Neco 2022/2022 Chemistry Questions and Answers

(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

(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

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(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)