Your message is too short
Your message is spammy
» »

[2014 WAEC EXPO] CLICK HERE FOR CHEMISTRY ANSWERS






Verified CHEMISTRY OBJ:
1-10: DAAABDADCA
11-20: BABBBDCCAC
21-30: CDABDDACBC
31-40: BBDCCCCDBA
41-50: DCABADBABB

(1ai)
Ionic bond is the bond that
exists between metallic and
non-metallic atoms.

(1aii)
I ionic bond,
II ionic and covalent bond
+1y2+2x-2*3=0
(1bi) Na2S2O3, 2+2x-6=0
2x=6-2
X=4/2=2
:. the oxidation number of
sulphur in Na2S2O3 is 2

(1b)
Na2S2O3, 2+2x-6=0
2x=6-2
X=4/2=2
:. the oxidation number of
sulphur in Na2S2O3 is 2
(1c) Faraday's first law
states that the quantity of
substance deposited or
liberated during electrolysis
is directly proportional to
the amount of current
passing through the
electrolyte at constant
temperature an pressure at
a given time i.e Q=it
(1d)
i. SO3
ii.NaO
(1e)
i. Exothermic
ii. Endothermic
(1f)
i. It turns blue litmus paper
red
ii. It turns red litmus to red
(1g)Efflorescence are
substances when exposed to
the atmosphere, they lose
their water of crystalization
to the atmosphere.
(1h)
(i.) It is used in making
typewriter ribbon
(ii.) for making shoe polish
(iii.) For making carbon
black
(1i)
i. It is used for hardening of
oil
ii. It is used in breaking
heavy hydrocarbon light
fraction
iii. It is used in producing
esters for making soap
(1j)
Recall,
m= MQ/nF,
where m= mass of sliver
deposited.
Ag^+ + e^- --> Ag
n=1,
m/M=No of moles =
10920/1*96500
m/M=No of moles = 0.113
moles.
====================
PLS NOTE THAT ^ MEANS
RAISE TO POWER.
====================

(2ai)
This can be defined as a
substance formed when an
element gains or looses
election

(2aii)
Isotopes: these are element
that exist with the same at
its first election, in the
valence shell while
aluminium while aluminium
retains two hence
magnesium will become
more stable than aluminium.

(2bi)
(I)
12Mg= 1S^22S^22p^63S^2
13AL=
1S^22S^22P^63S^23P^1
(II) This is because the first
ionization energy increases
across the period but
decreases down the group
(2bii)
Na2O,MgO, Al2O3
(2c) (1) Allotropes are
atoms that exist in different
forms but have the same
physical state (ii)Diamond
and Graphite (iii) -Diamond
is used for making
jeweleries - Graphite is used
for making lead pencil (2d)
2Nacl+H2SO4 --
>Na2SO4+2Hcl (i) -The gas
turns blue litmus paper to
red -The gas is colourless
(ii)2Nacl+H2SO4--
>Na2SO4+2Hcl (2e) Molar
mass(gmol^-1)=mass/
Amount. Amount in moles=
5.68*10^-3/28. Amount in
moles=2.03*10^-4moles.
Amount in
moles=0.000203moles. (2c)
(1) Allotropes are atoms
that exist in different forms
but have the same physical
state (ii)Diamond and
Graphite (iii) -Diamond is
used for making jeweleries -
Graphite is used for making
lead pencil (2d) 2Nacl
+H2SO4 -->Na2SO4+2Hcl (i) -
The gas turns blue litmus
paper to red -The gas is
colourless (ii)2Nacl+H2SO4--
>Na2SO4+2Hcl (2e) Molar
mass(gmol^-1)=mass/
Amount. Amount in moles=
5.68*10^-3/28. Amount in
moles=2.03*10^-4moles.
Amount in
moles=0.000203moles.

(3ai)
coagulation sedimentation
filteration disinfection (3aii.)
magnesium ion and calcium
ion (3aiii) -it wastes soaps - it
causes furring of kettles and
boilers (3aiv) -crude oil
spillage refuse and sewage

(3bi) water is a polar solvent
which is an excellent solvent
for electrovalent substances
like NaCl but oil is non-polar
which has little affinity for
hydroxyl ion (3bii) -as
dehydrating agent -as an
acid -as oxidizing agent (3ci)
oxygen and hydrogen
chloride (3cii.) I-NaOH II-
because it is soluble in
water


(4ai)
I. Anode is the positive
electrode through which
current enters the
electrolyte
II. Cathode is te negative
electrode through which
current leaves the
electrolyte
(4aii)
I. Nacl--> Na+ Cl-
II.
Alcl3--> Al^3+ cl-
Al2O3-->Al^3+ O2-
(4aiii)
This is because the reaction
of sodium with water is very
explosive and violent but
that of aluminum is not
(4bi) This is because one
substance is undergoing
oxidation and another
reduction at the same time.
(4bii)
Cr2O7^-5 --> Cr^+2
Cl- --> Cl2.
(4biii)
Cr2O7 + Cl^- + 7H2O --> cl^
+
Cr2+ + 14H^+ + Cl2
(4c)
Current(I)=6A
Time(t)=1hr30mins
= (1*3600)+(30*60)
=3600+1800
=5400s
Quantity of Electricity(Q)
=I*t
=6*5400
=32400coulombus
m=MQ/nF
Al^3+ + 3e^- --> AL
n=3
m= 27*32400/3*96500
m=3.02g
Mass of aluminum deposited
is 3.02g










Related Article


Posted By DUBAI4YOU On 12:50 Mon, 14 Apr 2014

Comments:0 || Views:

Comment
Name







.....................

Please LOGIN or REGISTER To Gain Full Access To This Article