(29) ***
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fixed items by rendering them incapable of performing the services for which they were intended and thus set the maximum, limit on the assets' economic life.
1992 中學會考預習專欄
(2) Obsolescence:-
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MILL & DALE PRESS
Accounting (7)
It refers to the effect of innovations and technical
improvements
the economic
life of existing assets.
(3) Inadequacy-
報日僑等
line method of depreciation is
that each ·year, of service absorbs on equal portion of Depreciation acquisition cost.
per year is thus computed as follows:-
Depreciation per year
Acquisition cost- Estimated disposal value Estimated year of use
(2) Reducing Balance Method:-
五期星日二廿月一十 ( 一九九一)年十八國民壁中
This method, sometimes called diminishing balance method or fixed-percentage-of- declining-balance method, is ́used to" reflect the fact that some fixed assets yield either a greater quantity of service or more valuable services in early year of use. Under this method, a fixed percent is written off from the reducing balance of the asset account (ie. the net book value) each year, 50. as to reduce the asset to the estimated disposal value at the end of its life.
Example:-
be
Acquisition cost of y
the machine
Estimated disposal value
$10,000 $2,000
Estimated year of use
5 years
The
Depreciation may recognized as the effect of growth and changes in the scale firm's -operation in terminating the economic life of assets.
Year End Adjustment -
M. T. Kwok
Depreciation (1)
of
A. Definition:
a
(4) Passage of time:
As time goes by, the assets! value may decrease .. and depreciation reflects such fact. (5) Depletion:-
It refers to the reduction in market value of using the fixed assets.
Depreciation may be described as the portion of the cost of fixed assets that is deducted from revenuc for asset services used in the operations of a business. It is closely linked, to the concept of business income. Since part of the service potential of depreciable assets is exhausted in the income → generating process each period, the cost of these services must be deducted from. income. (i.e. depreciation) in measuring periodic (1) Estimated year of use.
profit and loss
B. Causes of Depreciation:-
(1) Wear and tears:-
Depreciation may result largely..
from...
physical deterioration due to operating use. · These physical forces terminate the usefulness pf
1992 中學會考預習專欄
明德出版社一
Unit 4:
MILL & DALE PRESS
Chemistry (7)
-W. LAM
Revision Notes an Redox Reactions
Electronic Theory of oxidation Reduction Reactions
a
1.1 Oxidation is the loss or removal of one or more electrons from chemical species such as an atom, a molecule,
an ion or a radical.
1.2: Reduction is the gain or acceptance of one or more electrons by a chemical species. oxidation e.g. Na
Fe2+ oxidation
reduction
Cl2+20
Na*+e
201
reduction Fe2+
1.3 An oxidant is a chemical species:
which accepts or gains electrons during a reaction, ...it is an electron acceptor.
1.4 A reductant is a chemical species which donates or loses electrons during a reaction, ile., it is an electron donor.
1.5 Since
oxidation and reduction reactions always occur together, sa they are more commonly known as the redox reactions. Hence, a redox reaction is, according to the Electronic Theory; 'one' which involves a transfer of electrons from a reductant to an oxidant.
e.g. (a)
oxidation(loss of e)
̧3*{aq}+1 ̄(aq) →Fe2+(aq) + 112(aq)
reduction(gain of e")
(An electron is transferred from an Iion to Fe2+ ion.)
(b)
oxidation(loss of e")
Zn{s}+Cu2+(aq) →2n2+(aq)+Qu{s}
reduction(gain of e)
(Two electrons are transferred from
a Zn-atom to a Cu
Question 4.1
;2+
ion.)
State the oxidants and reductants in
the following redox reactions.
(a) 2Fe(s)+3C12(g) → 2FeCl ̧(s).
{b} Mg(s)+2H*(aq) -- Mg2+(aq)+H2{g}
- {c} 2Fe3*{aq)+Sn2*(aq) → Fe2+(aq)
· Cu2+(aq)+2Ag(5)
(d) Cu(s)+2Ag*(aq)
--
2. Oxidation Numbers
are
+Sn" (aq)
2.1 Oxidation number's
directed numbers assigned to the elements, whether free or rombined, according to some simple rules. The sign of an oxidation number must always come before the number.
