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WAEC SSCE - Chemistry - 2022

Question 1 Rapport

Electropositivity of elements across the periodic table normally

remains constant down the group

increases across the period

decreases across the period

decreases down the group

Détails de la réponse

The electronegativity of elements across the periodic table generally follows a pattern. Electronegativity refers to the tendency of an atom to attract electrons towards itself when it is part of a chemical bond. However, in this question, we are discussing electropositivity, which is the opposite of electronegativity. Electropositivity refers to the tendency of an atom to lose electrons and form positive ions. It is essentially the measure of how easily an element gives away its valence electrons. In general, as we move across a period from left to right in the periodic table, the electropositivity of elements decreases. This means that elements on the left side of a period are more electropositive compared to those on the right side. This trend can be attributed to the increase in effective nuclear charge (the positive charge felt by the outermost electrons) as we move across the period. The stronger pull of the nucleus on the valence electrons makes it harder for the atoms to lose electrons and become positive ions. On the other hand, as we move down a group in the periodic table, the electropositivity of elements remains relatively constant or even increases. This is because the valence electrons are located in higher energy levels that are farther from the nucleus. The increased distance reduces the attractive force between the nucleus and the valence electrons, making it easier for the atoms to lose electrons and become positive ions. Therefore, based on the explanation above, the correct option is "decreases across the period."

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Periodic Chemistry (Nigeria Only)

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Question 3 Rapport

A white precipitate was formed when BaCl $_{2 (a q)}$ was added to an aqueous solution
of a salt X. The precipitate dissolved in dilute HCl with rapid effervescence. Salt X is likely to contain

SO

_{4}

^{2 -}

ions

SO

_{3}

^{2 -}

ions

NO

_{3}

- ions

SO

_{3}

^{2 -}

- ions

Acids, Bases And Salts

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Question 5 Rapport

A hydrogen chloride gas reacted with oxygen gas to yield water and chlorine gas. The mole ratio of the hydrogen chloride gas to water is

1:3

2:1

3:1

4:1

Détails de la réponse

4HCl + O $_{2}$ -- 2Cl $_{2}$ + 2H $_{2}$ O
4:2 = 4/2
2:1

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Stoichiometry And Chemical Reactions

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Question 6 Rapport

The atom with the electron configuration
1s $^{2}$ 2s $^{2}$ 2p63s $^{2}$ 3p $^{6}$ 3d $^{10}$ 4s $^{2}$ 4p $^{4}$ is in

period 4, p- block

period 3, p- block

period 4, d- block

period 3, d- block

Structure Of The Atom

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Question 7 Rapport

Consider the following reaction equation:
CaO + SiO $_{2}$ ------> CaSiO $_{3}$
Silicon(IV) oxide is acting as

a basic oxide

a reducing agent

an acidic oxide

an oxidizing agent

Periodic Chemistry (Ghana Only)

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Question 8 Rapport

If the molar mass of X(HCO3) $_{2}$ is 162 g mol $^{- 1}$ , determine the relative atomic mass of X.
[H = 1.0, C = 12.0, 0 = 16.0]

40

48

61

101

Kinetic Theory Of Matter And Gas Laws

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Question 13 Rapport

Consider the following reaction equation
N $_{2}$ (g) + 3H $_{2 (g)}$ ; ∆H = -92kJ

shift the equilibrium to the right

increase the yield of ammonia

decrease the amount of hydrogen

decrease the yield of ammonia

Kinetic Theory Of Matter And Gas Laws

Chemical Kinetics And Equilibrium Systems (Ghana Only)

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Question 20 Rapport

If 50 cm $^{3}$ of a saturated solution of KNO $_{3}$ at 40 °C contained 5:05 g of the salt, its solubility at the same temperature would be
[KNO $_{3}$ = 101]

1.0 mol dm-3

1.5 mol dm

^{- 3}

2.0 mol dm

^{- 3}

5.0 mol dm

^{- 3}

Détails de la réponse

V= 50cm $^{3}$
Mass= 5.05g
Relative molecular mass of KNO $_{3}$ = (39+14+(3*16)) = 101
Convert 50cm $^{3}$ to dm $^{3}$ which is
1000cm³ = 1dm $^{3}$
50cm³ = 50*1/1000
= 0.05dm $^{3}$
Moles = mass/ molar mass
= 5.05/101 =0.05mole
Solubility= mole/volume
Solubility=0.05mol/0.05dm $^{3}$
Solubility=1.0mol/dm_ $^{3}$

