(35)育教 日十二月十年巳己曆寬
1990 中學會考預習專欄
et,
MILL & DALE PRESS
Accounting (7)
Year End Adjustment
A. Definition:-
M. T. Kwok
-
- Depreciation (1)
Depreciation may be described as the portion of the cost of fixed assets that is deducted from revenue Sor asset services used
the in operations of a business. It is closely linked to the concept of business income. Since part of the service potential эт depreciable assets is exhausted in the income - generating process each period, the COSL of these. services must be deducted from income (i.e. depreciation) in measuring periodic profit and loss.
8. Causes of Depreciation:-
(1) Wear and tears:-
Unit
Depreciation may result physical
Largely from acterioration due to operating use.
These physical forces terminate the usefulness of fixed items by rendering them
B
1990 中學會考預習專欄
明德出版社
MILE & DALE PRESS
Chemistry (7)
R. Chu
4: Revision Notes on Redox
Reactions
1. Electronic Theory of oxidation
Reduction Reactions
1.1 Oxidation is the loss or removal of
more electrons one or
from 'a chemical species such as an atom, a molecule, an ion or a radical.
1.2 Reduction is the gain or acceptance or one or more electrons by a chemical species.
e.g. Na
Fe
oxidation
2+ oxidation
Na-e
+e
reduction
201
reduction F2+
1.3 An oxidant is a chemical species which accepts or gains electrons during a reaction, i.e., it is an electron acceptor.
1.4 A reductant is a chemical species which donates or loses electrons during a reaction, i.e., it is an electron donor.
1.5 Since
oxidation and reduction reactions always occur toge liter, 50 they are more commonly known as the redox reactions. Hence, a redox reaction is according to the Electronic Theory, оле which involves a transfer of electrons from a reductant to an oxidant.
e.g. jaj
oxidation loss of e")
Fe3*(an)+1"(ac)--Fe2+(aq) + H2(aq)
reduction gain of e)
(An electron is transferred from an
1 on to Fe2+ ion.)
(b)
Oxidation(less of el
Zatsi+cu2*Log)-Zn2*(aq)+Gals?
reduction(gain of e)
(Two electrons are transferred from
a in-tor to a Cui
incapable of performing the. services for which they were. intended and thus. set the maximum limit on the assets' economic life,
(2) Obsolescence:-
It refers to the effect of innovations and technical improvements оп the economic life of existing assets.
(3) Inadequacy:-
Depreciation may
be the recognized as
effect of growth and changes in the scale of a firm's operation ia terminating the economic life of assets.
(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.
C. Factors Affecting
Calculation:-
(1) Estimated year of use.
報日僑華
line method of depreciation is that each year of service absorbs on equal portion of acquisition cost. Depreciation per year is thus computed as follows:-
Depreciation per year
Acquisition cost Estimated disposal value Estimated year of use
Example:-
Acquisition cost of
the machine
: $10,000
Estimated disposal value: $2,000
Estimated year, of use
: 5 years
Depreciation per year
$10,000 - $2,000
दु.
= $1,600.00
(This amount of depreciation will be treated as expenses in the Profit and Loss Account, 1
五期星日七十月一十(九八九一)年八十七國民中
(2) Reducing Balance Method:-
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. (.e. the net book value) each year, so d3 to reduce the asset to the estimated disposal value at the end of its life.
The pre-centage rate) is
(depreciation
1
-
Estimated disposal value Acquisition cost
(n= years of use)
* In.
examination.
depreciation
rate
the is
frequently supplied.
Depreciation
Year Depreciation Depreciation
Accumulated
Net Book Value of the machine
$
$
Example:-
1'st 1,600
1,600
2 Rd
1,600
3,200
|10,000-3,200-5,800
Acquisition cost
:
3 rd
1,600
4,800
10,000-4,800-5,200
Disposal value
:
$2,000 $250
4 th
1,600
5,400
110,000-6,400-3,600
Estimated year of use
:
3 years
15 th
1,600
9,000
|10,000-8,000=2,000* |
3
250
Depreciation rate = 1 -
2000
3
1 - J8
(2) Cost of fixed assets and
disposal value.
(3) Method of calculation.
D. Simple Methods of Calculating
Depriciation:-
(1) Straight-line:- The
distinguishing characteristic of the straight-
more electronegative atom is given a negative 0.N. and the less electronegative atom а positive D.N., e.g. in the compound ICI, the O.N. of Cl is -1 and that of I is +1 because chlorine 15 electronegative than iodine.
more
(4) Hydrogen is usually given an 0.4. 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 0.N. of -1.
(6) Oxygen, being the second most electronegative element (next to fluorine) usually has an Q.N. of - in all its compounds except the peroxides, where 105
-1, and the fluorine oxide fe_0) where its 0.N. is +2.
