真四第張三第
日二十月十年午丙膺聯
WAH KIU YAT PO
育敎僑華
中文中學會考試題預習專欄
日
1英文中學會考試題預習專欄
化學科
违
數學科
二四)
·喬仲强·
3.1
a. Density
(1)
・第三次預習試題解答
KOM CHEMISTRY ’(i)
11.由100呎高之塔頂T,測得手地面上两奌A. B 之俯角依次 為2525'及3% 47% LATB :42-15 求A,B間距離
b
Solubility
in water
(解)設如下图:TC裝塔高,則
42°15"
¿CAT=25°25'. LCBT=32*47'
(平行线間之內錯角相等)
ArtsCAT.
AT=100 cac 25*25′— 2 $3 (0£),
C
Art ACBT BT=100 cac32°47=184·7 (9%)
依餘弦定律在△ATB中
AB ATH BI~ 2·AT· BTCH LATB
- <= 2332 + 184-7 - - 2x233 × 184-7 cos 42°15′
· 54290+34120-63700
24710
AB÷124715157.2
答:AB距離1572呎
3147
B
4
A
NO. LOG.
466 2.6684
1 184-7 2.2664
cos 42"is" T. 86936
63700 4.8041
2.(a) 設在右图中口為單位园(Unit Circle即半径)單位長度】
A cos2x=1-2 sințe.
(註):ZAOP=2x(国心角等於同弧
c. Combusti› bility
Supporter
of combus • tion
Action on heated met callio axides
f. Action with ethylene
action with
heavy metale
Hydrogen
Least conse of the three gases; light
est substance known
Least soluble of
Cha three gáséð
Burnia with a pala blue flame in air with the Cormation of water
Does not support combustion
Reduces metallic
such as cupria
oxide to metallig copper
OuO+H2aCu +Hyu
Reacts directly
with ethrylene 20 ordinary tempera-
#
【四)
王錦釗 ·
· Oxygen
Densest of tha threo gases. 16 times denser than hydrogen
NOBL 9510Die of threa gasde....
ני
Nitrogen
slightly lear dense than
BET
oxygen; 36 times denser than hydrogen Leas soluble . than oxygen sbut very slightly more soluble than hydrogen
Wat combustible Not combustibl
Supports comvuь- pues not sup- tion; a glowing part combus - eplint will burst tion.
into flame when
put in a jar of, abrzen
年六六九一魁公年五十五國民革中
The carbon monoxide is removed by passing the mixture of the two gases through an ammoniaca) solution of cuprous chloride and the hydrogen is then collected in gas jara y
over water
111)Small quantities or nyorogen are obtained in the laboratory
by the electrolysis of water water acidulated with dilute sulphuric acid is put into a Hoffman's. Voltameter: When ⇓ direct current is passed through the volḥamster. hydrogen i is given off at the cathode, and oxygen, aɩ the anode This method is not suitable for collecting the gas lo gṣa Jara
be von wow viac nyarogen disuses more rapidly then) air by means of the following experiment
The apparatus is set up as shown in the diagram, Initially the liquid levels in cubes A and B were at the same height. Since nydrogen diftuées out through the forous pot faster than air difluses in, the pressure inside the porous pot falls Consequently the water level rises in tube B and falls in Lube A
POROUS
! POT
No action!
No act 100
BIR
Oxidises ahylene
NO BILION
to form acetalo- enyde
tures to form ethane Car "02" 24. CHO
Platinum black is
used as a catalyst
C2
C2H6
No. act200
豐
·园角之倍】
故在A0MP中.
MP
sin 2x=== ОТ
MP
B
M
cos 2x = CM
OM.
ስ
OP
在AAPM中,
联PA,則&APM=x(同∠A之餘
Action with pyrogallo)
No action
sinx
MA
1: 80ion
AP
halogens
ZA BPM.
sin X
Q.2
__MA2
MP
ΑΡΣ
BP
~
去分母,去括号
加比定理
AP+ BP
=2-20max (2) cos^2x+sin2x=1)
試証:
+4(1-1)
Farm basio oxides will form the with Calcium, mag-nitride if the nesium ron lead; metal burns and copper
brilliantly E JM8+N2M832
Is absorbed by an alkaline solution
Combines caly of pyrogallol chloride
wt. copper oxide
Wt. or copper • boat
No action
No acte
No action
323
-300 110
2002
Wt of green removed by nydrogen wtfesorption Lubes (before)
kt. of absorption (tubes{a£uar}
Wt
of water formed
% of oxien in water
Lab 34 B
• 47.02 m
• 2.68
3. Hydrogen may be obtained from water by-
OF the action-of metals on; water. (11) · iNe-Action of non-metals on water; -{i}} the electrolysis of water
2.68 88.8 %
(1) We can conveniently collest a small quantity of hydrogen
by the action of calcium on water This method would not
be suitable for collecting a few gaa jers of the gas
test tube files with hot sund beaker
Wales
A few places of
are dropped: into the water and the test-tube is placed so that the hydrogen. liberated rises into the test-tube.
