TIBHAR Bλ#A=#028I WAH KIU YAT PO
$%$# 5%S
1969
・
報日僑華
四期星日五十月五年九六九一屦公年八十五國民華中
̇育教僑翻
Solutions
Both sodium acetate and sodium carbonate are salte formed from strong bases and weak acidé. When they are dissolved in water, hydrolysie occurs, resulting in the formation of a Large amount of hydroxyl ions, which turns. Litous blue.
2Na
2820
201
28
《一九六九年中文中會考試題預習
數學科
(廿八) 喬仲强•
第廿七次預習題解答
(1)三角形之边各為207,417,及480 不許用表,証
明其中一角恰為60.0
(註)因:ABC為任意三角形,故其中一角
笃60°者則以角必不為最大或最小
化學科
ቲቲ )
·王錦釗。
CHEMISTRY (27)
Which of the following substanoee turn iitrus blue?
ammonium chloride
II sodium acetate
III sodium carbonats
IV sodium chloride,
sodium nitrate
Which of the following alternatives
III only
I and II only.
C II and III only
I, II, and V only.
II, III, and IV only
Which of the following substances era. oster?
#CO.
(molecules)
The formation of molecules of carbonio acid
(which is a weak acid) means that there is an
excess of OH fons in the solution.
A similar reaction happens when sodium acetate 18 dissolved in water.
CH COO.CH is an ester formed from acetio
acid CH3COOH and ethanol C2H2OH.
HCOO.CE, is an ester farmed from formic acid. ECOOH and methanol CH OH.
ethyl acetate:
dCOOK
CH2OH
HOOO.CH
methyl formate
3. DCH CH
KH CHO
CHCH acetaldehyde
CH2C
CB COOH
#133 a=207,
=417, C-480.
依餘弦定律
atc f
480
480-47
cor B=
zac
2X207×480.
99360
480
E.D.
現
12849+230400–173.
2x207x480
#k B=60°
ac
5=2 (207 +417+480)=553
-f-552-417=135
1552x1354=隻.
207×480
cos 30718=30° tk 8=60° ($)
(2) A, B 第一河岸两奌相距200碼;
ZCABR LCBA 494 47°R 72: $19.
(解)設如右图△ABC中,
C = 180° ~ ( 47° + 72°) = 61°
依正弦定律
sinß
AC-
ABsin B Ain C
200 sin 72°
AC = sin 47°
NO
LOG
II CEO. CH
III CH GOO.
IV ECOOCH
CH-COOR
Which of the following alternativés is corréat?
ITI only
BI and II only.
C III and IV only.
D II III and only
E All except I
If one mole of ethyl alcohol is oxidised by one molecule of oxygen under suitable conditione, the main product of reaction is
carbon monoxide.
carbon dioxide
acetaldehyde
acetic acid.
E formaldehyde
When a current of 2 amperes is passed through an electrolyte for 80 minutes, I c.o. (at N.T.P.).
of oxygen gas are liberated. If 1 Faraday may be taken to be 96,000 coulomba, what is. „X? ̧
A 22,400
B 11,200
C. 1,120
acetic acid-
D The gram-molecular ut. of
The gram-molecular vol. of 6
-12 grams
- 22.4 litres
The gram-equivalent wt. of 02 The gram-equivalent volume of 05
(at ..P.)
8 grams.
- 5.6 litres
(at M.T.P.)
Volume of 02 (at N.P.P.") liberated W 1
Faraday of electricity
If current
and time
80 minutes
A
5.6 litres.
J
2 amperes
4,800 sess.
Quantity of electricity - 9,600 coulombs
t.T.P
Thus volume of 02 liberated at
litres
X #st. A ACD &
CD= AC sin 47°.
15.9
答:河寛 159
(D) 117 AD-hcot 47
BD-h cst 72° 200
cot 470 + cot 720
以解之
200 2:3016
sin 727978z
sin 47° 78641 2.1433
sin 61T941 8.
∙159 12.2015
(3) W2 # A B C D & * AB=12.8 BC=116, CD=76,3
DA=74, RAWAC-110 BD ZE
(解)在△ABC中,依餘弦律
Cot of ==
11.6 +11.0 -- 12.8°
2X116×110.
1346+121-163.8
2552
=0.3597
2=68055
Cos pin 11.02+76274a
9/8
255.2
查表得
XA ACD
查表得
== 42°14′
Z± ABCD 4
12.8
2x110x74 167.2
124 -0.7404
< BCD === || 1° 9'
BD*— 116"+ 76 — 2x 166 x 76.com,
D 560
E 280
Double decomposition will take place in aqueous Solution except when
one of the products is a gas
one of the products in insoluble in water one of the products formed is not ionised one of the reactants is insoluble in water both reactants and products are ionised in Water-
The following compounds yield a common gas on besting, except
A
ammonium nitràte
B
lead nitrate:
Csilver nitrate
D sodium nitrate,
E
zino nitrate"
7. Ordinary common salt gets 'wet' easily becaus
contains calcium chloride as an impurity contains magnesium chloride as an impurity. sodium chloride is deliquescent
A
B
DBodium chloride is efflorescent
E sodium chloride is hygroscopio
8. Which of the following reactions is not affected
by light?
