1978-01-05 — Page 27

LITR

頁三第張七第 · 日六廿月一十年巳丁屦夏

WAH KIU YAT PO

報日僑業

豪苔白露長亭。秋茲锺峯和尚说。

TH

三來

年路

三姝媚(和爽啖齮長城自運 》

*E-RSA-EKEISEN RES

錦漪 ̇⊊喑頭碗染,綠畫呵譯。淡雅姿杏,恭帝

四期

日五月一年八七九一公年七十六國民華中

華鹃件主席,齐孤士王,她去年在歐洲各的展出 未能忘。肺朝秋尙好,把酒再被律。 浪淘沙(取陽》

尚會陟召(從山】繞,劉記 朝。思親時

「夢迢迢,眼底欢光撩客堵!無限號→

成

崇

D to C:

育教僑華

ZnCo

Zno + Con

D to E

ZnCO2+2H01-ZnCl2+H2O+CO2

Eto Fi

ZnCl2+2NaOH➡Zn(OH)2+2Nacı

1978

F to Go

中社會考試題當專欄

Zn(OH)2+2NaOH Na Z

Zn(OH)

C to A:

200 + 00

公司

化學

十四 朱宏林

Zn + CO2

Zn(OH)

Chemistry (14)

Solution to 19

(a) X+H2SO4*

X50

2NaOH+H2SO 50+2H20 (b) Amount of acid originally

present is

50

x 0.5 = 0.025 moles 1000 According to the equation 2NaOH+K2SO →→ Na2 50+

04+20,0

amount of excess acid

x no. of moles of NaOH

required for neutra- lisations

16

1000 1.25). moles

-0.01 moles

amount of HSO, used up

by the metal 18

(0.025 - 0.01)moles. -0.015 moles

But, according to the fol- lowing equation

I + H250 → XSO4 + H2 No. of moles of metal X = no. of moles of H250

0.015 moles

0.015 moles of metal X have a mass of 0.975 € mass of 1 mole of metal

20.975

0.0158

=

65

Hence, the relative mass of metal X 18 65

fi

(b) According to the equation X + B250 →→ XSO #2 1 mole of X will liberate 22.4 dn of hydrogen gas at s.t.p.

the vol of H, liberated by 0.015 moles of I is

22.4 1.1000 x 0.015 em3 336 em

Formula mass of XS0=161 Formula mass of Na250 -142

Wt. of

XSO obtained is

4

0.015 moles x 161 g mol

2.415 &

Wt. of Na,504

16

obtained 18

** (1000 x 1.25) x 142

0.01 moles x 142 g mol

1.42

€

total amount of sulphate is

2.4151.42 g 3.835 €

Solution to Q. 20

:(a)

A is zine;

B is zine nitratej

C is zino(II)oxide;

Dis zinc carbonate

E is zino chloride solution;

F is sodium hydroxide;

G is sodium zincate,

(b) A to B:

Zn+4HNO3 →→ Zn(NO3)2+2NO2+2H2O

B to C1

22n(NO3)22

B to Di

Zn(NO3)2+2NaHCO ZnCO+CO2

+2NaNO3+H20

C to G:

Zn0+2NaO+H

A to Ga

Zn+2NaOH+2H2O-Wa Zn(OH) +E2

(c) Gas X 18 nitrogen dioxide,

Its solution is acidic, thus turning the litmus paper red.

(d) White precipitate of zinc

hydroxide is seen which will dissolve in excess ammonia solution due to the formation of a soluble complex ion. ZnC1+2NH →→Zn(OH),+2NH, OH

2

+4NH.

Solution to 0.21

8

Ba(NH) 1

+20H

Gopper, being wore electro- positive than silver, is Dore reactive than silver, thus displacing silver from its salts. The white crystalline solid is silver crystal,

Cu(®)+2Ag*(aq)→→Cu2+(aq)+2Ag The copper(II) ions formed Imparts a pale blue colour to the final solution.

(11) At first, carbon dioxide rem

acts with lime water to fori insoluble calcium carbonate which appears as white preci- pitate.

2+201

However, when an excess carbon dioxide is passed, calcium" hydrogen carbonate is formed which is soluble in water, so the water becomes clear again.

