真四第張七第 日十初月一十年辰丙夏
WAH KIU YAT PO
郭日僑華
1977中學會考試題預習專欄
明德社主编
四期星 e) The reaction would be
too slow at room perature.
育教僑 日十三月二十年六七九一公年五十六國民中
The flat-portion in- dicates
0.05g hydrogen and
化學(八) 朱宏林·
Answer
CHEMISTRY (8)
to 0.11
a) at 100°C; the value is
51.49
b) 30g at 70°C and 10.5g
at 30°C.
/
c) With the reference to the data in (b), to make 130g saturated potassium chlorate solution at 70°C requires 100g water and 30g potassium chlorate.
Mass of potassium chlorate in 50g solu- tion saturated at -70°C is
7000, 15500
30g X
1304
Mass of water
11-547
(50 -11.54)g
38.460.
But, mass of potassium. chlorate that can still
be dissolved in 38.46g. water is
b)
10.5 x 38.46g
4.003
4.04q.
Mass of crystals obtained of cooling 50g solution, satu- rated at 70°C, to 30°C is
(11.54 - 4.04)g -7.59
Mass of water is:
(120 45)g = 75g.
Mass of potassium- that can be dissolved by
75g water at 40°C is
75
༢༠༠-.
40G X 200 Mass of potassium chloride crystals. deposited will be (45- 30)a 159 3012 + 6XOH
5KC1+KC103H20
No. of moles of C12
absorbed is,
5.76dn
X 1
mole
24
3
am
0.21 moles.
ii) Formula mass of KCI
=39+35+5 = 74.5 Formula mass of KC103
39+35.5+3X16
122.5
According to the equa- tion in (i), 3 moles of Cl2 gives 5 moles
of KC1 and 1 mole of
KC103.
Mass of KC1 formed by 0.24 moles of Cl2 is,
(74.5 x 5) x 0.34
29.8g.
Mass of KC1D,
КС103
-formed
by 0.24 moles of C12
is.
即可完成 全身健康檢查
(122.5 X 1).X
= 9.8g.
iii) The 100g water will
become saturated with the 9.8g Potassium chlorate at 28°C)
because the solubili-
ty of potassium ch- lorate is 9.8g at..
28°C. So, on further cooling, Dotassium chlorate crystals will Come out:
iv). At 10°C, 100g water.
can dissolve 5.2g potassium chlorate. and 31.8g potassium chloride; hence, on cooling the solution- to 10°C, no crystals of potassium chloride are obtained but the mass of potassium · chlorate crystals to
be obtained is
(9.8 - 5.2)g.
= 4.69.9
c) The method employed is
called fractional
crystallization.
e) The solubilities of
•A.
Dotassium chloride and potassium chlorate at 40°C are respectively 40g and 149.
Maximum mass of Dota-. ssium chloride that' can be dissolved in 50g water at 40°C is
50g
409 X 1009
= 20g
and that of potassium. chlorate is
14g X
50g
= 79. .100g.
Since the mixture con- tains 20g potassium
tem-
f) This is to dry the
oxygen and sulphur dio- xide gasses because sulphur trioxide reacts explosively with water so the whole apparatus. has to be dry.
a) This is to protect the
sulphur trioxide from the moisture of the atmosphere.
h) The substance is called vanadium (V) oxide,
B.
V
The sulphur trioxade is collected in solid state. at is white, silky. needle-like crystal.
a) i) The energy is libera-
ted as the reaction is exothermic.
ii) AH for the reaction
is -95kJ-mole
-1
or
-90kj-ecuationTM,
b) 1) The yield of sulphur
trioxide would be decreased as the for- ward reaction is
exothermic.
ii) The yield of sulphur
trioxide would be in-. creased as the to tal number of molecules of reactants is. larger than that of Product.
c) This is because sulphur
trioxide reacts violently or explosively with water and forms a mist of acid drops with water vapour; but it dissolves smooth- ly in concentrated sul- phuric acid to form a liquid called oleum.
chloride and 20g potassium d) 502 (9)+H, SO
(so (1)
Chlorate, so, from the above calculations, it is found that all the potas- sium chloride are just. dissolved but (20-7)g = 139. potassium chloride remains andissolved. Fiswer to 0.12
a) The sign
indicates
that the reaction is reversible, i.e. it does not go to completion in either directions and
will finally reach an equilibrium » ...
b) 1) The catalyst can
speed up the rate of forming sulphur trioxide.
ii). The catalyst has no
effect on the amount of sulphur trioxide formed as it does not affect the com-. Position of the equilibrium mixture.
c) This is to increase the surface area of contact so as to increase the rate of the reaction concerned.
d) The reaction is
exothermic, thus pro-
ducing sufficient heat to keep the catalyst at the optimum tempera- ture.
