HONKAR atua=+f7PEZ WAH KIU YAT PO
!XXC-
HPO
4
+
+ POT
4
4010 reacts with water to give phosphoric acid which partly
橋
ww
dissociates to provide H which turns blue litmus red.
L975
中學會考專欄
(h) HI + H2O →→→→ 1150+ X
blue
(1) NECO
2Na*
•CO + H2O
+ OH
OH
:堅道書主編
化學科(十)·陳湛杰。
CHEMISTRY (#10)
Solutions for Chemistry (#9)
Multiple Choice Questions
1. (E) 2. (C) 3. (0) 4. (0) 5. (B).
Gonventional Questions
1. (a) Since there were originally
0.1 mole of HCL in 300 ml which is now diluted to a volume of 6. litres,
resulting solution Molarity of resulting = 0,186 = 0.0167
Molecular weight of 101 36.5 Concentration of resulting solution 0.0167.x 36.5
6.096 mm per litre
(b) Molarity of HC1 in a solution
of 2 moles of acid in 0.5 11tex
of solution = 2 +0.5 = 4
weight of
the
Since concentration molarity x molecular weight Since the molecular H01 is itself a constant, comparison of the molarities of the solutions is equivalent to that of the concentrations. Thus the first HCl solution is 2M/4M 0.5 times that of the second HCl solution in concen tration.
(é) In solution (1)
molarity
In solution (2), molarity
1000
10.1 x 2500.4.
Thus solution (2) is the more
dilute solution.
(a)(1) Jilution
factor
1000
2504
0.025
Thus normality of diluted solution = 0.14
Since one liter of N. qolution contains 1 ga equivalent
Thús 250 ml of 0.18: solution contains
01 x
= 0.025 gm
7009quivalent weight
Since the scoond solution: is obtained from the firet one by, dilution it should contain the same number of equivalents i.e. 0.025
(e) Number of equivalents in 50mi
50 3M HOL = 3 x 1000 0.15 Number of equivalente in 20 ml: 0.5M H01 ± 0.5 × 1000 = 0.01 Total number of equivalente in combined solution of (50 +20)
70 ml is (0.15 +0.01) - 0.16. Normality of resulting solution
=2.286
2 0.16 x
2. (a) NaOH
H2O
OH
Ae OH is an excess over that of H,
the litmus turns blue.
(b) HCT HO
017
The hydrogen ion (hydrated)
turas litnus red
{e) N@20 + H20
2NaOH
Na+ + OH Reaction of Na0 results in fomation of NaOH whose so- lution in water turn litmus
(a)
blue
H+
**
+
HCO3
2+
When 00, dissolves in water,
the weak carbonic acid is. fomed which partly dissociates to yield hydrogen ions. However the amount is so small that the neutral litmus ia turned from purple to only faint et H2O
(e)
+
3ND,
with time Fe**++ Fe(OH)3+ 3H← The ferric ion.
is not very stable in water, with time the ferric nitrate is partly hydrolyzed to the insu~ luble hydroxide with corre sponding amount of hydrogen ior 1bemted. Thus the solution
a slightly acidic and the turns pale red,
Na+
+
(f) Nagr
Br the Nat and Brions are quite stable in solution so that there are
no excese hydrogen or hydroxide
ions with the litmus retaining
the routral purple colour.
(8) P4010+ 60
H PO
4H3PO4
H
litmus is turned red.
H.0
As carbonic acid is a weak acić and NaOH a strong base, the solution is alkaline turning red. Litmus blue
HO
(3) NaH PO-> Na+ + H2P0
4
HP + HPO
4
B..
報日僑華
え表感
肉中的:小皮每支持细绒翰
中央のお茶不完全藥的級 述,那一项才是正確的答案:小缺少 葉 葉狭まの葉釘、葉
托葉和禁刮;菜缺少托菜和葉 精心缺少葉柄,托葉和乘封。
(5)葉綠或葉尖處生有許多小孔液 些小孔做:A氣孔B冷剂,葉
H; HamNER, E. PIERR
成針、
養為業的變態:南瓜的粉駭 Bebé, cxIEJATEG
HPOH + PO
4190
H2+20
Litmus, retaine
H,0
H2 PO+OF neutral colour
I:
越少
(2)
+ .003
HCO3 + H2O H2CO + OH HCO is very weak thus KICO, solutions turna red litmus blue
H2O
(1) NaHS > No.
