報日僑
“五期星 日九廿月一十(一九九一)年十八國民華中( 30 )
機大
透大
之 李 能有
工
生
平
大陸軍事工業企業
生產民用產品萬種
- 交通不便, 業多數地處山區
人矚目的成 就。 i已經取得了令 民經濟建設上 ‘製能力轉到唠惠
歉已能生產一萬多粗民用產品。一九九一年中 菜,掌握國外市場動向。例如我州的軍工企業 世界和平運動的新領域。目前中國軍事工業企 支持軍工企業到沿海開放城市和經濟特區揜企 平利用國防工業技術方面所取得的成績開創了一品,以了解國外市場的需求,另一方面又镜極 據中國國防科學工業委員會透露:中國和,方有外價渠道,積極組織出 上百種產品已進入國際市場 不靈,當地政
大一小的實依靠技術,開發、生產的高級豪華型旅游
一兵骨一百萬的同時,相應地削測了軍工生產規 僅窗口企業的產值即達四億元。江西省軍工企 ... 據透露,中國政府在一九八五年決定诚 產品出口產值已達二千多萬美元。一九八九年
卷、利軍一據
門的
衛軍工千
精”南石製品上市
有關資料證明 - 中國把强大的第工生產研
了滿足國內市場的需要,可工企業也生產了一 大批鍠工和家用器產品,如照相機、洗衣 大陸航空專家發出强烈呼籲 星托車的產量已佔全國生產總狂的百分之六十。了它們的技術水平和發展後勁。 機、電視撝-電冰箱、摩托車等,其中各種似 擴大了中國軍工企業的生存空間,而 汁等國家重點項目和重大裝備的以圈、研製、爲 北美等國家和地區,一九八八年民用產品外貿 門海承擔了莱山核電站,北京正負傘子對掾機 八種進入國際市場 - 產品遠銷歐、東南亞、 舶、吃車、火車車廂、發電設樑產品,工部 元。山西省軍工企業生產的民產品,包有三十 產值超過三千六百萬美元。 中國軍工企業的產品進入國際市場,何在
航空工業陷入低谷
國家必須大力扶持著洗洗出道
在發展問津。
航空專家們現代國家實力的主要 中整作用的密集的高科技產業,是 .骡方面起着工業是知識密集、技術 中铟航空工. 專家們指出,航空
展中的地位和作用進行再認識,軾正把航空工業放在應有的地位 "_依靠科學技術進步,才 航空工業的「頂蜃設計∫(宏餓規劃 ),應對航空工業在國家安全和發第一生產力」的政策, 家要加强對一只有落實「科學技術是
「鞋中國航空工業存在的url,打到达正使中國航空工業
振興之途?航空航天部,
戰攻家召開了「科學技 第一生產力」的保 科學技術研究院本月組 才,是落實一科學技術用,而大部份成果無人
術是第一生產力」的專 說,與吉良家們對此選 深圳假人民幣
業企辦就口用地 山工就了
受尤深。他們說,現在
中國的民航和空軍,人人自不穩定。航空科研 姊家們認爲·花展 科技隊伍不合理,科技 搜和航空工業的未來,一七、八年,而成果獎發 年深圳市發現的人民幣假鈔比往年明顯增多。
·們寄希望於中國航空科 很辛苦,有的項目長達
今年明顯增多
深圳市每年均可發現人民幣假鈔,但是今 連中國銀行深圳分行出納威的不完全統
消除需要航空工業部門下來,平均每人可能只
·一生產力]-涉及到思 礎上、管理上、決策上一 风o. 离家说- 堅持 實,中國航空工業和西 商反映了深圳市社會上貨幣流通中人民幣假鈔 合,四方鼎力,八面來 審知微子政策的落 共收繳的人民幣假鈔總數的一半多 - 從 | 個侧 策、支持、理解和配主性、大鍋阪等無不影 假鈔總數的百分之八十三點五,是過去十年來
發展的劳頭離以持續之上是八四年版的、窩機製印假另
和落實「科學技術是第 方發國家相比,在蒸明顯增多,中價實的假人民幣面籍分
段時間由於種種原因, 很大發展,但是後來一待解決的問題。據最近部門在收免國庫券中,發現了部份假國庫券, 期間,中國航空工業受品化,加速科研成果的印,溼案印製粗糙 - 膰次不亮,正面梅花底線 到國家高度重視,獲得 推廣應用 - 亦是一個亟 秋由不同顏色焿合,被綫不范晰。另外,有關 持。在第一個五年計, 科技成果的工程化與隨張,紙質比真鈔紙質鬆軟,而且救類,無水 想觀念、發展戰略、政和人才等方面差距較 中五十元面額和一百元面額的人民幣假鈔多是 對楊凝方面,雄 囓要政策扶持。”的一九六五年版的擔元辦人民帶假鈔,均是1 策導向、資金投向、人大,要開小這個差距,電子掃描、拓印和攝製店印的。最近首次發現 、重視、玄米:與會專家說,加速顶70824116(號碼),其特點是紙張爲普通紙 一元分塗改爲五十元的假券。
台灣中小型企業
·中國「七五一簪家 學重點科技攻關課癌之一
成衝擘。.... 增加一倍,而今年經濟衆氣雖然回穩,但新竹 的科學數據證實:晚婚生 幣快速升值 - 勞工普遍缺乏,又對中小企業造 有的孩于整格發育狀況明 與告業註飽或歇業家數高達七千餘家,比前年作的大型科研活动以詳盡 於受到經濟景氣不振的影響,去年台灣中小型的科學決策。這項歷時八
·經濟發展中所佔比近正在下降。他並透露,由邊結婚的政策是恰到好處 .接受立法院質詢時指出,台灣中小企業在台灣門提倡的婦女二十三歲以 最近台灣铯凌部中小企業處處長王党民在一告表明:中國計劃生育部 ...顯優於非晚育的孩子。.