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e) The ordinary concentrated
sulphuric acid aan be obtained by diluting the oleum with sufficient amount of water. That is
H2520, (1)+H2O (1)
2450 (1)
Answer 50
A
a) 488 cm3 of CO at
is equal to,
3
488cm 22400cm
3 X 1 mole
0.02 moles.
there are 129 carbon in 1 mole of CO2.
Mass of carbon atoin in 0.02 moles of
is,
202
129/ X 0.02 = 0.249 Molécular mass of H.
= 1 X 2 + 16 = 18.
there are 29 hydrogen atom in 1 mole (or 18g)
.0.
Mass of hydrogen in 0.45g water is
0.459
29 X
= 0.059 189
Hence, 0.37g of P con- sists of 0.24g carbon,
(0.37 -0.24 - 0.05)g
0.08g oxygen.
mass of c:mass of H
: mass of o =0.24 0.05 0.08 Let x=no. of moles of C, y=no. of moles of H, z=no. of moles of 0,
x y z
Then,
0.24 0.05. 0.08
2
1 16 =0.02 0.05:0.005 -4:10:1
The empirical formula of
is C.H
560сл of Pat's.t.P. weighs 1.85g.
1 mole (or 22400 cm)
at s.t.p.) of weighs,
7.85g X
3
5600m 22400cm
= 749. 3.
The molecular mass of P is 74
Now, let the molecular formula of P be
(C4100)
Then,
P
n
molecular mass of
= n_X (4X12+ 1X10 + 16X1 )|
74n 74
n
1..
The molecular formula of r is CH,
C4H100
·4002
The equation is, .C.H (0+60)
04 100+ 602
d) Alkanols
The Possible structural. Formuläe are:-
H H H H
H-C-C-C-C-OH
ii) H
1.
ET
Η Η Η Η
HHHH
-C
F HH-OR H
H
H-C-H
H
H
iii) H-C-C-C-OH
41
iv)
H
H-C-H
1-butanol)
2-butanol
2-methyl -Propan- -1-01
2-methyl- ProDan- 2-01
that the amount
of precipitates formed remains constant. The reason is that all the chloride ion has all been used un in forming the precipitate c) From the graph,
it is
a)
seen that the volume of 1.OM silver nitrate solution required.”
moles of silver nitrate are re- quired to react completely with
3
?cm
3
1000cm
-X 0.5)moles
of metallic chloride Lo, or moles of
silver nitrate to
react completely with 1 mole of me-
tallic chloride is,
3 1.000
3 moles
The ionic charge on me- tallic ion is +3. The formula of the metallic
chloride is MC13.
7) 1). The 100
11
3.
"resent
arc
3+
M
ND
and cl..
are
Phc one resent
M3, NOTM, and Ag**
3
a) when both jasses are dry. there is no reaction. When water is added, reaction take places at once. Yellow particles of sulphur are deposited
on the sides of the gas- jars.
2
2H2 S() + SO
so2(9) 2H2O(l) + 35(s)
odine is liberated by chlorine. Then iodine will react with starch to impart a blue colo- ration.
Cl2(9) + 2K1 ̄(aq) 2KC1(aq) + L2(aq). 12(aq) + starch
blue complex.
when sulphur is heated in concentrated nitric
acid, it is oxidized by conc nitric acid to sulphuric acid. When the filtrate is treated with calcium chloride solution, white precipi- tate of calcium sulphate
is observed. Reddish- brown fumes of nitrogen dioxide are also observed
$ (s) + 6HNO2 (1).
\2 SD4 (aq) +6ÑO2 (4)+2H2O(1)
TH
H
OH
H H
H-C-C-H
H-C-С-OH
H H
cyclo-
butanol
Answer
C H
a) Ag (aq) + C1 ̄(aa)
AgCl (s)
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