2019 NECO Chemistry Essay & Obj Question Answer – June/July Expo

NECO 2019 Chemistry Essay and Objectives ANSWER

100% Correct and Verified NECO 2019 Chemistry Essay/Theory and Obj Answer.

NECO 2019 Chemistry Obj
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NECO 2019 Chemistry Essay/Theory 
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INSTRUCTION: answer any 4 questions in all
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(1ai) Ammonium Chloride -> Sublimation
Ink -> Chromatography
Iron filling -> Magnetic

(1aii)
Examples of Physical changes are: Change of matter, Evaporation of sodium chloride
Examples of Chemical Changes are: Burning of firewood, Rusting of Iron

(1bi)
(i) They are good reducing agents
(ii) They formed basic oxide
(iii) They formed electrovalent bonding

(1bii)
Aluminum tetraixosulphate (vi)is prepared by the action of hot concentrated tertaoxosulphte (vi) acid on aluminum oxide
Al2O3 (s) + 3H2SO4 (aq) ( Al2(SO4)3(aq) +3H2O(l)
Aluminum tetraoxosulphate (vi) is moderately soluble in water.

(1ci) PICK TWO:
(i) Hydrogen Sulphide
(ii) It is dense than air
(iii) It is very poisonous
(iv) It burn with a pale blue flame
(v) It is weakly acidic to litmus

(1cii)
Pb(NO3)2 + H2S PbS + 2HN03

(1ciii)
Pb(N03)2aq + H2Sg, ( PbSs + 2HN03aq

(1civ) It turn black due to the formation of black lead (ii) sulphide.

(1di)
Water gas contains flammable gases whereas the producer gas contains both flammable and non-flammable gases.

(1dii)
(i) Diamond ( Octahedral in shape
(ii) Graphite Hexagonal in shape

(1e)
(i) for manufacturing of ammonia and fertilizer
(ii) It is used as a refrigerant and also used to shrink metal parts
(iii) It is used in grinding substances that are too tough to grind at normal temperature
============================
(2aii)
Avogadro’s law states that equal volumes of all gases under the same temperature and pressure contain the same number of molecules

(2aii)
no2=?

2SO2(g) + O2 —> 2SO3(g)

2 : 1 : 2

VSO2 = 7.50dm^3

nSO2 = V/Vm = 7.50/22.4 = 0.3348moles

2moles of SO2 —-> 1 moles of O2
0.3348 —-> x

x = 0.3348/2 = 0.1674moles

nO2 = 0.1674moles
np = n × L = 0.167 × 6.02 × 10^23
np = 1.008 × 10^23molecules
np = 1.01×10^23 molecules

(2bi)
(I) NaCl —–> Valency of chlorine is 1
(II) NaO2 —–> Valency of oxygen is 2
(III) Na3N ——> Valency of Nitrogen is 3

(2bii)
– Formation of black soap
– Formation of glycerol
– It helps in the precipitation of large biomolecules such as protein

(2ci)
– Na2O
– MgO
– CaO

(2cii)
Because they react with water to form acids

(2ciii)
– Reaction with sulphur
S(s) + O2(g) —> SO2(g)

– Reaction with metals
4Na(s) + O2(g) —-> 2NaO(s)

(2di)
Ion is any atom or group of atoms which possess an electric charge

(2dii)
TABULATE PLS
ION : H^+, F^-, N^3+

NUMBER OF PROTONS: 1,9,13

NUMBER OF NEUTRONS : 0, 10, 14

NUMBER OF ELECTRONS : 0, 10, 10

(2ei)
Monosaccharide are simple six carbon sugar having the formula C6H12O6 eg glucose and it’s isomers are fructose and galactose, both obtainable from fruits and honey

(2eii)
– fructose
– galactose
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3ci)
Pollution is the introduction of contaminants into the natural environment that cause adverse change. Pollution can take the form of chemical substances or energy, such as noise, heat or light

3cii)
Pollutants, the components of pollution, can be either foreign substances/energies or naturally occurring contaminants.
+++++++++++++++++++++++++++++
(3ai)
Chemical industry are companies that produce industrial chemicals

