NECO Chemistry Questions and Answers 2021 | Theory and Obj

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Questions and Answers for NECO Chemistry 2021. In this post, I will provide you with free access to old Chemistry objectives and theory repeated problems. You’ll also learn how NECO Chemistry questions are constructed and how to respond to them.

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NECO 2021 Chemistry Answers are now being loaded…

NECO Chemistry OBJ Answers for Today…

Today’s Chemistry Essay Answers are as follows:

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

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

(1aiii) (i)An element’s attribute is a periodic function of its atomic number.
(ii)Elements are arranged in the periodic table in increasing atomic weight order.

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

1ci) According to Faraday’s first law of electrolysis, the chemical deposition caused by current flow through an electrolyte is directly proportional to the amount of electricity (coulombs) carried through it.

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

(1di) Using H2SO4 H+ SO4-1 H+ OH- A+ 1di) Using H2SO4 H+ SO4-1 H+ OH- A+ 1di) Using H2SO4 H OH + e —-> anode OH -\s2OH+(aq)+2OH(aq)—->2H²O(s)+O²(aq)

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

(III)phenol (non-electrolyte)

(1ei)\s(i)mercury

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

(2ai) Basicity is the number of hydrogen ions that can be replaced in an acid (2aii)
(I) —> 3; (II) —> 1; (III) —> 2; (IV) —> 3; (V) —> 1; (IV) —> 2; (V

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

(2bii) (i)Light (ii)Temperature (iii)Reactant Nature

(2ci) Compile a S/N table; (I) (II)

Methyl orange and phenolphthalein are indicators.

In acid, there are two colors: red and colorless.

At the end, the color is orange and it is colorless.

Yellow and pink for the foundation color.

Strong acid and weak base, weak acid and strong base, strong acid and strong base

(2cii)

1s2,2s2,2p3 (i)Nitrogen

Fluorine —> 1s2,2s2,2p5

Aluminium —> 1s1,2s2,2p6,3s2,3p1

(2di)

(I)Hydrogen gas escapes.

(II)The purple color fades and becomes colorless.

(III)It results in the formation of a black carbon residue.

(2dii)

(i)It can be used as a quick source of energy.

(ii)it is used in the confectionery industry.

(4ai) (I)Burning necessitates heating, whereas corrosion does not. (II)Boiling takes place at a specific temperature, whereas evaporation takes place at any temperature.

(4b)

A concentration solution is one that forms when a big amount of a chemical dissolves in a little amount of water.

(4bii)

(i)in the electrochemical series, the ion’s location

(ii)concentration ion ion ion ion ion ion

(iii) type of electrodes

(4ci)

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

(4cii)

Teraoxosulphate of aluminum (iv)

(4ciii)

1,2,3,-triol (i)Propane

(ii)Salt of potassium

(4di)

Tabulate.

-Detergent from Staples-

(i) In hard water, it does not develop scum.

(ii) They are non-biodegradable.

-Soapy detergent – (i)firms scum in hard water (ii)is biodegradable

(4dii)

(i)RCOOH

(ii)ROH

(4diii)

V1=300cm³.

P1=760mmHg

P2=800mmHg,

V2=?

Using the formula V1*P1 =V2*P2

300*700=V2*300

V2=300*760/800

V2=228000/800

V2=285cm³

(4div)

It has a +3 change.

(4dv)

+Al3+ (Aluminum ion)

(5ai)

Coal and coke are two types of coal.

(5aii)

NO2 nitrogen (iv) oxide (I)acidic

NO nitrogen (ii) oxide (II)neutral

(5aiii)

H2SO4/H2O CO HCOOH (g)

22.4 dm3 CO = 46 g HCOOH

X = 600g of HCOOH

CO(s) is created in the amount of 600*22.4/46=2.92dm3.

(5bi) (i)To standardize an acid or basic solution

(ii)To calculate the purity and impurity of an acid in a base.

(5bii)

(I)density

(II)solubility

(5ci)

2NaCl2(aq) + Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe(OH)2 Fe (s)

Sodium chloride and iron (II) hydroxide are the primary products.

(5cii)

(I)FeCl²

Iron (II) chloride (II)

(5di)

(I)it has a somewhat higher density than air.

(II)it is water soluble to a degree.

(5dii)

Rust occurs as a result of the development of hydrated iron (iii) oxide when exposed to oxygen. In other words, when it rusts, it turns a reddish brown color.

