1966-04-06 — Page 22

第二第張六第

日六十月三年牛芮膺夏

WAH KILI YAT PO

英中會考試題預習專欄化學科(十四)王錦釗一九六六年度

CHEMISTRY (14) ·

1 (s) The fundamental particies of which an atom is thought to be

built up are the electron, the proton and the neutrou

Particl

Proton

Neution

Float Pan

(b) Electrova)ency

Mass, relative to

that of the hydro gen ion

I

l L/1645

Slectrical charge

unit positive charge

No charge unit negative charge

In general each atom tends to surround itself with a stable 'octet' of electrons The 'octet 'nɛn be achieved by a gain or loss of electrons Such atoms as will react to form compounds by electron transfer are themselves

converted into ions. Commounas formed by electron transfer are called electrovalent compounds. For example, the electronic structure of the calcium atom is 2.8.8.2 while that of the chlorine atom is 2.8.7. In the formation of calcium chloride, the two electrons in the outer shell of the calciul atom move into the outer shells of two different chlorine atoms The calcium atom and the two chlorine atoms now have the electronic configuration of argon.

The calcium atom has last two electrons and become a positively charged ion, while the chlorine atoms have each rained an eletron, and hecome neretively charged ions..

ÉLECTRON TRANSFER

QUANG INAMICAL COMBINATION

中中試題預習專欄

數學科 ( 十四 ) ·喬仲強·

第十三次預習試題解答

1.(a)地图上以11碼4吋表200哩,試求地图之比例尺.卷 地图上佔面積 3.6平方吋,該地的真是面積着干方哩?

(石)14吋:200哩=400吋:200哩

-400 of: «o × 5280 × 12 of

=1:31680(比例尺)

其比例尺亦可寫為1吋:咖哩,故面積之比為了方吋裡 即地图上1方时之面積,蹇際上為最哩

放 3.6方时面積,表示為 36女方哩=0.9.方哩. 答:比例尺為1:31680,真是面積0.9嘿

(6)美金76元可兑港幣446.5元、英鎊34鎊可兑港幣

544元,間英鎊 28元鎊,可兑美金多少?

(解)英鎊|鎊合港幣4元;又港幣1元合美金5 故英鎊282鎊合港幣2825元,

合美金28245*

34

(注)本題亦可用連鎖比例解之,其式如下:

:768元《答

-282鎊

2828 $44×76

港幣544元

X

-美金76元

=768.

美金1元

吳34鎊

港幣4租房.5元

2.(a).

.但由銀質鑄成之直锥体,底半狂12吋,斜高20 吋着此园錐体是由每個直径冷时,原古吋六銀幣熔釤 而成,問需用銀幣于相?

(解)直园錐体体積v=

CHLORINE ATOM

Ca +

{2+}

a

(282)

Covalency

CALEM ATOM

Ca**

[2.6.81

→

(2.8)

CHLORINA ATUM

In the case of covalency, the 'oclet' is achieved by sharing electrons. There is no question of transfer of electrons, as both atoms are electron accepters Electrons are not actually gained or lost by the atoms involved. For example, in the formation of ammonia, three hydrogen atoms share their single valency electrons with the nitropen atom, which in turn, shares its five valency elections with the three hydrogen atoms This leaves each atom with a stable

electronic configuration. There are three covalent bonds,

each formed by 2 electrons, one from a hydrogen atom an one from the nitrogen atom

Bu42

(e) It is know that atoms of the same element can have different atomic weights, although the atoms have the same chemical properties The explanation is that this is due to the rou that each element has atoms with diflerent numbers of neutrons, which causes the stoms to have different weights: An element possessing this property is said to show isotopy, and the etoms of different weights aru said to be isotopes of the element.

A good example of isotoby is given by chlorine whose atomic weight 1s 35-455

The

There is an isotope whose atomic weight is 35.

nucleus ionsists of 17 protons aid 18 neutrons. There. are 17 electrons if the atom.

Chlorine has a second isotope whose atomic weight is 37

The hucleus consists of 37 protons and 20 neutrons.

also 17 eloatrons in the atom.

