NECO Chemistry Questions and Answers 2021/2022 (Essay & Objectives)

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.

GET THE FULL QUESTIONS AND FREE ANSWERS HERE

GET THE FULL QUESTIONS AND FREE ANSWERS HERE

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.

NECO 2021 Chemistry Answers Loading.

Today’s NECO Chem OBJ Answers…

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

GET THE FULL QUESTIONS AND FREE ANSWERS HERE

GET THE FULL QUESTIONS AND FREE ANSWERS HERE

(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

GET THE FULL QUESTIONS AND FREE ANSWERS HERE

GET THE FULL QUESTIONS AND FREE ANSWERS HERE

(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

——————————————————————————————————————-

The following NECO Chemistry questions are questions to expect in the 2021 NECO examination.

1. The minimum amount of energy required for effective collisions between reacting particles is known
A) Activation energy
B) Bond energy
C) Kinetic energy
D) Potential energy

2. The bond formed between H2OH2O and H+H+ to form the hydroxonium H3O+H3O+ is
A) Dative
B) Covalent
C) Electrovalent
D) Ionic

3. An element XX forms the following oxides X2O,XOX2O,XO and XO2.XO2. This phenomenon illustrates the law of ________.
A) Conservation of mass
B) Definite proportion
C) Mass action
D) Multiple proportion

Download:  Prince Kaybee shares photos from “Hosh” visuals

4.. How many moles of oxygen would contain 1.204×10241.204×1024 molecules?
NB: Avogadro’s constant (NA) =6.02×1023=6.02×1023
A) 1
B) 2
C) 3
D) 4

5. Which of the following statements about solids is correct?
A) Solid particles are less orderly than those of a liquid
B) Solid have lower densities than liquids
C) Solid particles have greater kinetic energies than those of liquids
D) Solid particles cannot be easily compressed

6. Which of the following apparatus can be used to measure a specific volume of a liquid accurately?
A) Beaker
B) Conical flask
C) Measuring cyclinder
D) Pipette

7. The general gas equation PVT=KPVT=K is a combination of
A) Boyle’s and Charles’ laws
B) Boyle’s and Graham’s laws
C) Charles’ and Graham’s laws
D) Dalton’s and Graham’s laws

8. The spreading of the scent of a flower in a garden is an example of?
A) Brownian motion
B) Diffusion
C) Osmosis
D) Tynadal effect

9. Propane and carbon (IV) oxide diffuse at the same rate because [H = 1.00, C = 12.0, O = 16.0]
Options
A) They are both gases
B) Their molecules contain carbon
C) They have the same relative molecular mass
D) Both are denser than air

1O. The energy which accompanies the addition of an electron to an isolated gaseous atom is
A) Atomization
B) Electronegativity
C) Electron affinity
D) Ionization

11. A sample of hard water contains some calcium sulphate and calcium hydrogen carbonate. The total hardness may therefore be removed by
A. boiling the water
B. adding excess calcium hydroxide
C. adding a calculated amount of calcium hydroxide
D. adding sodium carbonate
E. adding magnesium hydroxide

12. During the electrolysis of copper II sulphate between platinum electrodes, if litmus solution is added to the anode compartment,
A. the litmus turns blue but no gas is evolved
B. the litmus turns blue and oxygen is evolved
C. the litmus turns blue and hydrogen is evolved
D. the litmus turns red and oxygen is evolved
E. the litmus turns red and then becomes colourless

13. The reaction between an organic acid and an alcohol in the presence of an acid catalyst is known as;
A. saponification
B. dehydration
C. esterification
D. hydrolysis
E. hydration

14. The IUPAC names of the compounds CH3COOH and CH2=CH2 are respectively;
A. acetic acid and ethane
B. ethanoic acid and ethene
C. methanoic acid and ethylene
D. ethanol and ethene
E. acetic acid and ethylene

GET THE FULL QUESTIONS AND FREE ANSWERS HERE

GET THE FULL QUESTIONS AND FREE ANSWERS HERE

15. If 30cm3 of oxygen diffuses through a porous pot in 7 seconds, how long will it take 60cm3 of chlorine to diffuse through the same pot, if the vapour densities of oxygen and chlorine are 16 and 36 respectively?
A. 9.3 sec
B. 14 sec
C. 21 sec
D. 28 sec
E. 30.3 sec

16. When heat is absorbed during a chemical reaction, the reaction is said to be
A. thermodynamic
B. exothermic
C. isothermal
D. endothermic
E. thermostatic

17. When large hydrocarbon molecules are heated at high temperature in the presence of a catalyst to give smaller molecules, the process is known as
A. disintegration
B. polymerization
C. cracking
D. degradation
E. distillation

