NECO 2021 Chemistry Questions Answers (OBJ And Essay) – June/July Expo

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NECO Chemistry Questions and Answers 2021. I will be showing you past Chemistry objectives and theory repeated questions for free in this post. You will also understand how NECO Chemistry questions are set and how to answer them.

The National Examinations Council (NECO) is an examination body in Nigeria that conducts the Senior Secondary Certificate Examination and the General Certificate in Education in June/July and December/January respectively.

NECO Chemistry Objectives and Essay Answers 2021(Expo)

The 2021 NECO Chemistry expo will be posted here on 28th July 2021 during the NECO Chemistry examination. Keep checking and reloading this page for the answers.

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Today’s NECO Chem OBJ Answers…

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Today’s Chemistry Essay Answers:

(1ai)
(i)graphite
(ii)diamond

(1aii)
(i)animal charcoal
(ii)carbon black

(1aiii)
(i)The property of elements are a periodic function of their atomic number
(ii)Elements are arranged in the periodic table according to the order of increasing in their atomic weight.

(1bi)
Periodicity can be defined as the trend or recurring variation in element properties with increasing atomic number.

(1bii)
using; mole = no; of atoms/avogadro’s constant
0.5=no; of atoms/6.023*10²³
no; of atoms = 0.5*6.02*10²³=3.012*10²³atom

1ci)
Faraday’s first law of electrolysis state that the chemical deposition due to the flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.

(1cii)
2O²^- + 9^e —>2O²
no; of electron = 4
Q=20300C
G.M.V =22.4dm³.
F=96500C
Mole=Q/n,f
Mole=20300/4*96500
Mole=20300/386000
Mole=0.05mol
Recall; =vol/G.M.V
0.05=vol/22.4
vol=0.05*22.4
vol=1.12dm³

(1di)
Using H²SO⁴
H+ SO⁴^-¹
H+ OH^-
A+ Anode
OH —-> OH + e^-
2OH+(aq)+2OH(aq)—->2H²O(s)+O²(aq)

(1dii)
Tabulate
-Electrolyte- (I)teraoxosulphate(iv)acid
(II)Ester

-non electrolyte-
(III)phenol

(1ei)
(i)mercury

(1eii)
(I)no; of electron in Y =16
(II)no; of mass number =16+18=34
(III)sulphur

(2ai)
basicity can be defined as the number of replaceable hydrogen ion in an acid

(2aii)
(I) —> 3
(II) —> 1
(III) —> 2

(2bi)
(i)Concentration
(ii)Temperature
(iii)Pressure

(2bii)
(i)Light
(ii)Temperature
(iii)Nature of reactant

(2ci)
Tabulate
S/N; (I), (II)

Indicator; methyl orange, phenolphthalein

Colour in acid; red, colourless

Colour at end point; orange colourless

Colour in base; yellow, pink

Suitable for; strong acid and weak base, weak acid and strong base

(2cii)
(i)Nitrogen —> 1s²,2s²,2p³
(ii)Fluorine —> 1s²,2s²,2p⁵
(iii)Aluminum —> 1s¹,2s²,2p⁶,3s²,3p¹

(2di)
(I)Hydrogen gas is liberated
(II)The purple colour turns colourless
(III)It leads to the deposit of black residue of carbon

(2dii)
(i)It serve as immediate source of energy
(ii)it is used in the manufacture of sweets.

(4ai)
(I)Burning requires heating while corrosion does not
(II)Boiling occurs at a certain temperature while evaporation occurs at all temperature

(4b)
A concentration solution can be defined as a solution formed when a large quantity of a substance dissolves in a little volume of water

(4bii)
(i)position of ion in electrochemical series
(ii)concentration ion
(iii)nature of electrodes

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(4ci)
Al²(SO⁴)³=(27*2)+(32*3)+(16*12) =54+96+192=342glmol

(4cii)
Aluminum teraoxosulphate (iv)

(4ciii)
(i)Propane – 1,2,3,- triol
(ii)Potassium salt

(4di)
Tabulate.
-Soaples detergent-
(i)it does not form scum in hard water
(ii)they are non-biodegradale

-Soapy detergent-
(i)It firm scum in hard water
(ii)They are biodegradable

(4dii)
(i)RCOOH
(ii)ROH

(4diii)
V1=300cm³.
P1=760mmHg
P2=800mmHg,
V2=?
Using; V1*P1 =V2*P2
300*700=V2*300
V2=300*760/800
V2=228000/800
V2=285cm³

(4div)
Its change is +3

(4dv)
Al³^+ (Aluminum ion)

(5ai)
Coal and coke

(5aii)
(I)acidic — NO² nitrogen (iv) oxide
(II)neutral — NO nitrogen (ii) oxide

(5aiii)
HCOOH. H²SO⁴/H²O CO(g)

46g of HCOOH = 22.4dm³ of CO
600g of HCOOH = X
X= 600*22.4/46=2.92dm³ of CO(s) is produced.

(5bi)
(i)To standardize a solution of an acid or base
(ii)To determine the percentage purity and impurity of an acid of a base

(5bii)
(I)density
(II)solubility

(5ci)
FeCl²(s) + 2NaOH(aq) —-> 2NaCl²(aq) + Fe(OH)²(s)
The main product is sodium chloride and iron (II) hydroxide

(5cii)
(I)FeCl²
(II)iron (ii) chloride

(5di)
(I)it is slightly denser than air
(II)it is slightly soluble in water

(5dii)
Because on exposure to air of rust due to the formation of hydrated iron (iii) oxide. In other words rusting it changes to reddish brown f

BMS, [18.11.20 07:04]
[Forwarded from Horlajoy]
laky powder is formed with new properties and irrversable permanent change.
Fe(s) + 3O³(g)+ XH²O(s) —> 2Fe²O³ XH²O(s)

(5diii)
Mas of dry hydrogen =35g
Mass of dry hydrogen + oxygen vapour of a compound= 440g
Mass of organic vapour of the compound = 440g-35g=405g
V.D of the vapour =mass of vapour/mass of equal volume of H²
405/35 =11.51 ≅ 11.6
V.D = 11.6
R.m.m of the vapour =V.D *2
11.6*2=23.2

(6ai)
(I)Efflorescence
(II)Isotope
(III)Isomerism

(6aii)
Kipps apparatus

(6aiii)
(i)Temperature
(ii)Enthalpy change value

(6bi)
Polymerisation can be defined as the arrangement of smaller nuclei to form a large nuclei

(6bii)
(i)Addition polymerisation
(ii)Condensation Polymerisation

(6biii)
OH^- =4.583r10^⁵
Since [H^+] [OH]= 10^-¹⁴
[H^+] [4.583*10^-⁵]=10^-¹⁴
[H^+]=1*10^-¹⁴/4.583*10^-⁵
[H^+]=0.22*10^-⁹
[H^+]=2.2*10^-¹⁰moldm³
Since; PH= – logH^+
PH= – log¹⁰ 2.2*10^-¹⁰
PH=0.34+10
PH=10.34

(6ci)
(I) —-> carbohydrate
(II) —-> R-OH and R-CHO

(6cii)
(I)Brass composition; copper and zinc

-uses of brass-
(i)it is use in making hammers
(ii)it is used in application where low corrosion resistance is required.

(II)steel composition; iron and carbon

-Uses of steel-
(i)it is used on roofs
(ii)it is used as cladding for exterior walls

(6ciii)
(i)Fermentation
(ii)Preparation from ethene

(6iv)
This is because there are on molecules in that can accept protons

NECO Chemistry Questions And Answers 2021 (Paper 2)

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