日一月四年八七九一脑公年七十六國民華中 育教濟經
皇务短煎外身際頭
華僑經濟
今年內戲院生意尙穩好
【本港藝院生意保持爆好|添一支不容忽視的生力一對本衪影片的光埠亦有 今年以來本港的電影市場,將用。這意味苦內地新片
戲院商會現並無加票價打算
假期增多最有助力 業內競爭漸見激烈
「高貴的打算。
院的生活,除了本
戲院商會現時尚未有,爲激烈了。 理事長江祺給表示,戲|在東南地區亦甚受歎一激烈的了。
·港九新界洒院商會,不但在本港亶座,而且「塲的競爭,將會是空前 -業內競爭頗爲激烈,畢,而業內的競爭就一打擊力。目前本港製片 他攢稱:內地簪片,在不满色,今後電影市 一公司甚多,台灣方面亦
·美經濟指標有改進
六期星
眼的假素
本月汽車銷創紀錄
一類的兩個多月來假期甚
日復活節,加上裹假及
+-8324• KB -
零售銷值較去年增
「不俗。
·期例假,對校院生 視家梁發展類快,但第一、五二一檙增加三六|啊,這可能放慢了政府 二。〔) 商務部 在建築上的開支。 (四) 江氏規稱:煞然電|紀錄,比去年同期二11|係反快出寒冷天氣所影 一今年以來“院生意已漸
年宮民生菌滋長有核 月十一日至廿日銷譟 有刺激力。總的來說, ;(一)美新汽車在三支劑按,會使一九七八 三〇一、九一六棚的少,但又稱激減少可能 (特)約消息一同時相信最近報告的開
的-當然-吸引橋案的最C,三點,輕季塑性攜,石油市塲塔勞報告,經 讓視的觀衆是有所不同,三百萬美元,或相當於入口激堀,在最近毎序 | 並不大,事實上遭影與一個,零售銷售祜少四|煤口和工逸已造成石油
「我在視对接,一至三月十八日爲止的經濟合作發組織稱类
國五天組勞在1
.五%,上週就堆加四 網外加二、00
楊業近來的一款含 將提高進口關稅
福 美對民用對講機
池,頗能迎合觀眾的計,預料三月各嵇濟指数會 那麽耱偉共三六百萬
郭日僑
頁二第張七第日四廿月二年午戊膰夏 WAH KIU YAT PO
「菜——
那些响本
大打漿倒 *片遭藥陜
稅率百分十五有效一年 “日影响最大次爲台韓港
的戲院生意發望降,正 在談到對今後數月
統週二要求國會把外國 息;白宫宣佈,卡特
民用莜段對講機進口 目前日本供應英國
關稅提高百分之十五, 他說,民用波段對講機的進口的百分之八十五,其餘
百分之十四史中華民國
「假開始,戲院生意將置 將有蝎激作用。而至著
出裘示,本燕某一院线,把進口關稔提高百分之一率百分之六。
. 邅項將使日本 易委員會所地藏的提高百分之十五,在接備的 這項花比美國國際質、進口隔鸯在第一年提高 幅度小得多。該委員會 兩年每年降低百分之三
·然後恢復至目前的
、韓國及香港供源。
·提據卡帶的建盤,
俄四人劍後,那些名人 他說,由於內地打
這項進口限駹將在
·卡特邆拒絕該委會今後兩週內卡特發佈聲
東京股市普上揚
出口股較爲突出
受歡迎的「楊門女將」 財三姐」,另一將大場,成交非常活躍。 一套的大旗瓣片將陶,收市時,獲利9
·一出現,因而未以全日。 日一來將推出。以後,一套大分別,市况蘭桜上揭|九。 |近已火緻香港-「劃」出色,因而刺激大市上. 登上舞台。此種風氣衰 突出 -而投資股出現 大受歡迎的影片則推新|-因而一般出口股較爲 打入冷宮,而在文革前對日圓的事阽爲碼升 發闔的緤椒戲已全部被日,東京股市由於美元 午後市况與早市無
本田汽車:五八二
Aj8月卅一日收市價
費再:九一〇。 HREE--1100
星辰表:四一四。
最吐
義下既工;大五。
三菱地所:八三
地產:娀五台
第一套新戲來說,爲再收入一千萬股。 -缺亦隨這種趨勢而改,收五四四七突。 絕,從早時上映的典·成交法數共七億四
·一些影業公冢的製作路斯指數升三十六四人氣
大成郡設:二四九
|都超過百萬以上。因此
新股成
新市场指散升一點
新力:一六九。
,在這種懶况下,今一〇六,收四〇七點八四
日本石油:六九五
21 1.5
0.5A (Ang.)