與母親分娩年的研究報
: 二王覺民說,去年關門的七千多家中小型企·透項研究是由湖南醫
營運難歇業者衆 清新生兒報格音方
台灣未来的經濟發展中,中小企業仍然扮演極 教授張寶林婚夫婚發起一 降,而博統性產業在海外投資者亦以中小型傳 健單位的近五百人次參與協作研究 辛 大陸投資持續成長已達二千多次,投資的行業新生兒的體格發育狀況進行調查登記,結果照 醫. 王覺民透露,近幾年來,台灣中小企業去一萬八千七百二十四例妊婦及分娩的單胎活產 [ 以勞力密祟的三次加工爲生,如製鞋業、電示:小於或等於二十三歲的非晚婚婦女生下的 七億五千萬美元,他認爲,這種產菜外移現 標均比大於二十三歲的晚婚滑女生下的新生兒 工、自行車、塑膠製品等行業,總投資金額達 新生兒,其證重、身溉、頭面、胸圍等發育指 重要角色,但是中小型企業所佔比重正在下主持工作的,先後有全國各地七十七個磐療保 他們對南方七省、自治區的十二個城市的
象,値得警惕。
十八國民華中, 有教息訊岸兩日四廿月十年未辛曆夏
華
廣告
巢七工以大
小字卅四個
上十元
但也不應過於晚育。
1.通過對發育指標的綜合分析·專家們認 育最優。另外 - 他們遵提醒人們;不要早育, 爲婦女二十四至三十四歲之間分娩的孩子, 祖。尤其烱於二十歲以下的姉女生下 網格發直飛好,其中尤以二十四至二十九歲生
下的新生兒的六項發升指標都明顯優於非 「五個城市的二萬四千一百五十個新生兒及其母 觀的狀況進行更全面深入的調查研究。結果進 一步鹃示:二十四歲至三十四歲晚婚頴婦女生
這是20年的百分比卻比
七稿惡登 - 每段(大字四個,
·本報分類廣告·歡迎各界來. 過一個位,现多出字數控一
小字卅四個 } 定爲一個位. 每天收廿二十元,刊牟如超
*供午膪有意碴 458
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1992 中學會考預習專欄
明德出版社: MILL & DALE PRESS.
Chemistry (8)
Unit 5 Revision notes.
calculations
Basic concepts
明表果成究研學醫陸大
歲九廿至四廿 期佳最育生為
1.7 The mass of one mole
constituent particles
of.the.
of! 习 substance is often known as its
molar' mass (M). Its unit is
9
e.g. M(Mg) 24 g mol
0.4% 1000
0.01 mol
1.13 A Molar solution of a substance is one that contains one mole of the substance dissolved in 1.dm of the solution.
M(H250g) = (1x2+32x1+16x4).
98 g.mol M(MnOq^) = (55x1+16x4)
-119 g mal
2.
11: LAM
onchemical
1.8 Thus, for a substance X,
n(X) Mass of X in. grams
M(X) in g mol
e.g. M(CaC)) = 111 g mol
̇n(Catiz)
in 2.22 g of CaCl,
Calculations on Chemical Formulae.
2.1 The empirical (simplest formula of a compound show, its constituent atoms and their simplest integral ratio. imp e.g. Ethane has an empirical
formula of. CH, indicating that it is composed of C and -H. atoms in the ratio 1:3.
of substance shaws athe actual number af each kilop atoms in minis molecules..
2.2.The molecular formula.
e.g. Ethane chasaw the
formula of Cg
Thus," for
covalent
mole
of a
gas rust contain the
X + H2SOA
XSO
mol
mol
same number (the Avogadro' number) of molecules.
