1977-11-11 — Page 27

育教偶 南三第張七第日一初月十年巳丁墓

fi & IG *

化學

1978

中學會考試題預習專欄

Chemistry (6)

Solution to 0.7

またこ

朱宏林

Gas G is sulphur dioxide. Cu+211,50 CuSQ4+2H20+SC

4.4

(c) sore of the black lead(17)

oxide have turned white.. The substance formed is the white powder of lead(II) sulphate.

(a) FbQ + $02 →→→ Pb304 (0)

(e) It is because sulphur dio-

xide is an acidic gas so it

can be neutralised by the sodium hydroxide. (*) It is because concentrated

sulphuric acid is a much stronger oxidising agent when it is hot; so, by heating, it can oxidize copper more readily. (g) It is the anhydrous copper

(11)sulphate.

(h) (1) Fo because the lead

(II)sulphate formed, being insoluble in water, forms an over- coat on the unreacted leadmetal, thus stop- ping the reaction. (11) Yes; the equation is

Zn + 2H2S04

ZnS04+2H20+502

(1) A blue solution is obtain ed due to the presence of the hydrated copper(II) Long.

Solution to Q.8-

Na CO2+2HCT

+HC1→→ Na C1+H2O+CO,

(11)Tet y g be the weight of

the anhydrous Bodium carbo- nate in the 3.80 g mixture. Then, the wieght of sodium hydrogen carbonate in the mixture is (3,80-7) 6.

*- formule mass of sodium

carbonate is 106

nos of moles of

moles

Na2CO3

Similarly, the no. of moles of sodium hydrogen carbon- ate is (3.80 y moles

84

Now, according to the fol- lowing eqaution

a2CO2+2HC1-2NaC1+H2O+C©2 no. of moles of HCl requi- red to react with Ba, CO

- 2 x no. of moles of Na,Coz

(2 x 106) moles

moles

And, according to the fol-

lowing equation

NaHCO3+ +HC NaC1+H2O+ CO2

no. of moles of HC need-

ed to react with MaHCO no. of moles of NaHCO

3,99.-2

84

moles

the total amount of HO required to react with the mixture completely

3.80- is (+ 300

moles :84 But, the actual amount or KCL used

sed up is

50

1000

x 1.2 moles

0.06 moles

Hence, we have the follow- ing equality:-

3.80.-

+

2004. 64y+201.4-53y 31y= 65.72

2.12

Hence

2.12

3.60 and

of

=0.06 267.12

Yan CO2 is

100 -=55.8.

of FallCoz is

2.12)

* 100

Copper is less electropos itive than sine but more electropositive than silver (as indicated by their re-. lative positions in the Activity Series), so it can- not displace zino out from zine nitrate but can dis- place silver from silver nitrate solution. The silver netal displace out appears as silvery-grey, needle-like Cyrstals.

Cu + 2Ag

Cu 2+ + 248

WAH "KIU YAT "PO

(11)The hydronium ions in a sol- ution are responsible for all the common acidic pro- perties. In either pure wa ter or dry sulphur trioxide crystals, there is no hydro- nium ions so the dry litmus paper does not turn red. However, on mixing the two, sulphur trioxide. reacts with water to form hydronium ions which thus turning the lit-: mus paper red.

SO2 + 3H20 -

Solution to 0.9

2130 + 50

2

(a) The gas X 18 hydrogen chlo-

ride.

(b) Na©1(a)+H2SO

NÁH$O+HC1

(c) The bond present is covalent

bond.

Conduction of electricity through liquids or solutions. depends on the mobile ions. In pure water there is no or little ions, so the cur- rent is not flowing. (ii) Then hydrogen chloride is dissolved in water, it rem acte with water to form H0 and C1 ions which are mobile, so a current is now flowing thus lighting up the electric bulb. As the concentration of ions in-. creases, the bulb becomes brighter and brighter.

+ Ca HC1 + 120 —☀ (iii) The gases observed around

the electrode £ is oxygen and that, around electrode I is hydrogen.guit (iv) At the electrode L

4011-

20+02+4€TM

At the electrode is 21+ + 2e7 → lly

(v)ccording to the equations-

00

(iv), 1 hole of elect rons will liberate (1/4)

ole of oxygen gas or (1/2) mole of hydrogen gas. So, the ratio of volumes of O2 to liberated by the

2 same anon

of electricity

is

White precipitates are seen

in both test-tubes, In tube A H2O + SO2

-2+

Ba + SO

In tube B

H20 +

24

2H+

SOA

Baso (s)

2H +

2-

BaS0, (8)

Be + 503.

(11) The white precipitates in

the tube A remain but that intube B dissolve on heat-

ing because they can react

with the hydrochloric acid.

2+

BaSO2->Ba2++SO2+H2©

經濟與公共事務(六)

孔繁盛

Economic and Public Affairs

(6)

Describe the trade restric- tions often imposed by countries on international trade. What are the diffi- culties Hong Kong encount- ers in her exporting trade?

To encourage development of domestic industry and to protect existing industry, a government may establish tariffe, quotas, boycotts, exchange controls and other barriers against imported goods. The reasons for such barriers may be econo- mic and political, and they are usually welcomed by local industry.

1. Tariffs: A tariff is a tax imposed by a govern- ment on imported goods for the porpose of raising revenue or protecting home industries from the compe tition of foreign-produced goods. Tariff rates are b based on value or quantity, or a combination of both.

2. Quotas: Countries may also impose limitations on the quantity of certain commodities imported during a specifio period of time. These quotas may be applied to imports from certain

郭日僑華

countries or from all fore- ign countries in general, Quotas are set for a varie ty of reasons, the most im- portant one being to pro- tect domestic industry and to conserve foreign exchan ge.

