1978-02-17 — Page 25

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

育教僑華

=

B

& Deze Preis:

化學

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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