Ca ZH2Q •\Ca(OH)2= ▼ 2
A
larger quantities of hydrogen may be obtained by passing. atṣam" over rad-hot iron contained in a hard÷glesa (preferably silica) combust 100 Lube. The "combust:100 tube la heated by means of a strong bunsen flane so as to keep the iron red-hot The hydrogen liberated le collected in gas jars over water
3Fe + 4 H2:0
-
14. ng
(11)nydrogen can be obtained by passing steam over coke
heated to while heat... The coke is nested in a muffle
furnace and the product obtained is a mixture of carbon monoxide and hydrögön
GUARD
WATIGE
A
amitimah LareL
OF LIQUID IN BOTH HAMS
D4 Ways by which iron may be prevented from ruating are **.
follows:-
(a) Painting: -- This is the commonest method.
ships, motor cars ste
from rusting.
Steel Bridges. are painted to prevent them
(6) Grease or tør - ver is used for bridget; grease or 011]
is used for tools.
(-) Salvanising » Iron sovevo por vulduange are gazranjbeg to prevent the iron from rusting. The iron sheets are dipped into molten zinc which forms a protective layer over the iron.
(a) Electroplating - many iron articios aIN CONTINU
layer of metal, such as tin, chromium or nickel to prevent them from rusting The shining netal parts of moter car are pieces of iron which have been chromium. flated.
(-) Plastics - zne sursace of the iron is spraysu misu
layer of some plastis material to prevent it from rusting.
(f) Stainless steva - adron 18 provantes : rom pusting by
alloying it with a metal such as chromium, the allos being known as stainless steel, m
Questions for next
How is hydrogen produced on a large ach) 67 - How, and under what conditions, does nydrogen react witne (a) oxygen, (b) sulphur, (c) chlorine, (a) black.ooppaa“
okide?
Salta my om pruparen sy (3) neutralisation of an acid } and an alkali, (b) double decomposition, (e) direct synthesis, Illustrate these thires mothods by describing} the preparation of (1) pure, dry trystals of potassim. nitrate; (15) pure, dry calcium sulphate, (111) anhydrous ferric chloride. Describe and explain what you ses nappening when crystals of copper sulphate are heated in a test-tube which is held horisontally. Kake à list of time suits in everyday uss. Give their chemical names and formulas, and state what they are useag for
`Name of Salt Chemical Mana Forma Use
LIBRARIES
•
MA`z MP" — ( 1-cm 22)2+(ain 2x)* (8° 8¥17−1).
AB*
4 nin'x=1-2002X+ cm*2x + sim2%
+ cos 2x == 1-2 sin x
(8)不用对數法,求下式之值:(可用极式表示)
n2 (F) + 4 (tan &π+CO3π}
(AE) 13 Ian&T=
= ten (1-4) = -tan 4.
cos 315 —cot (24+7}~~¢HT
sin (I) + $ { tam? 1+00831).
3. (a) ik 0 3 2 1.
(答)
(4) sin (180o 8) am air 8,
{ü) cos ( 180°~0)= - Cos 8
to to + LAOP=180"-6
XORITE, LA PARPMLDA, M
Jsing = CMP (& MP3x^])
MP ор
COS BOM OM (OM 43 11)
L& 0.31 #33 op' JE L AOP # L POMEÔ
X15 P′3] P′M'LOA BJ sin8= M'P'}
典口均為正值、
COS B=DM! & MPL
@st.s0PM art. Aop'M' (-413 — 4412),
= [MP]=]N'P'"], 10M=10M |
12 MP=MP OM=-OM' ('#ta Hz).
sin (180 -8)= sino, {(cos (1800)—cos
Q.E.D.
the 2 sind - 3 cos 0-3-0 (7# # 0°3 180°).
•{A}} = sin^0=100, MX
2 (1-crs*0)~30088-3-✪
化簡
分解
W
2 cas*0 + 3 0058+1=0
(2 cos 0+ 1) (200 @ + 1) ==== 1
cos ==
cos = 0 = 120°
事
=-1, §10=180°
·A·0=120 & 180°
4. **»*#*=B4: cot (sin☺) - cot' (sin 8+1)=tan (3)
A
(14) #2 cot (sino) = A cot (sino+1)=B. tau! (#)=0.
