1985-03-18 — Page 23

節樂音育 頁三第張六第日七廿月正年丑乙曆夏

· 1985

中學會考試題預習專欄

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(廿六)

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Solution to Q.37

(8)

(i) The substance is

ammonium chloride,

The equation is NH; (g) + HCl(g) →→NĖ, C1 ( 3 )

(ii) The ring is formed

at a place nearer to Q,

•white ring

(iii)This is because

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

than the latter). As

a consequence,

ammonia meetg

hydrogen chi ride

at a place nearer to 0.

(iv) The electronic

structure of ammonium chloride is:

(v) I would dissociate to form ammonia gas

and hydrogen chloride gas again.

NH, C1(6)-NH; (g) +H¢1 {g} (b)

(1) A is sodium sulphite. (2) B is sulphur dioxide.

(3) NS02+2C1 --

2NaC1+Sg

(4) The reducing property

of 50 as potassium

(ii).

dichromate is a typical oxidizing agent.

(1) When dry nitrogen and

hydrogen gases are passed over a beated catalyst such as iron powder, ammonia gas is obtained. N2+32NH

(2) Hydrogen gas is burnt in oxygen gas to give water vapour.

22+02

2120.

(3) When a mixture of

hydrogen and chlorine gases is exposed to diffused light (or when hydrogen is burnt in chlorine gas) hydrogen is chloride is obtained.

(iii)

2+C12HC1

(1) Ammonia dissolves

readily in water and ionizes to give ammonium ion (NFIL

and hydroxide ion (OH) 0 the

WAH KIU YAT PO

solution is alkaline...

+011

NH3+120

(2) There is no effect. (3) Hydrogen chloride

dissolves readily in water and is completely ionizes to give hydronium jone (0) and

chloride ions (C17).

HCI+H20 30 ci

Solution to 9.38

(i) Nitric acid is rather readily decomposed on 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. (ii). Concentrated 50, being an acidic substance, readily Deutralizes ammonia which is an alkaline gas. Thus, the former cannot be used to dry the

latter..

(ii)When a piece of

fresh aluminïum is put into dilute nitric acid, its outer surface is readily oxidized

forming a tough coat

of aluminium oxide, which being. resistant to

the

nitric acid, thus

preventing any

報日僑華

·期星

日八十月三年五八九一年四十七國民華中

maximum mass of H3PO4 obtained from 4.1 g

mol

of P

HA

c4.1

and (KC105) in.

mixturce

1.97-yo

122.5

According to the equation

further attack of

the metal by the acid..

(iv) When lead(Fr) oxide

is added into sulphuric acid or hydrochloric acid, it becomes conted with a layer of insoluble lead(II) sulphate or "... lead(II) chloride respectively, thus stepping any futher reactions.A

Phosphorus is a rather reaction non- me tal and combines readily with oxygen and many other elements, 80, it oçcura naturally as its compounds ratber than in the free... state.

(ii) The other allotrope

"white

(iii).

is called phosphorus).

(1) The formulae, are

PC13 and PC15 respectively.

(2) The oxidation number of phosphours is 13. in PC1, but +5. PCI

(3) Phosphorus(III)

chloride molecule has a 3-D Atructure 38 whown below:

(iv)

(1) Phosphorus is

oxidized by the nitric acid. (2) This is because red

phosphorus is more stable though less reactive at room temperature while whereas phosphorus tursts into flame spontaneously on coming into contact. with air.

(3) It is because

nitrogen dioxide gas produced is harmful to us...

(4) There are three kinde

of salts: sodium dihydrogen phosphate (Nall,PO); disodium

hydrogen phosphate (Na2HPO) trisodium phosphate (NPV).

(5) It is to filter off

any unreacted phosphorus,

(6) M(P)=31 g mol-1

M{H2PO,)=84 y mol According to the

equation... P+SINOPO2+5NO2+H20

1 mol of P gives

1 mol of 11,PU

-10.85 g Solution to Q.39

(i) Magnesium oxide would

be obtained in the boat A and copper powder in the boat B. (ii) Mg+C0g0+CO CO+Cuo→Cu+C02

(iii)The limewater" would

turn milky as carbon dinxide is present.. in the exit gas X.. (iv) Yes; because the

exit gas X may contain unreacted carbon monoxide which in poiṣunous

in nature,

(1) No; because carbon. monoxide cannot. reduce zine(II). oxide.

(2) Yes; lead metal would

be obtained. Pb0+Co→→Pb+C02

(b)

(1) KC10, 2C1+302 (11) Ag*+01→→→→AgC1 (iii)N(KC10 ̧)=122.5.8

..: moi

M(KC)-74.5 g moll M(AgCl)-145.5g mo17

(KC1) in mixture«yg (KC10) in mixture. (1.97-y) £

(KC) in mixture

Now,

2 mol

2mol Imol

1 wol

n(KC) formed on de composing KC103 n(KCI) totally present

n(AgCl) formed

2.87 14351

-0.02 mol

According to the equation

·Ag+C1′′ →AgCl(s):

Amplimol

(KC1): reacted #n(C1") reacted -nagcl) obtained -0.02 mol.......(2) Equation (1) and (2),

we have

1.97-2.0.2

745+122.5

Om solving the equation, y=0.745

by mass of KC, in the original mixture

0.745 1.97 =37.82

x100

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届七世第

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(左)與其老師 一級說好獨奏,冠軍姚張莊(右)、季軍說

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I

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