最習垴段的學
報日僑華
床,期定 應接,些氏佛業的
頁二第張六第 日二十月一十年寅甲歷要
A BEZ BE
WAH KIU YATİPO
(嶄
腐校
生由批示 學盤
業營業範 生委粗
出
示校工後心爲前旅 由華 五育。
BET
西日樂 云效員又生滨 經
、全,燐將兩西縣 育示 往人
FERRAGER DER DA
夏季
堅送及聖類斯學生義工
明在城門水塘清潔示範
《特凯丷返盛民政 西區民政署、西區葵是人
to maintain metabolism at a high
the. body temperature of
mammals must be evenly maintaified
豬敢 我合然 有話要
鹽 取活除,宿們:
在第
致书司負
店
本,敢、強和做賽期 年
送
*
CA THE AP
人
at the correct optimum. This is executed independently of external environmental influence, but, it was initiated by extemal factors...
surroundin cold, i temperature it at a suitable
That is to say, whe
L975
medium or a
regulates its'
中學會考專欄
by main
maintaining
:堅道書院主編:
生物科匣(八)·廖國置。
溫習内容,免疫的意我要素
抗原抗體免疫的類型3.接 種 病能油碗天花的原理4烧 苗熊抗生素
**WATI PERS
的特異成份鵝A.沉澱装5抗毒無
CASA, J
HOLE BUSUKSES
FUME, A. #DAR 6-23 conto. MÉR & MA
A豬蛋白脂肪)無糖漿 25 FRA FORD,
B: trosk * <. saa kotid ME
白喉病人的體內;
III問答題:
法
童牛痘
#SM, A IRS & A
(注)試述牛症能勘天花的藏酒味 病毒免疫的理由:
ZIMBA, ISƦmA.016.
VEE <DETE (BYA D
Date, dein BA, B.S
KBEN. 1 BÍLM1 598 1 305
-
雨子枚落宜環境」 有成新的菌絲體..
Roma
-
强者、消费者以及分解者香楼道
- MARTELE.
·化響方法以造成一傻不適宜防生: 南生活繁琐的環境緒病症。
荷花物受细菌或微纳米 蔡用,由根雜的化合物變成米無 用的化合物的現象她貌畋
BERLIN. --B 代的生物而來。如果沒有生的地 宜,絕對不能產生下一代的奶、安
如哩化磨或其他方法将他殺死,便 ̇拖不能繁殖下一代
消毒,消毒法进给多用物理和化學
法:热用衣物用焚燒法,正常的嵴
樂病菌於病人的
..
when
.the
is
range in order to enable the proper -funct
nctioning of the whole body.
we can say that the ultimate advantage of regulation of
of body temperature is to maintain life.
Water is an essential of proto- plasm, which must always be satu- reted' with it. It permeates all tissues and its presence and in- Compressibility enable them to resiet enormous pressure exerted
upon the body by the atmosphere. By maintaining constant water.
So
content within the body, the main-
taining of osmotic equilibrium
between the cells and the extre– cellular fluid is also kept constants &
As all tissues are
bathed in aqueous solutions, it also help to maintain constant
emotic pressure of plasma. No physiological reaction can take place in ita absence. Thus we can. that to maintain a constant Bay. rete of physiological activities in the cells throughout the body, a constent water content must be maintained.
ture.
b. (i) The skin plays an important part in regulation of body temper
If the body temperature is high, blood vessels dilate (vasodilation) thus more blood. flows near the surface and aà a result of convection and radiation, more heat, is lost to the surround –. ing medium.
In this WEY temperature is maintained constant.
the body
Another way by which heat is lost is by the increase rate of sweating If
more sweat is lost the evapo—” ration of sweat helps to remove. heat from body by the
Latent heat of vapore of
But
In some animals, a way to reduce body heat is by shedding.
hedding the fur
or hair in summer thus reducing the Insu lation on the surface of the skin.
if the body temperature is Low, the reverse would have to occur. In other words, blood Vessels would contract (yaso constriction) in order to reduce or decrease the flow of blood to the surface, thus less heat is lost by
action and
fon and radiation. By
By hair. rection or the "fluffing up" of
convac
hair or fur or feather, more air is trapped in between to fom a
thicker insulation so
90 heat is képt "in". Sometimes, in aone animals, they grow more hair in winter to provide
thicker winter coat. And the temperature is low, less sweating occurs 30 18ss heat is Lost from the body.
