郭日僑華
理
育教僑體頁三第張六第 日十三月二十年乙
工學院首次開放日將
人士蒞臨終雄。 生會推行姝辦工作。李格致博士又表示歡迎各界 是由學生自己做,或钻不餓,我相信过天開放日 一定舉淋成功。」李博士囑咐教職員凼力協助學 學院開放日的主動精神。他說一次部份工作都 理工院長李格式上稱讚學生舉辦首次理工
·土元博士員主持 舉行,由校董會主席 1将在紅磡理工學院 「將於二月七日正午十 開日開幕儀式 學生會舉辦。. -WELEDANDE 湖及太古兩校舍舉行 於二月七、八日在江
五期星
日十三月一年六七九一年五十六國民中
院學工理
放開次首
行舉日入及七月下
信由你 堪舆師一番理論,眞是信不
園,河幅及獅獅等。 ,開放日游且包括披露·音樂演俊,敞?
「我博士主持閉幕儀式。 開放對將於二月八日下午六時小陈,由李格
鼎盛,亦感英雄無用武門對面之山城松樹在十 影响之下,雖校内人才一開國以其势。現時校 而無去路,氣勢受阻,一現在崇基之火車站入口 正門面山背水,有日之限制,下大校門宜於 利,只是有四根石柱的:他說港非客觀環境 | 但以水火 - 對校園有一邪基女性,因其屬險。 浙起來煞氣太乐。新敬聯會較爲有将上
BARKENI ZE SKKA - SZEX RESKANCESE - PAKKEXAZR RESP(-3* N*E*K*·
們也指信風水之說,那有利,因崇基校園位於 (26)TER 2****KIMK
|中大校園照水塔
煞氣看來太重
英雄蠖屈難伸
脉炎燕
防水 以上爐林樱
對起的 面鮮粥 木山匯氣時
,在八卦中屬台席之位一大發。
€設校門面向四方 僑光。註接,中大定將
一頭發生火災,以免树木
四十五及五十七歲者。
,凡庸宇之人皆受其推 而如出生於陰曆六月,將 ,因現年廿一、北三、
WAH KIU YAT PO
更受影响。..
方童
鋒」。 一加棅柱;以擦一说牙之 | 方法可以將柱蒗麼不成 風水入虎牙之格,改善
·童氏謂谳莊似台燭, :至於校門之四柱,
童氏謂新及聯合
於聚之門,且姑安會 光大,目前至多只能列
到許多始料所不及的問 加研究,也許可以發揚 設與卜筮之°C,日後再 所致,但命理風水之
誠立,在衆黃中成長, 之,姑妄聽之。
所以過去十年來~泡遇 題。有人說紅先萬水欠
· (1)
按:中大在艱苦中
【作,亦不宜包括有關名次之任何資料。 特別小組委員帶,以處特殊困難問題,晚娘學
月放一切資料,應予保密。因此,吾人建於本校 評定而不穩記分。客人又露爲小學之或橫報告
• [}}><#EEQENIGEKEİ
發給家長之成橫報告書中,學生之成閡應以等摄
·細資料。爲枝等本身之利益計,正式記分表上所 三,一二、乎人不就學生知悉有關和分之年
於獲得分區委員會同意後,當可進行。
—例如哚用綜合教學法學校,於處理上需有 箇用於此類 校。挐校校長如說欲作任何改, 干伸縮拙。人相信上文所述之各項原則,亦可
研究取代中學入學試
— 三、對上課形式與一般不同之學校一
KK DOX (80 40 40
主爲成長成
撰小六々《撬第三! 立 證件將只能向分攝委員動能,而非由個別小學 自行發給。小半校長不得取錄無轉校實之康生 小六目的,學生於轉校時必須出示投劃。是項 變。好人建斷,爲防廠家長利用諸校而寧南面積 生不得兩度參加中學入學試辦法,基本上並無改 一成散,即實影响學位分配辦法之公允,故吾人 爲此種留級,于菜止。是項措施,與現行之举 長若得以安排其子女 小六日期取每校考試 戒获以分配學位之方法不致產生漏洞。例如,家 : 三、 四、吾人所關注者乃確保採用校內 及名次聯同計算之辦法,桀經瓢慎查核,始予决
罗僑榮
1976中學會考試題預習專欄
化學科 (十一),李國榮,
CHEMISTRY (11)
Answers to question. 1 and.
1A.
(1) Sodium nitrace or potassium
nitrate is also needed.
(ii) If sodium nitrate is used, the
reaction will be
2NaNO
3(s);
+
HASC4
Na SO + 2 HINO
in solutio
(iii) Partial decomposition of nitric
acid also takes place. 41NO
3(D
+ 4NO
12(6)
Nitrogen dioxide is a brown gan which readily dissolves in the acid making it a straw yellow 11- quid.
