1968
#
10463 英文中學會考試題預習專欄
-3 FEB 1968
貸数備藥
買三第張三躬
日五初月正年時代歷
CITY
WAH KIU YAT PO HALL
1968
大題額習專桶
化學科
(十四)
林錫復
複題十四'
化學科 (十四)
·王錦釗・
Chemistry (14)
機署項目 重要反應方程式 (二)
填-
空格並完成平衡所列方程式:
化合作用
'
(2) Na+ Cla t
(3) H2 * Q
(4) N2 * O, ment
(5) cor q
(6) SQ+H2Om
(8) NH3 * H2O -*
(五)分解作用
(1)硝酸鹽之加熱分解
2. KNO
6. Pb(NO3),→
C. AgNO
d. NHWND
(2)一般碳酸鹽及碳酸氫鹽加熱分解放
出
氣
a CaCO, →
6. NaHCO3
C. (NHƯ CÔ
3)凡氧化劑加熱放出
a. KCIO2 →
2. Cu502 5H2O 2
1. (a) Three important uses of hydrochloric acid are: (1) Cleaning metal surfaces before galvanising or electroplating. Iron rust is removed by
pickling' iron in this way.
(11) Preparation of glue from bones." Concentrated)
hydrochloric acid dissolves the calcium“ phosphate present in the bones.
(111) Preparation of drugs, dyes, and photographio
chemicals.
b) (1) A test for a chloride ion in solution is by treating the chloride solution with silver/
(M)*C1-"
goluble chloride
iitrate solution and dilute nitric acid,
The addition of nitric acid ensures that :he precipitate formed is silver chloride and not some other insoluble silver_salt/ which is soluble in nitric acid.)
in solution
Ag*NO3TM
silver nitrate solution
}
AgCl
silver ohloride;
(H)"NO2***
soluble metallic nitrate in volutior
A white precipitate of silver chloride is formed. This precipitate is soluble in ammonia solution, forming a solution of a complex salt
of silver, [Ag(NH3)2]* 01′′
(11) A test for a chloride ion in the solid state is by heating the solid chloride with concentrated sulphuric acid in the presence of manganese dioxide.
2 [M]
一魏(有例外)
b. Hgo
C. K2Cr2O,
(4)水化物加熱會.
b. Cason 2 H2O
(5)濃硫酸對水化物或碳水化合物等
作用
Q. HCOOH
H500
6. Hal2Ou insoy
C. Cratta 0, MASON
a. Cusly SH2OM
複習題十三解答
(T)觀察法平衡
(1) Pu Q1at H2O → H3PO4
A greenish yellow ges, chlorine, 18 evolved. This gas will turn a piece of filter paper dipped in potassium iodide solution brown, owing to the liberation of áodine.
(C12 2X1
golden
(c)
'Metallic
Radical
12,
Potassium Sodium
Colour of the flame
Lilac
Intense
•
----
3.
Calcium
4.
Strontium.
5.
Copper
6.
Barium
Tast
yellow
Brick red
2XC) + 12/
Apple-graen
Observa Lion'
(1) Appearance X was white
solid
(11) Solid & was Solid was'
posed to the
aar for a few
days
HONG KONG
Colour of the flam” when seen through blus. zla sa
Crimson Invisible
Light"green. Crimson (Green Green
A (Infarar
Solid I is a deliqui
surrounded by escent substange
a pool of
liquid.
六期鼻
日三月二年八六九一靨公年七十五國民中
(111) Waterwas]
A solution
added to X
(iv) Silver)
nitrate |solution_was]
added to solution of
was formed'
white precipitate was obtained.
(v) AmmoniaTM
The white"
solution was
ppt. dissolve
added to white to form a ppt. from (iv) clear
⚫lution.
́is soluble in water
Silver nitrate
solution will give ä
white ppt. of silver chloride with solutions. of soluble chlorides.Í Hence X may be a chloride
Equations
▼
[M]_C1_+ AgNO3"
ARC1] [M] NO
This confirms that'
the ppt, is a chloride,
because silver
(chloride will.
dissolve in amgonis solution to form_a complex_ion.
Equation:
*
NOT
"AgCl +`ZNH4OH
<[Ag(M{3}a]*C1"+2120
X is therefore a chloride which is deliquescent, Thus X might be anhydrous calcium chloride CaCl2 or anhydrous magnesium chloride MgClą, both of which are white, deliquescent solids..
,b)
NaCl
23+35.5 -58.5
+
AgNO3
300
AgC1 108+35.5 .-143.5
NANO.
