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(1a) A molecular formula consists of the chemical symbols for the constituent followed by numeric subscripts showing the number of atoms of each element present while structural formula consists of symbols for the atoms connected by short lines that represent chemical bonds.
(1b) (i) Screening effect (ii) Size of atom (iii) Nuclear change
(1c) (i) The reaction must proceed both forward and backward i.e reversible (ii) DG = O
(1d) (i) B (ii) It has the least ionization energy among the three
(1e) Graham’s law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of their densities provided other conditions remained constant
(1f) (i) Na2SO4. Mg(NO3)2 (ii) CaCO3
(1g) (i) Protein – Addition polymer (ii) Perspex – Condensation polymer (iii) Nylon – Condensation polymer
(1h) Atomic radius can be defined as the half distance between the nucleus of two covalently bonded atoms.
(1i) Ethanol has a higher boiling point than propane because it has stronger force of attraction than propane which has a weak van der waal’s force.
(1j) (i) It aids the digestion of food (ii) It aids the action of enzymes (iii) It controls the availablity of nutrients ==============================
(2ai) (i) They have the same method of preparation (ii) They have the same chemical properties
(2aii) (i) Ethane has a single bond while Ethene has a double bond (ii) They belong to different homogeneous series (iii) They have different molecular formula and structural formula
(2bi) (i)HCl (ii) – It is colourless – It has an irritating odour (iii) 2Nacl(aq) + H2SO4(aq) —> Na2SO4(aq) + 2Hcl(g)
(2ci) Hydrocarbons are compounds having hydrogen and carbon atoms only. Examples are alkanes, alkenes, alkynes etc
(2cii) (i) Coal (ii) Crude oil
(2ciii) C H 83% 100.83 83/12 17/1 6.92/6.92 17/6.92 1 2 Empirical formula = CH2
(3aii) Given, Va = 25.0cm³, Vb = 24.0cm³, Cb = 0.15mol/dm³, Ca =?, Na = 1 , Nb = 2 Using, CaVa/CbVb = Na/Nb Ca = CbVbNa/VaNb Ca = 0.15 × 24.0 × 1/25.0×2 Ca = 3.6/50 = 0.072mol/dm³
(3bi) (i) 2Mg(s) + CO2(g) —> 2MgO(aq) + C(g) (ii) This is because magnesium can reduce carbon (iv) oxide to black carbon and producing magnesium oxide
(3bii) Mg(NO3)2 M of Mg(NO3)2 = 24+(2×14) + (6×16) = 24 + 28 + 96 = 148g/mol % of N = 28/148 × 100/1 N = 18.9%
(3ci) (i) Reaction with KMnO4 (ii) Reaction with bromine water
(3cii) It turns them colourless
(3d) Boil the vegetable oil in sodium hydroxide solution, it breaks down releasing the organic acid and the alkanol. The process is known as saponification. The organic ester is immediately neutralized by the sodium hydroxide solution to form the sodium salt of the organic acid which is soap =================================
(4ai) (i) Graphite – Hexagonal (ii) Diamond – Octahedral
(4aii) Diamond us hard because of the strong force of attraction that hold them together and it can not conduct electricity because all the available electrons are used for bonding while Graphite is soft because it’s particles are in layers and can conduct electricity because not all the electrons are used for bonding
(4bi) (i) It helps to remove dissolved gases and oxidizes dissolved metals (ii) It helps to remove objects such as rags, paper, plastics, and clogging of downstream equipment (iii) It reduces particle concentration in water and minimizes the need for coagulation and flocculation
(4bii) (i) Calcium tetraoxosulphate (vi) (ii) Calcium hydrogen trioxocarbonate
(4biii) (i) Filtration (ii) Boiling
(4biv) It has a sweet taste and it builds the shells of lower organisms
(4ci) The ore (SnO2) is wasted out of the ground with water, powdered, washed and roasted to drive off oxides sulphur, arseric and antimony. The roasted one is now reduced by heating with coke in a reverberatory furnace and molten tin is tapped off from the bottom of the furnace. The tin obtained is impure and it is remelted on a slopping furnace in which the tin melts and run its moulds. The tin obtained is 99.9% pure
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Waec Chemistry Answers 2021 for 28th September 2021
Tuesday, 28th September 2021
Chemistry 2 (Essay) 2:00 pm. –4:00 pm.
Chemistry 1 (Objective) 4:00 pm. –5:00 pm.
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WAEC Chemistry Past Questions and Answers
Waec Past Questions and Answer
The West African Senior School Certificate Examination (WASSCE) is a type of standardized test taken in West Africa, mostly by students who wish to proceed to higher institutions. It is administered by the West African Examination Council (WAEC).
