Neco chemistry question and answer 2020

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Today’s NECO Chem OBJ Answers…
1-10: BBDDADEBCA
11-12: BDEABDBCEE
21-30: DEEECCCCEC
31-40: ABDAADCAAB
41-50: EDEDAEADDD
51-60: DBEBABBADA
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
(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
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