(iv) Rubber parts like joints or
stoppers.carnct be used since ni- tric acid reacts with rubber. (v) When copper metal is put into
the acis, oxidation redustic re- action occurs.
Cu(s)
+410
•Cu(NO3)2 +
in solution
B.
콘(4)
20(8)
+ 2010
2(g)
Copper (II) ion in solution is
The brown gas is
blue in colori nitrogen dioxide.
(1) The degree of dissociation of
an acid in aqueous solution is usually a good indication of its strength as an acid.
An acid which completely dissoci- ates in solution produces a high concentration of hydrated nydro- gen ion (H2O+). For example, hydrogen chloride in water under goes the following reaction to large extend.
HCI
it.
While weak acids like acetic acid undergo a much smaller degree of dissociation.
CH2000+ H20
30* + CH2C00 ̃
The population of hydrazed hydro- gen ion in this solution is much more scarce than that in a solut- ion of hydrogen chloride of same concentration..
(ii) Strong acids are good conduct-
ors of electricity due to the fact that the ionic collbration in a solution directly determines the conductance of that solution,
(iii)
a. 41INO
3(2)
2120 (g)
+
C(s)
2 (6)
+ 4NG
+002 (g)
Concentrate nitric acid can oxid- ize carbon to carbon dioxide.
b. 5HNO3 (1) +5(s)
220 (
GNO 2
+ H_SO
2 (6)
4(aq)
Concentrated nitric acid can ox- jdize sulfur to sulfuric acid.
(iv) According to the equation
INO + HaOH →→→ NaNO3 + + H2O..
C.
3
same amounts (in moles) of the two reactants will produce sodi- um nitrate in solution. On beat- ing sodium nitrate, oxygen is olved, and godium nitrite remains
2NN
2NaNO12+ 92
ev-
If heating is not long enough, we would expect a mixture of so- dium nitrate an nitrite.
(1) When calcium nitrate is heated,
oxygen and nitrogen dioxide #ra froduced, and white calcium oxide is left as solid residue.
20& (NO)
3/2(e)
20a0 + 4NO
2(g)
[(s).
When silver nitrate
(g)
is heated, both gases will also be observed.
but metallic silver will be left
as the solid residue.
·2Ag30
3(5)
NO 2
2(g)
2(e)
all nitrates are soluble
test for the golubility of the
unknown. Then D the solution
add a few c.c. of freshly prepar- ei ferrous sulfate solution and
Holding dilute sulfuric acid.
the test-tube at an inclined angle of 459, slowly but steadily pour. in concentrated sulfuric acid, a browa sibstance will be seen bet- ween the two layers of solutions if the unknown contains nitrate ion. The chemistry involved in this experiment is as follows:-
(ii) Volume of mercury expelled
= apparent expansion of mercury
0.444 c.C.
When the temperature is incre- ased to 8390; the new volume of mercury 50 +0.444
50:444 0.0. fraction of mercury expelled from the bottle.
0.444
50.444
0.009
ns.
Solutions for multiple choice 3. (B) is correct.
Bright surfaces are poor radi- ators. Delete A Cand E.
B, C and E indicate dull sur face is a good radiator, which are correct statements. But C and E are deleted. Therefore (B) is co
rect
is correct.
All the metal blocks are same mass, same weight, and same base. That is to say, all these blocks are exerting the same pressure or the ice blocki The borders of met- als of higher specific heat capaci: ties is warmer and so ice melts there.
5. (A) is correct.
The period of oscillation of a simple pendulum depends on the pen- dulum length but not on the mass of the bob or the amplitude of swing.
On hot day the pendulum length expands, so it loses time. 6. (D) is correct.
Mass is constant. (A) is in- correct.
Weight depends on mass and gravity. It is independent on heat W = ¤g
Both volume and surface area of the body increase when heated.
If volume increases, while mass remains the same this indica- tes the density of the body is dem creased. Therefore, (D) is correct. 7. (B) is correct.
7
Molecules moves faster and ex-. erts heavy bombardment on the wall of the container when the temperat- ure of a given mass of gas increas- es. This exerts a greater pressure in the space.
Question for next week
1.
95cm
Lead shot
-Bung (for insertting,
thermometer)
As the figure shown above, m kg of lead shots are put into a card- board tube which fitted with 2 stop- pers, the priginal room temperature. is found that 25°C.
The tube is then inverted ra- pidly 100 times so that the shots fall down vertically through 95cm. each time, and the new temperature is found that of 30°C.
(i) In order to estimate the spec-
ific heat capacity of the lead shots, what assumptions will you make?
(ii) Try to determine the specific
heat capacity of lead shots from the above records. (Take g-10ms-2) (iii) Give two reasons why we cannot
obtain a good results.
1976中學會考試題預習專欄
the apparent mercury
expansion of
Aris.
-
(50) (0.000148) (83 — 23). 0.444 c.c.
(1) NO
HSC
物理科 (十二)
3(aq)
·何輝明
-2
·HNO
3.(1).