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Solubility Of Substances (Ghana Only)

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Question 22 Rapport

A colourless gas with a pungent smell is evolved when dilute hydrochloric acid is added to a sample of a salt. The gas evolved could turn

acidified K

_{2}

Cr

_{2}

O

_{7}

solution colourless

acidified KMnO

_{4}

solution colourless

Fe(NO

_{3}

)

_{2}

solution green

Pb(NO

_{3}

)

_{2}

paper black

Periodic Chemistry (Ghana Only)

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Question 23 Rapport

What is the percentage composition of carbon
in Ca(HCO3) $_{2}$ ?
[ Ca = 40.0, 0= 16.0, C = 12.0, H= 1.0]

22.2%

14.8%

7.4%

3.7%

Kinetic Theory Of Matter And Gas Laws

Separation Of Mixtures And Purification Of Chemical Substances

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Question 24 Rapport

Which of the following gases is alkaline?

NO

_{2}

CO

_{2}

NH

_{3}

N

_{2}

O

Acids, Bases And Salts

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Question 30 Rapport

The IUPAC name for CH $_{3}$ CHCH $_{3}$ CHCLCHCH $_{3}$ CH $_{2}$ CH $_{3}$ is

2, 4 - dimethyl - 3 - chlorohexane

3, 5 - dimethyl - 4 - chlorohexane

4 - chloro - 3, 5 - dirnethylhexane

3 - chloro - 2, 4 - dimethylhexane

Détails de la réponse

- Recognise the functional group in the compound. This will determine the suffix of the name

- Find the longest continuous carbon chain that contains the functional group (it won't always be a straight chain) and count the number of carbon atoms in this chain. This number will determine the prefix (the beginning) of the compound's name

- Number the carbons in the longest carbon chain (Important: If the molecule is not an alkane (i.e. has a functional group) you need to start numbering so that the functional group is on the carbon with the lowest possible number). Start with the carbon at the end closest to the functional group.

- Look for any branched groups:
Name them by counting the number of carbon atoms in the branched group and referring to Table 4.6, these groups will all end in -yl.
Note the position of the group on the main carbon chain. If there is more than one of the same type of branched group then both numbers must be listed (e.g. 2,4 -) and one of the prefixes listed in Table 4.7 must be used. Important: If the molecule is an alkane the branched group must be on the carbon with the lowest possible number.
The branched groups must be listed before the name of the main chain in alphabetical order (ignoring di/tri/tetra).
If there are no branched groups this step can be ignored.

- For the alkyl halides the halogen atom is treated in much the same way as branched groups:
To name them take the name of the halogen atom (e.g. iodine) and replace the -ine with -o (e.g. iodo)
Give the halogen atom a number to show its position on the carbon chain. If there is more than one halogen atom the numbers should be listed and a prefix should be used (e.g. 3,4-diiodo- or 1,2,2-trichloro-). See Table 4.7 for a list of the prefixes.
The halogen atoms must be listed before the name of the main chain in alphabetical order (ignore di/tri/tetra).
If there are no halogen atoms this step can be ignored.
- Combine the elements of the name into a single word in the following order:
branched groups/halogen atoms in alphabetical order (ignoring prefixes)
prefix of main chain
name ending according to the functional group and its position on the longest carbon chain.

Lisez la note de cours sur Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Chemistry Of Carbon Compounds (Nigeria Only)

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Question 33 Rapport

If 5.0 g of marble reacts with 25.0 cm3 hydrochloric acid, which of the following combinations has the fastest reaction rate?