The precipitate formed is stiver chloride, which is obtained by reaction between sodium) chloride and silver nitrate. There is no reaction between sodium nitrate and silver/ nitrate.
Wt. of NaCl reqd. to produce 143.5 gm,” of AgC,
☐ 5815 gm.
Wt. of NaCl reqd. to produce 2.87 gu." of Agc3.
2,87
58.5 - 58.5 x
ga. *. 143.5
50
#t, of mixture of NaCl and NaNO
Percentage of NaCI in_the_mixture
questions for next week (15a;
–
- 2 ge
58.5
50 x 2 x 100 - 58.5
1. (a) Write an account of the extraction of sulphur,
by the Frasch process.
Describe briefly two methods by which sulphur is obtained as a by-product in industry, What happens when sulphur is heated slowly in a test-tube until it boils? Accomt for these changes:
What do you understand by allotropy? State
briefly how you would make two crystalline
forms and two non-crystalline forms of sulphur.
LIBRARIES
Cut HNO2? Cu(NO3)2+ NO+ Ho
# a Cu+bHNO, →→ © Cu(NO3) + α NICHO 考慮cau.
as e------
6
右方H為に
覆核
* PuOμ*6 M20-#4 H3PO4
左方右司
16 16
[M] 12 42
(2) NaCl + H2SO4 -* Na» SQμ ↑ HCI
2
1
因右方Mazi
28e/2 右方S I'
*. 2NaCl + H2 SOμ → Në 2 SQμ † RHCI
代入
d
0000
20+d
6c+d=3
40 = 3
Cù + 2 HNO2 Cu(NO3), + {NO + H2O
* 3 Cu + 8 HNO3 →→→ 3 Cu(NO3)† 2 NO+ 4 H2O
此法較繁但必可得
覆核
|左
右方
百便但對較繁
Ne 2
2
的方程式,則會屢試 不合無所適從
Cl 2
Z
左方右右
H 2
2
1
出結果適用於較繁
程式.
a 3
3
H
O
4
8
24
24
: (两代数法平衡
(1) Pb(NO3)→→→ Pb0+ No2+ O2
(四)原子價變運法平衡
a Pb (NO 2) → 6 PbD + © NO2+ d'O2
考慮P6.
N:
૨ વનાનો છે
任設一值予其未知數之一,最好選取 各式中最常見且馬上可代入算出别 值而不必解联立方程式的。覆核
* Q = 1, #ok¥ b=1
(3)44 c = 2
在氧化還原反應中,一元素之原子價增 加必有另一元素之原子價減少,且增減 之總数必相等。
3) H2S+ KMnQy+ H 2 SOμ→→ K2SO4 + MnSO2
TH2OT S
H2S被氧化為S; S Se
2
[左右右
Lez
2
BP: 5H2S+ 2 KMnO2
{N 4 4
55+2Mn504
KMing, & Mn504: My M
Ø×5+©×2; 55=+2Mñ">55°+2Mn?
+ H2SO4
(觀察得)3
5H2S+2KMnO4 + 3. H2SOμ
K2S04+2Mn50, +8420+55
生産品H2SO4,MASON 中之S是由 M2SO4 供给,三者之S箱+6價 並無增滅故不在12中考慮
(?) K2C10ƒ? HC12 MCI+ C/C/3+ HaQ+ Clá
AGO Craig Cr C 'HC被氧化成以
·C/°--(2)
'U)+3(2). Cr+301 Cr+3€/°/
KACAO, +3 HCl → CrCl2+ Cla
Cha!.
Kalno,+ 6HCI #2 [rCl2+3Cl
SHCL 2KCI+TH&C I (ii)
*. K2CO2+ 14 HCI →2KCI +2CrClz #78607308 注:(i) (iii)為用觀察法決定之次序
AHCI中有6HCI是供给CK用原字樓
Kai,
1-1 43 0,3 8 HCI Z
Crc/s的原子價不變皆為-5
(3) H2S + SO2 → H2O+S
MAS被氧化成S
So, #RS:
૨૮+().
职
25*
+2
5+4 -4 50.
* » མཁ
·25′+ 5*,
2 H2S + SQ2+ 35
+2H20
2 H 2 S + SO2 − 2 10 +35
此法平衡較繁之方程式甚有價值上面
歉例看似步驟麻煩實際上所有步酵箱
可在
或心中進行而不必寓出只將
最後結果填上便可。
代入(3)鲜之得d=
· Pb(NO3) ☆ Pb0 + 2No02+ ± 02 [0\ 12|12
• 2 Pb(NO3)2 P60+ 4N0+02
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