(3aii)
– Favourable climatic conditions
– It’s nearness to the source of raw materials
– The availability of space to store raw materials

(3b)
(i) It is aldose and it contains alkanal (-CHO) group
(ii) Brick red precipitate is formed
(iii) Reducing sugar possess a free aldehyde (-CHO) or ketone (-C=O) group while non reducing sugar a free aldehyde or ketone group is lacking

(3ci)
(i) Pollution is the release of unwanted materials into the environment which cause adverse change
(ii) Pollutant is a substance or energy introduced into the environment that has undesired effects or adversely effects the usefulness of a resource

(3cii)
(i)Plastic
(ii)Polythene
(iii)Pesticides

(3ciii)
VC2H2=?
MCaCa=3.2g
MCaCa=40+(2*12)
=40+24
=64glmol
n=m/m=3.2/64
n=0.05mole of CaC2
CaC2+2H2O —> Can(OH)2+C2H2
0.05mole of CaC2 —-> 0.05mole of C2H2 because the mole ratio are equal
NC2H2=0.05moles
n=V/VM
VC2H2=n*Vm = 0.05*22.4
VC2H2=1.12dm³

(3d)
(i)Na2CO3 —-> soluble
(ii)K2CO3 —-> soluble
(iii)ZnCO3 —-> insoluble
(iv)(NH4)2CO3 —-> soluble

===========================
(4ai)
Sublimation is the process whereby a solid changes state directly to the gaseous state without passing through the liquid state upon heating

(4aii)
Iodine and Ammonium chloride

(4b)
(i) He may have conducted the experiment in a partial vacuum
(ii) Addition of impurities like salt
(iii) T°K = 78 +273 = 351K

(4ci)
(i)presence of moisture
(ii)presence of oxygen

(4cii)
4Al(NO3)3(s) ——> 2Al2O3(s) +3O2(q) + 12NO2(q)

(4ciii)
(i)sols and gels
(ii)Aerosols
(iii)Emulsion

(4d)
Using CAVA/CBVB =nA/nB

0.122*20.50/CB * 25.00 = 2/1
CB = 0.05002moles/dm^3

Using :
Gram conc = molar conc * molar mass
Molar mass of hydrated salt = 6.203 * 1/0.05002
=124g/mol

Now;
106 + (18x)= 124
18x = 124 – 106
18x = 18
x = 18/18 = 1
SaH was NaCO3. H2O

(4e)
(i)the metal is potassium
(ii) period 4
(iii)group 1
(iv)2k(s) + 2H2O —-> 2KOH(aq) + H2(q)
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(5ai)
Standard electrode potential is the potential difference between an element and a solution of its ions under standard condition

(5aii)
I. Oxidation takes place at hydrogen electrode

H2(q) —–> 2H^+(aq) + 2e^-

II. Reduction takes place at copper electrode
CU²+(aq) + 2e ——> CU(s)

Overall reaction
CU(s) + 2H^+

(5bi)
TABULATE
-COMPOUND (Fes)-
(i) It can not be separated by physical means
(ii) The composition is fixed
(iii) Homogeneous
(iv) Can be represented by a chemical formula

-MIXTURE of Fe and s-
(i) It can be separated by physical means
(ii) The composition of a mixture can vary
(iii) Homogeneous or Heterogeneous
(iv) Can not be represented by a chemical formula

(5bii)
6P , 13Q and 15R
P=C
Q=Al
R=P
(I) Q is metal
(II) R is a non-metal
(IIII)Q and R belong to the same period .

(5ci)
V1= 300cm^3
P1= 410mmHg
P2=?
V2= 130cm^3
P1V1=P2V2
P2= P1V1/V2
P2 = 410*300/130=946.2
P2 = 946.2mmHg

(5cii)
2CuO(s) + C(s) —-> 2CuO(s) + CO2(g)

H2O(g) + C(s) —-> CO(g) + H2(g)

(5ciii)
(I) 20L and 17B
L= calcium
B= chlorine
(II) Calcium chloride
(III) Electrovalent combination
(IV) It is used in desiccators as drying agent for all gases except ammonia
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COMPLETED.

GOODLUCK

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