Marble chips and 2.0 mol dm

^{- 3}

HCl

_{(a q)}

Marble chips and 2.5 mol dm

^{- 3}

HCl

_{(a q)}

Powdered marble and 2.5 mol dm

^{- 3}

HCl

_{(a q)}

Powdered marble and 2.0 mol dm

^{- 3}

HCl

_{(a q)}

Rates Of Chemical Reaction

Chemical Kinetics And Equilibrium Systems (Ghana Only)

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Question 40 Rapport

The electron configuration of carbon atom in its excited state is

1s

^{2}

2s

^{2}

2px

^{2}

2py

^{0}

1s

^{2}

2s

^{2}

2px

^{1}

2py

^{1}

1s

^{2}

2s

^{1}

2px

^{1}

2py

^{1}

2pz

^{1}

1s

^{1}

2s

^{2}

2px

^{1}

2py

^{1}

Atomic Structure And Bonding

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Question 41 Rapport

What number of moles of oxygen would exert a pressure of 10 atm at 320 K in an 8.2 dm $^{3}$ cylinder?
[R=0.082 atm dm $^{- 3}$ mol $^{- 1}$ K $^{- 1}$ ]

0.32

1.52

3.13

31.25

Détails de la réponse

P = 10atm, T= 320K, V= 8.2dm $^{3}$
PV =nRT
n = P * V/R * T
10 * 8.2/0.082 * 320
= 82/26.24
n = 3.13

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Kinetic Theory Of Matter And Gas Laws

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Question 53 Rapport

a) (i) Define the term Avogadro's number.

(ii) If 2.30 g of an oxide of nitrogen, x, contains 3.01 x 10 $^{22}$ molecules, calculate the molar mass of x.

(iii) Deduce the formula of x. N, =6.02 x 10", N =14.0, O = 16.0]

(b)(i) Describe briefly what happens when each of the following substances are added to water:

(I) CCI $_{4}$ ; (II) SiCI $_{4}$ ,

(ii) Explain briefly why the reactions in (a)(i), (b)(i), (I) and (b)(ii) (II) are different Study the diagram below and answer the questions that follow.

(c) Study the diagram below and answer the questions that follow.

Open photo

(i) What is the set up used for?

Réponse

a) (i) Avogadro's number is the number of particles (atoms, molecules, ions, etc.) present in one mole of a substance. It is approximately equal to 6.02 × 1023 particles per mole. (ii) First, we need to calculate the number of moles of x in the given sample: Number of moles = Number of particles / Avogadro's number Number of moles = 3.01 × 1022 / 6.02 × 1023 Number of moles = 0.05 mol Molar mass of x = Mass of x / Number of moles Molar mass of x = 2.30 g / 0.05 mol Molar mass of x = 46 g/mol (iii) The empirical formula of x can be determined by finding the ratio of the atoms present in it. Since the molar mass of x is 46 g/mol and it contains nitrogen and oxygen, we can assume that the formula is NxOy. The ratio of N to O can be determined using the atomic masses of N and O and the given molar mass: Molar mass of NxOy = (N × 14.0) + (y × 16.0) = 46 N + y = 3.29 Since N and y must be integers, the closest possible ratio is N:O = 1:2. Therefore, the formula of x is N2O. (b) (i) When CCl4 is added to water, it does not dissolve as it is a nonpolar substance and water is a polar solvent. When SiCl4 is added to water, it reacts with water to form HCl and SiO2, which precipitates as a white solid. (ii) The reactions in (a)(i), (b)(i), (I), and (b)(ii) are different because they involve different substances and chemical reactions with water. In (a)(i), an oxide of nitrogen is being analyzed for its molar mass and formula. In (b)(i), two substances that are insoluble or react with water are being added to it. In (I), Pb(NO3)2 is being added to NaCl solution to precipitate PbCl2. In (b)(ii), SiCl4 is reacting with water to form HCl and SiO2. (c) (i) The setup is a simple distillation apparatus. It is used to separate a mixture of liquids with different boiling points by heating the mixture and collecting the condensed vapors. The mixture is heated in a distillation flask, and the vapors travel up the fractionating column, which contains several trays or packing material. The vapor with the lower boiling point condenses on the trays or packing, and the vapor with the higher boiling point continues to the condenser, where it is cooled and collected as a liquid. This allows for the separation of the components of the mixture based on their boiling points.