B
combination between chlorine and hydrogen combination between chlorine and methane decomposition od dinitrogen tetroxide
.D. decomposition of silver bromide
E photosysthesis,
9. Which of the following malte, do not form hydrates!
I
II
Calcium chloride
III lead nitrate.
copper sulphate
IV
potassium sulphate
V.
sodium chloride
Which of the following combinations is correct?
A
II only
B
I and IV only
C II and V only
D. I III and V only
E III..
IV and V. only
10. In a certain experiment 1 c.c. of soap solution was required to give a permanent lather with 25 c.c. of distilled water, and 15 d.q. of soap solution were required to give a permanent lather with 25 c.c. of a given sample of water. After boiling the water first, only 9 c.o. of soap: solution were required to give a permanent lather with 25 c.o. of water.
What is ratio of temporary hardness to permanent hardness in the water?
2.+ 3
B. 34
C
3:5
4 13
E
4.1
560 #1.
5. B When BaCl solution is added to Na,504
solution, say, the following reaction coeursi BaCl2
- BaSO +2NaC1
white ppt.
But suppose potassium chloride is added to sodium sulphate solution, no reaction takes place for the liquid mixture contains a mixture of iona, which do not interact with each other.
Ci
2Na
504
6. A All the other nitrates will yield oxygen on
heating. T
7. B
*.g. 2Pb(NO3)2
2NaNO
- 2P30
+
4802
2NaNO3 ZNA02 02.
But ammonium nitrate yields nitrous oxide on heating
NHNO3 NO
2820
Pure sodium choride does not get "wet"1.0. it is not deliquesgent. Common salt is impure @odium chläride. One of the main impurities is magresium chloride which is a deliquescent substance.
8. C All the other reactions are greatly affected by ligat. For instance, hydrogen and chlorine will combine slowly in diffused qunlight, but a mixture of the two gasem will combine with explosive violence when exposed to direct sunlight.
9. E
Dinitrogen tetroxide will not decompose by exposing it to sunlight..
Calcium chloride forms a hydrate CaCl.61,0 and copper sulphate forma a hydrate Cu30.58 0.
The other three salts form crystals which are not hydraten:
10. De Volume of soap solution required to redőve
"temporary bardness" in 25 ml. of water
(15 9) 0.0. 60.c.
Volume of soap solution required to remove "permanent hardness" in 25 ml. of water
(9-1) cic.
= 8.0.0.
टस
Ratio of temporary hardness to permanent hardness 6.18
34.
-134-6+57.76 +281], 6×76 cos: 68°5)".
=192.36+63 62
NO.
209
=256.0
2
10.30/0
M
1.06450
7.6
0.8808
BD=1756.0-
*: BD=16.0
cos 68°51′ 7.55734
63.62 1.8036
(4)在4000呎高之飛機上,測得地面上A,B两地的俯
·角容為35°10及2730A在飛機之正西,而B在北 偏西14.求AB踩離
(解)設ㄗ為飛機的位置,
C為其在地面上射影
AAC-4000 cot 35°10′
4000× 1.4193
5677 (28)
BC-4000 cot 2.7°30
=4000 X 192} 0
-7684 (%)
#SABC, LC=90°-14°= 76°
AB2 = 5 6 7 7 2 + 7684–2×5677 × 7684 004 76°
-322 3,0000+ 5904,0000-2111,0000 -7016,0000
AB 7016,0000 = -8376.
C:AB相距8376呎
(5)(a)試利用面積公式
sin (A+B)=
14000呎
2
0.3010
5677 3.7541
7684 3.8956
cos:76° 7.38376
2.111×10 7.3244
A=Labsinc, $**g
== sin A'cos B+ cos A sin B.
(SE) 47 CD LAB,
CD=b sinA.
X ABCD
ACD
AACD.
AD=&cos A
CD=a sin 8,
BD= a cos B
ADX CD=1bers Ax a sin B ab cos A sin 8.
ABCD=/ BDXCD = = a cor B & bain A= £ absin Acos B,
X^ABC=&ab sinc = £ab sin [180°—(A+B)]
=abain (A+B)
$1 sin(A+B) == Ain A Cas8+ coś A sin 8. [3x$1
(E) (a) 11, £ A=B, 4] sin 28 — 2sin8 cos B.""
依正弦定律:我
再依法老律
"simb
Rac
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