(ii) The kidneys are
are known as the oano-regulators, that is,
19, it
regulates the osmotic pressures of
the pl
plasma,
sma, lymph and fluid in the tissue, cells.
Thus the work of the
kidneys is to correct any variation in the water content of the body If
large amounts of water is drunk,
plasma becomes too dilute,
cours, and more water
excreted thus a
a watery urine. If the body loses a large Amount of
water, such as through sweating, the plasma becomes too concentrated, then the kidneys
remove less Water and a less
copious and more concentrated urine
is excreted.
N.B. Actually, how much water ia
absorbed from the glomerulus
filtrate and pass back to the blood.
is controlled by a hormone ADH (anti-diuretic hormone
aecreted by
pit the tutt
itary gland which in turn was stimulated by an area of the If the brain, the hypothalamus. blood passing through this area of the brain is too concentrated, mox hormones is secreted and thus less water is removed from the kidneys. In conclusion, we can say that the kidneys play an important part in homeostasis of the body.
Brat
e. The function of the capsule of the kidneys is to filter off from the blood plasma of the glomerular capillaries a fluid which is known as provisional urine or capsular urine or glomerulus filtrate,
the tubulas of the function of the
return Keys is to reabsorb and to the blood plasma of the capillary network those substances which the body cannot afford to lose.
1975
中會考專欄!
冷藏乾烧
長炒環境或
兩類細菌,使食如不需生
熱鬧丁度專標
□永爍。
*
Solution for the last issue
1. Readings taken from the experi-
menti
生物科(八)
BIOLOGY (#8):
Answers to Review on Exore tion.
a
·e. - b
i-a.
b
a.
h
Q-14.a. What advantages are gained
in a mammal through regulation of bady temperature and water content?
b. Describe the part played in amammal by-
(i) the skin in the regulation of
body temperature.
(1) the kidneys in the regulation
of water content.
c. Distinguish between the
funotion of the capaule and the
tubule of the kidney.
Explanations:-
a. Mammals are homofotheric (warm-
blooded) animale, "ha important
chemical actions (metabolic processes) taking place in the animal's body are enzyme actions and proceed best at an optiman temperature (about 4020T In order
物理科(八)
PHYSICS (8)
Additiona 10 50 100 150 200 250 Load()
Scale Teaú=||
ing (mm)
120140160 180 200 1220
K is a constant
Upthrust, U = k + W
where w is the additional load
(ii) To determine the S.C.of.the
liquid
Theory and calculation
With additional load win the tibe
.(3) Upthrust Uk
With additional load w, in the tube
(2) Upthrust U2- W2+ Subtract (1) from (2),
AU =AW
where is the radius of the tube
fis the density of the
liquid
h is the difference in the
depth of the tube sub- merged P.
cocote
-0.8
2. By Archimedes Principle, the
upthrust experienced by a body immersed in a liquid is equal to the loas in weight of it in the liquid and is equal to the weight of the liquid displaced by it.
Weight of water displaced by body
its lose in weight in water * 75 gf p
Weight of oil displaced
its lose in weight in oils 49 gf Wt: of alcohol displaced
ite loss in wt. in alcohol=60gf
Vt. of oil displaced. S.G. of oil of water displaced
= 0:65-
9.0.of alcohol-wt. of alcohol disp.
wt.of water dispd.
500-3
(b)]
U
UW+P,
F= UW
0.8-0.6
= 0.2 kgf
Force required to sub- merge it completely in dil F
Upthrust of oil acting on it U Uwt. of equal volume
of
(0.8 x 0.6)/0.6
Wt. of bail = 0.6 kgf
When released, a force. of (U-W)g newtons acta on the ball, and (U-V)gma
a 0.2 x 10.
10
日
Its acceleration is
3.
me
Questions for this week
1. (a) Distinguish between mass
and weight.
A body of mass 10 kg is trans- ferred to the surface of the moon. Calculate.
its
的按类。
在湛江下選擇一些注易的動作再遇到一些触发性 力的活動,和希望把技術練習扽熱,所以他們 _何低證:第一,每個兒童都喜愛需要運用|
(5)
weight, (ii) the force required to
give it acceleration of 5ms (iii) the force required to keep
in uniform motion on a horizontal plane with a coefficient of friction 0.4 (iv) the additional power need-
ed to keep the body with the seme speed up an ineline of 1 in 20 to a height oh 10 min 20. seconds Assume that the gravitational acceleration of the moon is 1.8 m/s/s.
