JAMB UTME - Chemistry - 2019

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Question 2 Report

How many alkoxyalkanes can be obtained from the molecular formula $ \mathrm{C_4H_{10}O} $?

4

2

1

3

Answer Details

Alkoxyalkanes have a general formula of R-O-R', where R and R' are alkyl groups. From the given molecular formula C4H10O, we can see that there are four carbon atoms, so the longest possible alkyl group is butyl (C4H9-). To form alkoxyalkanes, we need to attach an oxygen atom to the alkyl group. This can be done in three ways - by attaching the oxygen to one of the terminal carbon atoms (forming a primary alcohol), by attaching it to one of the central carbon atoms (forming a secondary alcohol), or by attaching it to the carbonyl carbon atom (forming an ester). So, we can obtain a maximum of three alkoxyalkanes from the given molecular formula. However, we need to take into account that there are different isomers possible for each type of alcohol or ester, depending on which carbon atom the oxygen is attached to. Therefore, the correct answer is (at least) 3.

Read lesson note on Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Read lesson note on Acids, Bases And Salts (Nigeria Only) (WAEC)

Read lesson note on Chemical Equilibria (JAMB)

Acids, Bases And Salts (Nigeria Only)

Chemical Equilibria

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Question 3 Report

The electronic configuration of element Z is $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^1$. What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.

Z

_{3}

(SO

_{4}

)

_{2}

ZSO

_{4}

Z

_{2}

SO

_{4}

Z

_{2}

(SO

_{4}

)

_{3}

Answer Details

Z = 1s

^{2}

2s

^{2}

2p

^{6}

3s

^{2}

3p

^{1}

∴

We have Z

^{3 +}

and SO

_{4}^{2 -}

The reaction : Z

^{3 +}

+ SO

_{4}^{2 -}

\to

Z

_{2}

(SO

_{4}

)

_{3}

.

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 4 Report

Sulphur exists in six forms in the solid state. This property is known as

Isomerism

Allotropy

Isotopy

Isomorphism

Answer Details

The property of sulfur existing in six different forms in the solid-state is known as allotropy. Allotropy is a phenomenon where an element can exist in multiple forms, called allotropes, that have different physical and chemical properties but are composed of the same atoms. These different forms arise due to differences in the arrangement of atoms or molecules within the substance. In the case of sulfur, it can exist in multiple solid-state allotropes, including rhombic, monoclinic, and plastic sulfur, among others. Each of these allotropes has a different crystal structure, melting point, and other physical and chemical properties, even though they are all composed of sulfur atoms. Allotropy is a common phenomenon observed in many elements, including carbon, oxygen, and phosphorus, among others.

Read lesson note on Non-metals And Their Compounds (JAMB)

Non-metals And Their Compounds

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Question 5 Report

In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm$^{3}$ of 1.25 molar tetraoxosulphate (vi) acid?

25cm

^{3}

10cm

^{3}

20cm

^{3}

50cm

^{3}

Answer Details

Equation of reaction : 2NaOH + H

_{2}

SO

_{4}

→ Na

_{2}

SO

_{4}

+ 2H

_{2}

O

Concentration of a base, CB = 0.5M

Volume of acid, V $_{A}$ = 10cm $^{3}$

Concentration of an acid, C $_{A}$ = 1.25M

Volume of base, V $_{B}$ = ?

Recall:
$\frac{C_{A} V_{A}}{C_{B} V_{B}} = \frac{n_{A}}{n_{B}}$ ... (1)
N.B: From the equation,
$\frac{n_{A}}{n_{B}} = \frac{1}{2}$

From (1)
$\frac{1.25 \times 10}{0.5 \times V_{B}} = \frac{1}{2}$
$\frac{12.5}{0.5 V_{B}} = \frac{1}{2}$
25 = 0.5V $_{B}$

VB = 50.0 cm $^{3}$

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Stoichiometry And Chemical Reactions

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Question 6 Report

Consider the reaction

$ \mathrm{A(s) + 2B(g) \rightarrow 2C(aq) + D(g)} $

What will be the effect of a decrease in pressure on the reaction?

the equilibrium will shift

rate of forward reaction will increase

rate of backward reaction will increase

rates of forward and backward reaction are not affected

Answer Details

Given: The equation below

A $_{(s)}$ + 2B $_{(g)}$ → 2C $_{(a q)}$ + D $_{(g)}$

Since we have a higher number of moles of gaseous species on the LHS, i.e 2, a decrease in pressure will favor the forward reaction.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

Read lesson note on Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Read lesson note on Periodic Chemistry (Ghana Only) (WAEC)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Periodic Chemistry (Ghana Only)

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Question 7 Report

A solution X, on mixing with $ \mathrm{AgNO_3} $ solution gives a white precipitate soluble in aqueous $ \mathrm{NH_3} $, a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain

Al

^{3 +}

, Cl

^{-}

Hg

^{+}

, SO

_{4}^{2 -}

Pb

^{2 +}

, Cl

^{-}

Ag

^{+}

, SO

_{4}^{2 -}

Chemistry Of Carbon Compounds (Ghana Only)

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Question 8 Report

What volume of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.

