報日僑華
六期星
-10A
to 0.1 mole thus consist- ing of 0.1 atoms.
(11) 2 g of 00 is equivalent
to 1/22 mole thus con- sisting of (1/22) M molecules.
育教僑華
育教僑頁三第張五第 日七十月九年巳丁夏 WAH KIU YAT PO
(iii) (2) liquid can con- duct electricity, becaurs it consists of mobile ions in liquid state while : coniste of covalent mole- cules only..
學能推理練習專欄
(a) and 3 have similar openi
cal properties both have
(111).
WENDWA
one electron in the valenc shell.
文字推理練習十三
選出下列有题最適當的答案,並在它的下面
ALIMBEN. 4 GEURES
PRAHA
GAA CHA
R6 WATTEN F
•HÆGHANI SE DE PES
***BÆNK. MELTHERMAL KAH 的露,往往能的雾纪量的終身 我細括父避術
AMSE
E#*
***MAGPAK
CRAA
Anik
ELFGA
COCHET HE54. SE
EMALE 29 mg 19 -
--
*****K ETA MA
***ME IS 19°
以前
的工作。多年来依然凝段:瀟瀟碟綠地過日學
AST
B总线
C&#
DRE
作者認為彼得是
ASTRUK
CHW**
·EPA
同學們認為作者没有變,是指他的
A装纸
CŁA
年齢
上大裁寫的小说
A黎明
ERER
[]衣飾
K L M X J A PLAT
到鬥小失大
(e) Laso of F=17+17=34 alu ass of P=17+18=35 anu
lass of an 'averaze ato of 2 is
34x404 75x50
*** (40+ 60).
54.
the relative atolic
Lass of clouent is
182 of an averne aton
xas of an a
Solution to 0.5
(a) (1) X(0)2-2HC1→→XC12-2H,0
Nað" + ÏÇı➡NaCI÷H2O (11)
The amount of HCL origin- ally present is
50 1000
x5 moles
0.25 moles
According to the following equation
NaOH + HCl -> fac +H, O
no. or moles of IC1 present in 25 om diluted soln.
no. of moles of NaOH re-
quired for neutralization
.20
1 moles
1000
0.02: moles
total amount of excess
HC1 in the mixture is.
0.02 moles x
250 25
= 0.2 moles
Hence, the amount of HC1
used up by ammonium sulp- hate is
0.25 moles – 0.2 moles 0.05 moles
But, according to the
following equation
X(011) 42101XC12+2}2O
12
no. of moles of X(OH),
no. of moles of HCI used:
2
= 0.05 moles
0.025 Boles
·(2).
Let the relative atomic mass of X be y, then the formula mass of X(OH), is (y + 34).
Hence, no, of moles of the hydroxide in 3.05 g solid.
3,05
moles
(y+34)
·(2)
Equating (1) and (2), we
have 3.05
(+34)
34
=0.025
3.05 0.025
.122
88
Therefore, the relative atomic mass of X is 88.
Gas X is hydrogen and liquid is water. (ii) Zn 2 n
1978
Solution
中學會考試題預習專欄
(a)
HOLERAETE ARA
[(1)
Dole Pic
化學(四)
朱宏林
Chemistry
Solution to Q.4
(a)
(i) the element C consiste of
mozatomic/molecules.
(11) the element is a halo-
rün.
(111)he element wil re-
act most readily visi chlorine.
(17) The element B is the
(0)
strongest reducing erent and the elements ahorn, heel ent, will form
an ion of C by losing is two valency electrons.
(1) The foule of the hydride
(ii)
e Toril of the olde of 13 Co.
(i) the electronic structure
02 X 16
The electronic structure
H
(11) the bonding in is ioic
G. électrovalent an. tit in covalent.
(111) Add a little liquid v
to anhydrous conper(11) sulphate or cobalt(IT) chloride, the former will trn from white to blue while the latter.. will
turn from blue to pink if the liquid Y is water.
(iv). Mg or Pe can be used as
3
(v).
they are above hydropen in the Activity Series. Fb, though being above hydrogen in the Activity Series, is not used in this experiment because the lead (II)chloride. formed in the reaction is insoluble in water thus stopping the reac- tion.
Redox reaction was taken place in the combustion. tube.
H2
Cuc H20+ Gu (vi) Only Fbo can be reduced
by hydrogen.. (vii) Cone HASO, is used to dry the hydrogen gas. (viii)From the equation below.
#2 Cu +
H2O
Cho+
79.5 (or 1 mole) of Cun gives 63.5 (or 1 mole) of copper metal. ..mase of copper obtained
on reducing 7.95 g Cuo
7.95 & X 79.5 63.5
6.35 €
(b) 2 g of hydrogen is equiva-
lent to 1 mole which now consists of N molecules.
·(1)..
2 of neon is equivalent
6 of bromine is equiv alent to 1/40 mole thus consisting of (1/40) H molecules,
2. (iv) 2 g of 00, is equival-
ent to 1/30 mole thus consistir og 1/30 dons. (b) Fotassium chloride is com-
posed of K and C1 ions,
in the solid state, but, these ions are held strong- ly together by electrosta- tic attraction, so they are not mobile nor they allow electrons to flow through. However, in the aqueous solution, the ions are fres and mobile, so they can now conduct electricity. According to the following ionic equations:-
(c):
11 and
Ma
+ 2e Mg/ it is seen that i 1 mole of electrons (or 1 faraday) of electricity) are passed through, 1 mole or lithium metal but only hole or agresiui metal will be deposited. So, when the
te amount of electricity
in pood, the relative
sces of Li to lig will be 7(x-24).