(b) A mass of 5 kg and specific gravity 2.5 is suspended, complete. ly immersed in water, from a spring balance attached to the roof of a lift. What will be the reading of the balance
а
(i) if
if the lift is ascending with a unif
iform speed of 10 m/s (ii) if the lift is ascending with
an acceleration of 2-4 m/s/s (iii) if the lift is descending
with an acceleration of 2.4me? (iv) if the lift is falling freely
under gravity?
2. (a) You are provided with a long capillary tube, a metre-rule, and Bone mercury. How can you deter- mine the atmospheric pressure. with these apparatus. (b) A metal cylinder contained 10 litres of gas at a pressure of 1 m of mercury Some gas escaped and as a result, the pressure of the gas in the cylinder dropped to 0.9 m of mercury. Find the volume of the gas escaped. The atmospher
ic pressure is 0.75 m of mercury
and the temperature inside and
outside of the cylinder is. the
same.
(c) To determine the S.G. of.
lead shot by using a density bot-
tle, the following data are obtain
ed:-
Wt. of bottle filled with wat57%
of lead shot - 28.5 g Wt. Wt: of bottle with lead shot and
water in it 84.7.g
史協會專標
化學科(八)、陳湛杰。
CHEMISTRY
Test tube
Scale in m
Liquia
Additional load
Multiple Choios
TO. (B)
Load shot
(1) Graph of depth against addi-
scale in m
tional load
星
195
2130
additional load
Weight of the tube +lead shot-
(#8)
SOLUTIONS FOR CHEMISTRY (#7)
Conventional Qusetions
B) 5. (D
T. Sodium hydroxide can be pre- pared industrially by the electro- lysis of aqueous sodium chloride. ((trine). In order to obtain the hydroxide, two cangamonts in the electrolytic cell may be used. In the diaphram ac11, the anode and cathode are separated by a dia- phragm, preventing the mixing of products. In the merury coli, the Gethole is of meroury, so that sodium amalgam is formed first. This latter is typified by the
日四月一年五七九一公年四十六國民華中
六期星
效各傭育中
地位厚 當學期
作性有如餘師落。榭理學麴、半活學 ,讓之許何人外的
水沙老興 高腰 會將失分純建多配。老行政升,
古代重而的兜 于冥席各
然情 算是果把配副
∵番便面·要的貢
製程追劇實商除會學致
而熊式有
便 方管 地只 功激的關
∵莱·能鍆·始譔蘡所通匪 音僑胞們、圖以常均
Kellner-Solvay sell, with carbon anode and mentury cathode. During electrolysis of saturated brine, the following reactions occur:-
At anode, 01-
201 - 01 At outhede,
Nat+6 Na
Na Na F+ àH ↑ Two Factor govers the liberation of Modica instead of hydrogen (the dual electrolytic product) at the cathode. First, the very high over-potential of hydrogen at a mercury cathcde ising ite discharge potential consider bly. 20 th it is difficult to be libirsted.
119- Secondly, soud wa selves in the mercury to give a dilute solution, the electrode potential of such a dilute amalgar is so low that sodium ions can be discharged. Th- sodiumalgan is withdrawn from the sell and treat- ed with water to give sodium hyl oxide (caustic anda) and byl regen. Reactions of sodium hydroxide: ((a) Aluminium. As, this is only weakly electropositive, is reacts with sodium hydroxide to form zal, sodium ajunirate envolv ing hydrogen:
2A1 + NaOH + 21,0
2N0A10,347
(b) Phosphorus White phosphorus dissolve. 14 bodium hydroxide to give sodium Zypophosphite and phos- phine gas:
+ 3NaOH + 3H2O
=3NaLP02+ Pä2?
(c) Chlorine Chlorine dissolves readily in sodium hydroxile 2012 tion, however, different products are formed depending on the pond tions used. In the cold with ex- cess alkali, chloride and hypo- chlorite ions are formed:
20+ C2201010 + 820
and chlorate La the hot,
iona are formed:
EOH 4301
010,
+3820
This latter can be interpreted as a disproportiomtion of hypo- chlorite at the higher temperatur
3010201 + 0103-
(a) Sulphur Sulphur reacts slow- ly with strong alkali to give-a mixture initially of sulphite and aulphide:
35+ 60H
25
+503
However, &s the sulphide ion can be attached to additional atons of sulphur (estenation), various polysulphide ions are also formed, 0.5, S
S
Similarly, the sulphite on eadily adds on a eulphur ator to give the thiol- phate fon S
2. There are two kinds of harness of water, tempomy and permanent hardness. Temporary hardness is caused by the presence of the bi- carbonates of calcium and magnesium, while permanent haru- nega is due to the presence of the -sulphates of calcium and magnesiumD Temporary hardness may be removed by!.
