1979-06-07 — Page 40

頁四第張十第日三十月五年未己腐宴

WAH KIU YAT PO

報日僑華

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Chemistry.

The solutions suggested

are very often only one

of the possible answers to the questions concerned.

The four mistakes are corrected as below,

(1)

the tap Bis

missed.

out

(2)

tripod stand

(3)

(4)

Min 02

NAG1

(used a round bottomed

flask & heat, has

to be applied)

Then draw a diagram to show the correct set-up.

Manganese(IV)oxide is an oxidizing agent, oxidizing the hydrogén chloride. (ii)The white fumes are due to

the moisted hydrogen chloride gas.

The equation is:-

NaC] + II 50,NaHSO,

(iii)Water in flask B is to

dissolve any escaping hydrogen chloride from flask A.NG

FIC 1

Concentrated sulphuric acid in flask G is to dry the chlorine gas..

(iv) Iron (11)chloride solution would turn from pale- green to yellow as

iron (11)ions are oxidized to iron(III)ions by chlorine gas

(v) The red fumes are brozine

vapour. This is because any hydrogen bromide formed is further oxidized by concentrated sulphuric to bromine. Nabr

Na!{S0, +BBr

21Br

The order of their

reactivities is..

X>Z>R>T> @

(ii)Q might be mercury. When

00 (mercury(II)oxide) is heated, it changes from orange to deep-red in colour and finally

decomposes to rive oxygen and mercury vapour which condenses as silvery tiny drops on the cooler part of the test-tube 20g0—72llg. + On

(iii)The carbonate of X

not decompose on heating. (iv)The equations are:

XNO

26(103)on?2Z0 + }X0g+Ug (v) The simple cell is se us shown below:-

beake

porous .pist

2. (u)

T TSUL

solution

(1). a is the smallest in.

group 1,

(ii) j is the most reactive

non-metal.

(111) h and i can form ions

with

-24 It is the family of

(iv)

(v)

Key;

charge

alkali metals,

It possibly has

tetrahedral shape

an atom of f.

an atom.

Tetrahedral

(vi) The formula for the

compounds formed by g and e ia

€32 and that

for the compound formed

by g and J is fiz.

The compound

have a higher melti

point because e 352

Lonic and the ions held together by the strong electrostatic force while gåz is covalent and the molecules are held together by the weak intermolecular (or

van der waals) forces only.

(vii)The probable formula for

the oxide of 1 is 10,

and the bond is covalent in nature. This is because bóth

1 and 0 are

elements in the group VJ

The purpose of

galvanizing iron is to prevent iron from rusting (ii) The watch glass is to

prevent the splashing of the solution and of the acist,

(iii)This is because “zinc reacts first and the reaction is fast and moderately vigorous. When all the zinc have

reacted, iron starts to Tract and the reaction less vigorous

and more slowly because iron is less reactive than zinc. (iv) Excess sodium hydroxide.

is used because, besides precipitating the zine

Vontae, it has

to convert the zinc hydroxide to zincate ions.

The precipitate turns. from green to reddish-

The solid residne is iron(III) oxide,

·

4Fe (011) g+0,−>?F¢g0z+4Hy0 Formula mass of Feg0z×160 Mass of iron in 1.20g of

obtained

56 x 2

160.

weight of zine

in the sample of galvanized iron,

41.0

16%

x: 100%

The total volume of gas collected at the end of the experiment ів 60см3. (2) 0,9 mlautes have

elapsed when half of the total volume of gas is liberated,

(3) When three-quarters of

the reactant have. reacted, 1.7 minutes

have elapsed,

(ii)The reaction rate is

fastest at D.

四期星

(11).) Initially, the slope is steep, indicating a fast rate of réaction, because. the concentration \ëf«?« hydrogen peroxide is high; At the intermediate stage the slope is less steep because the concentration of hydrogen peroxide Finally, decreases.

the curve levels out because. ell the hydrogen peroxide have decomposed and no more oxygen is giver off. (iv)The equation for the

reaction, is:-

2011, 0, (2 q )→211, 0(1)+0,{8})

No. of moles of

liberated when all

have decomposed

60 24000

0.0025

(1)

According to the equation. (1) above

no. of moles of l¿0ŋ

present in 50.0cm3 solution.

2x110. of moles of oxygen obtained

0.00253

1:05

concentration of hydrogen peroxide solution

0.005 1000 mo)

0.1 mol dm

(1) Peanut oil is unsaturated

consisting carbon-carbon

double hond(s), given conditions,

hydrogenation

Under the

place

(1.e. hydrogen adds across the carbon-carhon double bonds) to give saturated compounds.

(11) Nickel powder acts as a

catalyst.

(iii)At room temperature,

peanut oil is a viscous liquid while the produc will be à semi-solid

The electronic

configurations of X and Y

2.8.2 and 2.8.7 respectively. Both X and Y tends to attain the stable electronic configurati lons of noble elements, the best way for X to do so is fo lose 2 electrons, thus having

an electronic configuration similar to neon (2,8); while the best way for T to do this is gaïn 1 more electron, thus having an electronic- configuration of argon (2, 8.8). Since X tends to loose electrons, so it is a reducing agent while Y tends to gain electrons so it is an oxidising. agent.