The resources below on Chemistry have been provided by WAEC to assist you to understand the required standards expected in the Chemistry final Examination. Students’ performance in the examination under review was done by the Chief examiner Questions.
The contents in each WASSCE Chemistry question paper (for a specific year) are usually similar from one country to another. Questions on the WASSCE Chemistry section may be specified to be answered by candidates.
The standard of the paper was good and did not deviate from those of the previous years. The questions were straightforward, unambiguous, and spread to cover most aspects of the Syllabus. The rubrics were clearly stated.
However, there has been so much significant improvement in the overall candidates’ performance compared with those of the previous years because of the Past question that student engages themselves.
WAEC Chemistry Past Questions 2020 PAPER 2 (ESSAY)
SECTION A 1. Compound A consisting of carbon and hydrogen only. The compound was found to contain 80% carbon by mass. (a) Calculate the empirical formula of compound A using the data above. (b) The relative molecular mass of compound A was found to be 30. Use this information to deduce the molecular formula of compound A. [H = 1.00 C = 12.00] 2. (a) When calcium oxide and coke are heated in an electric furnace, the products are carbon (ii) oxide and calcium carbide (CaC2), write the equation for this reaction. (b) The addition of water to calcium carbide leads to the formation of calcium hydroxide and ethyne. Write the equation for the production of ethyne. 3. Calculate the percentage by mass of silicon tetrachloride. [2 marks] 4. Ammonia, NH3, and phosphine, Ph3, are the hydrides of the first two elements in group 5. (a) Draw a dot and cross diagram for the ammonia molecule. [2 marks] (b) Sketch and explain the shape of the ammonia molecule. [3 marks] 5. The first ionization energy of chlorine is +1260KJmol-1. (a) Define the term first ionization energy. (b) State and explain the general trend in the values of the first ionization energy for the elements across the period, sodium to argon in the periodic table.
6. An aqueous solution has a pH of 4.0. (a) (i) What is the hydrogen ion concentration of the solution? (ii) What effect will it have on litmus paper? (iii) Which of the following salt solutions would have the same effect on litmus? Give a reason for your answer. NH4Cl(aq); NaCl(aq) ; CH3OON(aq). (b) (i) Differentiate between a fine chemical and a heavy chemical. (ii) Name two sources of air pollution. (iii) Suggest one way of reducing air pollution in cities
WAEC Chemistry Past Questions SAMPLE QUESTIONS PAPER 1 (OBJECTIVE TEST)
1. Isotopes of the same element have the same number of
A. Protons, neutrons, and electrons. B. Protons and neutrons but a different number of electrons. C. Protons and electrons but different number of neutrons. D. Neutrons and electrons but a different number of protons. 2. Which type of chemical bond is formed by the transfer of electrons? A. Covalent B. Dative C. Ionic D. Metallic 3. The concentration of an aqueous solution is 5mg dm-3. Determine is the concentration in parts per million (ppm). A. 500 ppm B. 50 ppm C. 10 ppm D. 5 ppm
4. Consider the following species: H, H+, H-. What is the number of electrons in each of the species respectively? A. 1, 0, 2 B. 0, 1, 2 C. 2, 1, 0 D. 1, 2, 0 5. Two electrons can occupy the same orbital if they have different A. Angular momentum quantum numbers. B. Magnetic quantum numbers. C. Principal quantum numbers. D. Spin quantum numbers. 6. What mass of NaOH is required to make 250 cm3 of 0.10mol/dm-3 solution? [Na = 23, O = 16, H = 1] A. 1 g B. 4 g C. 8 g D. 16 g 7. Which of the following substances is not a hydrocarbon? A. Benzene B. Butane C. Ethyne D. Ethanamide
8. A substance which ionizes completely into hydroxonium ions is a A. Strong acid. B. Strong base. C. Weak acid. D. Weak base. 9. Which of the following solutions is able to resist changes in pH when small amounts of an acid or a base is added? A. Buffer solution B. Neutral solution C. Saturated solution D. Supersaturated solution 10. Protein is a polymer formed from the linkage of A. Amino acid molecules. B. Fatty acid molecules. C. Glucose units. D. Monosaccharides.
