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· BEANS ·6*RE-
Math Dale Signs
化
(vi) According to the equations
1978
012
+ 2e
中學會考試題預習專欄
2H + 2e
明德社主編
+ OH
學十八 朱宏林
Chemistry (18)
Solution to Q.25
(a) Pb
2+
+ + 21 →→ Pbl2(s)
(b) The minimum volume, as
read from the graph, is 4.5 cm.
(c) The minimum amount of
lead (I1)nitrate solution
0.1 moles X
4.5 1000
0.00045 moles
(4) According to the equation
(e)
Pb2+
+ 21 — Pbiz
amount of I ions present in 6 cm of iodide solution
amount of Pb2+
ions re-
quired for precipitating all the iodide ions
0.00045 moles x 2.
0.0009 moles
amount of metallic iodide
in 6 om of 0.05M solution
= 0.05 moles x.
= 0.00030 moles.
1000
and, there are 0.0009 goles of iodide ions in 6 cm of
solution. metallic iodide,
1.e. 0.0003 moles of meta- 1lio iodide give 0.0009 moles of iodide ions..
1 mole of metallic iodide give 3 moles of iodide. (r) The most probable formula
for the metallic iodide is XI.
Since the iodide ion carries 1 unit of nega- tive charge, so the metal X must have a charge of +3.
(6) Formula mass of Pbl .461
According to the equation in (a), amount of FbIz
2+
=no. of soles of Pb -0.00045 moles
wt of the precipitate
461 g 0.00045. 0.20745 G
Solution to Q.26
chlorine is discharged at the anode because chloride ion, 01, is an anion. (11) Hydrogen gas; because H
is selectively discharged.
- C12 (iv) No. of moles of Cl2 dis-
(111) 201
charged is
moles
22.4 22400 =0.001 moles
+
2e
According to the equation in (111),
amount of electricity used = 2x (0.001) faradays
= 2 x 0.001 x 96500 G
= 193 C
the time is 1930 seconds
.*, the magnitude of current
193
amperes
1930. 0.1 amperes
(v) The discharge of H at the cathode will shift the fo- llowing equilibrium
H20 - H*. +
OH
to the right. This results in an excess of hydroxyl
ions, OH Bo the solution
5
becomes alkaline.
no
of moles of OH
=no. of moles of. H*
discharged
2 x no. of moles of H
- 2 x no. of moles of of
= 2 x 0.001 moles
0.002 moles
Let y I be the molarity of the sulphuric acid. According to the equation
2. +20H
H2SOA no. of moles.
Ꭹ *
25 1000
of acid
moles
x no. of moles of OH
0.002.
0.001 moles.
25y
1000
0,001
0.04
So, the acid has a molarity. of 0.04M.
(vil) Amount of C1 originally
present is
no. of moles of NaCl
100
1000 5 moles
0.5 moles
Amount of C1
discharged
= 2 x no. of moles of Cl
= 2.x 0,001 moles
= 0.002 moles
.. amount of excess 01
-(0.5-0.002) moles
= 0.498 moles
Let v cm be the volume of silver nitrate required. According to the equation Ag →C1→→ AgCl (B) then, no. of moles of AgNO
no. of moles of a
in excess
0.498 moles
5x
moles
1000
5x
1000
= 0.498
x= 99.6
The volume required 1s 99.6 cm.
(b) (1) 2Fe + 3012 2FeCl3
Conditions:
(2) dry chlorine
strongly heated iron
(11) PbCQz+2 HNO3 -> Pb (NOz)2 + H2O
+ 002 Fb(NO3)2+NaCl→→→PbCl2+FanOz
solid
(iii) By electrolysing sopper (II)sulphate with copper
cathode. 2+
Qu +2e
+
Cu(s)
(iv) CH, COH + 2(0) CHC
COOM CH2 CH2OH CH, OOOH H2O
+ CH3COOC2H5
prolonged heating with cone H30 as catalyst
Conditions:
Solution to 0.27
neutralisation reactions:
(1)
H2S04+2NaOH Ne2SO4+2H20 HCl + NaOH . ) Nac14,0 – (2)
\precipitation reaction:-
BaCl2+H2S04
+H2SO-BaSO +2HC1 —(3)
(ii) Formula mass of H
H2SO4
# 98
=233
Formula mass of BaSO According to the equation (3), weight of H2SO4 in 25 cm of S
= 98 g x
0.262
-233
233
-3 4.408 g dm
0.262) x
concentration of H2SO in S
(98 x
4
1000 25
dm
the outside pressure. The
water index will be pushed
up out of the tube, OR, if the tube is long enough, the water index may dis-
appear completely as the water in the index vapuri-zes. 24】i) Heat transferred to
al mode
wavelength
21 - 2 x 0.8
344
Frequency
the alcohol
*.(4)
No. of moles of NaOH used
35
0.15 X moles
1000
0.00525 moles
But, according to equations
(1) and (2),
no. of moles of NaOH required
2 x no. of moles of H.