+HS
HS
HS + H20 — H2S
он
HS is a very weak acid thus a solution of NaHS turne litaus blue
H2O
(m) £50
HSO
250
Sulphurous acid is a weak acid, thus K2SO solutions tums red Litmus
A. HO
(n) NHNO3 NHÀNG,
Ammonium hydroxide is a weak base, thus a solution of NH NO in HO turns blue litmus red
Questions for Next Week
The questions in this issue are related to 3. Organic Chemistry study your notes and text book before attempting the questions. Multiple Choice Questions
1. Elementel analysis of compound
I shova a composition of C, 54.5% and H, 9.1% by weight, At N.T.F. conditions 1 litre of the vapour weights 2 gms. Compound X is (a) (glo (b) CHO (0) CÔ
(d) CH2O2 (e) none of the above
2. When othylalcohol is mixed with
excess amount of concentrated sulphuro sold at 18090, the major product formed is
(a) ether (b) acetylsne
enethane: (d) ethylene
(e) alcohol is stable in acid
compound
3. Which of the fol CROH 7
is an isomer of CH,
(a) CH2CH2 CH2CH ON
(b) GHCHOHOH (c) CH2 CH2CKO
(a) CH2000H (4) CH2ON CH2
4. The solubility of C
greatest in
(a) HO (b) CHCH2OH
Bromine, water
C2HD-CH
5. Which of the following compounds
will decalorize potassium permanganate solution in the
presence of dilute sulphuric acid? 2(a) CH2OH (b) CH2 = CH2 (c) H
(a) CH2000H
(e) CH COCH
Conventional QuestionA 1. What do you understand by
homologous series"? Illustrate your answer with suitable examples Describe two chanical testa each for distinguishing the following pairs of organic compounds:- |(a) (CH3)300H and CH2OH2OCH2CH3.
(b) CH2CO•O•C2H, and C2H CUCCH2 2. Explain briefly, with a diagram
of the apparatus employed, the nothed of determining the per contages of carbon and hydrogen ir an organic compound.
0.185 g of gave on com-
an organic compound and bustion 0.440 22
0.225 gm. of water;
Vapour
density of the compound being 37.
What is the molecular formule Of the compount, and what structural fomulae ere possible for it?
生物科(十)・本國翼、
‘她文
1. HELHARM, 2. Biedrska.
練婆道:選擇題
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TEADY
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生物科(+)
15.迷落禁巡程。 冬天會變成红色?小弑索出第
P. 3. 12. 彈球纲瘰莲。 三上期答案:
IS-CIA AIS. DIACja.ci. acta; xo, AB, TAIC 2.8.
立填-
:光胶芽、横孩城礼
Řeš / AIR, J. ÞİR. Hh查根系指报 臌大根菜
AR CAR, THIR· 寄生憾水根;P柱根。
4.製造食物營養繁殖:
問答題:艺
2743
11⁄2 4 x
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12: 34 4 w ma
drp
世梅
手葉蒸
AMBERT test
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五、根之功能约有下列表獵
S. 343
T. 512
18。在73955中,3的值是6的值的是倍?
P: 0.5 Q.5倍 R. 504
T500% PQRST
□□□□□ 19.下列那一個分數的數值與0.15相同?
S. 55
BICZOGY (#10)
Answers to 2.16
b. A:
a. D: 11:30 am 12:2012) 6:30 am. Dat no time during the
experiment.
02:30 pm 6ph r-sched
highest point.
2. B: race is detemind My food ont
temperature. EB130 Kin
sive an account of how eponse to stima dan secur in (1) Anize ba (b) a lowering Flert; and (2) a umm21:
Explanation:
All living things are capable of raking a response wher, acted upon oy bimus. recestor is any
art of an organism which is
apable of being affected by s
e, and an effector is any
(apable of making a response. an Ampete, any part of thej: body can act us a receptor or az an effactor. Au. he know, the bod; of
hanging shape by means of the flow of
ch resulted in protoplasm which numerous projections or pecudo pod- je protrading and withdrawing. such a pseudopodium, vomés into contact with a grain of sand, it will.withdraw, had projections but
But
Amoola fa
to constantly sh
out in another direction. when an Amoeba comes into contact
particles with food t
as my
*
કે ફ
11.