(ii) Apply P = 12R
the power consumed by I (0.5)2(50)
= 12,5W (Ang.)
(d) The circuit is
From: (1)
x
X 20
X 80
Y = 12a
K2
(*) If K, and
(Ans.)
are closed,
the left half of the circuit. will be shorted because the current will just flow through
resistor Y and the lake re-
sister only and therefore
reading of the ammeter A
will be zero.
育教僑
Date Presi
化學
1978
華
【中學會考試題預習專欄
品扌务品
(廿六 ) 朱宏林
Chemistry (26)
Solution to 0.37
(1) HCOONa+NaOHN CO2+H2O
The product la water vapour.
(11) CECH_COONa+NaOH-Na2CO3
The product is ethane;
(iii) CH
Solution to 0.39
Solubility of 2 at 15°C
× 100 g = 25 §
8.5
3410
Solubility of Z at 40°c.
75-40 40
ox 100g=87.5
(ii) The difference in solub- (87.5-25) g· ilities
62.5 B
+
(Lii) Amount of Z2 which has to be
added in order to maintain the solution to be saturated
250 = 62.5 8.x
125 =:130 6
Mass of iodine in Y-0.508.g
Mass of oxygen in Y
= (0.668
0.160 s
0.508)
molar ratio of 1:0 in Y
0.508 0.160
+CH3CH
127--
·16-
+ NaOH Na Coz
+CH2=CH¿
- 0.004
0.01
25.
the formula of Y is 1,0.
2X
(1)
In the above figure, the' equi- valent resistance R for the
·branch
(e)(i) Neither Ly. nor. Iy will
be lighted
育教僑華
The product is ethene. (b) The IUPAC name for the com-
pund Y is
2-chloro-4-methyl-hex-2-ene (1) The bromine solution is
decolorized.
(11) The product is called
2-chloro-2,3-dibromo-
4-methylhexane,
CH3G=CCHCH3
given by
(2)
(ii) Only L1 will be light-
ed.
If bath K and K are clos-
Mass of iodine in 1.336 g
1.336 0.668
Substitute (1) into (2)
1, bath diodes are aborted,
學能推理練習專櫚
1.016 g
X
maas of iodine in X
= 1.016 g + 0.245 g
= 1.270 g
mass of oxygen in X
(1.590 – 1.270) 0.5208
molar ratio of I: 0. in
1.270 .0.320
X
when balanced
预味
Br. Br
CIB C25
(iii) It is an addition
reaction. Solution to 0.38
A is potassium chloride; Bis oxygen gas.
(ii) Formula mass of A=74.5
Formila mass of KC10=122.5
(1) No, of molea of KC103
募
2.45
= 0.02 moles 122.5
(2) No. of moles of A..
= 0.02 moles 74.5
(3) No. of moles of B
672
22400
(ii) From the above results, 王七 is seen that 0.02 moles of KC10, on heating gives 0.02 moles of KC1 and 0.03 moles of oxygen gas.
8
That is, 2 moles of KC10. gives 2 moles of KC1 and 3 moles of 0.
Hence, the equation for the
·reaction: "should be
2KC10
10 - 2KC1 + 30,
Time = 60 x 12 + 52 = 772" sec
∴ quantity of electricity used
= 0.25 x 772
=193:coulombe
193
96500
faradaya
= 0.002 faradays
(ii) Amount of ji liberated.
.0.216 0.002 moles
108
**0,002 faradays liberate
0.002 moles of M
.to liberate '1 mole of N
requires 1 faraday.
(111) Amount of N liberated
t
0,059
59
0,001 molas
6.0.002 faradays liberate
0.001 moles of N ..to liberate 1 mole of N
requires 2 faradays.
127
∙16
0.02;
-0.01
32
the formula of X is 102.
(11) The equation is
10102
(3) Add dilute. sulphuric acid to
each of the substances. Copper(IT)oxide will react: with it to give a blue solu- tion of copper(II) sulphate while carbon remains unchanged.
(11) Add acidified potassium perm
manganate to each of the liquids.
Methanoic acid will, decolo rize it due the redox react- don between the two while. ethanoic acid does not react with it at all.
(iii) Pass each of the gases into
Bome bromine wolution in:
tetrachloromethane,
Ethene will decolourize the solution while ethane has reaction with it at all..