(7) The algebraic sum of the 0.N.'s of all the atoms is zero for a neutral compound but equal to its electronic charge for a polyatomic ion.
a
(8) The R. O any atom in chemical Species composing of two or more elements may be obtained by first assigning a reasonable 0.N. to each of the other elements in the species.
2.3 It be noted that the same elementnight have different oxidation numbers when contained in different chemicals species, e.g. carbon has a 0.N. of +2 in CO but
+4 in CC
the
*
5
10,000-1,600-9,400
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,
+4
92+02
+
+6
- 2503(502 is a reductant. J.
502+2H2S →2H20+35(502 is an
+4
0
oxidant.)
This implies that the oxidants and reductants are actually relative terms and their applications depend on the relative strengths of the
substances concerned as oxidant or reductant. In the example above, it is obvious, that sulphur dioxide has a reducing power stronger than oxygen but weaker, than hydrogen sulphide, so, it reduces oxygen but is reduced by hydrogen sulphide. However, powerful oxidarits and reductants
as always act Such: Only
the weaker oxidants and reductants can act as both oxidants and reductants according to the substances with which they react.
general, the metallic elements are usually the reductants because they lose electrons readily to form the non-metallic cations while elements are often oxidants as they accept electroCreadily to form anions The more electropositive the fetal the stronger will be its reducing power, and the more electronegative the por metal, the stronger will be oxidizing power.
5. Table of Some Common Oxidants and Reductants
Oxidants
Acidified potassium permanganate (KMnO4)
Alkaline potassium permanganate Acidified potassium
dichromate (K2Cr207)
Usual Changes In Terms Of Electron Transfer
8H* + 5e". M30+
+ Ze
+ @
ལ
14K* + 6e-
02-
201
- Mn 2+ + 44 20
MnO2
+ 40H
2Cr 3+
+
개20 세계대
Mno
+
2H20 + 3e.
Cr207
Oxygen
Halogens (e.g. Cl2)
2072
12
2H4504
.504
and
Hot conc. nitric
acid Acidified peroxi
4HNO3
+ Ze
+ 2N02
-ZH2O
SHNO 3
+ бe*
Question 4.2 Assign ата O.N. to each of underlined elements, the following chemical species. (a) Call (1) Na202
(b) A1203 (e) MnD
(c).
3. Change in Oxidation Numbers
Oxidation-Reduction Reactions
3.1 An atom is said to be oxidized if its oxidation number is increased or becomes more positive during a reaction. The atom 15 thus à
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.
increase in 0.N.
(oxidation)
0
1
Na
Na + e
decrease in 0.N.
(reduction)
V
C1
+ Ze
-t 201
3.3 A redox reaction is one during which there is a change in the oxidation number of one or more atoms.
Cu2+ ion.)
Question 4.1
State the oxidants and reductants the following redox reactions. (a) 2Fe(s)+301(g) → 2FeCl3(s)
in
e.g.
Hot conc. sulphuric
acid
Dil nitric acid
Reductants Hydrogen (H) with heated metal oxide
Hydrogen sulphide.
(H2S)
Moist sulphur dioxide (SC2) or sulphites
(503
Neutral or alkaline hydrogen peroxide Most metals (alone or in dilute acids,
+ 2e
-6ND 3" + 4H20 + ZNO
Usual Changes In Terms of Electron Transfer
H2 + 02--
+ 2e
H2$ 24+ + 52-
S + 2e
Z-
SO2 + H2O → H2SO3 = 24* + $022-
· 2420 + 02+ 2e
2-
-5042- H20
24+
- Ze
H202
+ 20#"
Na
Na + e"
e.g. Na, Mg, etc.)
Carbon with heated metal oxide
C+
C+
02- 202-
CO + 2e
CO +
+ 2e
12
+ *
-1 +1-2
0 -2-1
H ̄ + {H} 20
+ e
4+
+Sn*** (aq)
increase in 0.N.
(oxidation)
TiniII) ion in acid
medium
Sn2+
4+
Sn
+ 2e
Hence, in
directed
decrease in 0.N.
(reduction)
Carbon monoxide (CO) with heated metal oxide Iodide ion (17) in dilute acid
Iron(II) ion in acid medium
50%
Depreciation
NetBook Value x 50%
Year end
Depreciation
Accumulated Depreciation
Het Book Value
i st 2 nd
$ 1,000
1,000
3 rd
500 290
1,500
.. $2,300-1,000×1,000 $2,000-1,500
$
500
1,750 $2,000-1,750
250*
1
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 machine
at last year end
$5,000
$3,400
Value of machine
at this year end :
The depreciation of this year
- $5,000 - $3,400
= $1,600
- END -
to multiply equation (3) by 2 and equation (4) by 3.
(3) × 2:2A1-2A13+ +.68"...(5) (4) x 3:35n2+ + 6e-35m...(5) Step 3
the
the
Adding up the equations (5)
to and (6) eliminate electrons, we have following balanced equation.