Algae present in the water, a nusve o pseudopodia flow round the food particle a rog of water,
a tood
Frun this ve can see that Anoeta shows diffor antial irrita.
It can lity. respond to sauli, and any vari- ation in the stimulus caucus a
The stimuli different response. must be conducted in some way through prot toplaym and nust also be perceived over the whole of its surface.
ths
(b) A flowering plant
Gan
respond to a variety of stimula, vete:11M resulting in the performance of simple reflex movementa. It may
日八十月一年五七九一公年四十六國民華中育教僑華
be a taxis gr movement of a free -living plant aither rectly
towards or avey from the source of the stimulus, or, & tropie o movement often a bending of one or more organs of a fixed' plant either directly towards or away from the sore of the stimulus.. Unlike the Amoeba, only part of the lowering plant respent to stimli. They are the apices, mely the tip 2 sten and roo
· Let in the meristemitic cells. ua: consider one such respouse→ the response to light or photo- travism.3tens are, moetly posi tively phototropio aut grow towards the source of light. Roots are not sensitive to the direction of light, but some are negatively phototropic and grow. away from it. Terves are mostly diephototropic, that is the petiolee hold the lanira at right angles to the light source,
Experiments conclude that the
ahout tip is sensitive to the stimius of Light, the cells of this region ser rute, and discis.rge auxins (plent hormones) which
travel from the tip to the zone of elengation where they cause co-
rid resulting
celoration of However, the same
in parvature.
stimulus of
light
un mot hig
auxing to accumulate illuminated side but since
arorcentration of Euxins in roots
100
X 100
A simple Constant-volume Gas-Ther mometer.
Dry air is contained in bulb B which is connected by..rer. row tubing to a mercury pressure gauge, and the volume of the air. ia kept constant by moving the open tube until mercury level on other side, reaches, a mark AT If the bulb is surrounded by ver of 1000 air in the bulb expats, The open limh has to be adjusted. such as to keep mercury in touchi with the mark A. The mercury level
the
on both tube is recorded.
The pressure in the bulb 15 P100.
P100 h where H is the atmospheric pressur and h is the respective difference. in levels of mercury.
Temperature
of surrounding
०६
Boiling point 100
Melting point. ter and the roote
inhibi
side
Crows fatto:
growth, the luminated
curves away
from
the
a
The simplest route
ervous impulse
respords to a 3-mulus is
Pressure. dt air in bul
H+ 100
of water
3
H ́+ Do Por
of ice
which in
x
Hh p
called
efiex are, and the action is
Ich notion.
as a
It occurs through the spinal cord,
vit wch or ithout the awareness of the brain. Let us consider an
is example of how response effeited.. The withdrawing of the arm if the hand comes into contact with a hot obsecf. A nemre ending in the hard in stimulate,
૩
sensory neurons, conveys this 1ervous impulse along a asusery nerve to the central nervous system where to is analysed. · A motor neurone conveys the respondi along a motos ne we to the organ which will react, in this case it is the bicep muscles of the Rim which will contract, withdrawing the arm from the hot jesti, 2.18
a. What as meant by irritability? b. Give 3 examples in plants: e. How does plants respond to the
inilateral stimulus of (i) ligh (ii) gravity (111) water (iv) chéndol (v) touch d. Deze ar experiment to
illustrate any ONE of the above- * Je spots28,
explanation:
Answer to (a) and (e) are quite. straight forward. What you must. remember, when answering, is to Keep to the question and only
As for produce what is asked for. part (b), here is where you often
Many candidatea, instead go wrong.
asks
of simply giving the name of plants and the type of response shown, tend to waste too much time in describing the hour and why plants a certain manner. react in a As for part (a), the question for an experiment to illustrate just one type of tropic response so you must remember that and select the one which you have done and therefore can give an accurate account of
Try answering this question on youd
OWN
物理科(十)·陆永恒。
PHYSICS (10)
1. (a) A simple Constant-pressure
gas Thermometer.