[Na = 23, N = 14, O = 16]

1.7 litres

0.420 litres

0.588 litres

2.35 litres

Answer Details

To calculate the volume of a solution, we need to use the formula: moles of solute = concentration x volume First, let's find the number of moles of sodium trioxonitrate (V) in 5g of the solute. The molar mass of NaNO3 is: Na = 23 N = 14 3 x O = 3 x 16 = 48 Molar mass = 23 + 14 + 48 = 85 g/mol The number of moles of NaNO3 in 5g is: moles = mass / molar mass = 5 / 85 = 0.0588 moles Now, we can use the formula above to find the volume of the solution: moles of solute = concentration x volume volume = moles of solute / concentration volume = 0.0588 moles / 0.100 M volume = 0.588 litres Therefore, the correct answer is 0.588 litres of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Stoichiometry And Chemical Reactions

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Question 9 Report

The velocity, V of a gas is related to its mass, M by (k = proportionality constant)

V =

\frac{k}{M}

V =

\frac{k}{M^{\frac{1}{2}}}

V =

k M^{2}

V =

(\frac{k}{M})^{\frac{1}{2}}

Answer Details

Recall:
V =

\sqrt{\frac{3 R T}{M}}

∴ V \propto \frac{1}{\sqrt{M}}

V = \frac{k}{\sqrt{M}}

V =

\frac{k}{M^{\frac{1}{2}}}

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 10 Report

Elements in the periodic table are arranged in the order of their

atomic numbers

metallic character

mass number

relative atomic mass

Answer Details

Elements in the periodic table are arranged in the order of their atomic numbers. The atomic number of an element is the number of protons in the nucleus of an atom of that element. The elements are arranged in order of increasing atomic number from left to right and from top to bottom in the periodic table. The elements in each row, also known as a period, have the same number of electron shells, while the elements in each column, also known as a group or family, have the same number of valence electrons. This arrangement makes it possible to predict the chemical and physical properties of an element based on its position in the periodic table. Therefore, the correct answer is: - atomic numbers

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 11 Report

2-methylprop-1-ene is an isomer of

3-methyl but-1-ene

2-methyl but-1-ene

but-2-ene

pent-2-ene

Answer Details

2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene. An isomer is a molecule that has the same molecular formula as another molecule, but a different arrangement of atoms. In this case, 2-methylprop-1-ene has the molecular formula C4H8, and so do 3-methyl but-1-ene and 2-methyl but-1-ene. The difference between these three molecules is in the arrangement of the carbon and hydrogen atoms. 2-methylprop-1-ene has a branched structure with a double bond between the first and second carbon atoms. 3-methyl but-1-ene is also a branched molecule, but the double bond is between the second and third carbon atoms. Similarly, 2-methyl but-1-ene has a double bond between the first and second carbon atoms, but it has a different branching pattern. On the other hand, pent-2-ene has five carbon atoms, so it has a different molecular formula than 2-methylprop-1-ene. Therefore, 2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene, but not of pent-2-ene, because it has the same molecular formula and a different arrangement of atoms compared to the other two isomers.

Read lesson note on Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

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Question 12 Report

When ammonia and hydrogen ion bond together to form ammonium ion, the bond formed is called

electrovalent bond

co-ordinate bond

ionic bond

covalent bond

Answer Details

When ammonia and hydrogen ion go into bonding, they form ammonium ion by combining with a dative/coordinate covalent bond.

Read lesson note on Oxidation And Reduction (JAMB)

Read lesson note on Atomic Structure And Bonding (JAMB)

Oxidation And Reduction

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Question 13 Report

If acidified Potassium Dichromate(VI) ($\mathrm{K_2Cr_2O_7}$) acts as oxidizing agent, color changes from

yellow to green

orange to green

yellow to red

orange to red

Answer Details

Potassium Dichromate (VI), when it is acidified, acts as an oxidizing agent. When this happens, the color changes from orange to green. This is because the orange color of the potassium dichromate is due to the presence of Cr(VI) ions, which are oxidized to Cr(III) ions. The green color that is produced is due to the formation of chromium(III) ions. In this reaction, the dichromate ions are being oxidized, which means that they are losing electrons, and the chromium ions are being reduced, which means that they are gaining electrons. The transfer of electrons causes the color change from orange to green.

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 14 Report

Which important nitrogen-containing compound is produced in Haber's process?