2A1 + 3542+ 2A13+ + 35n
Question 4.4
Balance the following redox reaction
Fe2+(aq)+C12(aq) → fe3+(aq)+C1 ̈(aq)
Solution to Q.4.1
(a) fe is the reductant as it has lost
electrons.
C1 is the oxidant as it has accepted electrons.
(b) Mg is the reductant.
oxidant.
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)
.. 0.N. of Ca + (-1) x 2 = 0
..0.N. of Ca = +2
(b) 0,N. of 0 = -2
.'. O.N. of AT + (−2) x 3 = 0
.. 0.N. of Al - +3
(c) 0.N. of F = -1
.. 0.N. of 1 + (-1) x 5 = 0
.. Q.N. of 1 = +5
(d) 0.N. of Na = +1
O.N. of 0+ (+) x 2 = 0
.'. 0.N. of 0 = 1
-3
+4
-1 2(Ag)(C1)
+2
-
4{N} {0)+6H20
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(b) Mg(s) <24(19) — My2"(ag) (4,
2+
(c) 25e3*(ac)+5n2+(aq) → Fe(aq)
(d) Cuts)+23*(aq) -- Cu?*;ac)+2Ag(s)
2. Oxidation Numbers
2.1 Oxidation
are
numbers numbers assigned to the elements, whether free or combined, according to some simple rules. The sign of an oxidation nunder must always come before the number.
Two physical
water reaction.
Question 4.3
is oxidized while H-atom reduced in the above
State whether the elements underlined are oxidized, reduced or neither, in the following reactions. (a) ZAO+CT - AgC1
There is
to
be
(b) 4NH3+502 — 4NO+6H1⁄2Ð
significance attached to these arbitrary numbers it any circumstances.
2.2 Rules for assigning the Oxidation
Numbers (0.N.):-
(1) Atoms in the 'elementary (free) state are given an O.N. of zero. (2) The 0.N. of a monatomic ion is simply the charge on it. A cation has a positive 0.M. while an anion has a negative one, s.g. the 0.N.'s of Na', fe
and Cl are +1, +3, and -1 respectively.
(3) In a polyatomic structure the
(c) SQ2+0H ̃ ̄ — HSQ2
(d) Zn(OH)2+2NaOH → Na2Zn(OH)q
--
(e) 102+5I ̄+6H* → {2+3H2O
4.
Relative Strengths of Oxidants and Reductants
4.1 According to the Electronic Theory
of redox reactions, a reductant is an electron donor and an oxidant is
an electron acceptor. The more
6. Balancing Simple Redox Reactions
Balancing redox reactions 15 rather complicated. However, simple redox reactions can be balanced quite readily by using the fact that the total number of electrons transferred in each of the half-reaction equations of a redox reaction (i.e. the oxidation reaction and the reduction reaction) must be the same.
e.g.
(1) Balance
Fe(s)+Cu2+(aq)-Fe2+(aq)+Cu(s)
Step 1 Write down the two half- reaction equations.
- મુ
-Fe2+
Cu2+ + 2e
+ 2e .....(1)
Cu .....(2)
+ 4e
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.
€u2+
Fe Cu
(ii) Balance
-Fe + Cu
Al(s)+Sn2+(aq)—A13+(aq)+5n{s)
Step 1
Write
down the two reaction equations.
A1 -AT + 3e.....(3)
Sn.....(4)
3+
Sn2+
+ Ze
Step 2
half-
In order to make the number of electrons the same in the equations (3) and (4), we have
(e) 0.N. of 0 = −2
..0.N. of Mn+ (-2) x 4 = -1
.. O.N. of Mn-1 + 8 = +7
(f) 0.N. of K +1 and 0.N. of 0 = -2
.. (+1) x 2 + 0.N. of C + {-2} x 3 = 0
.. 0.N. of C+62
Solution to Q.4.3
0
{a} 2(Ag)+(CT)2
.. C1 is reduced as its O.N. has
decreased from 0 to -1.
(b) 4(N)(H)3+50%
.. N is oxidized as its O.N. bas
increased from -3 to +2.
+4 -2 -2+1 +1 +4 -2
{c} ${0}2+[OH]~~
(H S (0)33
.. 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
{d} Zn(OH}2+2{Ma}{OH)
+1 +2-2+1
(Na)2Zn(OH)4
.. Zn is neither reduced nor oxidized. This is not a redox reaction as none of the atoms has its 0.N. changed.
+5
(e) 103
~+51"+6H*
0
12+3H20
.. I in 10 is reduced as its 0.N.
has decreased from +5 to 0.
Solution to Q.4.4
Fe2+(aq) → Fe3+(aq) + e .......(1)
*
201 (aq) ......{2}
Cl2(aq) + 2e
~(2)x1; 2Fe2+(aq) → 2Fe3+(aq)+2e ...{3)
(3)+(4): 2Fe2+(aq)+Cl2(aq)
委實
師。
~~ 2Fe3+(aq) +201 ̄(aq)
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