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Free Chemistry Waec Expo Answers

IMG 20180410 WA0008 Free Chemistry Waec Expo Answers

1ai)
Fermentation is the process in which a
substance breaks down into a simpler
substance. Microorganisms like yeast and
bacteria usually play a role in the fermentation
process, creating beer, wine, bread, yogurt and
other foods.
(1aii)
Zymase
(1b)
-The strength of an acid
-The PH of the solution
(1c)
Fe+H2SO4->FESO4 + H2
H2SO4=10cm^3
Concentration=1mol/dm^3
Amount=Vol * Concentration
=10/100*1
=0.01mol
1mol=56g
0.01mol=56*0.01
=0.56g
Mass of unreacted=5-0.056
=4.44g

continuation(1e)
A catalyst increases the rate of a chemical by
swinging it’s activation every.
(1f)
(i) CH3CooH + NH3 —->CH3CooH2 + H2O
(ii) Ethanoamide
(1g)
(a) It doesn’t corrode easily.
(b) it doesn’t react with contain inside.
(c) It can stand for a long period of time.
(1h)
(a) Sweet production industries
(b) perfume industries.
(1i)
(a) E
(b)
(1j)
(a) it increases the volume water in ocean and
seas.
(b) It causes increase in rainfall and also
corrode painted surfaces
2ai)
Collision theory states that the rate of a
reaction depends on the rate of collision of the
reactant molecules. Hence effective collision
determines the rate of the reaction because it is
the collision that leads to the formation of a
product.
(2aii)
When the rate of collision increases, the rate at
which molecules collide with each other will
also increase thereby making the kinetic energy
of the molecules to also increase. The
temperature will also increase because
temperature is a measure of average kinetic
energy.
(2bi)
Draw your diagram
(2bii)
C2H5OH + 3O2 —->2CO2 + 3H2O
1mole of ethanol = 3moles of O2
2.5moles of ethanol will require 3 × 2.5/1 =
7.5moles of O2.
but 1 mole of gas = 22.4dm3
7.5moles = 7.5 × 22.4 = 168dm3 of O2.
(2Ci)
Esterification is the formation of an Ester by the
reaction between Alkanol and an acid.
(2cii) Two uses of alkanols
(i) They are used as solvents for cellulose
(ii) They are uses in making perfumes and
cosmetics.
(iii) They are used for quick drying of paints
and nail varnishes.
(2ciii)
Sodium Ethanoate
(2di)
Tin
(2dii )
This is because the galvanized plate is
corrosion-resistant. It has a protective coating
which prevents further oxidation of the metal.
4ai)
DRAW THE DIAGRAM
(4aii)
Na2SO3(aq) +2HCl(aq) ->2NaCl(aq)+H2O
(l)+SO2(g)
(4aiii)
Dry the gas evolved by passing it through
H2SO4 and collect it by downward delivery
(4aiv)
Because it is denser than air
(4bi)
It states that there is a mixture of gases which
do not react chemically together,then the total
pressure exerted by the mixture is the sum of
the partial pressure of the individual gases that
make up the mixture
(4bii)
T1=12C=285K
P1=690mmHg
V1=30cm^3
T2=273K
P2=760mmHg
V2=?
Using general gas law
V1P1/T1=V2P2/T2
V2=V1P1T2/T1P2
=(30*690*273)/(285*760)
V2=26cm^3
(4ci)
Cl2(g)+H20(l)->HCl(aq)+HOcl(aq)
2HOCl(sunlight)->2HCl(aq+O2(g)
(4cii)
Litmus turns red because of the presence of
acid HCl
(4di)
Chlorine->Bromine->iodine (increasing boiling
point)
(4dii)
Because water has two lone pairs of electrons
as compared to ammonia which has one
3ai)
H H H H H
| | | | |
H-C-C-C-C-C-C=C-H Heptene
| | | | | |
H H H H H H
(3aii)
The sixth member of Allene C7H14 =
7(molecular mass of carbon) + 14(molecular
mass of H)
= 7(12) + 14(1)
= 84 + 14 = 98gmol-¹
(3aiii)
-Cracking is the breaking down of higher
molecules into smaller molecules while
reforming is the change in the functional group
or activity of the compound.
-In cracking more than one product or
component is formed. But in reforming one
product can still be gotten.
(3bi)
Enthalpy of Neutralization change is the heat
change which offers when one mole of H+ from
an acid reacts with one mole of OH- from an
alkali to form one mole of water.
(3bii)
Weigh about 100cm³ of dil HCL(aq) and that of
KOH(aq) of equal volume and put each into a
glass calorimeter. The temperature of the two
Solutions are taken. The mean volume of the
temperature of the acid and base taken. Then
quickly transfer the alkali to the acid, turn and
take the final temperature of the solution.
Record the mass of the mixture. Find the
temperature difference.
Total heat covered = mass × specific heat
capacity × temperature change.
(3ci)
I – copper
II – silver
(3cii)
Because silver is below copper in the
electrochemical series.
(3ciii)
Oxidation – anode

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Updated: April 10, 2018 — 4:16 pm

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