There is to be
no physical significance attached to these arbitrary numbers it any circumstances.
2:2 Rules for assigning the Oxidation
Numbers (0.8.):-
(1) Atoms in the elementary (free) state are given an 0.N. of zero. (2) The 0.N. of a monatomic ion is simply the charge on it. A cation has a positive D.N. while an anion has a negative one, e.g. the 0.N.'s ", Fe3+ and Cl are +1, +3. of Nat. and -1 respectively.
(3) In a polyatomic structure the
G. Factors Affecting
Calculation:-
(2) Cost of
fixed disposal value.
Depreciation per year
$10,000-$2,000
$1,600.00
(This amount of depreciation will be treated as expenses in the Profit and Loss Account.)
pre-centage (depreciation. rate) is
Estimated disposal value Acquisition cost
In
Depreciation
Year Depreciation Accumulated
Depreciation
Net Book Value of the machine
(n years of use)
examination, depreciation frequently supplied,
Pate
the is
$
1st 1,600 12.nd 1,600 assets and 3rd 1.600
1,600 3,200
4,800
th
1,600 1,600
6,400
8,000
10,600-1608-8,400, 10,000-3, 200-6,800 {10,000-4,800÷5,200 10,000-6,400-3,600 10,000-8,0002,000*
(3) Method of calculation.
D. Simple Method's of Calculating
Depriciation:-
(1) Straight-line:- The
distinguishing
characteristic of the straight-
more electronegative atom is given a negative Q.N. and the Tess. electronegative atom В positive 10.N., eigi in the compound ICI, the ON. of C1 is 1 and that of is +1 because chlorine is more electronegative than iodine.
(4) Hydrogen is usually given an, D.N. of +1 in all its compounds except the metallic hydrides, in which its 0.N.is -1.
(5) Fluorine, being the most electronegative element, is always given an O.N. of -1.
to
(6) Oxygen, being the second most electronegative element (next fluorine) usually has an O.N. of -2 in all its compounds except the peroxides, where its of 1
and the fluorine oxide la Diwhere its 0.N. is +2. **
(7) The algebraic Sum of the on. of all the atoms is zero for a neutral compound but equal to its electronic charge for a polyatomic
(8) The "pl" sapy atom in a chemical species, composing of two or more elegents may be obtained by first assigning a reasonable 0.N. to each of the other elements in the species.
2.3 It is to be noted that the same
element might
different have oxidation numbers when contained in different chemicals species, e.g. carbon 0.N. of +2 in CO but +4 in Be
Question 4.2 Assign an underlined elegants chemical species.
(a) Call2 ..(b). (d) Na202 (e) Mn0)
each of tho the following
(c) F5
3. Change in Oxidation Mumbai's. Oxidation-Reduction Reactions -
3.1 An atom is said to be oxidized if its oxidation number is increased For becomes more positive during a reaction. The atom is thus a
reductant.
3.2 An atom is said to be reduced if its oxidation number is decreased or becomes more negative during a reaction. The atom is thus an oxidant.
e.g.
Na
increase in 0.M.
(oxidation)
+1 Na e
decrease in O.N."
(reduction)
C12
201
3.3 A
one
during
redox reaction is which there is a change in the oxidation number of one or more atosis.
e.g.
If estimate is accurate, the net book value of the machine would equal to estimated disposal value after the machine had been fully depreciated.
readily a chemical species donates electrons, the stronger a reductant. it will be. Similarly the more.. readily a chemical species accepts electrons, the stronger an oxidant it will be. However, since donating electrons and accepting electrons are opposite processes, so, a stronger oxidant must be a weaker reductant and vice versa.
4.2 The same chemical species can be a reductant in one reaction but an oxidant in another depending on with what substance it is reacting
5.
+6
+4
-25031502 is a reductant
502+2H2H2O+35{SO, is an
oxidant.)