Lisez la note de cours sur States Of Matter (WAEC)

Lisez la note de cours sur Chemical Equilibra (JAMB)

Détails de la réponse

a) (i) Avogadro's number is the number of particles (atoms, molecules, ions, etc.) present in one mole of a substance. It is approximately equal to 6.02 × 1023 particles per mole. (ii) First, we need to calculate the number of moles of x in the given sample: Number of moles = Number of particles / Avogadro's number Number of moles = 3.01 × 1022 / 6.02 × 1023 Number of moles = 0.05 mol Molar mass of x = Mass of x / Number of moles Molar mass of x = 2.30 g / 0.05 mol Molar mass of x = 46 g/mol (iii) The empirical formula of x can be determined by finding the ratio of the atoms present in it. Since the molar mass of x is 46 g/mol and it contains nitrogen and oxygen, we can assume that the formula is NxOy. The ratio of N to O can be determined using the atomic masses of N and O and the given molar mass: Molar mass of NxOy = (N × 14.0) + (y × 16.0) = 46 N + y = 3.29 Since N and y must be integers, the closest possible ratio is N:O = 1:2. Therefore, the formula of x is N2O. (b) (i) When CCl4 is added to water, it does not dissolve as it is a nonpolar substance and water is a polar solvent. When SiCl4 is added to water, it reacts with water to form HCl and SiO2, which precipitates as a white solid. (ii) The reactions in (a)(i), (b)(i), (I), and (b)(ii) are different because they involve different substances and chemical reactions with water. In (a)(i), an oxide of nitrogen is being analyzed for its molar mass and formula. In (b)(i), two substances that are insoluble or react with water are being added to it. In (I), Pb(NO3)2 is being added to NaCl solution to precipitate PbCl2. In (b)(ii), SiCl4 is reacting with water to form HCl and SiO2. (c) (i) The setup is a simple distillation apparatus. It is used to separate a mixture of liquids with different boiling points by heating the mixture and collecting the condensed vapors. The mixture is heated in a distillation flask, and the vapors travel up the fractionating column, which contains several trays or packing material. The vapor with the lower boiling point condenses on the trays or packing, and the vapor with the higher boiling point continues to the condenser, where it is cooled and collected as a liquid. This allows for the separation of the components of the mixture based on their boiling points.

Lisez la note de cours sur States Of Matter (WAEC)

Lisez la note de cours sur Chemical Equilibra (JAMB)

States Of Matter

Chemical Equilibra

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Question 54 Rapport

(a)(i) Define the first ionization energy of an element

(ii) Consider the following table and use it to answer te question that follows

Element	Li	Be	b	C	N	O	F	Ne
Atomic number	3	4	5	6	7	8	9	10
1st I.E/kj mol $^{- 1}$	520	900	801	1086	1402	1314	1681	2081

Explain briefly why the first ionization energy of B is less than that of Be despite the fact that the atomic number of B is greater than that of Be.

(b) When Titanium chloride was electrolysed by passing 0.12 A current through the solution for 500 seconds, 0.015 g of titanium was deposited. What is the charge on the titanium ion?
[ IF= 96500 C, Ti= 48.0 ]

(c)(i) Aluminium can be obtained by the application of electrolysis. State the electrolyte which yields aluminium on electrolysis.

(ii) Name two major factors which would favour the siting of an aluminium smelter in a country.

(d)(i) Define the term paramagnetism.

(ii) Consider the following ions: 24 $^{C r^{2 +}}$ , 24 $^{C r^{6 +}}$

(I) Deduce the number of unpaired electrons in each of the ions.

(II) State which of the ions will have a greater power of paramagnetism

(l) Give a reason for the answer stated in (d)(ii)(II)