Q.6 What is the 'volume of oxygen
required to burn completely gaseous
每
mixture consisting of Now 1.3 g
40 cm of methane (CHg): and. 45 cm)
of carbon monoxide (CO)?
Q.7 50
Cm3
of a mixture of methane (CH4) and ethane (CH) required exactly 115 ofoxygen for complete. combustion: What is the composition by volume of the original mixture?.
0.8 12 cmd of a gaseous hydrocarbon..
requires 36 cm3 of oxygen for complete combustion, giving off 24
of carbon dioxide. `Calculate its molecular formula.
1.3
22400 448
x = 65
22400 cm
at's.tp. 448 cm3 ats.tp.
Q.6 N.B. Write a separate equation for the combustion of each component...
CHg(g) + 20g{g) - CDy(g) + 2H2O(1).
2. vols (by Avogadro's
1 mol
-2 mol
ie. 1 vol
40
?cmi
Val of Oy required. by CH4
2 x 40
80
200(9) + O2(g)
2002(g)
2 mo}
1 mol
Soluti
Fems
QUE FOR
Xas of Na200 ̧•101,0 = 286
23x2
1.e. 2 vols 45 cm3
of Na=
16.1
x45
% of C
% of 0.
100 = 16.8
The relative atomic mass (A) of an element is defined as the mass: afan.,"averager: atom of the element relative to the mass of an atom of carbon-12, taken as 12 exactly.
A of an element
Mass of an "average" atom of the element Mass of an atom of x12
carbon-12
It is a ratio and hence has no units.
The mass of an "average" atom of an element is the weighted mean of the masses of all its isotopes, 1.e., taking both the isotopic masses and proportions ·înto account.
e.g. chlorine exists as two
35
Isotopes, 37C1 and 351, in
the ratio. 3:1.
Thus, A. (ci}
35x3+37x1 3+1
35.5
1.2 The relative molecular mass (M)
1.4
of a substance is just the sum of
the relative atomic masses of all the atoms
in contained
molecular formula.
̇e.g.M(H250g):= 1x2+32x1+16x4
0.02 mov
The molar volume of a substance is the volume occupied by one mole of its constituent particles at a particula temperature and
pres
Moldr olume Molar mass of X
Density of x
1.1Đ A11 gases have the same molar volume. at the same temperature and pressure.
The palm volume of gas is about 224 at st.p. but 24.0 do at 'temperature and:
its
(e.g. n(02)
The formula mass (Fn) of an ionic compound is just the sum of the retative atomic masses of all the atoms: contained'in its empirical (simplest) formula.
me.g•F(CuSO•5H2O)
#63+1+16x4+(2x1+16x1)x5.
#249.5
F (Cr2oy
52x2+16x7
216
The male is defined as the amount of a substance which contains as many elementary entities of the substances as there are carbon atoms in exactly 0.012 kg (12.00 g).of carbon~12.
The elementary entities refer to any type of particles that may represent
the
chemical constitution of the substance under investigation, e.g. atoms, molecules, ions, electrons, etc. n(x). denotes the amount of substance X in moles. e.g. n(20) = 0.2 mol.
1.5. It has been found that 12.00 g of
carbon~12 contains carbon
6.02x1023
atoins .
refers
6.02x1012
This number,
is called the Avogadro number denoted as L. Thus, a mole often
the to
Avogadro number
of any particles particula. kind, e.g. atoms, ions, molecules, cars, etc.
1.6 One mole of a substance is just its relative atomic mass (for mass
atoms), relative molecular
(for molecules) or formula mass
(for ionic compounds or ions),
expressed in grammes.
圖文傳真
5594238
ic pressure.
gas. K..
Volume of X at s.t.p.
they
at s.t.p.
72 cm 24000 cm
0.003 MOT
one
substance X.
Molecular Formula of X
nx Empirical Formula of X- where n is an integer.
the Q. Calculate
percentage composition by mass of hydrated sodium carbonate (NaC0, 100).
Q.2 A hydrate has the. following
percentage composition by mass: 20.25% iron, 11.5% sulphur, 23.00% oxygen and 45.30% water Find its empirical formula.
Q.3 Hydrated magnesium sulphate (MgS0 xH20) consists of 48.9% by mass of the anhydrous salt, what.. is the value of x?
3.
ar volume of gas.
at sit.p.
0.4
oxygen gas
1.11 The concentration of a solution is commonly expressed in terms of the ̇riumber of moles of the solute in on cubic decimetre of the solution. It has the unit mé) dm-3
This is also known as its molarity and denoted as M.. Thus, for a solute X,
( Concentration,
of X
·in ol mor
Molarity)
Amount of solute in moles "Volume of the solution in dm3
in
of
its
e.g.21.2g of anhydrous sodium carbonate is dissolved water to give 400 cm solution. What is concentration?