品

There are

POdinary chloride consists of about 75% of the 35-1sotope.

and 25% of the 37-isotopes to a humid at

Q21a) When commercial iron is exposed to a humid atmosphere for a ahom

time, it soon becomes covered with a reddish-brown film which 16 called rust Rust usually contains ferrous oxide, ferrie oxide, carbon dioxide and water.

The rusting of iron will only take place in the presence of both air for oxygen) and water, Iron will not rust in air alone, nor will fron rust in water alone. Pure iron does not rust

If carbon dioxide is presenty dron rusts more rapidly, but if an alkali 19 present, iron will rust more slowly than in ordinary air.

There are a number of methods used to prevent iron from rusting.

(3) Pointing. This is the commonest method. Steel bridges, ships, motor cars, etc,, are pointed to prevent them from rusting::

lii) Using grease or tar -- Tarte used for bridgesy: grease or

011 is used for tpole.

(111/Gal van teing-- Sheet iron la dipped into molten zing which

forms a protective surface over the iron.

liv] Electroplating-- The tin on tin căng is put on in this

(v) Plastics-- The surface of the iron is sprayed with a layer

of some plastic material, which provides a protective surface against hist

(vi) Stainless steel-- Stainless steel which is an alloy

consist lig mainly. of: Iron and chromfum is being ised increasing quantities.NA

(B)Mast methods or preparing.ferase chloride give the aqueous

ebiut lon of the substance, Pure anhydrous ferric chloride cannot be obtained by evaborating an aqueous solution of ferric chloride, because ferric chloride nydrolyses in

FeCl3 520 Fe(OH), * 3HGI

When ferric chloride solution is heated to dryness, the ferric hydroxide decomposes into ferric oxide and water.

2Fe(OH)3 = €25

Oy + Hz O

A sample of oure anhydrous ferric chloride is obtained by passing ast reum of dry chlorine over some hot iron wire: Are combustión cube The wire glows and black crystals of Fergie chloride are obtained in the cooler parts of the tube

Cuestions for next week.

2FeCTz

What: weight af zinc oxide can be obtained by heating 50. gh. of 2n Zn(NO3)2 · 6H_07 What Is the least: weight or sulihuric acid containing 80 per cent by waluht of HS01 that would be reoudrea to dissolve the zinc oxide

12. 48 cc. of 0.12 N sulphuric acid neutralson Barcely 30.0-CI DI

potassium hydroxice solution. Name the salt formed, and calculered (a) the normality of the potassium hydroxide solution, (b) the number of gm, of tiesium hydroxide in 1 litre of the origina) solution, (c) the number of ga. of the salt, corrent to two decimal places, forsed by neutralising the original 56 c.c. of Fotassium hydroxide. SAN

flask wiphed 75. 34 mm. when evacuated and 79.60 g. when filled with a certain gas at atmospheric temperature and pressure, Thá volume of the flock was 1230 cc. and the room temperature and pressure throughout the experiment were 16,0 and 745 min., Calculate

he molecular weight of the gas.

依畢氏定理 龙:

202-122 = 16-(0)

V

768折(立方吋

(立方吋)

32

而每一銀幣之体犢开片长:

故旺需個數為768÷4=768#x= 819元

答:

8192

(七)甲乙丙三人,作一哩單車競賽中勝乙 160 設丙之速度海時 20哩,来甲乙二人之速度。

(解)丙20秒可走,20x誌哩=4

新故甲走1哩時丙走 一憂哩,因時間相同,故距離與

速度成正比

:

又甲走1760碼時,乙走 1600 码

1x20

~x=

1760:1600=22

You 1600x 12 xx=2071

答:甲海時 224哩,乙每辕20分哩.

3.(0) 某人以現金每股85元買股票面額100元之股票10.0.

股,一年後分得 股息

| 股:90元售出全部

(解)股票本金鸡

每股息红利~100 股票買賣藉利 共獲利

獲利百分率

((按票面計),並特股票按每

0元

¥100= 2000元.

