1986-03-24 — Page 30

(3) When a mixture of

WAH KIU YAT PO

hydrogen and chlorine

gases is exposed to diffused light (or

頁二第張八第日五十月二年寅丙歷夏

21+02

1986

中學會考試題預習專欄

化學(廿六)

明德出版社

朱有理 提供資料

Revision Exercise.

for.

Cert. Exam. (1986)

CHENISTRY (26) REGINALD V. CITU

MILL & DALE PRESS LTD,

Solution to 9.37

(1)

The substanceia. ..ammonium chloride.

The, equation is

• NHL 3 ( 6 ) + HC 1 ( g ) →→→→ NH, C1 ( s)

(11) The ring is formed

at a place nearer to 0.

White ring

(iii)This is because

ammonia diffuses faster than bydrogen chloride (as the former is lighter.

than the latter), As'

a consequence,

ammonia meets. hydrogen chloride

at a place nearer. to 0.

(iv).The electronic

structure of

ammonium chloride

(v)" It would dissociate to form ammónia gas and bydrogen chloride gas again.

NB, C1 (3) ~ NII, (g) +h€1 (8)

8

(1) A is godium sulphite, (2) B is sulphur dioxide. (3) Na,80,+2HC)

2NaC1+502

(4) The reducing property

of Sly as potassium

dichromate is a

(ii)

typical oxidizing agent

1) When dry nitrogen and hydrogen gases are passed over a heated catalyst such as iron powder, ammonia, ges

(2) Hydrogen gas is burnt

when hydrogen is

burnt in chlorine:

gas): hydrogen is

chloride: is obtained.

+1 201-

(ii) (1):Ammonia dissolves

readily in water and ionizes to give

ammonium ion (NIY *) and hydroxide ion (on), so, the solution is alkaline

NII+HONII,*+OH” (5). Tvarogen coffee

There is no effect.

dissolves readily in water and is completely. ionizes to give hydronium ions (H500).and. chloride ions (Cl HC1+0

10

Solution to 0.38

i) Nitric acid is rather readily decomposed op exposing to light, Brown bottles can absorb part of the sunlight, thus slowing down the decomposition of nitric acid. Thus,

it is usually stored in brown..bottles..

(11) Coed

5043

being

àcidic substance, readily neutralizea amonia which is an alkaline gus. Thus,'the former cannot be used to dry the latter

(iii)When a piece of

fresh alumi ́m á um put inte dilute nitric acid, its- outer surface is readily axidized forming a tough coat of pluminium oxide,' which being resistant to the nitric acid, thus preventing any

further attack of the metal by the acid

(iv) When lead(II) oxide

is udded into sulphuric acid or hydrochloric acid, it becomes contact. with a layer of tak insoluble lead(II) sulphate or

(b)

lead(II) chloride respectively, thus stopping any further reactions.

Phosphorus, 18: '0

rather reaction non-

detul and combines

"readily with oxygen

∴and many other

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occurs naturally as

its compounds rather tuan in the free state.

(ii) The other allotrope

(立)

is culled white

phosphorus! -

(1) The formulae ave

PCIy and PC15. respectively.

(2). The oxidation nubber.

of phosphours is +3 in PCI, but +5 in PC15

(3) Phosphorus(III)

(x)

chloride molecule bas a 3-0 structure as shown below:

(1) Phosphorus is

oxidized by the nitric acid. This is because red phosphorus is more stable though: Less reactive at room temperature while whereas phosphorus bursts into flame: spontaneously on coming into contact with air-

(3) It is because.

nitrogen dioxide gas produced is harmful

報日僑華

(4) There are three kinds

of salts: sodium dihydrogen phosphat好 '(NallyPO); 'disodium

hydrogen phosphate (Na,HPO); trisodium phosphate (Ne,PO4).

(5) It is to filter off

any unreacted phosphorus.

(6) M(F)=51_g_molTM"

gmol. According to the equationAR

1 mol of P gives

maximum mass of U FO

obtained from 4.1

of .p

82

-10.83 R

Solution to 0.39

8

Magnesium oxide would

be obtained in the

boat A and copper powder in the boat B

(ii) Mg+CO2→MgO+CO

CO+CuB→Cu+COg.