x no. of 12304
+no. of moles of HC1
2 x (4.40825) no. of
98 1000
0.00525 moles
-+
moles of
HC1
rio. of moles of HC1 in 25 cm -(0.00525. 0.00225) moles
0.003 moles
Formula mass of HC1 36.5
ctation of HC1-
4.38
x 0,003 x
g dm
1000 25
dm
A is ammonium chloride;
B is ammonia gas;
C is hydrogen chloride;
D is silver chloride.
(11) AgCl(s)+2NH ̧→→→ A6 (NH3)2
(iii) NC1+NaOH->NaCl+NH +F20
4
C1+ SONH HSO +HC? NH CA + ACNO AgCI+NE HO2
物
理汁
魯榮家
十八
PHYSICS (18)
Answers to Exercise 9
1.(a) Let the cross-sectional
area of the tube be A, the unknown temperature be T'C
and the absolute zero1
températurë according to
these datas he
Since,
Therefore,
280A
ec
(constant
pressure)
385A 100 + 9
322A
T +
280A
385A 100 +
2801 266.67
280 266,67
T = 40
(ang.) 322A
322
T + 266,67 (Ans)
· the : unknown temperature
is 40°C (ii) Since @ = 266.67
the absolute temperature obtained ia -266.67°C. (b)(i) If it is replaced by coloured water, the new! reading will be greater than that obtained in (a), ft is because when the temperature rises from 0°C to 40°c, the increase in s.v.p. of water pushes the water index fur-
ther up. (ii) At steam point, L will
increase because the total
pressure in side the tube
is the sum of the s.v.p. of
and the water at 100°C
pressure of the trapped air which is greater than.
(ii) leat lost by the calam rimeter and water
={m_o_x_m_c) (8 - 8)
(6)Assume no heat is lost to or gained from the surroundings
+ m_ c ) (H) -
10
Usually, the experimental
value of specific heat capa- city of alcohol is lower than the value as calculated from the above expression because
heat is usually absorbed hy
the cooled alcohol or mixture
from the surroundings and measured is higher which results a lower value 3. (a)
1.5M
As shown in the figure
above, N represents the nodes
total number of
nodes
7
(ii) wavelength
=
1.5
x 2 (iii) Apply
The frequency
20
0.5
= 0.5m (Ans)
40 Hz.
(iv) the wavelength
344
- 8.6m (Ans.) 40
(b) For vibrating strings
frequency & →
Length .Let 1 and L' be the ori- ginal and the final length
of the wire.
366 366 -
H
.60
the fraction increased·
(621 -1)/1
(Ans.) (c) Pipe A
Pipe B
λ
(ii) Let the length of pipe
H be L m.
...om
L 0.8m (Ans.) (iii) Pipe B at its fundament-
215Hz (Ans.):
advantage is that
end correction can be eliminated easily by just finding the first position
of resonance
(ii)
T
L+C
Let the velocity of sound in air be vms. and the end correction of the
(7)= (Lübe
$
c)
be
(1)
Substitute the result of the experiment into (1)
x 320
. (2).
0.17+ c .(3)
(2)- (3)
x.320
0.9
4 × 480
(b) Since
345.6 ma
ma-1 (Ans.)
味瓜
Where is the frequency of the stretched wire, Lis the length of the wire, T is the tension and a is the linear density of the wire.
It d and rare the density and the radius of the wire
then,
and hence,
t
2L
W d
(i) Let the fundamental freg-
uencies of X, Y and Z. be
and
f ̧(=30012), £y
pectively.
1
300 =
2L(0.2)
rea-
(1)
8000w
.. (2).
· (3)
21(0,4), X00 T
f, * 2L(0.2) 9000π (2)/(1).
150 (Hz) (Ans.)
-
(3)/(1)
B000
x 300 9000
283(uz) (Ana,)
(ii) the wavelength
330
300
* 1.1m (Ans.)
(iii) It is short radio wave.