★教僑瑕頁三第張五第日四廿月一十年辰丙膳 WAH KIU YAT PO
1977中學入學試試題預習專欄:
物理
智慧社主編
高榮家
Physics (10)
Answers to exercise
a)i) Heat energy required
to raise the tempera-
tare of the calorimeter to 100°C/
0.5 X 420 X (100–25) 15750 J
Heat energy required to raise the tempera- ture of the water to 100°C
252000 J
10.8 x 4200 x (100-25)
Power of the gas stove 252000 + 15750
8 X 60
267750 J/sec.
480
557.8
J/sec.
ii) Additional heat energy
required to boil the
water away com/letely
0.8 x 2260000 J
1808000 J
Time required
1-808000
557-8
sec.
3241 sec.
54 min. (Ans.):
b) Let the specific latent
heat of fusion of ice be
L J kg
mass of water
0.16kg 0.06kg
0.1 kg.
mass of ice
0.1845kg
- 0.160kg
0.0245 kg.
Heat energy lost by water
and calorimeter
(0.1x4200+0.06X420) (35-
13.8) J
1.06 X 420 X 21.2J
9438.24 J
Heat enercy gained by ice (0.02451, )+(0.0245x4200
X 13.8))
((0.02451 ) + (1420.02)) J
9438-24 -0.02456+1420.02
0.
C5751 - 8018.22
8018.22
0.0245. 327274-3
The specific latent heat
of fusion of ‡ce is 327274.3 Jkg
For both convention, when mis plotted against v, straight line of slope
will be obtained
ence, from the graph,
Slope
5m
f =
2 0.4m
Am
=
f = 0.2m
=5m-1
(Ans.)
The wave length for
the fundamental note
of the transverse
wave in the string
2 X 2m = 4m
Since the frecuency of
the fundamental note
emitted is 40 Hz.
the velocity of Pro- Pagation of the trans- verse wave in the string
4 X 60 240m/sec. (Ans) ii) The linear density of
the string
0.05kg 0.025kg/m
2m
V =
when v = velocity of the transverse wave in string
T= tension in the string u = linear density of the
string
240 =
T
10.025
T = (240)2(0.025)
1440 N. (Ans.)
The wave length of Fundamental note heard by ear
velocity of sound in air frequency of the note 330
40
.8.25m
(Ans.)
wsing dirtovece
iii) Let the object distance be a, image distance
be v and the focal length be f, magnifica- tion be m
Real-is-Positive,
multiply both sides by v
V
V
=
+1
Since
m=
mid
V
m
+
New Cartesian convention
Since
44
iv)
1
น.
1
V
L
=
+
ū
-
multiply both sides by v
1
1 + m
Since the pipe opened at both ends
. L
=
즐
郓日僑單
ii) According to the equa-
tion
2H
iii)
+
+ 2e
2f of electricity
are required to liberate 1 mole
(or 22.4dm at s.t.p.) of hydrogen gas.
Quantity of elec- tricity required to liberate 22.4cm hydrogen at s.t.p. is
2 x 22.4cm
22400cm3
0.002.£:
0.002 X 96500 coulomb/f
193 coulombs
*time used
(3 X 60 +13)sec.
193 sec.
the magnitude of
current is,
193coulombs
193 sec.
A
iv) According to the equa-
tion
Cu + 2e →→→→ Cu
1 mole (or 63.5g)
Cu is deposited by 28
Now only 0.002f electricity has been passed,
le)
the mass of Cu deposi- ted on Q is,
0.002£
63.5 X 2f
0.06359
Hence, the mass of elec- trode Q has increased by 0.06359.
i) Both hydrogen and chlorine
gases are present in gas- syringes X and Y.
ii) The two volumes are the
same, i.e., 44.8cm-
The length of the pipe iii) The masses of electro-
x 8.5m
4.25m. (Ans.)
if the organ Pipe is. closed at one end then
19
waves fenoth the vide
"Fendtn"
wave length
4 X length of the Dipe
= 4 X 4.25m -- 171.
the frequency of the fundamental note emitted
330
= 19.4Hz.
the frequency of the se- cond over tone from the
organ Pine closed at
one end
= (5) (330
= '97 Hz.
(Ans.)
化學(十)·朱宏林·
Chemistry. (10)--
Answer to 0.15
a) At P: Cu 0: Cu
R: 201→→→ Cla
2+
+2e Ca
+ 2e
S: 2H + 2e
H
2
b) Chlorine gas is collect-
ed in gas-syringe X and hydrogen gas in gas- syringe Y.
c) The rate of electrolysis
in cell B can be in- creased by
d)
i) increasing the voltag
(or current),
ii) increasing the concen-
tration of comper(11) sulphate solution, iii) decreasing the dis- tance between elec- trodes and 0.
1) The volume of gas in
gas-syringe x is ecual to that in gas- syringe Y, i.e. 22.4
des and Q after the experiment are the same as those before the experiment. That is, there is no change in their masses.
(f) The litmus solution turns blue because the solution left behind in cell B after the electrolysis consists of sodium hydro- xide which is a strong alkali..
g) The intensity of the blue coloration remains un- changed because amount of copper(II) ions discharged at Q is equal to that formed at, so
in that amount. of Cu
the solution remains the same.
Answer to 0.16
4
A.
a)i) CH1(g) +
202 (9) - 2420
CO2(g) + 2H2O(l)
ii) 200(g) +
2002(g)
b) According to the equation
(i);
1 mole of CH
reacts with
2 moles of 0 to give 1. mole of CO2
to
i.e. 1 vol. of CH reacts with 2 vols. of give 1 vol. of CO2.