(ii)The compound formed

between X and Y has an electronic structure as shown below:-

LOLOLOL

(iii) The

on is:-

>JY HYO

Y has an oxidation of in Y and +1 în YO

be

(iv) White precipitate would seen because ammonia combines with IT to give the compound

monically which

insoluble in

nature and

methylbenzene.

(i) Copper wire acts

catalyst,

The other metal is platinum. (ii)The equation is

ANIL + 50, 4NO கர்

The liquid in U-tube C is water which is formed in the above reaction and condensed on cooling.

日七月六年九七九一瑟公年八十六國民華中

Brown gas in flask Dis nitrogen dioxide.

It is formed when nitrogen monoxide, NO, comes into

contact with atmospheric oxygen.

ív)The while fumes are

amronium nitrate and Apino Dim nitrite in fine powders

育教備藥

(iv) C is propanoic acid,

CH,CH2COON.

The equation is:-

CH ̧CÊ¿CH2 ©H+CH,CH2C

GIT.

The reagent required is concentrated sulphuric

acid.

2NILz+I90+2NO,➡➡NHI, NO, +MH, NOŋ| (v) Conditions:

(v) Iron(II) sulphate solution

turns from pale green to dark-brown because of the formation of the complex Fe30, NO.

(b) no, of woles of §,50, originally preseut in 500 cm3 solution

8

500

1000 * 0.05.

0.025

According to the equation below

2NaÐII+H2SO¿ →→→Na2SO2 +20,0

of meles of H2SO2 eft unused by NH

of Nab

1⁄2 x no, of moles required for neutralization x 500

25

500 25

.01.

amount of H2SO4

nentralized by NH,

(0.025 - 0.01) 0.015 moles

According to the following equation

> (NH)250,

of moles of NHI

nally present in the

of moles of neutralized by NH

x no.

2 x 0.015.

volume of Nil, ga present at 25

atmosphere

24 dm x 0:03

.72.

720 cm-

The gas is ethene. CIL, CH, (I→→→CR2-CH,

and

(ii) Sodium hydroxide solution

is to removed any acidic gases such as sulphur dioxide. (iii)The bromine solution in

C is decolourized (ie, from orange red to colourless).

The product is 1.2-dibromoethane.

BrBr

(iv) The acidified KMn0

solution Dis decolourized (1.e. from purple to colourless).. because á redox reaction takes place whereby ethene

is bridized and Mn0, iun

is reduced to Mn CH2=CH2

5. (a)

(0) from

(1) A is propan-1-ol, CHCIL¿CH, OH.

(ii) Amisomer of A is

propan-1-ol

H-C

ons

Acidified potassium dichromate or acidified potassium permanganate can be used to convert A to

(1) excess aqueous, sodium

hydroxide or

potassium hydroxide (2) heat under reflux.

When D (n-propyl propanoate) is hydrolysed with an alkali, A

(propan-1-01) and a salt of C, which is propanoic acid are obtained. Propan-1-ol, can be obtained by fractional distillation of the reaction mixture. When all the propan-1-ol have been distilled off, a mineral acid, e.g. dilute sulphuric acid, is added to the residue to rec the propanoic acid from its salt and the mixture is then fractionally distilled to obtain C. The gas liberated is sulphur dioxide ii)The bromine water is

decolourized (i.e. turn from orange-red to

8

calourless),

recover

SO2+2H20+Br2→→→Í2SO2+2HBr (ii))The acidified potassium

dichromate solution turns from orange to green iD colour.

(iv) The rate is greatly

8

slowed down because part of the sodium sulphite -ozidized to sodium

sulphate by the atmospheric oxygen. Dissolve the solid in water and add acidîfied barinm chloride solution into the solution thus obtained; a white precipitate of barium sulphate will be seen the sulphate ions are present.

2

2+ 50, ̄(aq)+Ba2 *(aq)

blue (or

Solution A has a hinish green) colour. (ii)NII+ and K' will be present

in the filtrate BU

The presence of K+ can be detected by a flame test as it will impart a lilac colour to the bunsen flame. The presence. Nill, can be detected by boiling the filtrate B

with some dydroxide and, amrionia gas will be evolved which can be recognised by its smell or alkaline nature to litmus

paper. (iii)Precipitate C contains

magnesium carbonate, MgC), and copper(II) carbonate,

The Lonic equation is

2-

_C03 ̄ +20—П2Ø+C0,

Solution D contains magnesium ions and copper(II)ions. If

ampionta solution is added drop-wise to the solution D, white precipitate of magnesium hydroxide and bluish-white precipitate.

of copper(II)hydroxide are observed.

Mg

2+(aq)+201− (aq)

Mg(OH)2(S)___ Cu ́ ́(aq)+20H (aq)

€u(011), ($)

When excess: ammonia

solution is added, white precipitate of magnesium.

hydroxide remains

unchanged whereas

copper(II)hydroxide dissolves to give an intense blue colour

Cụ(ou),(s)+N

→cú(NH3)2*(oq)+2011TM (aq)