11. The bond formed between H2OH2O and H+H+ to form the hydroxonium H3O+H3O+ is
A) Dative B) Covalent C) Electrovalent D) Ionic
12. The minimum amount of energy required for effective collisions between reacting particles is known as
A) Activation energy B) Bond energy C) Kinetic energy D) Potential energy
13. 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 proportions
14. 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
15. 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
16. Which of the following apparatus can be used to measure a specific volume of a liquid accurately?
A) Beaker B) Conical flask C) Measuring cylinder D) Pipette
17. 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
18. The spreading of the scent of a flower in a garden is an example of ______
A) Brownian motion B) Diffusion C) Osmosis D) Tyndal effect
19. Propane and carbon (IV) oxide diffuse at the same rate because [H = 1.00, C = 12.0, O = 16.0]
A) They are both gases B) Their molecules contain carbon C) They have the same relative molecular mass D) Both are denser than air
20. The energy which accompanies the addition of an electron to an isolated gaseous atom is
A) Atomization B) Electronegativity C) Electron affinity D) Ionization
Chemistry-Obj!1BBDDADEBCA11BDEABDBCEE21DEEECCCCEC31ABDAADCAAB41EDEDAEADDD51DBEBABBADACOMPLETED.Love You All. Subscribe For Your NextSubject…..====================================(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 constant0.5=no; of atoms/6.023*10²³no; 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 —>2O²no; of electron = 4Q=20300CG.M.V =22.4dm³.F=96500CMole=Q/n,fMole=20300/4*96500Mole=20300/386000Mole=0.05molRecall; =vol/G.M.V0.05=vol/22.4vol=0.05*22.4vol=1.12dm³(1di)Using H²SO⁴H+ SO⁴^-¹H+ OH^-A+ AnodeOH —-> 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)TabulateS/N; (I), (II)Indicator; methyl orange, phenolphthaleinColour in acid; red, colourlessColour at end point; orange colourlessColour in base; yellow, pinkSuitable 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=760mmHgP2=800mmHg,V2=?Using; V1*P1 =V2*P2300*700=V2*300V2=300*760/800V2=228000/800V2=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 CO600g of HCOOH = XX= 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 flaky powder is formed with new properties and irrversable permanent change.Fe(s) + 3O³(g)+ XH²O(s) —> 2Fe²O³ XH²O(s)(5diii)Mas of dry hydrogen =35gMass of dry hydrogen + oxygen vapour of a compound= 440gMass of organic vapour of the compound = 440g-35g=405gV.D of the vapour =mass of vapour/mass of equal volume of H²405/35 =11.51 ≅ 11.6V.D = 11.6R.m.m of the vapour =V.D *211.6*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^- =4.583r10^⁵Since [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^-¹⁰moldm³Since; PH= – logH^+PH= – log¹⁰ 2.2*10^-¹⁰PH=0.34+10PH=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=====================================Completed =========================== CHEMISTRY OBJ 1-10: ABDCCDBDCD 11-20: ABCDDDCDBC 21-30: CABDCCCBAD 31-40: BCBDABCACC 41-50: CABDBACCBD °°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°′°°°°°°°°°°°°°° 1ai) Entropy is a measure of degree or disorder or randomness of a system unit is JK^-1 mol^-1. 1di) _Under Electrovalent Compound_ 1. They are solid at room temperature and do not vaporise easily. 2. They have high melting point and boiling point. 3. Elctrovalent Compounds readily dissolve in water and other polar solvent. 4. They do not dissolve in non – polar solvents such as benzene, ether, etc. _Under Covalent Compound_ 1. They are gases or volatile liquids because their molecules being electrically neutral are not bound by strong attractive force. 2. Covalent compounds readily dissolve in non – polar organic solvents such as benzene etc. 3. They have low melting and boiling points. 4. Covalent Compounds do not conduct electricity. 1dii) 1. High melting and boiling points. 2. Good Conductors of heat and electricity. 3. Relatively high densities. 4. Sonorous. 5. Malleable that is can be hammered into sheets. 1diii) 1. Nature of reactant. 2. Surface area of the reactant. 3. Presence of catalyst. 4. Presence of light. 5. Temperature of the reaction mixture. iii) Gay Lussac’s law states that the volume of gas which take part in a chemical reaction bear a simple whole number ratio to another end to the volume of the products, if gaseous, when measured at constant temperature and pressure. H2(g) + —— 2HCL(g) 2ei) Hardness of water is due to the presence of dissolved Calcium tetraoxosulphate(VI), Magnesiumtetraoxosulphate(VI) and Calcium hydrogen trioxocarbonate(IV) in water. 2eii) _Under Temporary Hardness_ 1. This type of hardness is caused by the presence of dissolved calcium hydrogen trioxocarbonate(IV) Can(HCO3)2. 2. It causes furring of kettles and boilers and also stalagmites and stalactites. _Under Permanent Hardness_ 1. This type of hardness is caused by the presence of calcium or magnesium ions in the form of so lube chloride. 2. It can be removed by using chemicals. 2eiii) 1. Addition of caustic soda. 2. Distillation. 3. Addition of Washing Soda.
WAEC Chemistry PAST QUESTION AND ANSWERS FOR 2021/2021
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