10cm3 of CH will
3
react with 20cm of 02 to give 10cm of CO2
Similarly, according to the equation (ii),
.*2 moles of CD reacts with 1 mole of to
02
PAI JOE JE
日三十月一年七七九一腊公年六十六國民體中
b) Molecular mass of H50
X 0.230cm)
200cm
3.
40cm
30cm
3
∙10cm
the vol. of CO,
similarly,
d is, 10cm3 +20cm3
formed
=30cm3
c) The initial volume of the
mixture
= vol of CH2+vol.of co
vol.ofair
(10 + 20 + 200) cm3
230cm
3
The final volume of the mixture
= vol.of
2
+ + vol.of co2
1.of unreacted o
(200x0.8+30+10) cm3
=200cm
3
the change in volume is (230-200)cm3
30cm 3
d) % of N, in the final mix-
to give 2 moles of CO2,
3
20cm3 of CO will react with 10cm of 2 to give
20cm3 of co
CO2*
Hence,
is.
vol. of
used up,
3
+ 20cm
10cm 30cm3
the volume of unreact- ed oxygen is,
B.
3
is,
2
15,
160cm 200cm
% of CO
30cm-
3X 100% 200cm
X 100% = 80%
in the final mix-
15%
% of in the final mix-
3 10cm ture is,
200cm
3× 100% = 5%
a) A is hydrogen chloride, B is hydrochloric acid. d is chlorine gas, D is silver chloride.
b)i) Add concentrated sul-
Dhuric acid to the con- 'mon salt.
ii) Add solution B onto
potassium permanganate crystals..
c) i) NaCl(s) + H12SO4 (1)→→
NaHSO (s) + HCl(g)
4
ii) 16HCl(ac) + 2KMnO (s),
2KCl(aq) + 2MnCl2(aq) +8H20(1) + 50120)
d) i) A turns it red as A is
acidic
ii) B also turns it. red. iii) C turns it at first
red and finally bleaches it white.
e) i) The precipitate, D,
dissolves in excess. aqueous ammonia. This is because the Precipitate, D, has. reacted with ammonia to form a complex ion which is soluble in water.
AgC1(s) + 2NH2 (aq)——
[Ag(NH ̧), ]*(aq)+C1 ̄
(aq)
ii) The precipitate, D, turns from white to violet and finally to black.
This is because ne silver chloride is decomposed by light and the silver ion is slowly reduced to silver.
+
Ag + 1e
Ag(s)
C. Formula mass of iron (111)
oxide,
is Fe203.
56 X 2 + 16 X 3 = 160.
1 mole of Ee
Fe203 con-
tains 2 moles of Fe
contains
i.e. 1600 Fe 203
112g Fe.
the mass of iron ore required to obtain 1 kilogram of iron is,
*kg X
160 160 X
85 112
= 1.68kg Answer to 0.17
A.
a) i) 2NaOн· 4. (NH) SO
Na2S04 +2NH3 + ii) H2SO + 2NaOH Na2SO4 + 2H20
4.
98
the concentration
of sulphuric acid is 19.6
98 moles/dm3 0.2M
No. of moles of H, SO
2504 in
12.5 cm solution is,
12.2mole 0.2. X
1000
= 0.0025 moles.
According to the equation
(ai) in (a)
2 moles of NaOH is neutralized by 1 mole
of H2S04
No.of moles of Naон neutralized by 0.0025 moles of HSO
is, $2 4.
0.0025 moles X 2
0.005 moles
But, the original amount
of NaOH present is,
0.5 X
50 1000
moles
0.025 moles
ist
Amount of NaOH used up by (NH)So (0.025-0.005)moles
= 0.02 moles
4
According to the equation (i) in (a),
No.of moles of (NH4)2SO4
2
X no.of moles of NaOH
x 0.02 moles
0.01 moles.
Mass of (NH)90 in
the mixture is, 0.01 X 13291
1.329 (where M.W. of
(NH) 50 is '4'2 4 132)
Hence, the original mix- ture consists of 1.329 ammonium sulphate and (2.74-1..32)g = 1.429: sodium sulphate.
c) i) The white precipitate
is calcuim. sulphate.
ii) Ca2+(aq) + So
caso (s)
iii) No. of moles of (NH) SO4 = 0,21
(aa)
moles.
No.of moles of Na
1.42
motes.
142 M
= 0.01
No.of moles of H50
0.0025 moles.
** NO. OF moles of SOA
ions is,
(0.01+0.01+0.0025)moles.
0.0225 moles
2-
Since 1 mole of so
SOA gives
1 mole of Caso
4
and the formula mass of Caso is.
B.
136.
Mass of Caso that can be obtained as, 0.0225
X 136 = 3-069
+
+ sng
2-(aa)
S0219)
2 b) The tube consists of
acueous sulphur dioxide (or sulphurous acid) 1) The Potassium dichromate
solution turns from orange to green. Aqueous sulphur dioxide has reduced Potassium dichro- mate to chromium(111) salt which is green and itself been oxidized to sulphuric acid.. 11) The bromine vater,
which is brown, is decolourized due to the reduction by aqueous sulphur dioxide which itself is oxidized to sulphuric acid. iii) The clear solution.
turns milky due to the apparance of yellow particles, insoluble- in vater. The Darticles are sulphur Powder. in this case aqueous sulphur dioxide is re- duced by hydrogen sulphide which is a stronger reducing agent than sulphur dioxide.
+H SO
S
2425
H2SO3
2
30+ 35(s)
c) White Drecipitate is
observed. The Precipitate is barium sulbhate be- cause sulphuric acid is formed by the oxidation of sulphurous acid.
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