Neco Chemistry Obj & Essay Answer Now Available

Neco Chemistry Obj & Essay Answer Now Available

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*CHEMISTRY OBJ*

1-10: CEEEABCDEC

11-20: ACBCBCDCAB

21-30: EAECBCDBCB

31-40: BBBACDDBDD

41-50: ECEEDACBDB

51-60: CCEBDBBBBD

(1a)

Ammonium chloride —> sublimation

Ink —> chromatography

Iron filling —> magnetic separation

(1aii)

-Example Of physical Change

(i)change of matter

(ii)evaporation of sodium chloride

-Example of chemical change

(i)burning of firewood

(ii)rusting of iron

(1bi)

(i)they are good reducing agent

(ii)they form basic oxide

(iii)they formed electro valent bonding

(1bii)

Aluminum tetraoxosulphate(vi) is prepared by the action of hot concentrated tetraoxosulphate(vi)acid on aluminum oxide

Al2O3(s) + 3H2SO4(aq)—->Al2(SO4)3(aq)+3H2O(i)

Aluminum tetraoxosulphate(vi) is moderately soluble in water

(1ci)

Hydrogen

(1cii)

(i)it is denser than air

(ii)it is very poisonous

(1ciii)

Pb(NO3)2+H2S—->Pbs+2HNO3

(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)

Given : 2SO2(aq) + O2(g) –> 2SO3(aq)

Volume of oxygen, O2 required = 1/2*7.50dm³

=3.75dm³

No of moles required = 3.75/22.4moles

=0.1674moles

No of oxygen molecules required = 0.1674×6.02×10^23

=1.008×10^23molecules

(2bi)

(I) NaCl —–> Valency of chlorine is 1

(II) NaO2 —–> Valency of oxygen is 2

(III) Na3N ——> Valency of Nitrogen is 3

(2bii)

(i)Formation of black soap

(ii)Formation of glycerol

(iii)It helps in the precipitation of large biomolecules such as protein

(2ci)

(i)Na2O

(ii)MgO

(iii)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)

(i)fructose

(ii)galactose

===================================================

(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)

============================

*BONUS ANSWER*

(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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