頁四第張十第日三十月五年未己腐宴
WAH KIU YAT PO
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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)