18. The pH of four solutions W, X, Y, Z are 4, 6, 8, 10 respectively, therefore
A. none of these solutions is acidic
B. the pH of Y is made more acidic by addition of distilled water
C. Z is the most acidic solution
D. W is the most acidic solution
E. X is neutral

19. When each of the nitrates of Potassium, Magnesium and iron is heated,
A. all the nitrates decompose to their oxides
B. the nitrate of magnesium gives the nitrite and oxygen
C. the nitrates of iron magnesium and iron give the oxides
D. the nitrate of iron gives the nitrite and oxygen
E. the nitrate of the magnesium is not decomposed

2O. Which of the following metals cannot replace hydrogen from water or steam?
A. Sodium
B. Magnesium
C. Iron
D. Calcium
E. Copper

21. small quantity of solid ammonium chloride (NH4Cl) was heated gently in a test tube, the solid gradually disappears to produce two gases. Later, a white cloudy deposit was observed on the cooler part of the test tube. The ammonium chloride is said to have undergone
A. distillation
B. sublimation
C. precipitation
D. evaporation
E. decomposition

22. Elements P, Q, R, S have 6, 11, 15, 17 electrons respectively, therefore,
A. P will form an electrovalent bond with R
B. Q will form a covalent bond with S
C. R will form an electrovalent bond with S
D. Q will form an electrovalent bond with S
E. Q will form a covalent bond with R

23. An element X forms the following compounds with chlorine; XCl4, XCl3, XCl2. This illustrates the
A. law of multiple proportions
B. law of chemical proportions
C. law of simple proportions
D. law of conservation of mass
E. law of definite proportions

24. The oxidation state of chlorine in potassium chlorate is
A. +1
B. +2
C. +3
D. +5
E. +7

25. 10 When air which contains the gases Oxygen, nitrogen, carbondioxide, water vapour and the rare gases, is passed through alkaline pyrogallol and then over quicklime, the only gases left are;
A. nitrogen and carbondioxide
B. the rare gases
C. nitrogen and oxygen
D. nitrogen and the rare gases
E. nitrogen, carbondioxide and the rare

26. Which of the following statements is NOT correct?
A. The average kinetic energy of a gas is directly proportional to its temperature
B. At constant tempearture, the volume of a gas increases as the pressure increases
C. The pressure of a gas is inversely proportional to its volume
D. The temperature of a gas is directly proportional to its volume
E. The collisions of molecules with each other are inelastic

27. Zinc Oxide is a
A. Basic Oxide
B. Acidic Oxide
C. Amphoteric Oxide
D. Neutral Oxide
E. Reactive Oxide

28. When sodium chloride and metallic sodium are each dissolved in water
A. both processes are exothermic
B. both processes are endothermic
C. the dissolution of metallic sodium is endothermic
D. the dissolution of metallic sodium is exothermic
E. the dissolution of sodium chloride is explosive

Download:  Engineer creates Lobola – an automated pap cooker

29. The periodic classification of elements is an arrangement of the elements in order of their
A. Atomic Weights
B. Isotopic Weights
C. Molecular Weights
D. Atomic Numbers
E. Atomic Masses

3O. In the reaction between sodium hydroxide and sulphuric acid solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralise 10cm3 of 1.25 molar sulphuric acid?
A. 5cm3
B. 10cm3
C. 20cm3
D. 25cm3
E. 50cm3

NECO Chemistry Questions And Answers 2021 (Paper 2)

Don’t worry about this NECO Chemistry Questions And Answers 2021. All you need to do is to keep on refreshing this page for the 2021 NECO Chemistry Questions And Answers for this year. It will be posted here in few minutes.

Tips on How to Pass 2021 NECO Chemistry Examinations

The following guidelines will help you pass the 2021 NECO Chemistry examination with flying colours.

Have a Target and Work Towards Actualizing it 

You have decided to pass NECO Chemistry 2021 and I am sure of that. Now, the next thing you should do is set targets.

You have told yourself, “I will score A in Neco Chemistry 2021”, that’s not all. You need to plan on how to make it happen. Create a timetable and master plan to achieve your goals.

 Get the Recommended Textbook on Chemistry for 2021 NECO Examination

Normally, Neco recommends books for the examination. But apart from NECO Literature in English where certain novels are compulsory, you are free to use any good Chemistry textbook to prepare for NECO 2021 exam.

Some textbooks are more difficult to understand. If you have any topic you are finding difficult to understand, then get a textbook that will simplify the topics and make life better for you.

