37tle
CITY HALL
育数化事
真三集張三纬
日六廿月三年申戉壓式
WAH
SAT PO #
B
764
ngs #X DEEFTALITE Y RMA
地理科
(十七)
劉五領。
GEOGRAPHY (17)
27. Explain, with the help of diagrams, the follow-
ings:
(a) Days are shorter in Peking than in Singapore
during De gember.
409.
The earth spins on its axis once every 24 hours, giving rise to the phenomenon of day and night. The half of the earth facing the sun will be lit up by the sun's rays while the other half will be in darkness. Rowever, the earth's axis is not perpendi- cular to the plane of the ecliptic. Instead it is inclined to it at an angle of 664 that in June, the Northern Hemisphere is tilted towards the sun while in December. it is turned away from it.
SO
Peking lies on 40° and Singapore, near to the Equator. Owing to the inclination of the earth's axis and its revolution round the sun, the length of the day decreases with increasing latitude north of the Equator in December ( see diagram ). Consequa, tly days are shorter in Peking than in Singapore.
Overhead S$300.
1 offìlude of Midday
Sam is an* »
day
da crim
Great circles are frequently used in navin ` gation.
A great circle is a circle on the earth's surface whose plane passes through its centre The shortest distance between any two pointe on the surface is an arc of a great circle- So, great circle routes are frequm tly folm lowed in nazigation,
Ax8
- a great circle route AYO - O small circle Pouda A*B is Sherfartian AT&
(c) When flying from
Ans.
ONGK
3ombay to Manila, passon- gers are told to put their watches on,
As the earth rotates from west to eat, all merdians to the east of a given meridian have sunrise before that weradian. Local tines along these meridians are therefore ahead of the local time of the given meri.... dian. Manila ( 121°E) lies to the east of Bombay (73°E). Therefore, when travelling from Bombay to Manila ( from west to east) clocks are advanced.
SAN
Petation from
(d) The length in miles of 1o of longitude
increases from latitude 80° N to the Equator.
MI
The length in miles of 1” longitude at a certainïatitude can be found by dividing the length of the parallel by.
360 since every parallel has an angle or 360° at the centre. For instance, tăia length between C and D in the diagram is equal to 1/360 of the length of the, equator, and AB gquals to 1/360 of the length of the 80° parallel. However, the parallels are not circles of equal mag- nitude, but they get progressively
smaller away from the equator, is a result,
the length in miles of 1 longitude in-
creages from high latitudes to the equator,
23. (a) A plane takes 6 hours to fly eastwards
from Calcutta (90°E) to Sydney (150°E).
→
it takes off at 10:00 p.m. on tionday in Calcutta. When will it arrive at 3ydney by Sydney local time AASA as the earth makes one complete
rotation from west to east in 24 hours, it turns an angle of 15
que hour.
Distance in longitude degrees between Calcutta Und Sydney 150 90°
60°
-
Difference I time betweekvense
4 hours.
Calcutta.
Sydney lies to the e. Therefore, its local time ould be shead of the local time at Calcut ta When it is 10:00 p.m. on Monday in Calcutta, it is 2:00 a.. on Tuesday by Sydney local time.
llowever, the plane takes o hours for making the trip.
Therefore, time of its arrival at Sydney 8:00 am. Tuesday
(by Sydney local time)
$7:00pm
Local time of a place X is 5:00 Tuesday when it is. 11:00 p.m. Hopaay according to GHT. What is the
Ans.
六期
日四廿券二年八六九一座公平七十五磊茂章中
1968
題額
化學科 (十七)
林錦衡
複習題十七
定性分析與鑑別法(一)
(I)選擇(A)所列之氣体填於(B)適當括號
4.
(A) H2S, SO2, Cl2, HCI, CO, COI, NH3.