3. Exchange barriers: In order to conserve acarce foreign exchange and impro- ve balance of payment diff- iculties, a country may impose restrictions on the amount of their currency. which will be exchanged for the currency of another country. In effect, it rations the amount of curr- ency available to pay for imports. Exchange controls. can be applied to all comm- odities, or a country may employ a system of multiple exchange rates based on the type of import,

ure.

4. Other restrictions: Many other kinds of restri- ctions are also imposed on imports, such as regulations affecting the importation of harmful products, drugs and medicine, and literat-

Products must also: comply with all government standards set for health, sanitation, packaging, and- labelling.. Failure to com- ply with these regulations can result in severe fines and penalties. Requirement and regulations vary with w each country

Hong Kong encounters the following difficulties in her exporting trade:

1. There are increasing number of countries impos- ing quotas on her major export, textiles. This will inevitably limit the growth of Hong Kong's ex- porting trade.

2. Hong Kong have been facing keen competition from neighbouring countries such as South Korea, Taiwan, Singapore, and Philipines.

e. Local rising costs of production in the past few years inevitably leads to a general rise of prices of exports, which weakens the competitiveness of her exp-

orts.

f. Fluctuation of exchan ge rates in the world adds difficulty to the estimat-

ion of her exports' costs and revenues,

物理(六)

PHYSICS (6)

tational acceleration g

五期星 日一十月一十年七七九一座公年六十六國民華中

Since the blocks are initially

Hence

at rest

T.

Apply

TO

Hence the reading of the spring balance will then be 0 kgf.

2(a) The 10-kg block will

not fall if the friction- al force f between the block and the trolley is equal to the weight of the block,

0,48 where it is the normal reaction exerted

0.48-

the block

- weight of the block -100N

In order to obtain the norm- al reaction, the trolley has to accelerate to the right,

Let the acceleration be

R10a...

(b)

0,4(108) » 100

fig(a))

In figure

a 25 ma

Ans.

fig(c)

T - tension in the string

F2T

F

In figure (b).

(1)

➡ reaction of the floor

on block A.

weight of block A

= 800N

the block

cannot be lifted up. If the block is lifted up NO

Similarly, in figure (c)-

of the floor on the block B.

weight of the blockB - 48g - 48ON F➡ 480N

†(480)=240N

For block A

- 800-240 -560N

■ 0 (Ans,)

For block B;

480 240-240N = 0 (Ans.)

金榮家

(11) If F - BOON

T

or

- 400N Look A

Answers to Exercise three

Take

the upward

direction -

as positive.

Let the tension

in the spring of

the spring bilance be T, the neceleration of the elevator

be.

ea and the mass of the block be ma

weight of the block

W:

- та

(1)

(a) When a = 2.5

10kgf 10 x10N-100N Substitute

e into (1)

100 - μ(10)

(2.5)

100% - 8kg weight of the block

- 8 x 10 = 8ON: (Ans,) (b) If T - 6,4kgf 64N

Substitute into (1)

64 8(10) ► Sa

--208-

The elevator moves downward with acceler ation 2 maTM.(vr the elevator has an upward deceleration 2ma-2)

(c) If the elevator moves

upward with uniform speed, then 'a=0

From (1)

T T-

- 80N

-

Hence

8kgf (Ans.) (d) If the elevator moves

'downwards with uniform

speed, also, a=ü

T = 8kgf (Ans,) (e) If the cable wire breaks,

then the elevator will

fall down with accelera- tion equal to the gravi-

-800-400-400N

Still the block cannot be lifted.

1-0 (Ans.)

or block B, similarly,

480-400 - 80N

(111) If F

#2 - 0 (Ans.)

- 1440N

T 720N

For block A

- 800-720-80N

0 (Ans.)

For block B.

0 for the

NB tension T is greater than the weight of bloak no that the block will lose contact with the floor. Now the net force on the black Bis

-W.

720-480

- 240N

4882

5 ms

(Ans.)

2014

(a) Let the tension in the

cord be T and the accel- eration of the system be a

20gsine

·T 20

(1) + (2)

-

(T)**

T = 20a

20...

..(2)

20 - 20a

20gaine 20(10)(0,6)-20

- 20a -2

-80m

Jat2

time required

to travel 80m

- †(2.5)t2

- Bsec (Ans,)

(b) Substitute a 2.5 into

(2)

↑ – 20 × 20(2,5) T = 70N (Ans,)

4.(a) The centre of gravity

of the system is the centroid (intersection of medians) of the triangle formed by join- ing the centres ofA, B and C

(b)The method used in (a):

not applicable because the masses of the balls are not the same, Hence,

since the mass of B is the same as that of C therefore, the cag. of B. and C is at the mid-point D of BC.

Also, the mass of A is twice as that of B, in other words, the mass of A is equal to the sum of masses of B and C,

Therefore, the

the whole system is the mid-point of AD.

(a)he uniform lemina can

be divided into 3 portions. Each of them having equal mása. The cog. of the whole system can be obtained by the same method as used in. (a). That is, the cog. of the whole lenina can be obtained by 'finding the centroid of the triangle formed by joining the centres of gravity (G1, Grand G2) of the partions,

5(a) Let the distance between

the knife edge to the end hanging the metal block be x In figure 7a, by taking moment about the knife edge

Mg(0.4) ■ ******

■ Wx,... (1) where w is weight of the metal block in air.

In figure 7b, by taking moment "about the knife edge

Mg(0,35) - W** ***** (2)

where W is the weight

of metal block in water,

8

W!=

0.32-3

곱게

he relative density of the metal

(3)

Figure 7(e), by taking moment about the knife edge

Mg(0.32) ➡**.... where. W is the weight of the metal block in the liquid. (3) + (1)

WHOM

The relati

density of

the liquid

W - 0.8W

W

W

1.6 (Ans.)