WAH KIU YAT PO
wt. of C in the sample.
13.2 8 x 12 = 3.6 6
44
wt. of H in the sample
2 X
育敦儒龜食一第張七第
日一十月正年午戊展
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=
B
& Deze Preis:
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1978
【中學會考試題預習專欄
( 二十 ) 朱宏林
Chemistry (21)
tion
It is to try the cyren and sulphur dioxide gales because the product, salahur trioxide, is highly hygrosserie. (ii)latinum or vanadium.()oxide
can be used as the cltalyst.
(111) 2302(e)+9 (5) -→→250, (1) (iv). It looks lik yellow, silky
needles. (v)
It is to prevent moisture in air from entering into the receiver..
(vi) Disolve the sulphur trioxide
first in concentrated gul ph uric acid and then dilute the resulting solution called the 'olewn with 3 su table amount of water.
SC
4.3
+1024
over dissolvi tie sulphu trioxide dir etly into water. (vii) i olecular 1:33 of 202 = 80 Mobording to the equation.
7902903
2 males (or 2x 22400 cm at s.t.p.) of 30, are required to give 2 x 80 g of 303. „val of su2 required to form
is.
3x22400x
.CK
C
Mitrogen neither burns in oxy- gen nor resets with pyrogallol.
vol of nitrogen obtained.
final volume of the mixture 100 cm3
rvol; of oxygen left unused
reduction in volume of the mixture on passing through alkaline pyrogallol
(150 100) cm
= 50 cm
vol of oxygen used in burning. =(200)
50).cm3
= 15€ en
3.
According to the equation below
2 vols 1 vol
vol of hydrogen present in the mixture is
2 vol of oxygen used, up
2 x 150 cm
300 cm
Now, the deconvosition of fl
can be reprebuted as i. the
following equation:
nitrogen + hydroger
རྡ
2
200 cm? ∙100
21 cle- N chles
cules
8.1.6 × = 0.9 € 18
wt of 0 in the sample == (6.9 -13.6 – 0.9)5
= 2.4.8
Hence, molar ratio of C:H:0
3.6. 0.9 2.4
12
16.
=0.3 0.9 0.15
報日僑
· SP - S&P 14 MA
where is the wavelength,
and is an integer.
(ii) The condition for maxi-
mum destructive interferenc
at Pis
SQP
五期星
(c)(1)
Vir
Agord
入
3)Since the radius of curv-
ature of the concave mirror
2:6:1
So, the empirical formula' is Co.
is 50cm, therefore
the focal length f
25cm.
(ii) (1)
Pis hydrogen chloride, ILCI;
is ethanolo:adid,
(ii) The magnification = 4
.0
II CC
0-1
Ris ethene,
H
is ethanol,
(2) CH2CI+PC15
CH2CH2C1+ PCC1,
+ C1
CH101+2(C)CH COOH + H2O CH-COOK÷PC1-SON, COC1+FOC1+HC1
ai, OH OH 2 CH2 = CH2 + EO
(note that ethere has no react- ion with ammoniacal copper(I) chloride but ethyne.. CHCH, can react with it.).
Solution to 6.30
(a) A is iron(II)sulphate;
B is zine(II) nitrate;
C is ammonium hydroxide.
(b) D is zine(II)hydroxide; E is iron(II)hydroxide; Fis barium sulphate;
G is a complex of the
formula
His
zine ponA
K is zinc(II)oxide.
(1) _ 2n(NO3)2+2N!? OH-
(v) = 4 Jul
REALIS - POS IT (VE
Real image
Apply
+
u
Case
Apply
13
+
=
25.
31.25(cm) (Ans.) Vita Tage
+
125
18.75(cm) (Ane.)
V=-75cm
NEW CARTESTAN
Case I Real Image:
f = -25, N
Apply 7
+
-31.25 cm
-125 em
(Ans,)
Case II: Virtual image
(CH)2
21H
Apply
(ii) Zn(CH)2+2NI,→→ Zn(NH3)2
(111) F
FeSO +2Nri
+ 2011
2) Fe(CU) H2O+02 2Fe(OH)3.
2
FeSo +Ba(NO
(vi) Zn(0) + 20aHCO3
(vii) 2n003
物理
2nCO2+2NaMO2+H20+00
Zno +
PHYSICS (20)
Answers
CO2
魯榮家
to Exercise 10, Hi)Let f be the Incal length
of the convex lens. Since the magnification is 3, therefore,lvl
REAL-IS-POSITIVE
300 cm3
3N mole
u
= 10 cm
cules
V
·−30cm)
3 mole-
Apply
cules
+
con-
11
--10cm
-30cm
Apply
3jut.
−25, v =÷lį 11
-18.75 cm.
75 cm.
(Ana.)
Therefore, if the object is
placed 31.25cm in front of the concave mirror, a real, in- verted and enlarged image is formed 125cm from the mirror
on the same side as the object
and if the object is placed. 18.75 cm in frost of the
concave, mirror, a virtnal,
ereet and enlarged image is
formed 75cm from the mirror on
the opposite side as the obm jest. Both
cases are
magnification 4.
4.(a)(1)
bf
principal axis
opt ka!
'centre
(01)
Lens
S
(iii)
Apply
12
21
= 15 (cm) fans)
+
(by Avondro's principle).
2 mole- 1 mole- cules cules
Hence, 1 molecule of
sists of
nitrogen atom and 3
hydrogen atoms: So its formula
is Miz 1.8. xl and y=3.