O2, N2, H2, N2O
(B)(1)通入爽水中能全之
()
(2)燃燒生藍色火燭燒後注入石灰水 並不混濁,但若改以無水硫酸銅檢 驗容器硫酸銅變藍色.( )
(3)使高錳酸鉀溶液(硫酸化)褪色面
所得溶液混濁。()
(4)無色氣体能令湿石蕊試紙呈紅色 遇氨水生濃白煙,溶於水後加入硝 酸银溶液生白色沉澱(... ,令湿石蕊試紙轉紅色令高錳酸鉀 溶液褪色溶於水後加硝酸鉀溶液 生白色沉澱但再加盐酸則沉澱消 失.( >
(II)(A)氧化銅二氧化及硫化亜鉄暗黒色
固体,且不溶於水若加入藍酸加熱
(1)若生一種無色氣体,此氧体令硝酸铅 溶液生黑色沉澱則原物為. .** *
沉澱為
(3)若生黄绿色氧体此氧体会石蕊试纸
呈紅色後轉白色野原物為
若生藍綠色溶液無氧体産生则原物 是
(6)氧化鋅、氧化鎂與氧化鈣皆白色圃
先以鹽酸溶解之再以炤色反應試驗 若火猫呈磚紅色者原物是 仁者各加以氫氧化鈉溶液器生 色沉澱若加過量則其一沉澱溶解凍 一另一不溶物為.
IBPARIES
longa tuều
G.M.T. refers to the local time or the Greenwich (Prime) Heridian. Since the local time of the given place X is in fron of the G.M.T. for the reasons given in (a). I must be situated to the east of the Price Meridian .
Difference in time between the given place and the Greenwich Meridian = hours. Difference in longitude degress betWEDI
90° then - 6 x 15°
"Therefore, longitude of X = 90′′E
¿Ampw©je)'ara 00:01
Questions for next week
29. xplain, with the aid of diagruns, how çaon nasj Deen formed: block mountain. Fold mountain, residual mountain, volcanic mountain. Quote an actual example for each.
30. With the aid of diagrams, describe the formation
of three common features to be found in the lover courge of a river,
複習題+文解答ㄚ
(1) CuO + H2 → Cu + H2O
(2) ZnO + CO → Zn + CO2
(3) Fe203+ 3 CO→ 2 Fe + 3 CO2
(4) H2 Cr2O2+ 14 HCl →2 CrCl3 +2KCI
+7 H20+3 Cl2
(5) FeCl3 + H2S → FeCl2 + HCL+ S (6) 2 HNO3 + H 2 S → 2 H2O +2NO2+S. (1) 2 KMnO4 + 8 H2SOμ+ ¿Ò FeSou
→ K2SO4 *ZMAS04+5 Fe2(SO4)3
18120
(8) Cu + 2 H2SO4 -> CuSOμ + 2 H2O+ SO2 (9) H2S + H2 504 −2 H2O+ SQ2 + S (10) C + 4 HNO3 – Cu(NO3)*2NO2+2M2O (11)3Cu + 8 HNO3 (da) → Zlu(NO3),† 4 H2O_
+240
-
(12) Cut 4 HNU3 (CoxC) → Cu (NO3)2
* 2H3O+ZNQT. (13) SU2+ 2 HNO3 → H2SO2+ 2 NO2
(14)2H2S + SO2 →2H20138
(5) 5 H2SO2+ 2 KMnOù →→ K2 SO2 + 2 MÁSOL
† 2H2504+3 H2O
(16) 2 FeCl3 + SA Cl2- ~2 FeCl2+ SnCla
(17) 3H2S+ KəChÛz + 4 H2SOμ
→K2SOut Ch2 (5©n}z+7H2O+ 3 S
(18) X2 + 2 NQ2S2O3 + 2No25+ Na2SUO6 (19) H2O2+ ZAI + H2SOμ? K2SOμ+2#2O+łą (20) 2KM#Q + SH2S + 3H2SOu
→K2SO4 + 2MnSOμ + 8 H2O +55
(21) 5502 + 2 KMnQu + 2H2Q
→ K2SQμ † 2M150μ* 2 H2SO4
(22) 2 KMnO4+ JDKI + 8H3504
→ 6 #350μ† 2}#50μ ↑ PH2O + 512 (23) CÍ1⁄2 AZKOH ZKCI + KCIO + H2O
SKU + KUGTM 3H2O
+ H2O → HNG + HNO3