Now, M(Na2CO2) = 106 g mol-1
a[Na2CO3)
= 0.2 mol
106
Vol of solution = 0.4 dm?. concentration = 0-2
0.4
= 0.5 mol dm
(or, 0.5M)
-3
1.12 The amount of a solute contained in a known volume of its solution a' known concentration is readily found. by using the relationship below.
Amount of solute Molarity x vol. of
in males
solution in dm3
ㄓ
e.g. n(KC1) in 25 cm3 of 0.4M KCl
solution
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●每日下午五點半截藉
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(告類
Calculations on Chemical Equations 3.425.g. of Teadlla, Hi, IV} oxide is. reduced to: Jead using hydrogen gas. Calculate
(a) the mass of lead obtained, (b) the volume of H2 gas reacted,
measured at $.t.p.
Metal Bis in Group II of the iodic Table, 1.5.9 afrit is whead to some witate hydrolog acid, hydrogen gas is collected at st.p. and 0.2 g of the metal is left behind. What is its relative atomic mass of B..
Calculations on Volumes in Gaseous. Reactions
one
4.1 Gay-Lussac's Law of "Combining Volumes states that "when gases react, they do so in volumes which bear
to a simple ratio another, and to the volumes of the products. if gaseous, provided that all volumes are measured at, or corrected to the same temperature and pressure."
of H2O = 1DXDD x 100 = 52.9
19x10 286
Q.2 Mole ratio of Fe: S: 0.
1 'vol (by Avagädro's Law):
Vol: of 0, required by co =
=22.5cm
Hence, total volume of 0, required
80.+ 22.5
102.5 cm)
Q.7 M.B. Write a separate equation of
the combustion of each gas.
Let the volume of. CHg be x.cmd.
Then, the volume of C2HG 1s:/ (50,- x) cm3
20.25 11.50
0.36 0.36:
0.36
20
2H20
4: 7
1 mol
2 mol
its
empirical formula 15
i.e. 1..vol
x cm3
2 vols (By Avogadro's.
Law)
7 cm
Case of hydrates, the crystallization is
a single, separate
0.3 M(H_0) = tag mal
= 120 g mol M(504) M(MgSOg*xH20)=
= (120+18x) gmo1~1
% by mass of the anhydrous salt:
12
120+188
= 48.9
-x 100
(given)
迎X100 = 48.9.
= 7 (It must be
a whole
number.)
Q.4 M(Pbg0g) e.g. It is found that 50 cm3 of nitrogen reacts with. 150.cm3
of hydrogen to produce 100 cm3 of ammonia.
Thus, vol. of Naivo. of 2:vol.
= 50:150:100
= 685 g mol-!
N(Pb) = 207_j mol~1
Let the mass of Pb obtained be x.y and the vol. of Hy used be y m。
3Pb + 4120
3 mol
+
of NH3
1 mol
1:3:2 (simple whole-number
ratio)
art
equat
states 4.2 The Avogadros' Law
that equal volumes of all gases, at the same temperature and
pressure, contain
number of molecules. The reverse is also true, i.e. equal number of molecules will have equal volume, measured at the same temperature and pressure.
Avogadro's Law volume of gas
number of Molecules
4.3 The Avogadro's Law also explains
the why gases have
same nolar
volume at the same temperature and pressure.
This is because, one
1.e.685 g (4x22.4) dm3 (3x207) g Now 3.425 yin3
Vol. of 0 reacted with. CHg=2x cmd
20246
+
702
2 mo1
i.e. 2 vols
-4002620
7 vols (By Avogadro's
taw)
(50-x) cm3?
Val: of Oy reacted with C2H5
- 1⁄2 (50 - x) cm3
total volume of 0, reacted
2x+ 3 150-x)
115
(given) 1.e, 2x + 2 (50 x}#1.15
-
X. 40
Q.8 Let its molecular formula be C Hy
+ (x + 1)02 — XCO2 + H20 Gy
(x+ mo?
i.e. 1 vol
12
cm3
(x+4) vols x vol (By
12(x+) c
Avogadro's
Law) 12x CM3
Vol. of CO, farnied ' x 12x
XQ
X =
(a) x 3.425 (3x207) : 685
3x207
x 3.425
=
3.105
x = 2
(b)y: 3.525 = (4x22,4):685
4x22.4
* 3.425
= 24 (given)
Vol. of 0, reacted « t2tx + 1 -
= 36 (given)
1.e. 12(2+) = 36 (as x = 2)
y=4
= 0.448 din3 (or
448 cm3)
...
0.5 Let A of 8 be x.
Hence,
Mass of s reacted = 1.5-0.2 = 1.3g
僑菜
報日僑
(刋明)
(出) 【含】
its molecular formula is
(廣分) (告類)
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