(90元-85元)×100=500元 00元+500元=2500元, 2500

100% 29舌%(答)

(f)有甲乙丙三數,甲奖乙之比為4:3,乙兵丙之比為7:9.三數 之連乘積為127008. 求此三數

(解)甲乙 =43"設甲數為28x.則乙數為21x

乙:昞= 79 丙數為27元

5= 28:21:27 | 依題意 28元-21x27x=127040

197008

128X2/27

28x2=56

2/ × 2=42

56 42 54

4(a) § 4a+982 = 4al (訛)

=2.

This log za+3b- lega+·loqt

ab+9b2 = 16 ab

「鄉凡四百五十人。

·九龍城街坊會

設英文中學

南一段,鄭氏僻九 钚參加。 長有年,魏會早有小 各校友,屈時 歸命光,歷任該會首會長聞文戰爭邳通函 校歌染,旋鬋陳元素氣對會務推貸調育 洪孟平、徐維齡、楊柏南、茨菰、張善 城街坊殿钓會理事長超同學赴日本安造。,先行祝禱體經反培會長肉苕患,顯始,張炳南、陳阿珍、周生華、郭花道、 (國際可)九龍 舉行,同時跺黄媖 開會循宗裝惓式舉行鋼今職員,由於花 小康,司嫩葉類梁,理事長關絣生,滋養 弱,熱心至籍可點,費事,改擊醇,認會長黃佐時,殛王貞,秘葉 [骝漭上篇會是菜現一

【其中有發出方幕捕者, 一方面鼎力支授班校,,在後製模督感到。 科學校點收,並萬不會長分別報告交建 在發爾頓酒店舉行聯淼年會,在過杰取同 ,已先將甄自先行设梁由巫校長反高苕鸥唱江大學香港同學會,能(1月)晚 海 提名下,選出一九六六至六七年慶會長

倍,特可转各界與龍城人 食設法露磔基金,預辦理。 立辦委會,以質,可通知幹事陳光 ;最近將成,千設計如有紅花

【鸛理監事之使茼龍 林基督敦埸公 直以「建中學一件,赏析員及校友前往九龍 高雄,認爲可就原模加下午六時半,全體職 未能供子女升中月十七日(星期日 龍城區域內不少坊眾 又該會定期学太

日六月四年六六九一膺公年五十五回民華中

六 培正同醨總會

無士之然烈摂额,乍現培校友會 今日月叙粲 (國際 拮道女 選出新職員

于中寮在熙选·小学之

月份就縠于定今(4 何窕田街入號令所|校加者只見熱烈, 3日下午七時,在九中,各願學同學還 墁正同學們會太 樂紫工作盛橋行穿

| 杯與新任會長董落璁入左J

10

才孩得。

·唱校段及:呼,會長灣來明明絨被求會員 比赛资冠軍及攝影比賽冠軍,均爲伍籍 、 纯禽中,由閒學凌志編及張耕生韻淇

i其他節目對有同努之子女提任之女高音

哥舞 試。後選隨基督政大犘同學會上

·希照節目,本由筲韓供廠商之阿哥

支、張梵、周月亭、朱文嬌、鸡袄、

·受紀念銀杯,以免過去一年談會工作之

·座會長奶 網將校友杯交與新金長,並 田怡如,伍貫才、驗其良。選出後,患

等,于午夜由智杂的六入及驚厚照夫人抽,

無海慕敎育彩♯金而世之幸遇

選,

日本來闾畿,主流會長

秉椰夫人山峰,爾有那麼迷檢驗,二

华牌光衣機等共五十三項。荒會間季默

分解···

又X+9=45

·180.

36)(x+5)=0

去分母

【女士浩製,陳女士露"感,深受同學W獄, 當選入鸟如后: 前期廣州培遒校長,一款推荐邈任。本

會及高哲浆

道,在黑舘,剖朝坤人問禳

一方面共英懿師,一,所諾內娄亦與學校 【極連繫,禺意深長 對在路同學系醫TM 滬江同學會選出

新任會長輩厚

香港兒亨利院慶祝兒童额米特水版登滾歌無表演

會場一角。

20(X+q+X)=x(X+9).

X=36 # −5 (1678)

6. (a) BB4 { log4 = 4 (*) $41210 $574) ----

(解)由(1)式,两边取以0茑底之數

(2)

log x + log Y.