 Do not Skip Chemistry Examples and Exercise you Will Come Across While Reading: 

Many candidates are fond of skipping exercises and even examples while studying textbooks. In fact, we like notebooks so much that we could ask, “can I read my notebook and pass NECO Chemistry 2021?” Don’t be scared of attempting exercises in Biology. Face the challenges.

If you have any questions about the NECO Chemistry Questions and Answers 2021, kindly drop your question in the comment box.

GET THE FULL QUESTIONS AND FREE ANSWERS HERE

GET THE FULL QUESTIONS AND FREE ANSWERS HERE

 

CHEMISTRY OBJ
1-10: EECEBAEEBB
11-20: DDDBEACABB
21-30: CDECCDAAAB
31-40: BAECBBCDAC
41-50: CECCEBDBDC
51-60: CEBEEDDBAE

 

 

: *NECO 2021 CHEMISTRY VERIFIED EXPO*

 

 

*Chemistry theory*

-=-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-==-=-=-=-=-=-=-=-=-=-=-=

 

(1a)

-High temperature is required

-High pressure

-The presence of Catalyst

 

(1b)

(i) To avoid corrosion

(ii) To avoid rusting

 

(1ci)

Calcium oxide (CaO) is alkaline and Hcl is acidic. The two react to form calcium chloride and water; instead of dry Hcl, after that, it will end up with wet calcium chloride.

 

(1cii)

Concentrated H2SO4(aq)

 

(1d)

Conc HNO3(aq) reacts with a solution of FeSO4(aq). The Fe2+ was oxidised to yellow Fe3+ and brown NO2(g) was given off.

ie 2HNO3(aq) + 2FeSO4(aq) + H2SO4 -> Fe2(SO4)3 + 2NO2(g) + 2H2O

 

(1ei)

(I) ZnO(s) + H2SO4(aq) —> ZnSO4(aq) + H2O(l)

(II) ZnO(s) + NaOH(aq) —> Na2ZnO2(s) + H2O

 

1ei)

(I) ZnO(s) + H2SO4(aq) —> ZnSO4(aq) + H2O(l)

(II) ZnO(s) + NaOH(aq) —> Na2ZnO2(s) + H2O(l)

 

(1eii)

(III) Amphoteric property

 

(1f)

(i) They have the same number of atoms

(ii) They have different neutron number which makes their mass number to differ

 

(1g)

HclO – Oxochlorate(i)acid and Hcl

 

(1hi)

H2S is undergoing oxidation to S(s)

 

(1hii) loading……

 

 

*More loading……..* ✍️

✍️✍️

[7/28, 07:10] Obitiye Theophilus: 2aii.

*chemical properties carbon (iv) oxide*

– it has a density of 0.0019 g/cm3

 

– Critical pressure of 7.62 MN/m2

 

*chemical properties of Sulfur (iv) oxide*

– As a reducing agent – in the presence of water, SO2 acts as a powerful reducing agent.

 

– It turns moist litmus pink (being acidic)

: 4ai

– Enthalpy of combustion of a substance is defined as the heat energy given out when one mole of a substance burns completely in oxygen.

 

4aii.

– Isomerism is the phenomenon in which more than one compounds have the same chemical formula but different chemical structures.

 

4bii.

– Nitrous oxide (N₂O)

– Carbon dioxide (CO 2):

4bii.

– Nitrous oxide (N₂O)

– Carbon dioxide (CO 2)

TONY: CHEMISTRY-Obj!

1EECBBAEBBB

11DDDBEAEABB

21DBECCDAAAB

31BADCBACDAC

41CECCEADBDC

51CEBEEDDBAE

 

Completed!.

====================================

 

 

Chemistry-THEORY

(1ai)

(i) Base has bitter taste

(ii) Base is soapy to touch

(iii) Base turns litmus paper blue

 

(1aii)

The steps involved are:

(i)Shift conversion

(ii)Removal of carbon (iv) oxide

(iii)Steam reforming

 

(1aiii)

Cracking is the process by which heavier hydrocarbon molecule is splitted into two or more lighter molecules

 

(1aiv)

Thermal cracking

 

(1bi)

P ion = 1s²2s²2p⁶

 

(1bii)

Formula of chloride is Pcl2

 

(1biii)

Reducing agent

This is because it undergoes oxidation by losing electrons

 

(1ci)

Electron affinity is the energy released when a gaseous atom gains an electron to form a gaseous negative ion.

 

(1cii)

This is because it has lone pairs of electrons

 

(1di)

(i) By adding a catalyst

 

(ii) FeS(s) + 2HCl(g) —–> FeCl(g) + H2S(s)

 

Xg = ?