(a) ciling The uistelle bi-
iling carbonate la changed into the in- 301ible carbonate by heat:
Ca + 2HCO
→ Caco + 10 + H2O
(b) Slaked line. The addition of a calculated quantity of Flaked 1ime will cause the complete pre- ripitation of bicurbinate as the carbonate:
Ca (HCO
-Ca(OH)2 3/2 +
2caco + 2120 More limo is required to preci- pirate the magnesian bicarbonate since the magnesium hydroxide (ta magnesium carbonate is quite coluble), has to be presipilates:
Mg(HCO) 20a (OH),
= Mg(OH)2 + 20acb√+ 20 Both. temporary and perma: ent hari. ness may be removed by: (a). Sodium carbonate
The addition
of Eodim carbonate does not cause the purificà tion of water, there is merely a replacement of a mult velont ion by univalent ones whic has no effect on soap (e.g. Ca replaced by Ma
Ca (HCO3)2 + Na2CO3
Caso + Na,co,
3
(note that though CaSO is only alightly soluble in water, its solubility is enough to osuse significant hardness of water) (U) Jos exehange Ion exchangers
ugh 48 240life or a synthetic resin that possess giant anions that are static (.. an alumino- silicate ion of a three fixen- gional network of stoms) are known as cation exchangers. They Fossess the property that small ∙cations are free to move throug1:
the three dimensional network. Thus if originally the ion ex- changer de saturated, with. No one
and, then, shaken up with hard water these Na ions are exchanged with
iona of
the and Matter charge such 33
Ca
As the water sample possessing terpomary or perminent hardness is passed through the ion exchanger, the drained water senpls that filter out is free of hardnes
2001
TE
-+
+ dage ++ acca +
nhat When the sample of water con- taining sulphate and bicarbozate of calcium is titrated with acid with methyl orange as indicator, the weakly basic bicarbonate 1s neutralized:
f
de (HCO3)2
PH = Ca++
+220 + 200 Thus nos. of equivalents of Ca++ in 50 ml-18 0.06 x 2.5/1000 = 1.5 x
this in 100,000 ml there would ba C.3 aquivalente or 0.15gm ions. In tema of weight of calcium carbonate dar 100.000
& DESI
・放歌子
當午间除!
。印夢建其者二擔任講,作主有。蒙尔九推德 專營 設中 應 百教指者 还專任會小 近時行。
1. Curved
parts of water (1 ml weighs close
to 1gn for water), tempora
hardness is 0.15 x (40+12+3x15).
Tren of CAÇA. ner 100.000 parta ? of water by weight, 1.2. 15.
When 100 ml of water comple" is" mixed with sodium carbonate solu- tion and boiled, and then extmos. ed with water, the permanent hard÷ neas due to Caso in solution is transformed's
CACON, 504
Caso + Na CO2 whilst the temporary hardness ¿ue to Cs (HCO,), is converted to the carbonate under the condition of holling. Thus a. of equivalente of Ca for permanent hardness = equivalents of codium carbonate used excess of sodium carbonate.
11x0.1)/1000 1.e. (12.5x0.12-
per 100 ml is 2
FOT 100,000 parts of inter, the per nent harness may be expressed 210 x 100 = 100,000/100 parts of calciua
4x10 4. Thus 10.4 2. Ca*
20For the given sample of water, temporary hardness is 15 and permanent hardness 20 perts by weight of calöiur carbonte per 100,000 parts of water. QUESTIONS FOR THIS WEEK
The questions in this issue sre rated to 2.1. (ii) Trans- ition Metals; 2.2. Simple testa for ions. Study your notes ard textbook before attempting the question.
Multiple Choice Questions
1. Galvanised aron engeka nae coated with a layer of: (A)Zin (B)Lead (C) Chromiuri (D)Aluminium (x) Zi
be oxidized
2. Which of the fallowing caunst
hlorine: (A)Perrous ion (B)Copper metal (C)Antimony powder (D)Ferric ion (B)hosphorus
3. Concentrated hydrochloric acid is nacenary in flame tests for metal ions because;
(A) supporte combustion
(1) It does not suppert combustier (0)it is volatils:
(D)it makes the sample more. volatile
(E)it does not dissolve letinm 4. A substance A is a white Fowler. When dissolve in water, it produces a blue solutic. Upon acidification of the solution witl solution, áil. HCl, and the FC12
a white precipitate appears. What do you expect to observe if now <land nitrate solution ie added to the original aqueous solution of A?