This implies that the oxidants and reductants are actually relative terms and their applications depend on the relative strengths of the
Example:-
Acquisition cost Disposal value
50%
Depreciation Net Book Value x 50%
Year
end
1 st 2 ad
3 rd
Depreciation
$ 1,000 500 250
Accumulated Depreciation
Het Dook Value
$
1,000 1,500. 1,750
2,000-1,000 1,000. 2,000-1,500-
500
250*
2,000-1,750-
It equals to the estimated disposal value.
(3) Revaluation:-
The amount of depreciation
of the year may be calculated
simply by revaluating the value of the fixed assets at the end
of the year and comparing them with that at the beginning of
the year adjusted with any new acquisition.
Example:--
Value of machinë
at last year end
Value of machine
日七十月十年未
服時 54644
BM 54625
黼·黼:每日上午十時至下午五時半
即E
HE
圖文傳真:
55942.
省時又
每日下午五時半
誠聘校
九龍塘私立小 務員有經驗者
$5,000
at this year end. :
$3,400
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$2,000 $250
The depreciation of this year
$5,000
$3,400
Estimated year of use
3 years
*250 2000
Depreciation rate.
substances concerned as oxidant or reductant. In the example above, it is obvious that sulphur dioxide has a reducing power stranger than oxygen but weaker than hydrogen- sulphide, so, it reduces oxygen but is reduced by hydrogen sulphide. However, powerful oxidants and reductants always act as such. Only the weaker oxidants and reductants can act as both oxidants and reductants according to the substances with which they react.
a general, the metallic elements are usually the reductants because they lose electrons readily to form cations while the non-metallic elements-artepiteli oxidants as they. accept velect he readily to form anions the more electropositive. the metal the stronger will be its reducing power, and the
mdre electronegative the non-metal, the Bodizing be stronger w111. power.
Table of Some Common Oxidants and Reductants
Oxidants
Alkaline potassium permanganate
Usual Changes In Terms Of Electron Transfer
Acidified potassium permanganate (KMnO
BH+
+5e
410
MBO
24.0
+40H
14H+ 6e
102
20
Acidified potassium dichromate (K2Cr207).
Oxygen
Halogens (eig. C{2}]
Hot conc, sulphuric
acid
Hot conc. nitric
acid Acidified hydrogen
peroxide
Reductants Hydrogen (H) with heated metal oxide
Hydrogen sulphide.
(HS)
Moist sulphur dioxide (502) or sulphites
· (50327)
Neutral or alkaline hydrogen peroxide Most metals (alone
or in dilute acids,
e.g. Na, Mg, etc.)
Carbon with heated
with heated metal oxide
4HNO3 + 2e
2H
Usual Changes H2 +
HS 2H++
·6N03 + 4H2O + 2ND
ARIES
In Terms Of Electron Transfer
+ 2e
"S+ 2e
H2O H2SO 2H + SO22
50,
50
H202
+ 20H
· 2420 + 02 + Ze ̄
Na - Na* + eTM
+ 4e"
+ Ze"
metal oxide
に
c'+ 02-
CO + 2e"
C+
202-
CO2
Carbon monoxide (CO)
CO + 02-
+1-2
decrease in 0.N.
(reduction)
CO2
Iodide ion (17) in
0 -2+1
dilute acid
I
12
+e
increase in 0.N.)
(oxidation)
Iron(II) ion in acid .medium
Fe2+
-Fe
Tia(II) ion in acid
medium
Sn2+
+ 2e
Hence, H is oxidized while H-acou in water reduced in the above reaction.
Question 4.3
State whether the elements underlined are oxidized, reduced or neither, in the following reactions. (a) [2Ag+012 == 2AgOT
(b) 413+502
-
4NU+6H20
(c) $02+0H ̃— HSV3"
{d} Zn(OH)2+2NaOH → Na2Zn(OH) (e) 103 +51 +6н*
12+3H20
4. Relative Strengths of Oxidants and
Reductants
6. Balancing Simple Redox Reactions
reactions
is
Balancing redux
However, rather complicated. simple redox reactions
be Can balanced quite readily by using the fact that the total number of electrons transferred in each of the half-reaction equations of redox reaction (i.e. the oxidation reaction and the reduction reaction) must be the same.
e.g.
