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Neco chemistry question and answer 2020

Discussion in 'Education' started by Andy, Nov 17, 2020.

  1. Andy

    Andy Drug Lord Jr.Vip

    Neco chemistry question and answer 2020 will be shared here for free


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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
    ————————————————————————
    ——————————————-
     
    Last edited: Nov 18, 2020

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