報日僑華
WAH KIU YAT PO
QUESTIONS FOR NEXT WEEK.
KB + 160 - ALE÷14 KNO
頁二第張六第二日六十月二年卯乙歷
育僑華
史學會考專欄
化學科(十二) 陳湛杰
CHEMISTRY (#20)*
Solutions for chemistry #19)
1. The empirical formula of the organic compound may be calculated
Ca
ARMOTIOJ ET
Element:
-composition
At. vti
% composition
At. Wt.
changing to, whole
mumbai catio
Tous empirical formule je 62. the compound is treated with. When sodium, sodium chloride is formed. The andun of, glorile in gi fan equivalen a from 15g of crifinal compound is 1.5x0.620.0.0265
Amoun of 0.450, 450, required. for this titration.
0.0265
100% 1000-1.
58.0 ml.
For the vapour, density measurement the volume of vapout From organie, compound at NT.P. conti 5-
ions is 4102-762-2121. = 300 9mli Applying Avogadro's. Hypothesis the. number of gn.moles or was
of tite pound is 30049
This Molecular wtor compound
$390-9
1.5g
The molecular ət, calculated from the engiical formule C
= (3×12) + (6) + (2×35.5)) Thus the molecular formula corpoună" is the same as the
formula, ili Cl
2. Subetaine,
113 for the empiri
The substance is likely to be a bromide because aqueous polusions of bromides c insoluble silver brotide in the s présence of silver nitrate solution Silver bromide is pale yellow colour,
In dilute
nitric acid but schovihat soluble in AMGORIA. The lilur Clane is chara- cteristic of potassium compeinde. Thus A is potassium bromide and the reaction involved uge-
"(N.B. Zde, following sentence was
omitted in the question when hosNBÁ 7 gap is evolved that tums lime? watch milky The residue which remaine Lo yellow when hot but
to a white raver
folour when
-Subetance B As the powdery sutr stance yielded carbon dioxide upon heating it is either a bicarbonate or an inagluble carbonate. Since there is characteristic colour change for the residue during Temp erature tanges resembling that of zinc oxide, this is a likely possi-. bility."
This is further confirmed
by the fact that this residus dis- Golves, in'dilute anid, and give a white precipitate on addition of
ammonium sulphide. Pie conilunsa.
to be sine carbonate and the react- ions involved are:-
2100, - Z¤0 + Co, ↑
ZnCl2 + (MA) = 2:54 + 2NH CA
Subetance C The grain 'wid=cance sulphate since the solution " gives with acidified larium chlorie solution 2 whito precipitate. Il
fast
is also probably a Terrous salt, as ferrous Sults are grec The that with sodium hydroxide at yields a green precipitate strongly suggests such a conclusion. Upon rezgénue to siz, the green precipim tales tuvhed dark green and then
This is explained, by pro- greeely oxidation of ferrous hydroxide to the brown ferrie hydroxide. Thus ta ferrous sul phate and the reactions, are:-
FSC2NaOH = Fe(OH)+ &,50 14Fe (OR) + On + ?H2C = 4P¢ (OH)3.
FeSO+ Bac
Baso + Feci,
Subetande D An the substance. is of high density it is likely to be alicavy metal. This is further supported by the fact that it does not react with dilute pulphurie. acid, but however, 19 attacked by filute nitric acid.. Dre probably a metal between lead and silver ip
The the electro-chadical ceri93. fact the the solution obtained ives a black precipitate with
sulphide further confins hydroger that this is either lead, copper,
for silver. However, among mercury these four saly zate of lead will Είναι
a where precipitate with sodium hyd calds that is soluble in excess of alkali Thus lead. and the reactions are:
· 3Pb + &=No 2 = 3Pb(N), }; + 2NO ↑ + 4H2O P(NO)+ESPDS+ + 2HN();
P(OH) + NaOH - Ne
九六三正十 注特 意級
EEE 點時的
間片
+220
1. Explain what you understand
about oxidation and reduction, give,
examples of these reactions and indicate which ones are revengible.
Solutions of ZOI an 2.
26-Rail of standardised us follows, the acid neutralizes: 25. l. of the hydroxide and 41.261. of the hydro- xide, Lo end to titrate 200g of 4 benzoic acid, Benzoic acid has a molecular weight of 188.1 and near vỀ with FOH according to the following equstion,
In titrating the borzoic acid ther endpoint was 2536, en 2.50ml of NO solution was required to com plete the titution. Cazould to the molarity of the NC, aslution, and of the OH solution:
·0975:
中學會考專欄
堅道壽院主編
五期星
The total p.d. should be 0.5+0.4° = 0.0 volts
(c) for two wins in derise current of · Cy! A m
each bulb should have a plà... of 10.12 volte. For two pulbs. 14
series, requireg 0.24 veits.
4240 of 1.0V ic spied no rost. those two huts khách pre connect~
series, they share the p.