No. of FeCl2 = 127g/mole

No. of FeS= 88g/mole

 

Mole = Mass/ No. of Mass = 3.2/127 = 0.025mole of FeCl2

 

By comparison,

1 mole of FeS = 1 mole of FeCl2

X moles of FeS = 0.025 of FeCl2

X = 0.025mole of FeS

 

Recall,

Mass = Mole * molar mass

Mass = 0.025*88

Mass = 2.2g of FeS

X = 2.2g

 

(1dii)

(i)Chlorofluorocarbons

(ii)Ammonia

 

 

 

All will be posted keep refreshing

[7/28, 07:10] Obitiye Theophilus: 2ai.

(1) The particles in a gas are in constant, random motion

 

Download:  2021 NECO Geography Practical Questions and Answers,Expo/Runs

(2) The combined volume of the particles is negligible

 

(3) The particles exert no forces on one another,

CHEMISTRY-Obj!-

1EECBBAEBBB

11DDDBEAEABB

21DBECCDAAAB

31BADCBACDAC

41CECCEADBDC

51CEBEEDDBAE

 

Completed!.

====================================

 

 

Chemistry-THEORY

(4a)

(i)Enthalpy of combustion can be defined as the amount of heat evolved when one mole of a substance burns in air or oxygen

 

(ii) Isomerism can be defined as a phenomenon in which organic compounds are having the same molecular formula but different structural formula

 

(4bi)

NO2 – Acidic oxide

CO – Neutral oxide

Fe2O3 – Basic oxide

Al2O3 – Amphoteric oxide

 

(4bii)

(i)Nitrous oxide

(ii)Carbon (iv) oxide

 

(4c)

(i) Fractional distillation of liquid air

 

(ii) 2KNO3 —–> 2KNO2(s) + O2(g)

 

(NH)CO3(s) ——>2NH3(g) + CO2(g) + H2O(g)

 

(iii) No. of molecules = 4/16 × 6.02 × 10²³

= 1.5 × 10²³

 

(4di)

(i)Sodium

(ii)Potassium

 

(4dii)

(i)Carbon (iv) oxide

(ii)Carbon

 

(4diii)

(i)Hydration

(ii)Oxidation

====================================

 

(5ai)

Gay Lussac’s states that when gaese react, they do so in volume which bear a simple whole number ratio to one another and to the volume of their products.

 

(5aii)

2CO + O2 —-> 2CO2

2 1 2

100cm³ 70cm³ 100cm³

 

Total volume = 100+70+100 = 270cm³

 

(5bi)

(i)It is used in the production of other chemical substances

(ii)It is used as a drying or dehydrating agent

 

(5bii)

(i) ZnCO3 —–> ZnO + CO2

 

->Conditions<-

– High temperature

– High pressure

 

(ii) ZnO + H2SO4 —-> ZnSO4 + H2

 

->Conditions<-

– Catalyst

– Surface area

 

(5biii)

(i) Crystallization method

(ii) By heating the ZnSO4 with an alkali metal carbonate

 

(5c)

(i) This is due to the presence of extra electrons not used for bonding graphite

(ii) This is because sodium salts are soluble and double decomposition is used for the production of insoluble salts

(iii) It is an alkali substance ie possesses basic properties

 

(5di)

(i)Bitumen coal

(ii)Peat coal

 

(5dii)

Fractional distillation

====================================

 

(2ai)

(i)Molecules of gases are in constant random motion

(ii)The collision of gases is perfectly elastic

(iii)An increase in temperature leads to an increase in the average kinetic energy of the molecules of gases

 

(2aii)

(i)They both turn blue litmus paper red

(ii)They both dissolve in water to produce an acid

 

(2aiii)

Copper<Iron<Aluminium<Magnesium

 

(2aiv)

Copper

 

(2bi)

Activation energy can be defined as the minimum energy required for a reaction to occur.

