育教僑華頁一第張七第日九十月十年巳丁磨复 WAH KIU YAT PO
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擬試卷猁枪各一坐,薄刻行各時間爲四十五分鐘,內容及份量均携仿 NVERSAZNANDING CAKE ,增加對參加食盐孕对能力测劃的例心 - 特根某本月廿二日教署公佈, 日版 協助小學六年极考生在武林柏有-
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*HEERKREE+R+BATI 並送孔,日於本月(廿二日)分發一杯水力商村镇髻录演本一份,使 行致育局爲使應小六學生對行持行之爭視谋,有進一步 第一年只能力滚, - 童年在十二月六日 【星期三 } 下午二陵.
定本月三十日刊出
學能推理模擬試卷
正式发行之船民,上述兩科
期三季刊出 2
JAIZERENKE) VZELIN
本月三十日刊出之đ接就迷對動,仍由來向爲本版「升中民事
水枪,定期在本月三十日(E
二期星
a 三第三名:蕙理上午校
日九廿月一十年七七九一腊公年六十六國民中
29
初小月英诗集拼第一名錦田公會小學。第二名:公立
,
£ • DIN--JOERE WE IN 373–1
香港學校朗誦節特刊
育敦僑華
1978
中學會考試題預習專機|
明德社主講
化
學 (八) 朱宏林
Chemistry (8).
Solution to' 0.10
G is aniondej
1 is water;
X is nitrogen,
(b) (NH)2SO + Ca(OH)2
4
CaSO4 + 2NH ̧ + 2H 0
(c) Quickline(calcium oxide) can
be used to dry gas G.
(a) Reddish-brown copper powder
is obtained.
(e) (1)
Formula mass of (NH4)2SO4
is 132.
1.32 g of ammonium sulphate is equivalent to 0.01 moles. According to the equation in (b), 1 mole of ammonium gulp hate gives (2 x 22.4) am of ammonia (gas G) at s.t.p.
is
volume of gas G produced by 0.01 moles of (NH),50
2 x 22.4 dm x 0.01
dm3
0.448 dm
448 cm3
(ii).
According to the
low:-
quation be-
2NH2+300-N2+3H0+3Cu
2 vols.
1 vol.
.vol. of nitrogen (gas X)
2x val. of ammonia
x448 cm3
224 cm
(f) The solid is magnesium nitride.
3Mg +
14832
(g) The gas is ammonia.
MgzN2 + 6H20-3Mg(OH)2+2NH,
Solution to §.12
The yield will decrease; It.
is because a rise in temper- ature favours the backward reaction which is endothermic thus the equilibrium is shifted to the left.
(ii) The yield will decrease; it is because when the pressure is lowered, the backward re- action is favoured as the
reactants occupy a large vol- ume; so the equilibrium
18 shifted to the left.
(iii) The yield remains unchanged;
it is because a catalyst only speeds up the attainment of
the equilibrium but does not shift the equilibrium in any direction.
(iv) The yield will increase; it
is because the equilibrium is shifted to the right when the concentration of oxygen is increased.
8
A dark-blue solution is seen. Chlorine oxidises the iodide ions to iodine which then at once combines with the starch to form a dark-blue complex,
012 + 21 →→ I1⁄2 ★ 201 (ii) White precipitate is seen
when the first few drops of sodium hydroxide are added, The precipitate is zinc hy- droxide.
• Zn(OH)2(s)
2++ 201 Zn
However, when excess sodium hydroxide is added, the pre- cipitate is dissolved due to
the formation of a complex
salt, sodium zincate.
Zn(OH)2+2NaOH →→→ Ne, Zn(OH)
(c) P is ethanoic acid (or acetic
acid) and its structural for mula is
CHY LOH
Mg +20E COOH (CH_COO), Mg+H2
CH COOH+PC1CH COC1+POC13
HCL
The gas observed in (1) is... hydrogen while the white fumes
in (ii) is moisted hydrogen chloride gas.
Solution to Q.' 11
(a) Pb(NO3)2 + K2CQz
Pb00 (8)+2KNO
(b) Formula mass of lead(II):
nitrate=331.
..3.31 g of lead(II)nitrate
is equivalent to 251 = 0.01
moles.
According to the following equation
Pb(NO3)2+K2 CO->Phco,
no. of moles of PbC0z formed = no, of moles of Pb(NO3)2
= 0.01 moles
•*formula mass of PbC0z=307
wt. of the precipitate = 0,01 moles x 307 g/mole.
2.07 €
(c) Furthermore, according to the
equation in (a),
no. of moles of K,CO, used
up in forming the ppt. no. of moles of Pb(NO3)2 0.01 moles
But, the original amount of
KC
50
1000
present
x 0.25 moles
0.0125 moles
amount of K2CO
in excess
(0.0125 - 0.01) moles 0.0025 moles
Let y cn be the volume of
0.5M HNO required to react with excess KCO3.