Each rear pad. of ani, the current fiqwing is
(2)
be az in pral.el, they have those of 1.2v ACTORS 1. Ee bulb takes 0125A. That Pay total current of 0.5A draııın (row the nano, é currunt, e of 0.4A) is trenched aquadly Into two breary each bear C.2k, 116. p. équired, we used 0.24 through
lamang is tur
The
2(n) Ohia's Lar
FOR 2 given comic tor, at
constant, temperature, the current through it is directly proportionT
to potontiul difference across it.
1
物理科(二十)
BTR
and
PHYSICS (20)
pid. in veli
Current in. ampi
02 025
(a) The slope of a V-1 graph denotes the resistance of the bulk the graph plotted is not a straigh Tins
Its clope increscer, mean- ing that the resistance of the filament varies with temps reture. Its resistance increases ac Temperature riseb..
(b) For a resistor or 2 ohms, Sho graph should be e straight lin with slope of V
2.
when the bulb and the resispor fre connected in series with a current of 0.2 A flowing, the bulb should.. have & p.d. of 0.5V across it.
(Read from graph A) and the
realator takes z p.d. of 0.4v (from graph E).
Current in amp.
By Ohm's Law
where R is the equivalent resla- tance of the parallel combination.
Volte
(i) they obey Ohm's Law (streight line eritha)
(17) The slope of each graph
meosures the resistance.
4olins:
R2 ohms
R1.25 chas
(111) If R R and Rare
羅
series, and a current of 1.0A passes through them.
Tenuis à vid. of
都新華新莊
日八廿月三年五七九一层公年四十六國民華中
4V to Rond 1A to pass
ay to Band A to pass R
125V to vend i to pass R
pr a total of 7.257 to send A in passing through the cireult. The toul equivalent resistance
187-25 ohms. (iv) If they are connected in
rarallel and e pid. of 1 volt applied across it, the currents in R is 0, JA in R is 0.5A in Rs 18 0.8A
(a)
the total current will be 1.64. Their combined resistance in
B=0.625 olma
Let E be the evi.f. of the gell. The equivalent resistance for the
4- and 12-s resistog ie,
4x12
Roma Therefore,
P=0.4(2+3)
2.0 volta
The current flowing through 4-ohme
resistor ik HT=04X2 = 0,34
(e) Cost of electricity coneuved.
per day $1.80. Cost per "unit" $0.30: Amount of electricity consumed
1.80
per day 0306, units
= 6000 W - HIS
Power of the device,
6000
d
24.
250 watta
Mains voltage, V =200 volts. Resistance of device,
| V2 40000
R==
P 250
160. ohne
If the main voltage drops to 80% of 200 V, then, the power of the device becomics F
= 0.64
=0.64 x 250 160 watts Questions for this week 1(a) The diagram showa a auft
irori alectro-magnet and armature Show on suitable diagrams; the direction of the lines of force and Nand 6 poles whent (1) the armature is moved well
away from the magnet. (ii) the armature is fitted into
the gap
(iii) the armature is left in the
position shown in the dia
蘇士登查
C
lain what happens to the
商教偶率:
malesulee in the closed iron ring when the pair of switch
arme are moved upwards from thair present position to the position shown by the dotted lines: (c) The resistivity of the metal copper is 178 x 10 ohm-ċme. Explain what this statement meang (d) A steady current was made to pass through a solution of copper sulphate as shown After: 30 min. the cathode was found to have grained a mase of 1.485 g
copper electrodas
Fach has in 50 sq.co. Calculate: (1) the current flowing
(ii) the résistivity of the
solution.
(iii) the quantity of heat produ ced in the solution. Electrochemical Equivalent of cop per 3.3 x 10-* em per couloumb.
2. (a) State Faraday's Laws of
Electrolysis
·菲頓希爾
高天今
(b) You are given a sultable 9olution of silver salt and re- gaired to use it to electroplate a brass tray
Draw a labelled diagram of the
and circuit required. apparatus.
Do not describe the experiment, but explain any important pre- cautions that should be observed to give an even deposit of sil- ver that will not drop off..
(c) In the above experiment, you are given that the ever- chemical equivalent
0.001178 gm/coulomb and, that current of 4 amp, was passed. through the solution for 12.5 hours. while a voltmeter con nected from the anode to the cathode gave a reading of 20 volts.
Calculate:
the mass of silver deposgi- Plated.
(ii)
the resistance of the soltion.
(111) the cost of the electrical!
energy used if it was sup- plied at a rate of 50% per unit.
(d) Another electro-plating tank is connected, electrically in parallel with the first, and it is found that the second tank absorbs electrical energy st a rate of 100 watte. Calculate the resistance of the second tank, and the total: current now being shared bem tween the two tanks. (assume that the voltmeter reading remains at 20 yolts)
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