 

(2bii)

Catalyst lowers activation energy

 

(2biii)

Nikel

 

(2biv)

(i)Sodium trioxocarbonate (iv)

(ii)Hydrogen gas

 

(2ci)

(i)It reacts with metals to liberate hydrogen

(ii)It reacts with base to produce salt and water only

 

(2cii)

(i) 2KI + 2Cl2 —-> 2Kl2 + I2

 

(ii) 2Zn + Cl2 —–> 2ZnCl

 

(2ciii)

Cl2

 

(2civ)

Covalent bond

 

(2d)

(i) Butane

(ii) Propene

(iii) Ethanol

(iv) Ethyne

====================================

 

(1ai)

(i) Base has bitter taste

(ii) Base is soapy to touch

(iii) Base turns litmus paper blue

 

(1aii)

The steps involved are:

(i)Shift conversion

(ii)Removal of carbon (iv) oxide

(iii)Steam reforming

 

(1aiii)

Cracking is the process by which heavier hydrocarbon molecule is splitted into two or more lighter molecules

 

(1aiv)

Thermal cracking

 

(1bi)

P ion = 1s²2s²2p⁶

 

(1bii)

Formula of chloride is Pcl2

 

(1biii)

Reducing agent

This is because it undergoes oxidation by losing electrons

 

(1ci)

Electron affinity is the energy released when a gaseous atom gains an electron to form a gaseous negative ion.

 

(1cii)

This is because it has lone pairs of electrons

 

(1di)

(i) By adding a catalyst

 

(ii) FeS(s) + 2HCl(g) —–> FeCl(g) + H2S(s)

 

Xg = ?

No. of FeCl2 = 127g/mole

No. of FeS= 88g/mole

 

Mole = Mass/ No. of Mass = 3.2/127 = 0.025mole of FeCl2

 

By comparison,

1 mole of FeS = 1 mole of FeCl2

X moles of FeS = 0.025 of FeCl2

X = 0.025mole of FeS

 

Recall,

Mass = Mole * molar mass

Mass = 0.025*88

Mass = 2.2g of FeS

X = 2.2g

 

(1dii)

(i)Chlorofluorocarbons

(ii)Ammonia

====================================

 

Completed!. We Remain The baddest

Site Ever.

Uncle &TeamDragon

07085276140

: CHEMISTRY-OBJ

1EECBBAEBBB

11DDDBEAEABB

21DBECCDAAAB

31BADCBACDAC

41CECCEADBDC

51CEBEEDDBAE

Completed!.

====================================

CHEMISTRY-THEORY

(1ai)

(i) Base has bitter taste

(ii) Base is soapy to touch

(iii) Base turns litmus paper blue

 

(1aii)

The steps involved are:

(i)Shift conversion

(ii)Removal of carbon (iv) oxide

(iii)Steam reforming

 

(1aiii)

Cracking is the process by which heavier hydrocarbon molecule is splitted into two or more lighter molecules

 

(1aiv)

Thermal cracking

 

(1bi)

P ion = 1s²2s²2p⁶

 

(1bii)

Formula of chloride is Pcl2

 

(1biii)

Reducing agent

This is because it undergoes oxidation by losing electrons

 

(1ci)

Electron affinity is the energy released when a gaseous atom gains an electron to form a gaseous negative ion.

 

(1cii)

This is because it has lone pairs of electrons

 

(1di)

(i) By adding a catalyst

 

(ii) FeS(s) + 2HCl(g) —–> FeCl(g) + H2S(s)

 

Xg = ?

No. of FeCl2 = 127g/mole

No. of FeS= 88g/mole

 

Mole = Mass/ No. of Mass = 3.2/127 = 0.025mole of FeCl2

 

By comparison,

1 mole of FeS = 1 mole of FeCl2

X moles of FeS = 0.025 of FeCl2

X = 0.025mole of FeS

 

Recall,

Mass = Mole * molar mass

Mass = 0.025*88

Mass = 2.2g of FeS

X = 2.2g

 

(1dii)

(i)Chlorofluorocarbons

(ii)Ammonia

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°′°°°°°°°°°°°°°°

(4a)

(i)Enthalpy of combustion can be defined as the amount of heat evolved when one mole of a substance burns in air or oxygen

 

(ii) Isomerism can be defined as a phenomenon in which organic compounds are having the same molecular formula but different structural formula

 

(4bi)

NO2 – Acidic oxide

CO – Neutral oxide

Fe2O3 – Basic oxide

Al2O3 – Amphoteric oxide

 

(4bii)

(i)Nitrous oxide

(ii)Carbon (iv) oxide

 

(4c)

(i) Fractional distillation of liquid air

 

(ii) 2KNO3 —–> 2KNO2(s) + O2(g)

 

(NH)CO3(s) ——>2NH3(g) + CO2(g) + H2O(g)

 

(iii) No. of molecules = 4/16 × 6.02 × 10²³

= 1.5 × 10²³

 

(4di)

(i)Sodium

(ii)Potassium

 

(4dii)

(i)Carbon (iv) oxide

(ii)Carbon

 

(4diii)

(i)Hydration

(ii)Oxidation

Similar Posts:

Author: Admin Boss

Leave a Reply

Your email address will not be published.