(i) Balance
a
Fe(s)+Cu2+(aq)-fe2+(aq)+Cu{s}
Step 1
Write down the reaction equations.
Step 2
Since the number of electrons is the same in the equations: (1) and (2), so, simply adding up the two to eliminate the electrons, we at once have the equation balanced.
Fe + Cu2+ (ii) Balance
Fe2+
+ Cu
Al(s)+Sn2+(aq)→AT3*{aq)+Sn{s}
Step 1
Write down the two half- reaction equations.
معد.
A)
2+
two
half-
Sn
+ 2e
Step 2
Fe --
+ 2e" .....(1)
S
The more
Cu2+ + 2e
Cu.....(2)
4.1 According to the Electronic Theory
of redox reactions, a reductant is an electron donor and an oxidant
an electron acceptor.
+ 3e.....(3)
Sn .....(4)
In order to make the number of electrons the Satne in the equations (3) and (4), we have
$1,600
END
to multiply equation (3) by Z and equation (4) by 3,
(3) x 2:2A1
(4) x 3:35n Step 3
бе... (5)
+ Ge 35n...(6)
Adding up the equations (5). and (6) to eliminate electrons, we have following balanced equation.
2A1+35n
Question 4.4
the
the
Balance the following redox reaction.
-Fe3+(aq)+C) ̄(aq)
Fe2*(aq}+C)2(aq)
Solution to 0.4.1
(a) Fe is the reductant as it has lost,
électrons
C) is the oxidant accepted electrons.
(b) Mg is the reductant.
oxidant,
as
It has
H is the
(c) Fe3*(aq) is the oxidant. Sn2+(aq)
is the reductant.
(d) Cu(s) is the reductant. Ag(aq) is
the oxidant..
Solution to Q.4.2
(a) 0.N. of H-1 (as it is a metallic
hydride)- O.. of Ca +(-1) x 2 = 0
0.N. of Ca +2
(b) 0.8. of 0 = −2
0.N. of Al
·O.N. of AT
*
(-2) x 3
:= +3.
(c) 0.N. of F-1
•*. O.N. of I + {-1} x 5.=.0
„NO,N, of I=+5
\{d} ́O.N、 of Na = +1
.*. 0.N. of 0+ (+) x 2 =
.*. 0.N. of 0
(e) 0.N. of 0 = -2
.0.N. of Mn+ (-2) x 41
.*. O.N. of Mn = 18 = +7
(f) 0.N. of K = +1 and 0.N. of Q = -2
.*. (+1} x 2 + Ó.N. of C+ (−2) x 3 = 0
.. 0.N. of C+6 2 = +4
Solution to 0.4.3
0
(a) 2(Ag)+(C1)2 → 2(Ag}{C})
.". C) is reduced as its 0.N. has
decreased from 0 to -1.
-3
+2
(b) 4{N}(H)g+50, → 4{N}{0}+6H2O
N. is oxidized as its 0.8. increased from 3 to +2+.
+4 -2 -2+1 +1 +4.-2
{c} S{0}2+COH] ̃ → (H & (0)31°
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has
.". S is neither oxidized nor reduced.
This is not a redox reaction as
none of the atoms has its 0.N. changed.
+2 +2 +1
+1 -2+1
+1 +2-2+1
(d) Zn(OH)2+2(Na)(OH) - {Na}¿Zn(OH)4
.. Zn is neither reduced nor oxidized. This is not a redox reaction as none of the atoms has its O.N. changed.
+5
(e) 103
*+51"+6H*
112+31120
.". I in 103" is reduced as its 0.N.
has decreased from +5 to 0.
Solution to Q.4.4
Fe3+(aq) + e........(1)
Fe2+(aq) C12(aq) + 20°
w
201 (aq) .....(2).
(2)x1: 2Fe2+(aq) - 2Fe3+(aq)+2...(3) {3}+{4}; 2F@2+(aq)+Cl2{@q}
3+
— 2Fø31 (aq) +201 ̈{aq)
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