(iv) Add concentrated sulphuric
no
acid to each of the solids. Sodium chloride will give a white steady fumes which is hydrogen chloride while.
sodium bromine will give a
violet vapour which is
iodine vapour,
物理 (廿六 )
泰榮家
PHYSICS (26).
Answers to Exercise 13
30cm
wwww
to discharge 1 mole of M needs 1 faraday,
∴the charge on M La
to discharge 1 mole of N
needs 2 faradays, MTthe charge on M. Is
In the figure, when balanced,
(a)
(Ang',)
(i) apply tie principle.
Wheatstone Bridge
X:500 (Anan)
Albert will obtain 50.
(11) Apply V
0.99 (0.02)X
X = 49.5s (Ans.)
Bernard's, answer in 495
(iii) Again, apply V - IR 1.02 (0.02)x
X-510: (Ange)
Charlie's
Maker is 51
(b)(i) The result of Bernard
is actually the combined re-
sistance of X and the volt-
meter in parallel.
Let the resistance of the
voltse ter be
R
元
49500 (Ans.)
(ii) The result of Charlie's
method is actually the com-
bined resistance of X and
the ammeter in series,
Where
51 50+
RA
is the resistance
of the ammeter.
10 (Anm.):
(c.)(1) Let the current throu-
gh X be. I and the current
through R be I' in Albert's
circuit;
(1) Apply
power dissipated by the
8-ohm resistor
(0.1)2(8)
0.08W (Ans.)
(11) As shown in the figure below, let the equivalent.
resistan
ince of PQ and RS be
812
RAB
Pad across AB.
數字推理練習(十二)
文字推理棟習一
麵别的辨忘!
下列每赋有六個項目,其中四個是同類 的,其餘兩個跟進去個並不同類,試選出這兩 何
【例】(1)摭
12) (31
15)冰
A 1,2,
B2, 3-
C4.5
D3.6
E5.6
跳和蹣是足部的動作。答案應避E。
【例二】(1)電燈:(2)大把(3)蠟燭4)太陽:
(5)味驢(5)油燈
A1,37 B 2.4
D3.4E4.5
C4.5
題解:電燈、火把蠟燭、涵燈都是联明用 的,而且是人工的;而太陽是天然的, 暖爐是取骇用的。答案應避C。
下列每題有六個項目,其中四個是同到的。 其餘弟個跟這四個並不同類;拭選出還兩個項目: 並在它的下面燾一橫機。
B3, 5 C3, 6D4, 5 E5, 6
(2)
Wa(9):
1. (1100
{2}
13)
-(5) ・・・ (6) 367-
A1, S B2, 6
2. (1)
(2)
(3)
C3,4D3.6E45
(4)
(5)理
(6).
AZ, 4
3.1
(3)
(4)
(5)
(e.m.f. of the cell)
A2, 3 B3, -4
(51
CA.
5
{2}限:{3}布 (01
D2, 6. E1.
[41
A1, 2
(p.d, across the 8-ohm
resistor)
5. 01
A2, 4
2
1.2v.
-(0.1)(8)
6. (1)
{5}原料
P1, 4:
7. (1)清秀
120
(iii)
Where
Hpq
and
試
are the
resistance of the branches
P and HS respectively,
B24 C3, 4 D3, 5 'E 4; : 5. ́
(2) (3)
(41
.(6)
Bi 3 C2, 5 D3, 4 E46
(2) (3)** (4)塑膠
玻璐
Q2.5 R3. 4 S45 T4, 6
(2)游¢ (3)秀覺(4)艶魔
P1,2 Q1. 6 R2, 5 S2, 6, TJ, 4
8. WURG (2) (3) (4) j
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Q2, 5 R2, 6 S3, 4 T4, 6
(2)鼻于(3)耳朵(4) 【腰 (6) *#*
. P1, 6
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P1, 201, 3 R2, 6 $3, 5 T4, 5 10. (1): (2)廣播(3)看電視(4)寄信
[(5) K&M (6) 4T TK-
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(5) - (6)
E56
乘
12+18
200(Aoe。)
13..
: 21
I + I'm
I* (R+40) 1(50+80)=1(25+40)
I'
current drawn from the cell
(lv) If no change in the read-
inge of the ammetera then
no current passing through K. Hence
(1)
Also
X + Y = 20.
(2)
• I{X+80)
A 1, 2
(4) #BE
A1, 2 B1, 5 C3, 4 D3, 6
(2)款存 (3)**
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(5) 33 (6) AAA
A1. 3
B2, 4. C2, 5
D3, 5, E4. G
**:0 OC QE OD A 00 P OT 00 OA OC
S
A
B
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