1977-02-14 — Page 27

and finally to ethanoic acid- which gives a sour smell of the vinegar.

CHCH2OH CH

CH,CHOCH COOH

育僑華

育教僑華頁三第張七第:日七廿月二十年辰丙麿夏 WAH KIU YAT PO

(x-1)+iy . * )x+1)2+iy

報日僑

4

(x−1 )2+y2+(x+1)2+y2

= 4

(x−1)2+y2 = 4- (x+1)2+y2

1977中學會考試題預習專欄

明德社主編

附加數學

(十四)

2x + 1

岑俊彦·

Additional Mathematics(14)

Solation to Exercise.9

1. a

1

cose - isine

cose isine

cose-icine) (cose+isine)

coseisino.

2

cos Orsino

cose + isine.

b)+coso-sinë

Ans.

(1+cos@)+ising

1+cose+isino.

(1+coso)2+sin2o

1+coso+isine

2+2cose

1+(2cos2-1)+isin2 cos

2+212cos

2cos(costising)

cos ising

+

cos

0-

2(1+itan=) = 1 tan

2.

Let z

5+41 5+4

25+401-16

Ans.

Squaring both sides,

(x−1)2+y2 =16+(x+1)2+y2-

8

(x+1)2+y2

2

16+x2+2x+1+y2−8 (x+1)2+y2

-4x-16.

-8: (x+1)2+y2

x+ 4 = 2. (x+1)2+y2

Squaring both sides again,

བ ་

(x+4) 2

·4 (x+1)2+y2.

x2+8x+16 = 3x2+42 = 12

4x2+8x+4+4y2

the locus of z is an ellipse whose equation is 3x2+4y2

2

12

b) Put z = x + iy

x-i

z+i

x2+(y+1} i

Ans.

2

x2+(y-1)i

x2+(y+1)2 - x2+(y-1) 2.

+(y+1) 2:

= −1 + x2+(y-1)2 Squaring both sides, x2+(y+1)2

y2 + 2y + 1

= 1+x2+y2=2y+1

4y - 1.

x2+(y-1)2

Squaring both sides again, (4y-1)2 = 4 x2+(y-1) 2

· 16y2-8y+1 = 4x2+4y2-8y+4 12y2 4x2 = 3

the locus of z is a

hyperbola whose equation

4x

25 + 16

401

is 12y?

41 + 4 1

Now, - - +2

1. (cos120°+isin120o)=cis120o

and 1

13 2

1.(cos2400+isin240o }=cis240°

cis120° Cis240°

= cis(120-240°)

= cis(-120°)

1* cos(-120°)+isin(−120°),

Z

1 and arg z=-120°

2. b) Let z=

2. c)

Ans.

(i+1)(2+i)

3-1

(1+1) (2+1) ( 3+1)

(3-1) (3+1)

(1+31)(3+i)

9+1.

3+101-3.

1:0

· (cos90o+isin90°)

1 and arg z = 90°

Ans.

Let z

cos0+isino cos-ising

(cose+isine)(coserisine)

(coserisin☺)(cos +ising) (cos-sini) cosuri ing)

cos(0+0)+isin(0+0)

2

cos prsin2p

cos(+)+isin(0+þ) ·

z= 1 and arg z =

Ans.

3.2 3-2

z-61

онф

Let z=x+iy, we have

2 x+iy-2

=

x+iy-61

2 (x-2)+iy x+i(y-6) 2' (x-2)2+y2= x2+(y−6)?

Squaring both sides,

4 (x-2)2+ y2

=

x2+(y−6)2 ·

4x2-16x+16+4y2.

2

2

+ y2- 12ỷ + 36 3x2+3y2-16x+12y-20 = 0

Which is an equation of a Circle with centre (-2) the complex anumber represented by its centre

21

Ans

3

4. a) Let z =x+iy, we have

= 4

2-1

+

2+1

化學(十四)

Answer to 0.22

3

Ans.

c) A colourless gas with chara-

cteristic pungènt smell is

produced and it can turn the

moisted red litmus paper blue.