9] cot A=sino, cot Bang+1, tenC= # cot c=& 两边各取餘切函數,
By
12
cot [oot" (sin 0)-cot* (sin 8+1)]== cot [ tan' (#)]
cot (A-8)= cot c.
cot (A-8)= cot 8 cot A+ /
cot B-cat A
((sin 0+ 1) sing+1
簡
分解
(sin8+1) = 'sim✪
4 pin: 0 + 4 sin04 != 0
(asino +1)*
0=nπ+(-1)*ZT
: 0=n]+(-1)′′Z4
= 건II-(-1)*품.
5. 在任意四边形ABCD中,∠BAD之平分线及BCD之平
14 BD + − ‡ E, Q·] < ABC 27 R2 ADC 278#
亦交≤ AC之一点,
(Eku) AE, CE & BAD, 2B C D Z.
分角线,E在 BDE
() LABC, LADC <»åƒƒ'✯ à'
ACE.
(证明))說<ABC之分角线界AC交于F
联DF(两奌定一直线)
AE, CE & BLBAD, LBCD ŻDA3⁄4 (LA«)
AB
BC
器=器(4肉球細角,内神辺店
(代换)
WBLIC
龍二題:BF平分∠ABC,内角角线性矍定理
7 A2-AE (4*)
DC
OF平分ADC(分△~興部位成比例で直ち) By LABC, CADCZA & ACE.
Q.E.D. 6{a}园的切线,割线及国外线,典出心的距離着何?
(答)园的切线典园心的距離等於半狂,割线典国心的
距離小於半径,园外线典园心的距離大於半裸
(附註)上述的逆定理的能成立。
(陽)求作一已知园之外切三角形,使其三边各奖已知直线平行
(已知)p.gm直线及0园、
{求作)0园的外切三角形,使其感
边分别典中观三直线平行
Wit) 1.1.0 11 p. q. m biz
线分别交口园於P.Q.M.
2. 13 P, Q, MS $14% OP,
OQ,OM 的垂线並延長
21 A, B, C
3. AABC為求.
(証明)従略
M
(附註):如上图:遇PQM2作出线懂得之三角形,亦可適合
本题,故有两解
2.如遇PQ及M作垂线,听得之三角形,則。因為旁切闼.
7.(a)两平弦時截之弧有何關係?其逆定理成立否? (答)两平弦時截之狐相等
ABIICD, 91 A2=6Dð. 其逆定理亦成立,即“着A,B,D.C
«7 % £ £ ± ا, £ AC=BD.
ABIICD"
(8)求作-已知国之内接三角形,使其
三 峡已知三直线平行
Myrogallol
Alk Alaking orygald
(Zkm) f. g. m = 1 # ROB
(求作)0园的内接三角形,使各边.
分别典pg、批平转
(作法)1.作。园的適當弦PO
使典平
215 0 47 48 GRU J.
3 BY PR
4 13 034 0SL RP
5以口為心,os半往作同园. R
DA
7. 13 DIE OD 2 #102# A, B.
♪ 2 A 3 g < # # # 20B * C
9.联 BC,則ᅀABC 為好球 (証明)
第四次預習試題
1.(a)試述“餘藪定理並証明之
(セ) 多項式) ルーズ+ラズ
整除,試求a,6元值
2. $24✯ f(x)=x^-3x2+3x+ax+b Hi1⁄2 x2-3x+4q #
除試求 a,名之值,並求f(x)=0元一切根
3.(a) 何謂方程式之根?
£12x−(3m+2)x+12=0JF
4x2-(9m-2)x+36=0 /9-18
值并求出每一方程式之根以驗算之。
* m2
4(a)+6m y = π 14 (x 1£& −3x+5)
(B)於同一图中,再给 y=直线,此两图线之交类,疗
·求解之方程式為何?並求其根
(C)若x為是數,等之區域差何?(區域即之範圍)
5. 2 x 4. z 4
{2x-4=1
俱為塞數,求解方程组
4x+y=2xy=3′′
(† 7 x + x = 1 1⁄2 (a"-2ab+&4)~" — (a−b)2* (a+b)*
14 a, b 1 5 & asbro.
(更正)第二次預習試題解答中文第一題:x=(1+3玩
y=-1±32 74. A
(補-
)第二題討論“根式之主值”若在平方根時,应取亚 值,但在高次根(立方根或四次报等)時,則取正是數值 (* 1 1⁄2 v M 1P=2, FE =-2,而不取20
No comments yet.
Private notes are available after approval.