4(64)
(11) 2HNO
3(L)
+3H2SO
PHYSICS
Solution for last week
(12)
(a) Let the heat capacity of
calorimeter be H J/°C heat gained heat lost ((50) (4.2)+ H) (20-10) (100) (4.2) (26-20). H 42:
i.e. the heat capacity of the
calorimeter is 42J/0
(b) Suppose that the specific lat-
ent heat of fusion of ice is LJ/g
heat gained heat lost 100+ (10) (4.2) (14 - 0)
((150) (4.2) + 42) (20 14) 10L+5884032
I =
4032-588
10
= 344
the specific latent heat of fusion of ice is 344 J/g
(c) Assume that the specific heat
capacity of aluminium is CJ6°C Heat lost by that piece of alu- minium
(100)(405-64)C (J) = 341000 J Heat gained by the calorimeter and the water contained
42+((160)(4.2)) (64–14):
35700 J
heat lost heat gained
341000 = 35700.
01.05
the specific heat capacity of aluminium is 1.05 J/g°C
2(a) Assume that the coefficient of linear expansion of glass isd For a change of temperature A, the lengths of the rods become (b+b) and (a + ad AÐ): respectively.
the superficial expansion of the hole enclosed (a + ad ^ ☺) { b+bαA÷) -ab Similarly for the same change of temperature, the dimen- sions of the glass plate become (b+bα) and (a +ad@) respectively.
Therefore the same superficial expansion results.
(b) It is always true that the vo-
lume enclosed by a solid expan- ds at the same rate as would a. solid body of the same material as that of which the walls are composed. The re fore the two objects have the same cubical expansion.
(c) Let
R
=
A
V
coefficient of real expansion of the liquid coefficient of appare- nt expansion of the liquid
coefficient of cubical expansion of the con- taining vessel. Assume that the volume of the liquid is V
Real expansion of the liquid apparent expansion of the liquid + cubical expansion of the part of the containing vessel in contact with the liquid for a change of temperature
= V
i.e. →
R
(a)
=
7A +7x
+
(Q.E.D.)
(i) Making use of the result in (b).
coefficient of cubical expan- sion of the S.G. bottle coefficient of cubical expan- sion of glass. making the S.G. bottle
ww
0.000032/0
3 x 0.000008 Apply the result from (C), Coefficient of apparent expan- sion of mercury (coefficient of real expansion
of mercury)
- (coefficient of cubical expan sion of the S.G. bottle)
= 0.00018 - 0.000032
= 0.000148/°C
+ SO
4:(L) +6FeSO4
in
solution
3 Pe2 (504) 3+4H2+2NO(g)
in solution.
NO
+
(g). FeSO4
un solutior
FeSO NO (a brown compound)
4
(111) Ammonium nitrate when heated gives steam and nitrous oxide
2A
NH.NO
3(8)
(i) The reaction involved:
HC1 + N@OH -→→→→→
NaCl + H2O
(ii) Average volume of HCl used
- ( 22 - 1 +
+21.9+22.0)e.c.
22.0 c.C.
According to the equation above:
1 mole of HCl will neutralise 1 mole of NaOH
25
Mole of NaOH used()
Mole of HC1 used
25
(To00) (0.05) mole
Molarity of HC1:
25
= (+880) (0.05)/(2368).
-0.0568 (Molar)
0.05)mole
(iii) The acid should be held in a
burette
The alkali should be held in s conical flask.
(iv) The indicator should be added
to the alkalı in the flask.
B. Discharge of hydrogen gas invol
ves
217 de reaction:
+2@
H2(g)
2 moles of electron transfer can discharge 1 mole of hydro- gen.
Discharge of oxygen inmlves the reaction:
40 (sq) - 4e
2H20 (2)+02 {
Same amount of electron tranIS– fer as above can discharge only mole of oxygen.
Since hydrogen obtained in ex- periment was mole..
ты
Oxygen also obtained
(글) (구) mole
2200le
At. S.T.P. the volume of oxygen 22:41iters
32
i.e. at 21°C, 761mmHg, the volume
(224) (224) (769)liters
32 ..
= 0.743 liters
C. Molecular weight of Cu50 4-
= 159.5 Molecular weight of water-18 The weight of anhydrous salt left (15-5.4)g.
= 9.6gm.
Let x be the number of molecules of water in one molecule of hy- drated salt.
x18
159.5
5.4
X == 5 1.2. The formula is
Cu30
4520
D. When it is heated further, the
reaction becomes
Cu80
4(s) (uo(s)
+ SO
03(4)
1 mole of Cuso, produces 1 mole of copper oxide.
Mole of Cus0 4
9.
us ed
759.5
i.e. Mole of CuO obtained
Weight of Cuo
9.6
159.5
***
(molecular weight of Cudgm
9.6 159.5(79.5)m.
= 4.786 g.