So, the gas must be ammonia. 2NH C1(E)

+ Ca(OH)2(s):

|_ CaCl2(8)+2NH ̧(g)+2H20(1)

When the mixture is heated, ammotum chloride sublimes off as white fumes, Ammonium chloride dissociates on heat- ing to ammonia gas and hydro-. gen chloride gas which reform the ammonium chloride on cool- ing. Sodium chloride has no reaction.

NH_C1(s)

4

Answer to 0.23

NH,(s) + HCl(6)

a) P is potassium chlorate, 2 is

oxygen, R is potassium chloride, and S is silver chloride,

b) 2KC103 2KC1 + 302

Ag (aq) + Cl(aq)➡ AgCl (8)

Silver chloride can be disso- lved in aqueous ammonia as it can react with ammonia to færm a soluble complex compound.

AgCl (a)+2NHA(K)2]*ci"

soluble

e) No. of moles Q (1.e.oxygen)

- moles

* 72 cod 24000 cm

-0.003 moles

Let y be the formula mass of P, then, no. of moles of F in 0.245 g P 18 (0.245/y) moles. But, according to the follow- ing equation,

2KC103

- 2KC1 + 302

no of moles of O

no of moles of P

0.003Y

朱宏林·

0.245.

On heating, only potassium

hydrogen carbonate undergoes.

thermal decomposition,

2003 + H2O +

No. of moles of CO2 evolved

= 0.005 moles

120 24000 Formula mass of potassium hydrogen carbonate Is, (39+1+12 + 3x16) = 100 According to the equation in (a), 2 moles of KHCO give 1 mole of CO2

Amount of KHCO to liberate 0.005 moles of CO, is

0.005 x 2 moles

0.01 moles.

= 0/01 x 100 g

c) Both potassium hydrogen carbo-

nate and potassium carbonate can react with dilute HCL,

KHCO HGÌ –

KG1+H2O+CO2

K2 CO2+2801 - 2KC1+H2O+CO2

d) No. of moles of CO2 given off

360

24000

0.015 moles

According to the 1st equation in (c), 1 mole of 1

1 mole of

of KHCC, Gives

amount of Co, liberated by 0.01 moles of KHCO in the mixture is 0.01 moles the amount of CC, liberated by potassium carbonate is, 0.015 0.01 = 0.005 moles But, according to the 2nd equat- ion in (c), 1 mole of K,00, gives 1 mole of carbon dioxide; ..amount of K,CO2 to liberate

0.005 moles of 002 18,

0.005 moles ·

= 0.005 moles x 138 g/mole

= 0.69 g

(as the formula mass of potassium

carbonate is 138).

When the mixture is heated, the nitric acid vapour and reddish- brown gas of nitrogen dioxide are given off. Concentrated sul- phuric acid displaces off the more volatile nitric acid from its salt, KNO. Part of the nitric acid is decomposed by heat to give nitrogen dioxide, water and oxygen. KNO

KHSO

4 HNO3 2H204N02 02

+

+#2504 AHNO 3 b) Ethanol is oxidised by potassium dichromate first to acetaldehyde

B

3 x 0.245

2 x 0.006

0.003 0.245/3

122.5

Yes. It is because the formula mass of P as calculated from.

the formula KC103

一期星

1.32 : 1.98 2:3

日四十月二年七七九一瑟公年六十六國民華中

b) Hence, the empirical formula for

the oxide of M is M2Oz.

物理

(十四)·魯榮家

Physics (14)

1(s) (1) Let A he the cross- sectional area and ℗ be the temperature of the oil bath.

By Charles' Law

V

"therefore,

*

hence

sin iz MN/HO

sin r

As the ray is very close the principal axis CP,

therefore

MCNG, MQ<= {N}

Alo

the liquid layer

therefore

very thin

constant

NC

final vol final temp

initial: vol.

final vol.

initial vol. = 0.13A initial temp. - 273X final vol, 0.18A AN final temp

0.13A

273

(273 + 0)X

0.18A

273 + 0

9- 105°C (Ans.)

(11) Let 01 be the room temp.