12
物理(四) 魯榮家
PHYSICS (4)
ons in Exerci
10 ms
1,(a)(1)
hown in the force diagram above, take mument about A.
where
TOGE(AR)
is the horizontal
e required.
force
03 - 2 x 3 (m
OR MA
I'm
UB
24 -28990
F(1) = -100 (10)(4,899)
Fm 4299N (ans)
(ii) The minumin föree F
min should ingent to the wheel as shown in the diagram, be low
1009
tangent
Fmin
to the whe at print, & which is one of Mia extremes of diameter AUC
of the ecrcular wheel Taking moment about. A.
/Fmin(40) = 1002(XB)
mi, (19) = 1000(4,899)
489,98 (Ans.).
(b)(i)If, the force is applied
to the 10-kg block, then
10 kg 5kg
min
10kg
where H, is the force bêt- ween the blocks.
the reaction of 5-kg block an 10-kg: block;
Let the acceleration of the system he a mg-2:-
Y
F R
5a
(2
אל1
alno, F. solving from -(1) and (2)
we have
- • 5N (Ana.) (ii). Similarly, let the force
between the blocks be g.
and the acceleration of the system ba ́a”
15
ms
R
58
From which
2
日九廿月十年七七九一屦公年六十六國民華中
10N (ARN)
Note the readers should. understand the physical meaning that the values of the forces between: the blocks are different but the values of the accelera 1100 of the system are the same in. (i) and (ii), (i RB, but a
2. (a)
120kg.
As shown in the force diag-
ram above
frictional force exerted on 120kg block.
➡ frictional force exerted
on 150 kg block. Since sing – 0,6
e9s0 • 1- sin"8"
058 Henge, a apply Newton's 2nd Taw of mitïon:
120gsinë -f- T - 120a
120(10)(0,6)–(0,2)(120x)(0,8)
- T120a.
528 - T - 120a
150gs
1508
-+(150)(10)(0,5): -(0,5)(150)(10)(0,8)= 150a
T-300-
+ (2)
150a... (1)
228270a
-0.84.
the acceleration is 0.84-ms-2 (Ans,}
(ii) Substitute inte (1)
3.56N (Aus)
the tension in the cord. is 3.56N.
(1) The force diagram is.
shown be low-
where
A
Normál reaction acting on
the upper end from the
smooth wall.
2 the resultant force act ing on the lower end the ladder..
the vertical component
➡ the horizontal component
of Il
frictional force de tween. the ladder and the floor. weight of the Indifer. (11) Since the ladder i
in equilibrium, therefors,
(2)
Taking moment about B
W x 3
SW
x80
N
-0
From (3)
(3)
An's
3 x 160(10)
600N (Ans,)
The direction of the force acting on the upper end of the ladder is perpendicular to the vertical wall with magnitude 600N.
(iii) The frictional force t
is horizontal
From (1)
600N (Ana.)
(iv) From (2)
- V 1600N N2 The coefficient of sliding friction between the ladder and the horizontal floor
f
600
0.375 (Ans.) 1600
(7) The resultant R
2
+ Ng
= √600+ 1600
Also,
1708,8N
tane
(Ans.)
0.375
N2 -20°33'
g -
Ans: The resultant at the
end of the ladder in
1708,8N making 20o33? with the vertical,
v and ▼ be 4.(a)(i) Let
the velocities of the block A after the bullet emerges and the bullet as it emer- ges from block A respect- ively.
By conservation of energy 2 (2) 2 (2)(g)(0,2)
(Ang.)
(ii) By conservation of linear momentum
The initial momentum
(0,01)(2400)
- 24 kg ms-1. The final momentum
(2)▼ + (0,01)▼1 (2) (2) + 0.01×3
+0,01 v
24 - 4 + 0.01 1
2000 ms-1 (Ans)
Let be the velocity of the blockB and the bullet after the bullet hit the block B.
Since the bullet is embedded in block B, hence, by cons servation of momentum
(0,01)(2000)-(0.01+0.99)▼
y 20 mg-
By conservation of energy
}(0.01+0,99)(20)2 (0.01+0.99)gh
Take g 10 (
h- 20m (Ans.)
5. (1)As shown in the force
diagram below
The bouyant force B will apply at the mid-point of the part of the rod submerg- ed in water.
(11) W - weight of the rod
1.2g - 12N
weight at the end of the rod to be determined,
The volume of the rod.
2.43
(density of
water is.
a kg m
Volure of red submerged in water
2:
5
-
The bouyant force
weight of water displaced by the rod
density of water x volume of the rod submerged x g.
2.
- d x - x 10 - 20N (Ans,)
ང་
(111) Taking moment about-
the hingeờ
(Bsin0)(2,5) «(Wsin0)(3)
+(wwino)(6)
20x2,5 - 12x3 + 6w
= 2,333 N (Ans. (I)The downward force pro-
duced by the 120kg block
120gsing – fi 120gsine 120g(sing 120g(0,6 -120g(0.44).
528N (Ans.)
(0,2)(120gcose)
0.2cose 0.2x0.8)
(ii) The downward force pro-
duced by the 150kg block - 150gsinë – to
- 150gaine -u(150gcon8)
150g(sine
ucose) 1500(0,6 0.8u)N (Ans.) (b) The blocks will slide
down with the cord always stretched if the dawnward acceleration produced by the 120kg-block is great- er than or equal to that of the 150kg - block,
The minimum value of
Mis
528
1500(0.6 -0.8u)
150
From which
+
u 0,2 (Ana,) (c) If `u× 0.5 > 0.2 then the blocks will slide down
with the cord always stretch- ed. Let the tension in the cord be T and the accelera- tion of the system he
・a ms
120k
8.
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