Tit

log is

=

log 4 + log 10 == log 4+ 1

log 4+1-log x

log y log x ==

lag th

(3)式代入上式, Bagx (log4+1-legx)=lex4

去括号多項 分解

(log x) — log x log 4 ~ log X+ log 4=0

log x (log x-log 4)— ( log X-log 4)——

(log x-log 4) (lag X−1)=0

・0, a log x = log 4. X=4

log y = log 4 + 1 - log4=1= logicy=

Ai} log2=1==== log 10

3x-1=0.

y=4

xx(3),

代入(3), logy=log4 +1

log 4

=10

(b) iq_x+y=†.

特試

X4-18xy+

3p*g* ~ (p"sq")

Q.E.D.

•2(x+y)

Per il 16, 402+1228 + 9 tab, op (2013) ah 两边取對數 2 log 2a+3b= log a+ log b

Log garth Logan log b 除以2,

(註)本題中之口,須為正數

(f) 分解因式: (b-c)+(c-a)+(a-b)

(33) vx bit a f /k*—(b−c)2+(c-b)3+ (b¬b)3 =o__

故由餘數定理,知原式有 a-b之因子,又因原式為輪換式 故亦有(b−c) (c-a)2因子,而原式 笃三次同次式,故尚餘ż因 子必為常數危 ̇設原式=k (a-b) (l-c)(e=a)

a=a b-1, c- 代入上式两端,

(1~0)+(6-2)‍+ ( 2−1)3 — A (2-1) (1–0) (0–2)

*. $d=3(a+b)(b−c)(c-a'

=3

5.(a) 有等差級數,其首項為24;而第一,第五,第十一项 成等比級數,試求此级數(書寫至第十一

(解)設等差级数ż公差為d 則第五項為24+4d,第

+一項孟 24+10d, 依等比級數性質得

24 +44.

2474d 24+10d

24(24+10d)=(24+4d

576+240d=576+192d+16d

16¢-48d=0

3或d=0

(i) % d=3, 9|16 $ $ 24, 27, 30, 33, 36, 39, 42, 45, 43,

(ä)差d=0,則報數為

(答)

(b) 有一工程,由甲乙二人合 则如天可成,关分别由二人獨 做此工程,则完成之日數,甲比乙多9天,求二人獨做完工日數 (解)設乙獨做 x日可成, 則甲獨做需2+9日完成而每一日

一,但二人合作可完成全之程0 人分别完成全工程 +

(TE) + Chox of 4+q*=(x+y)+ (x−y}^;

(大+8

=4 (x*+y*+ zx*y*) =p*+ q*+2+8

R p2g*= (x+y) (x−y}= [(x+4)(x,y)]2=(x2y*)*

-(B)

右方= 5*q-(+*+Z*+2f*g*)

50x4

**y*+ y" (†}) 左式=方式

4(x2+ y2+2x^4) [WX(A)(B)\\^]

第十四次預習試題

J.(a)已知△ABC中,a边上中线ma=4.5cm, 长边上中线

C=40m, 求作&ABC(依尺寸作图,述作法,不须证明)

Q.E.D.

3cm,

△ABC中,AB=AC,延長BC至任一奌P. 試证 AP-AS"=RB-PC. (b)

2.(a) Nt A ABC中 LA笃直角,以AB為直徑作0园交BC於D通卫兵 作0回切线交ACE. 求証 AE=CE. 泥)銳角三角形ABC中

a=20 con

23.2 cm, R(外接园学弑

3. 銳角△ABC中,以ABAC爲边向形外作正方形ABEF 及 ACGH 由A作BC 垂线AD文之延线交FH柊M.試証

(a) FMSHM

4.(a) △ABC中,

(b) AM==BC.

435 &=11.91; c=906 #64.

f)設九尺分别为△ABC之内切图及外挂园半狸 s弟半周。

試計: (一)(一)(一)== :))

5.(a)解 3ww28 =+2cot26 (答案小於180°正角)

有一正角錐体,其高米6cm,底第一個边長60m之正方形 求其側面積及表面積

6.描绘+4=3pinx+2cosx(x取值由0°至70),並利用元以

解方程式