(iii)The limewater would

2 torn milky we carbon dioxide is present in the exit gas X..

(fv) Yes; because the

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exit gas X. may

contain unreactedTM-

carbonmonoxide which is poisonous in nature.

1) Noj. beniuse carhin monoxide cannot reduce, zinc(IT) toxide.

(2) Yes; lead metal would'

be obtained. Ph1+CO2b+CO2

(

{1} @kclu ̧ z> 21c1+30, (ii)AgT+C) - AgC1 (iii)M(KC10g)=122.5g:

Now,

01

\M(KC))=74.5 g mol ̈1 N(AgCl)=143.95 01- m(KC) in mixture=yg. m(KCT03) in mixture

= (1-47~y) B

n(KC1) in mixture

·#nd.n(K¢10g) t称:

mixturez

1.97-y

122.5 mol

According to the equation

KC103

Z mol

mvl

1 mol n(KCl) formed.con decomposing KC10, n(KCl) totally: present:

122.5

n(AgCl) formed

2.87

143.5

0.02 mol According to the equation AG+C)→ AgCl(s):

Imol mol 'n(KCl) reacted

(C1) reacted =n(AgCl) obtained:

0.02.mol.

:(2)

Equation (1) and (2),

we have

y1.97-y 74.5 122.5"

-0.2

On solving the

equation,

ym 0.745

%by mass 'of'KC1.

the original mixture

0.745. x100

37.82

物理(廿六)

明德 出版社

康樂

盧雄國提供資料

Revision Exercise

for

一期星

PHYSICS (26) HK.LO

日四廿月三年六八九一公年五十七國民華中育教僑華

MILL & DALE PRESS, LTD.

Example 3 (Multi-range

voltmeter)

The figure shows the internal wiring of a "two-scale" valtmeter whose binding posts are mari:ed 10V and 30V, The resistance of the moving. [coil galvanometer Gie

150. If a current of ImA through the galvanometer will cause it to deflect Tull scale. Find the resistances R and R

Solution:

30v:

When the switch is

switched to 10V-terminal,

10= (1 ×10−?)(E ̧).

R100000

(1)

(Aus.)

When the switch is

switched to 30-terminal.

30=(1x10~~) (R ̧ +H2 ) - • (2)

30-(1x10-3)(10000+Eg)

Rp=20000.n

Example

:(Ohmmeter) As shown in the figure, G is n galvanometer of internal

resistance 100. It gives full scale deflection- (ie, painting at, the mark "10") when a current of 10mA ia appliedi The battery is of en. SV, and negligihle internal resistance. Disin resistance such that when terminalen, and b are shorted, the meter reads full scale. When a and b

are open, the meter reads zero.

(a) Find the resistance

of n.

(b) When a and b øre

connected to an

unknown resistance H

and the-pointer

points at the mark

"6". Find the resistance of R

Solutions

(a)Since the meter reads.

full scale (i.e 10mA when a and b gharted.

* R-290 (ing.)

(b)When the pointer.

points to "6" the: current through G is JOMAX

-Oma

As shown in the figure,

3m(6x1075)(R+R+10

=(6x10-5)(500+H

.B-2000

(Ans.)

(D) Measuring unknown

resistance by Aumeter and Voltmeter. (i) High resistance

Low resistance

9.4 Magnetic effect of

current and electromagnetin induction

1. Right and rules

Using the right

hand the thumbprints in the direction of

the current, the

fingers then curl in

the direction of the field.

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field.

2. Magnetic fields

around current-

sorrying conductors

·(4) Straight wire

(ii) Circular

curren

(iii)Solenn

fie.

3. Electromagnetic

induetfon: (1)

Faraday's W The" feduced e.a.f.is proportional to

change of the magnetic flux linking the circuit,

Len's law: The induced currentag flows in sueli a direction as to

notion or change producing it. (ii) Fleming a right

and rule

(First finger)

▲ Field

current

Tinger)

Motion

(thumb)

current

Notion

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