Solution to: 5.29
8
(1) is 'downward delivery
of the gas or 'úpward displacement of air*.
(II) is upward delivery of
the gas or 'downward displacement of air', (III) is collection of the gas over vrter'.
(ii) The average relative mole-
cular mass of air is
(111)
80 x 28 + 20 x 32 80+ 20
28.8
(Note that the relative mole-
cular masses of N, and O are 28 and 32 respectively.)
+207
-15 (cm) (Ans,)
'NEW CAITES KAN
Therefore, the focal length
of the lens is 15cm,
(ii) The iris of the eye and
the retina of the eye corres-
pond to the diaphragm and
the film of R came ra (iii) The image formed in in front of the retina of the
́eye and a diverging lens is
required to correct this defect vision.
2(1) The condition for max- imum constructive interfer-
ence will occur at P is
Path difference
which gives
molecular
gases
mags
suitable method
0
32
(I) or (III)
NO,
46
(1)
1101
36.5
28
HS
34
(II) or (III) (1) or (III)
(b) Molecular mass of Co
Molecular mass of 1126 18
= 44
=
magnification =|||
Lens
Image
= 1
(Ans.)
日七十月二年八七九一腿公年七十六國民蕺中
(ii) It is because the obser- ver da not vertically ahove the object, In
otherwords, it is not
viewed vertically “above,
生物 二十 梁永華
BIOLOGY (20)
Unit 10: Organisms and
Environment
II Conventional Questions:
1. The following diagram is a simplified nitrogen cycle.
X
Animals
Flante
书
Fitrates
Mitrites
Bacteria in root nodules
Ammonium compounds
a. X is the greatest reservior
of the system. What should x be
blame the processes A, B
and C.
Give two ways by which nitrogen in animals returns to the soil.
d. state the importance of
bacteria in root nodules in the cycle.
What are the effects on plants and animals if process in the cycle is inhibited?
Ane.
a. X is the atmospheric nitro-
gen.
A: Denitrification
B Nitrification C: Absorption
1) The decomposition of the death remains of animals by putrefying bacteria.
2) The decomposition of
the nitrogenous wastes of animals by putrefy- ing bacteria,
d. They absorb atmospheric
nitrogen and change it to nitrogenous compounds so that they can facilitate: the entering of nitrogen into the cycle.
If process UC in the cycle is inhibited, plante would show poor growth at first and die finally since nitr- ates are important for pro- tein synthesis and chloro-
Animals phyll formation.
are consumers. The death of producers (green plants) results in death of animale.
2. There are two forms of tree-
trunk moth ( Biston betularia) in England, one of them has light-pigmented body and wings and the other has dark - pigmented body and wings. A countrywide survey has been carried out to find out the distribution of these two forms in the industrial areas of England where the bark of trees has become greatly darkened by industrial pol- lution and in the non-indus-. rial areas. The result is shown in the following table. Co. of Industrial| Fon-indus- moths.
trial
areas
counted
arees
Total 5392
3468
Light.
2080
31-23
moths
Dark
3312
2045
moths
The predators of moths are birda.
a. What is the possible select-
ing agent in nature that Cetermines the distribution of this two forms in such areas?
b. Which form of the moths is more naturally selected in 1) the industrial areas, and ii) the nonindustrial areas? c. Explain why the above moths are more naturally selected in such areas?
a. The birds.
b. 1) Dark moths,
ii) Light moths.
c. in heavily polluted indus-
trial areas. the dark motha
blend well with the dark
wooda whereas the light
ones can easily be detected
by the predator and be killed.
In non-industrial areas, the
light moths blend well with the light voods whereas the. dark ones are very conspi- cuous to their predator and bo killed.
3.The diagram shows a much
simplified version of the food web in a small fresh- water pond.
Green
Hater fleas
Fish
algae
es
Soluble
death
nutrients
Bacteria
Which component of the food web will provide the great- est amount of energy? Explain your answer.
blow will an increase in
number of fish affects inme- diately the balance of this pond, community?
Write the sequence of the food chain through which fish reta less energy from a cons- tant mass of algae,
d. Name two organisms that you
would expect to be present in greater numbers than the others.
e. Name two organisms that you
would expect to be present in less numbers than the others
Explain briefly the import- ance of bacteria to the food web.
Ans.
a. Green algae.
This is because of that algae. can convert light energy to chemical energy during photo- synthesis for energy supply of the other components in this food chain.
b. Ah Increase in number of
fish will result in a decre- ase of both water fleas and. tadpoles and leaving more algas in the pond.
c. Green algae
Water fleas. Tadpoles →→Fish
a. Green alrae and bacteria.
e. Fish and tadpoles..
Bacteria act as a decomposer in the food web. They break down the eliminated products and dead bodies of plants and animals into simpler con- pounds which can be used by the green algae again.
The following curve shows the population growth of yeast cells in a culture.
solution.
40. of
individuals
ІІ IIN
Time (days)
Addition of paramecium to culture medium Explain the change of population size at phase T JI,II and TV.
Ans,
1) Phase I: The rise of the curve indicates that the. yeast cells reproduce steadily to increase the number of individuals.
ii) Phase IT: The flattened part represents that the birth rate and death rate. balance each other The population, reaches an equilibrium.
iii) Phase III: The yeast popu-
lation drops since the paramecium population begins feeding onit.
iv) Phase IV: When the yeast
population falls so low that the paramecia suffer starva- tion so that less of them survive. With less predators the preys yeast cells) grow quickly and increase in number again.
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