x+2]

=味或

因此两值俱不能使分母為零故為真根。

(5)卷

試証:

at aft

(証)由己設式依反比得其

~ 2 + ( 4 ) = a+b

a+ab+b

(13) when an-aqued

a soluti on of . precipitate of

honia is added to

a brown gelatinous Pride is formed, which

ds insolble in excessor ammonia.

'`3NHOH_A:FeCl>

HONGKONG

Q. E. D..

3NH, C1 + Fe(OH) zł)

COPPARIES

Merric

hydroxide

(111) When an aqueous solution or ammonia is added "drop

by drop to copper sulphate solution contained in

a test-tube, at first a light blue grecipitata sopper hydroxide is formed:

„2NH2OH + CuSO1 = (NHL),30,*+ Cu(OH

gopper.

'If the ammonia Ls added until It is in excess, the precipitate will thesolve and a beautiful deep blu solution of a complex copper salt, cuprammoniuy sulphate, (Cu•4NHz) SOд H0 is formed

'vJWhen a gasjar of ammonia is inverted over a gas

Jar of hydrogen chloride dense white fumes of ammonium chloride are formed as a result of

chemical.combination between the two gases.

NH + HCl = -NH/Cl Thydrogen

chloride..

Suitable Drying Material,

H2S

Fused calcium chloride

Cl2

Fused calcium chloride'a.

cone. sulphuric acid)

NH3

Quicklime

ICO.

Fused calcium chloride.or] conc. sulphuric acid.

Questions?For Next Week!(24)

Q.1.How do the following behave towards,

(a) a solution of ammonium nitrite: (b) Solid ammonium sulphate: (c)

Solid ammonium dd chromate1.

[Chefactionfofi heari

2.2. (a) A salt M on being neated with caustic sodalgave) off a gas which turned red litmus blue.i

(b) Addition of silver nitrate solution to a solution]

of Min water gave a white, precipitate whichiwâg insoluble in nitric acid.

What is M7 Explain and give equational forīthe

· reactions

Q.3.Give lead nitrate crystals.describe now you could

obtain from t them:

(a) oxygen; (b) dinitrogen tetroxide: c/leadisulph ate; (d) metallic lead.

Describe the reactions whicn.take, place between nitrogen dioxide and (1) water, (ii), moist, sulphuri dioxide.

(註)本題亦可依下

依更比得哦

(6)描绘,y=(x+1)(x-4)图解(X取值由-2至+4),並利用之

以解 33x8-48+4=0(答案須準確至小數後一位)

(解)先製成下表:

14249

y== x(x+1)(x-4)

3. 42

012-24-24|| 012220

然後選擇適當之單位(维横軸均為一吋表一單位)描出各卖 依次連结之得·吉x(x+1)(x-4)曲线,因製表時當==-1及。

時,你有相同之值,故需再設...............

因绘曲线為5y=x338 15g+4=0.故於同一图中加终 即為求之根

之代入求解方程式得 (4)設0萬△ABC内任意点遇口作 DGHABEHI BC, FKII CA 若之直线,其交奌之横坐

EHFK

AB AC CA

由图中得 x=1.5 0.7或3.8(答)

(5)設如右下图: P笃△ABC 外接园之AB弧、 (註)由根與係數關係,可知方程式三根和*數之变号好上任意点 PELAB PF1BC(F在BC线上) 此值可利用之算

PF之延线交 ABC园於G. 試証 - EFI AG.

D

4£=== X(X+1)(x-4).

̇第十三次預習試題 {1) AB马d园直径,CD典AB交於P

過P作AB文垂线交ACAD(或其延线). 拎E及F試証CE DF共园

|2) 0為△ABC 外接园园心, AO延线!

交 BC柊KAHL BC交BC於H

"AH BH-CH...

AKL

BK CK

(3)以沈:△ABC之 三边向形外作諸 正方 ABDE. BCFG 及 ACHK. 则者三角形,

\AEK_BDG=CFH ¥IŤ (LBACŹ11r

H

(6) ABCD為任意四边形一直线

AB BC CD 及DA (或其延线)各於E

AE

LEB

DH GDWHAS