A pellet of mercury is put into
a short capillary tube with dry.
air trapped inside.
A half-etre rule is attached by the side. and is then put into a basin of boiling. water. The air column expands and its length is measured.
tube and its. attached is then insert- ed into some melting.
The air cauon is ice: cooled and contracts. The
length is then recorded
mm.
To measure the temperature of some liquid, the same process is re- peated and the air coulan is For all these measurements taken, the pressure is kept constant and is: the pressure of the atmosphere
Temperature
Length of air! column (mm)
Foiling point
of water Melting point
of ice
вду..
100°
0°C
X
$100
L
£
The centigrade temperature is de- fined on the scale of the thermo- meter as
心的一的一是多少?
4 5
325yt 日。650元 D. 1825 E. 292570
34.分解336質因數連乘式。
any
The centigrade temperature is: defined on the scale of this thermometer aa
X
700
*
Po
===
P100
- Pa
Proc. Po
100 PC
(b) Pressure of the trapped, ai? at 283K H 2 cm hg Pressure of the trapped atr at 373°C = H+22 cm Hg
(c)
By Charles Law (constant vol
Pressure
Absolute Temperature H+ 22
373
2
373(H 2)
==
283(H22):-) H(373283) 2x 373 +- 22x283
90H = 6972-
H-77-46
The barometric height is 7714 em of mercury
Wt. of balloon fabric-
Load lift -160 kgf Total weight of it =200 kgf.
Let be the mass of air. Density of air at T=290°K VIL
219-1.2 kg/m3 200 Density of air at 5000K 45 Since mass MVD, and the hot air balloon is open to the out side, the pressure is kept constant. at 750, com Rg. Therefore,
در
M/DM/D; DT = D?'
DDT 1.2
air
200 500 total weight of halloon 1,29
VD 200
0.51V 200.
-V 392 own; 2(a) (i) When brass is heated
through 1 deg.C. it will expand 18 x 10 of its original length. (11) 1 deg. C deg. F.
18 x 10-600 = 1.8x10"
=0.00001 per F
b) doer. of linear expr.
Coef. of cubical expn.- Y.
Length of each edge of a cune
Volume on
of the cube W。 = Temperature rise,
190
After increasing temperatura its edge = (1+x0)
its. volume
Therefore,
(1+23
(1+0) = (7, + 3)
·= 1 +34 + 35
The value of is very small: and aig negligible.
and, 3 α
(c) Coefficient of linear expu-
brass 18 % 10/9C Scale reading is correct only at a temperature of 0°C The reading obtained at 12°C.. is 76:52 cm which is incorrect It should be more than that v value as the scale itself ex- pands through this temperature' rise too. Therefore the reading should
be:
1 = ((1+α)
= 76.52(1..+ 18x1
76.54 cm
1
Questions for this week
1. Distinguish between the cool- ficients of real and apparent expansion of a liquid. Deduce an expression relating the density of a liquid at dire ferent temperatures with a doer ficient of expansion.
2
·盒牛肉乾便宜一,求牛肉乾的價值。
5
0.75
R. 90
9. 5x3105+ Y
1975
Y?
P. 3.
Q. 5
R. 7
升中試專欄
S.. 15
T. 21.
PQRST
智慧出版转生糖
3
21
10.
Y = ?
數學科(十)
Y+4
P. 4
驗
S. 46
112 Q..16. T. 64
R. 28
P.-
Q.
Ri
PQRST
20
200
2
1. 384-92-108
1
2
-0.5x0.3+0.2
B.:22
C. 38.
11.
S. 3
20
Ta Đa
E. 238
ABCDE
0.7
A. 0.3
B. 613
A. 10.6
B. 16.9
2. 789+6x21
A. 568
D. 4308
. E. 4560
3. 11x0.9+0.7
D 106 E. 176
C. 17.6
ABCDE
4. N-P-10 下面那一個算式是對的?
A. 10-P-N B. P-N-10 C1 N+10- P D. P-10-N E⚫ N-10 P:
5. / 280 的2倍是多少?
C. '865
A B C D E
A..9
A.