Réponse

(a)(i) The first ionization energy of an element is the amount of energy required to remove one electron from a neutral atom of that element in the gaseous state. (ii) The missing value in the table is 730 kJ mol⁻¹ for the first ionization energy of lithium (Li). Explanation for (a): The first ionization energy of B is less than that of Be because B has an electron in a p-orbital, which is further away from the nucleus than the s-orbital electrons in Be. As a result, the attraction between the nucleus and the outermost electron is weaker in B than in Be, making it easier to remove an electron from B. (b) The amount of charge on the titanium ion can be calculated using the formula: Charge = (Current x Time x IF) / Mass where IF is the Faraday constant and equals 96500 C mol⁻¹. Plugging in the values given, we get: Charge = (0.12 A x 500 s x 96500 C mol⁻¹) / 0.015 g Charge = 1.934 x 10⁶ C mol⁻¹ Since the charge on a single titanium ion is 4+, we can calculate the number of moles of titanium deposited by dividing the mass by the molar mass: n(Ti) = 0.015 g / 48.0 g mol⁻¹ = 3.125 x 10⁻⁴ mol Dividing the total charge by the number of moles of titanium gives us the charge on a single titanium ion: Charge on Ti⁴⁺ ion = (1.934 x 10⁶ C mol⁻¹) / (3.125 x 10⁻⁴ mol) = 6.188 x 10⁹ C (c)(i) The electrolyte which yields aluminium on electrolysis is molten cryolite (Na₃AlF₆) mixed with aluminium oxide (Al₂O₃). (ii) Two major factors which would favour the siting of an aluminium smelter in a country are the availability of cheap electricity and the presence of bauxite ore, which is the raw material used to produce aluminium. (d)(i) Paramagnetism is the property of an atom, ion, or molecule that has unpaired electrons and is attracted by a magnetic field. (ii) For 24Cr2+, there are 4 unpaired electrons, while for 24Cr6+, there are no unpaired electrons. Therefore, 24Cr2+ will have a greater power of paramagnetism. Explanation for (d)(ii): The power of paramagnetism is directly proportional to the number of unpaired electrons. Since 24Cr2+ has 4 unpaired electrons, it will have a greater power of paramagnetism than 24Cr6+, which has no unpaired electrons. This is because the unpaired electrons in 24Cr2+ are available for alignment with an external magnetic field, whereas the paired electrons in 24Cr6+ are not.

Détails de la réponse

(a)(i) The first ionization energy of an element is the amount of energy required to remove one electron from a neutral atom of that element in the gaseous state. (ii) The missing value in the table is 730 kJ mol⁻¹ for the first ionization energy of lithium (Li). Explanation for (a): The first ionization energy of B is less than that of Be because B has an electron in a p-orbital, which is further away from the nucleus than the s-orbital electrons in Be. As a result, the attraction between the nucleus and the outermost electron is weaker in B than in Be, making it easier to remove an electron from B. (b) The amount of charge on the titanium ion can be calculated using the formula: Charge = (Current x Time x IF) / Mass where IF is the Faraday constant and equals 96500 C mol⁻¹. Plugging in the values given, we get: Charge = (0.12 A x 500 s x 96500 C mol⁻¹) / 0.015 g Charge = 1.934 x 10⁶ C mol⁻¹ Since the charge on a single titanium ion is 4+, we can calculate the number of moles of titanium deposited by dividing the mass by the molar mass: n(Ti) = 0.015 g / 48.0 g mol⁻¹ = 3.125 x 10⁻⁴ mol Dividing the total charge by the number of moles of titanium gives us the charge on a single titanium ion: Charge on Ti⁴⁺ ion = (1.934 x 10⁶ C mol⁻¹) / (3.125 x 10⁻⁴ mol) = 6.188 x 10⁹ C (c)(i) The electrolyte which yields aluminium on electrolysis is molten cryolite (Na₃AlF₆) mixed with aluminium oxide (Al₂O₃). (ii) Two major factors which would favour the siting of an aluminium smelter in a country are the availability of cheap electricity and the presence of bauxite ore, which is the raw material used to produce aluminium. (d)(i) Paramagnetism is the property of an atom, ion, or molecule that has unpaired electrons and is attracted by a magnetic field. (ii) For 24Cr2+, there are 4 unpaired electrons, while for 24Cr6+, there are no unpaired electrons. Therefore, 24Cr2+ will have a greater power of paramagnetism. Explanation for (d)(ii): The power of paramagnetism is directly proportional to the number of unpaired electrons. Since 24Cr2+ has 4 unpaired electrons, it will have a greater power of paramagnetism than 24Cr6+, which has no unpaired electrons. This is because the unpaired electrons in 24Cr2+ are available for alignment with an external magnetic field, whereas the paired electrons in 24Cr6+ are not.

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