PC - 21

NQ = 14 - 21 - b

21

Similarly,

0.07 0.05 0.12

As figure shown,

30% (Ans.)

pa

(iii) Let the apparent

sivity »

금(소)

0,07 x 21 x 10′′

FO? x 105

1.4 x

10-x-1 (Ans.)

(b)

(1)

figure shown.

180

P+ q = 180

should be

3935.5+ 16 x 3

* 122.5

The value is the same as that determined in (e).

Molecular mass of ethanol is,

12 x 2 + 1 x 6 + 16.

= 46

92 g of ethanol is equivalent to 2moles.

From the given equation, it is noted that 1360 kJ is liberated when 1 mole of ethanol is burnt.

the heat change on burning 92 g of ethanol is, 1360 kJ * 2

= 2720 kJ

The heat is evolved because AH is negative in value. Answer to 0.24

W is nitrogen.

b) NaNO+NH C1- NH NO

NH NO

NH4NO2+NaC1 N2+ 2H2O

c) Liquid X is water.

Add a little of liquid X to anhydrous copper(II) sulphate or anhydrous cobalt(II)chlor- ide. If It is water, it will turn the copper(II) sulphate from white to blue or the cobalt(II)chloride from blue to pink; i.e., CuSO4 + 5H20

· CoCl2 + 68,0

CuSO4.5H20

·COC12 6H2O

.) It is to condense the water

vapour.

f) The substance is magnesium nitride which is white.

g) 3Mg(8)+N2(g) →→→ M ̧ ̧12(8) h) The substance is ammonium

nitrite. It is not used in actual practice because heat- ing it alone is dangerously explosive.

i) The gas evolved is ammonia.

The other substance is magnesium hydroxide.

j) Mg

B.

· 3Mg(OH)2+21Hz.

a) No. of g-atoms of metal Mis

100 31.6

1.32 g-atoms 52 No. of g-atoms of oxygen is / 31.6

= 1.98 g atomą 16 .*.g-atoms of Mg-atoms of O

and K are concyclic

+

•

.180°

180

* P + q

itute into (1)

D.

+ D) - A

(ii) For minimun deviation, Dit (i.e. d is the min- imum value of b)

nenee the ray passen synne- trically through the prism, therefore :

Since D =

n and

Hence

28-A

Also, from (2)

A... p + A 2p

Apply the Snell's Law

i

mini

sin r

(A+)/2+

A/2 sin(A+d)/2 si nk/2

2(a)As figure shown;

If the nnject and the image coinside; the ray must have incident on the mirror n0rm— ; ally and reflect back through tue centra. of curvature Ci

A the thin layer of li- quid is peared, the image

which is position is at q

cm below C

+

AB represents the object ly~

ing on the principal axis of the concave mirror and XY its image.

The image of point A

.

= 0.2m

0.150

12--1.15

V1 = 0.6m

hence, X is the real, zmage of point A and is át 0,6m from the mirror.

For the image of point.

0.2 0.1 - 0.3# -0.15

0:3

0.3m

0.15

Hence, Y is the real image of point B and is at 0.3m from the mirror,

size of the imag 0,6m2 = 0. 3m

magnification:

size of size of object 0.3 0.1

3 (Ans.)

(c)Since the image is enlar- ged, therefore it must be a converging lens, dut the image may be real or virtual. Far real' image

ME

v/u +4u

f = 0.16m

using

1/4u■ 1/0.16

u = 0.2m For virtual image

↑ 0.16m

using 1/u 1/v = 1/f

1/a- 1/40 - 1/0.16

u 0.12m Therefore, the object must be placed 0.2m er o.12m from the lens to produce a real or virtual image magnified

4 times linearly or 19 times superfically,

|經濟與公共事務(十四)

6. The ward system of the Council

The system was intro- duced in 1965 in order to maintain close contact with the general public.

The system divides the urban area into 10 distr- icts. Each district is under the care of '2 Coun- cillors, one appointed and the other elected.

There is an office for each ward and the public can make appointment to see the councillors of the ward.

7.

Financial structure of the Council

The Council has a share of 40% of the yield from rates in the urban area.

Fees and charges is the Council's second source of income.

In 1974 the Council received a once-for-all grant of $20 million from the government.