24 Caco (a) +H, O+CO2 Ca +21C

+2HCO3 (iii) As chlorine is more electro-

negative than iodine, so it' can displace iodine from the lodide by oxidation reaction. 01,(8)+21′′ (aq)-201. (aq)+I2 Iodine can alsolve in water. to a small extent to form a brown solution, However, as: more iodine is formed, much of it will not dissolve and settles at the bottom as dark grey solids.

(iv) Silver ion combines with

chloride ion to form silver chloride which appears as white precipitate. Ag* (aq)+01 ̄(aq)——AgC1(1) The precipitate is insoluble in acidic medium but it can. dissolve in excess amonia solution due to formation of a copplegion,

(b)

AECI(a)+2M, ) [AE(N)

Isotopes of an element are atoma having different mum- ber of neutrons, thus having different mass number but the same atomic number. They have similar chemisal properties. (11) The relative atomic mass of

europium is

151 x 1 + 153 x 1 1+ 1 152

(111) (1) It would be a metal

as it can lose two electrona. to form a dipositive lon.

(2) The formulae for its

oxide, chloride and

sulphate are respective-

ly are mo, Eucl, and EuSO

物理十四夔榮家

PHYSICS (14)

Answers to Exercise 7

(a) Lat the correct length.

he --X--

0006

+ 0.00002x30)

Ans: the correct length is

1.0006

(b) The coefficient of cubical

expansion of glass

- 3 x 0,000009

0.000027 e

Capacity of the flask at 100C

■ 50 x (1 • 0.000027×100)

50.135 cm

Let the volume of mercury required be V

V(1 + 0,00018 x 100)=50,135

50.135

1.018:

When ice is added

+0,2 x 4200

101,25-56700 67.50.

5. (a)(1)

(1)

Frequency f1/L (in m

(in Ilz:)

Heat

en up

252

4.20

(97.5 = 10)

300

5.00

73500 43.798 + 87.50

396

6.58

480

€8.00

Heat gained

• 18 x 336000

1 x 4200 × 10

105000 J

Assume no heat lost or gained, to or from the surroundings,

75500+43,75 87.5C 195000

43.758+87.50 - 31500

(2)

From (1)

38 m 1680 + 20 From (2)

2€

·(3)

(11) Since

720

(4)

and C from ($}&{4}

600

60

The slops of fang 1/L is

HE

The velocity ▼ of wave

49.2485 cm3 (Ans.)

(*) Apply the formula

Ang

The specific heat, sapa-

city of the metal sph- ere is 600 Jkg

travels in the wire.

density of water at

54 C

1000

+ 0,0005 x 50

1000

1.025

975.61 kgm

(Ans.)

Coefficient of enhical

expansion of gloss

3 x 0,000008

0.000024 °C

Volume of mercury contained

Capacity of the flask

1000(1 * 0.000024x50)

1001.2

1.0912

Mass of mercury in the -(975.61)x(1.0012×10/ * 0.97678 1g (ana,)

3. Let the specific heat

capacity of the metal sphere

-10 be s Jkg C and the heat capacity of the calori- meter be C J'c»l·

Hout given up by the metal sphere

0.5%(300 97.5) 101.25.

Heat gained

((0.2 × 4200 + C)(97,5–30) 56700 + 67.

Assume no heat gained or lost from or to the surround-

inge,

Бон

300m

shown in the figure above,

!1, P2 are the first and the

· 2nd positiona of the pin respectively.

When the liquid is not placed P, coinside with its image, that is, rays from the ob... Ject P to the mirror are reflected back along the original paths, therefore,

is also the centre of

curtă ture.

e. (The ray diagrams required is the above figure, (11)As shown in the figure;

the refractive index of the

liquid

n

#ini star

MN

MP.

MP2

MP

Since it is viewed from above and the layer of liquid coh- tained is very thin, therefore

MP PP 30 en

The slope →

slope from the

graph

480300

1

* 60

120 ma

the density of

(iii) Let the a

the wire be

the linear

(Ann

(where

A is the cross sectional ar

10-6a

72

120

- 5000(kgm) (Ana,);

If I n

T

- 30

w5-x

5000 10-3

The fundamental frequency

50

2(1)

- 50 Hz

5 x 10

Take the velocity of sound in air 340 mm

340 750

Wave length

6,8m (Aun.)

MI2

Sa

PPZ 30

-

30

6m 24am

1.25 (Ans.}

24