(4)No observable reaction
(B) White precipitat insoluble in vatar
(0) Mite precipitate soluble in watar
(D)Disappearance of the blue colour of the original solution. (E) Evolution of a brown ges 5. A solid X upon heating gives a colourless as witch on re- ignite a glowing splint." An aqueous solution of this solid in sodium hydroxide when heated Will produce a
which turns
red litmus blue. I may ba: (A)Sodium chlorate (B)Sodium chloride (C) Sodium nitrate (D)Ammonium nitrate (E)Amnorium carbonate
Tib Questions
the preperation of (a)copper sulphate crystals otating from malachite, (b). ferrous sulphate oryetale start-
(B)zine ing from iron wire, carbonate from metallic king, (d)pure lead monoxide from a mixture of lead monoxide ̈am2 ouprie cxide.
2.What would you expect to observė if you (a) dad ahlorine water to a colution of potassium bromide and then shaking the mixture with osrbon tetraoticride; (b) heat acdified ferrous tulphate solu- tion with nitric noid; (c) dop. crystals of oorger sulphate into concentrated sulphurie soil. What chemical explanations can you give. for the observations? Write aqua- tions where possible.
LO75
中高燴彥專欄
新數學(八)・謝國奧
MODERN MATHEMATICS Trigonometry III.
(#8)
Area Volume 3-D Problems
I. AISS
a. Triangle base x height
=ab sin 0-
- bo Bin A casin B
(a+b+0)
b. Rectangle length x breadth c. Parallelogram base x height d. Trapezium = (a+b)h
6. Quadrilateral y sin A
x and y are length of disgo nh
A is the angle Inoluded by
these two diagonals
f. Circle = p2
6. Sector = r2a =
3600 X 72
is the radian unit of angle subtended at the center of sector
h. Curved surface of cylinder
青敦僑
學課程均衡問題 另有三項專題|
·關教育司何雅明主件致小
小學教育研討會
九龍樂善堂舉行
演及討論會
surface
of
a cone stiri
1 is the slant length:
j. Curved surface of frustum
Til (a+b).
k. Annulus -7(R2 -|-2) 1. Surface of sphere
II. Volume:
4TER2
a. rectangler block area of
base x height
b. cylinder Tr2n
c. composing tube d. pyramid=
ocone. —
f. frustum =
E⚫ aghe re
bage area x height
TR
TIL. 3-Dimensionel Frölem
The angle of elevation /DAD B at the first point of observatior A, also /DAB and DEA = j are given, then fin
DAD sin gin(x+r)
DC DC
AB
=
DO
ÁB sin din⠀ [Bin(∞ + 1 }
The angle of elevation DAC = at the first point of observation A and the two angles (CAB = and
CBA are given,"
單
tax
AB tan Sain
sin(x +
A hall is 16ft long, 12ft wide and Sft high. Calculate the angle between the planes CAB, A BOD Let P be the mid-point of AB Then PO is perpendicular to AB also PN 18 perpendicular to AB /OPN in the required angle Fr. Eft
tan OEN
=
PR 62+
8
= 1.3333 OPN5308
Solution to exercise on equation 1 2 in 9 tan +1 tari+ 2sinf
་ ཌ་ 2cine COR
COF
Sin +25in € oogë
CO.
2sine 2sin e cosessin - cose 2ein e(sin e-cos e)-(sin 0-coa)|
(sin e
sin
- cos a)(28in ☺ - 1) = 0
cos 00 tan
@ 45° or 225
2sin 10 'ain Q=
= 30° or 150°.
1
aine - 2 - 5 cos &
20 - (1 - cos20) = 2 - 500s
+ 5008 - 30.
(2009.01) (cd3.0 + 3)-0
2cus1 =
430° or 3000:
cos + 3 = 0 (rejected}
2(0059-sinf) –
2(cos-1+0052e)
2008 8 - 1 = ♣
cos e
• = 369, or 330°, or 150°,
2108
4. 6COB = 1 ca 8
Ecosecos 1 = (3coa+1)(2008 - 2cos 1:0 008 0:
= 800 or 300°
3003
5. 48i
+1=0 000 =
e = 109°28′ or 250°321
12 sin2e
12312 tu e (681 +1)(2 sin Zain - 1 = 0
30° úr 1500 бsin 9 + 1 = C
10 é
sin o
Bin €
= 189°35' or 350°25 *