[18.11.20 07:04] BMS
[From Horlajoy] A laky powder is generated with new qualities and a permanent transformation that is irreversible.

2Fe2O3 XH2O = Fe(s) + 3O3(g) + XH2O(s) (s)

(5diii)

35 g dry hydrogen mas

A compound’s dry hydrogen + oxygen vapour mass equals 440g.

The compound’s organic vapour mass is 440g-35g=405g.

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Vapour V.D = mass of vapour/mass of H2 of equal volume

405/35 =11.51

11.6 V.D.

The vapour’s R.m.m =V.D *2

11.6*2=23.2

(6ai)

(I)Efflorescence

(II)Isotope

(III)Isomerism

(6aii)

Kipps’ device

(6aiii)

(i)Temperature

(ii)Enthalpy value change

(6bi) Polymerisation is the process of combining smaller nuclei to generate larger nuclei.

(6bii)

(i)Polymerization addition

(ii) Condensation Polymerization

(6biii)

=4.583r105 OH-

Since the [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

PH= – logH+ since

PH=-log10 2.2*10-10

PH=0.34+10

PH=10.34

(6ci)

—-> carbohydrate (I)

—-> R-OH and R-CHO (II)

(6cii)

(I)Composition of brass; copper and zinc

-brass uses- (i)it’s used to make hammers (ii) it’s employed in applications that require low corrosion resistance.

(II)composition of steel; iron and carbon

-Steel applications-

(i) It is used on roofs

(ii) it is used as an outside wall cladding

(6ciii)

(i)Fermentation

(ii)Ethene-based preparation

(6iv)

This is due to the fact that there are just a few molecules that can receive protons.

The following NECO Chemistry questions are examples of what you might expect on the NECO exam in 2021.

1. It is known how much energy is needed for effective collisions between responding particles.
A) Activation energy

B) Bond energy

(C) kinetic energy

D) Energy potential

2. The hydroxonium H3O+H3O+ is generated when H2OH2O and H+H+ form a link.

A) Nominative

Covalent (B)

(C) Electrovalent

(D) Ionic

3. The oxides X2O,XOX2O,XO,XO, and XO2.XO2 are formed by an element XX. This occurrence exemplifies the law of

A) Mass conservation

B) Proportions that are fixed

C) Massive mobilization

D) Proportions in multiples

4. What is the number of moles of oxygen in 1.20410241.2041024 molecules?

NB: Avogadro’s constant (NA) =6.021023=6.021023=6.021023=6.021023=6.021023=6.021023=6.021023=6.021023=6.021023=

A. 1

B. 2

C. 3

D. 4

5. Which of the following solids assertions is true?

A) Solid particles are less ordered than liquid particles.

B) Solids have a lower density than liquids.

C) The kinetic energy of solid particles are higher than those of liquids.

D) Solid particles are difficult to crush.

6. Which of the following apparatuses can be used to correctly measure the volume of a liquid?

(A) Beaker

B) Flask with a conical shape

C) Cyclinder measurement

(D) Pipette

7. PVT=KPVT=K is a combination of the general gas equations PVT=KPVT=K and PVT=KPVT=K.

A) The rules of Boyle and Charles

B) The laws of Boyle and Graham

C) The laws of Charles and Graham

D) The laws of Dalton and Graham

8. What is an example of the spread of a flower’s smell in a garden?

(A) Brownian motion

(B) Diffusion

(C) Osmosis is a type of osmosis that occurs when water is

(D) Tynadal impact

9. Because [H = 1.00, C = 12.0, O = 16], propane and carbon (IV) oxide diffuse at the same rate .0]

Options

They’re both gases, for starters.

B) They have carbon in their molecules.

C) Their respective molecular masses are the same.

D) They’re both denser than air.

1O. The energy released when an electron is added to an isolated gaseous atom is

(A) Atomization

(B) Electronegativity

C) Affinity for electrons

(D) Ionization

11. Calcium sulphate and calcium hydrogen carbonate are found in a sample of hard water. As a result, total hardness can be reduced by

A. boiling the water

B. adding excess calcium hydroxide.

C. adding calcium hydroxide in a determined amount

D. sprinkling with sodium carbonate

E. incorporating magnesium hydroxide into the mix 12. If litmus solution is supplied to the anode compartment during the electrolysis of copper II sulphate between platinum electrodes,

A. the litmus changes color to blue, but no gas is produced.

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