Then, according to the equat-
ion below
CO+2HNO
2KNO+H
no, of moles of
NO+H2O+CO2 needed
HINO
2 x no. of moles of excess
potassium carbonate
2 x 0.0025 moles
=0.005 moles
1000 x 0.5 moles
0.05 x 1000
100
0.5
Therefore, the volume of 0.5M nitric acid required to reagt
with excess KCO, is 100 em
物
理(八) 魯榮家
PHYSICS (8)
Answers to Exercise 4
1. (1) Let the tension. in
X be T and the tension in T be T2
When a load of mass 1 kg is added to X alone, the stretching force is LON, the extension of X is. 60 50 10cm. When the metal FQ is added, the extension of X is 55-50
5cm
=
= SN (Ans.)
Then a load of mass 1 kg
is added to Y alone the stretching force is 1ON, the extension in Y is 75 - 50 = 25cm. When PQ is added, the extension is 55 50 5cm
=
(Ans.)
22N
(11) Let W be the weight of the metal
W = 1 + 2
= 7N (Ans.)
(111) Let the distance of the centre of gravity of
from P be (cm).
By taking moment about F, we have
7x= 2(35)
x = 10(cm)
Ans: The distance of the
centre of gravity of
P from P is 10 cm.
2. (1) Take the direction
toward the right as +ve.
Total momentum, before impact
= 2(8) + 10(-1)
= 6 kgms-l
Total momentum aftes Impact
= (2+10)v = 12v
where v is the common
velocity of the box of
sand and the sphere after
impact.
By conservation of moment-
um,
21
The
required raulo,
- 1 : 2 (Ans.)
(11) Extension of the
spring in figure 1
0.23m-0.2m = 0.03m tension of the spring in figure 2
= 0.360 - 0.2m 0.13m
(Tension in figure 1)
(Tension in figure 2)
= 0.03: 0.13
313 (Ans.) (lil) Let T2 and T2 be the tensions in springs in (figure 1 and figure 2
ruspectively. U,, and un be the unthrusts on Can figurel and figure 2 respectively.
There 7 is the weight of
the cube C.
Fron part (1) uy From part (11) 1 =
Putting
these into (1)
(3)
solving for
2 from (2), ()
142 = 201
The relative density of
30
0.35 (Ans.)
4a)Total weight of oil
(Ans.) A
12v 6
+0.5 ms (ii) Total kinetic ene gy
Before impact
=
*(2)(8)2 + 1(10)(−1)2;
69J
Total kinetic energy after impact
= 3(2 + 10)(0.5)2
1.5J
Since there is no change
in potential energy before.
and after the impact,
mass of oil x 6
- density of oil
x volume of of:
= (0.75 x 103) x 6x104
4.5 (Ane.):
(b) Reading of the balance
weight of oil
weight of cylinder
≈ 4.5 + 0.1N 4.6N (Ans.).
(c) Total thrust
(pressure at bottom)
xbaue area)
= (density of oil)x (6) (depth of oil)x(base arca)
(0.75 x 103)x(10)x(0.1) x(0.01)
-
therefore the energy loss
in the collision is
697 -
1.50
7.5 (ans.)
(a)
= 67.5J (Ans.)
(iii) The energy lost in this im act will be diss-
ipated mainly in the form of heat and little in the
form of sound.
3(1) Let V volume of the
cube.C.
d = density of water Upthrust
weight of water displaced Upthrust in figure 1 =(v)ag
Upthrust in figure 2
ΣΕ
As shom in the figure, since the pressure acts normally to the surface in contact. The horizon- tal components will can- cel each other and the rosultant thrust vil act vortically upwards.
(e) The resultant thrust
(Total thrust)-(reaction
on flask)
= result of (c)
- result of (b) =7.5N
-
4.6N
= 2.9N (Ans,)
(f) Let a be the relative
density of the wooden cylinder and A be the cross-sectional area of the cylinder.
By law of floatation, weight of cylinder
- weight of oil displaced
(d x10) x0.06A x 6 =0.04A x(0.75x 103):
a = 0.5 (Ans.) (s) No, the same volume oil will displaced in
both cases, so that the liquid level will be the sane. Hence same pressure at the bottom and there- fore same thrust exert on the botton.
5(a) It shows that
(hair is made up or small
particles.
•
2. These senll particles are in constant ire- gular motion and never stop.
(11) When temperature rises,
the smoke particles nove nore vigorously. It is. because the air moleculos
more vigorously with rise. in temperature. They cause the smoke particles to
nove more vigorously when making collisions with them.
(b)(1) Volume of oleic acid in 0.03 cm3 of the solution
= 0.03% 1000
= 3 x 10-5cm3
Area of the patch of
oleic acid formed
(3)(10) 2
= 300cm2
The aproximate length of
a molecule
= thickness of the patch
=
10-5
300
107cm
10m (Ans.).
(11) The assumptions are (1) the oleic acid forms
mono-molecular layer on the water surface. (11) The oleic molecules
occupy the volume of the whole patch.
(3) Oleic acid is complete-
ly dissolved in methanol and the solution so form- ed is uniform
(4) Oleic acid does not. dis olved in water and does not evaporate.
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