12. 下列五個數中,那一個是質數?
B. 0.5
C.3
E 50.
ABCDE
B. 18 C. 21
D: 29
E. 33
A B C D E
ABCDE
13. 求最大二位偶數和最小三位數之盔。
A. 1. 6.2 C. 3 D. 12
E. 13
A B C D E
14.下列各數中,那一個不是24,36的公因數
?
A. 17
D. 34
6. 62-33
B. 18 C. 19 E. 36
A B C D E
›
- ?
P. 7
S. 55
8. 0.9x Z-0.36
P. 0.04
S. 3.24
Q. 0,4
9:2
a. 18 TA
R. 37
PORST
Z-7
R. 0.54
PQRST
A. 2 D. 6
8. 3.
C. 4
E. 8
ABCDE
□□□□□
ABCDE
PORST
26.
12x3 10-6
108+2 Y+2 P.0 a. 2 T. 8
}
11 1
Y?
R. 4
33.
PORST
C. 1300 ABCDE
27. W×17-W-17 W=?
16
P.
Q.
B-
16
17
18
5.157
-18
S.
T. 1
PQRST
17
POR ST
A. 2×32x7
c. 2 x3x7 É. 24 x2
B. 22 x32 x7 D. 23 x3x7
PQRST
□□□□□
39.晚上十時二十分睡覺,次性大脚三十分起
床,共睡了多少時間?
1
T. 22-
2
ABCDE
C. 675
15. 1640L«C»M»
A. B
B..40 C. 80 D. 160 E 240
16. SR75843085H.C.F.
P. 3 0.5 R. 10
S. 15
PQRST T. 150
7.下列工個數,何數的立方根不是壓斷?
P. 64 Q-125
R. 256
20. 26-400 - ?`
P. ****
0.5毫米 S. 6** R. 650
PORST T.6厘米200毫米
21.由20至50之間,質數之和是
A. 208
C. 220 B. 210 D. 228 E. 251
A B C D E
22,下列那一個數目最接近7007
A. 589 8. 598
A B C D E D. 697 E. 705
23.下面五個數目中,不同類的一個是 A. 91 B. 83 C. 61 D. 53 E. 11
A B C D E
24.某數的六分之一,比它的七分之一多3,
***
A. 42 8. 63 C. 72 D. 84
E./126
A B C D E
25.8要乘以那一個最小的數,它的平方很方
是整數?
8.2
A. 1
C. 3
D. 4
E. 8
ABCDE ``00000
28、佐4+10,19,31數下去,下一個數是多
0.46. R. 48 T. 53
PQRST
29.35至89之間,3與5的公倍數共有
P. 1
a. 21
R. 3 S. 4.
P Q R S T T. SH
·30.音樂室有座位七行,其中两行有座位八個 ,其餘每行有座位十個,音樂室共有座位
R. 60
PQRST
□□□□□ 31.柑115個分給甲、乙、丙三人,甲丙所得
·相同,乙比丙少5個,開乙得多少简? A. 3014 B. 350 C. 40 D. 45
37.茶葉一箱電20千克,隨溫箱茶葉的一比其
P. 79104
a. 7309
R. 8$10
3
A?
P. 44 $.50
35.
·是6的幾分之變?
1
3
A.
B
10
10
18
0.
P. 2261
A 231
ABCDE
7
P.55幅 5. 66
a. 5815 T.70個
36. 甲有疑是乙有
的一,若甲有欸140元
12
答
乙有欸多少?
(1) B (4.)
E
T. 8305
S. 8###205
40、膠布長30型,剪去6——再剪去12.5呎
3
·問餘下膝布多少?
言。23)1呎10吋
Q. 2310
C (3) A (6) R
PQRST
□□□□□
S.. 231950.
PQ R S T
P. 1000
S. 340
Q. 140. R. 240 T. 38072
PORSI
D (5)
(10)
T
E. 554
1
32. 18–7
4
1
A. 2
D. 162
B. 18-
,C. 126
9
E. 172
ABCDE
(13) (16) (195
一貫多少?
(22) D
5
P. 0
0.173
A. 2+
(25) (28
S. 4
5千克 PQRST
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