NH

_{3}

NO

_{2}

NO

HNO

_{3}

Answer Details

The important nitrogen-containing compound that is produced in Haber's process is NH3, which is also known as ammonia. Haber's process is a chemical process used to produce ammonia by reacting nitrogen gas (N2) and hydrogen gas (H2) under high pressure and temperature in the presence of an iron catalyst. The reaction between nitrogen and hydrogen produces ammonia as the main product, along with some nitrogen and hydrogen gases that do not react. NH3 is an important compound that is widely used in industry for the production of fertilizers, plastics, and other chemical products. It is also used as a cleaning agent, a refrigerant, and a fuel for engines. In addition, NH3 is an essential compound for life, as it is a key component of amino acids, which are the building blocks of proteins.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 15 Report

A cell shorthand notation can be written as $A / A^{+} // B^{2+} / B$. The double slash in the notation represents the

electrolyte

electrodes

salt bridge

solid - liquid interface

Answer Details

The double slash in the cell shorthand notation represents the salt bridge. A salt bridge is a component of an electrochemical cell that connects the two half-cells and allows the flow of ions between them. It consists of an inert electrolyte solution (usually a salt) that is placed between the two half-cells. The purpose of the salt bridge is to maintain electrical neutrality in each half-cell by allowing the flow of ions to balance the charge buildup in the half-cells. In the cell shorthand notation, the double slash "//" represents the salt bridge that connects the two half-cells of the electrochemical cell. The first half-cell is represented on the left-hand side of the slash and the second half-cell is represented on the right-hand side of the slash. The anode (where oxidation occurs) is represented on the left side, and the cathode (where reduction occurs) is represented on the right side. Therefore, the correct answer is option number 3: salt bridge.

Read lesson note on Electrolysis (JAMB)

Electrolysis

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Question 16 Report

Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?

Zinc

Manganese

Sodium

Iron

Answer Details

The metal that is most essential in the regulation of blood volume, blood pressure, and osmotic equilibrium is sodium. Sodium is a key electrolyte that helps maintain the balance of fluids in the body, including blood volume and blood pressure. Sodium ions are positively charged and are attracted to negatively charged ions, such as chloride (Cl-) and bicarbonate (HCO3-), which together help regulate the pH of the blood. Sodium is also essential for maintaining osmotic equilibrium, which refers to the balance of solutes between cells and the extracellular fluid. Osmotic equilibrium is critical for proper cellular function and is regulated by the movement of water and electrolytes, including sodium, in and out of cells. While the other metals listed (zinc, manganese, and iron) are important for various functions in the body, such as enzyme activity and oxygen transport, they are not directly involved in regulating blood volume, blood pressure, and osmotic equilibrium in the same way that sodium is. Therefore, the answer is not options 1, 2, or 4, and the correct answer is: sodium.

Read lesson note on Metals And Their Compounds (JAMB)

Metals And Their Compounds

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Question 17 Report

Which of the following represents the kind of bonding present in ammonium chloride?

ionic only

covalent only

ionic and dative covalent

dative covalent

Answer Details

Ammonium chloride contains both ionic and covalent bonds. In ammonium chloride, the ammonium ion (NH4+) is positively charged and the chloride ion (Cl-) is negatively charged. These ions are held together by ionic bonds, which are formed between positively and negatively charged ions. However, the bond between the hydrogen atom in the ammonium ion and the nitrogen atom in the ammonium ion is also a covalent bond. This type of covalent bond is known as a dative covalent bond, or a coordinate covalent bond, because the electron pair being shared is supplied by one atom only (the nitrogen atom in this case). So, the kind of bonding present in ammonium chloride is both ionic and dative covalent. In simple terms, ammonium chloride contains both ionic bonds between its positive and negative ions, and a dative covalent bond between the hydrogen atom and nitrogen atom within the ammonium ion.

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Question 18 Report

Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine?

addition of water, filtration, evaporation to dryness and sublimation

filtration, evaporation to dryness, addition of water and sublimation

sublimation, filtration, evaporation and addition of water

sublimation, addition of water, filtration and evaporation to dryness

Answer Details

The set of operations that will completely separate a mixture of sodium chloride, sand, and iodine is: - filtration, to separate the sand and iodine from the sodium chloride - evaporation to dryness, to concentrate the sodium chloride solution and remove any remaining water - sublimation, to separate the iodine as a solid from the remaining sodium chloride By using these operations, you can separate each component of the mixture into separate, pure forms. The order of the operations is important because each step must be done in a way that effectively separates the components and does not interfere with subsequent steps.

Read lesson note on Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

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Question 19 Report

An element Z contains 80% of $ {}^{16}_{8}\mathrm{Z} $ and 20% of $ {}^{18}_{8}\mathrm{Z} $. Its relative atomic mass is

16.2

17.6

16.0

16.4

Answer Details

R.A.M of Z =

16 (\frac{80}{100}) + 18 (\frac{20}{100})

=

12.8 + 3.6

= 16.4

Read lesson note on Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Read lesson note on Atomic Structure And Bonding (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

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Question 20 Report

Burning magnesium ribbon in air removes which of the following

(i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(iv)oxide?

i and iv only

i, ii, iii and iv only

i only

i and ii only

Answer Details

Burning magnesium ribbon in air will remove oxygen (option i) from the air, but not nitrogen (option ii), argon (option iii), or carbon dioxide (option iv). When magnesium burns, it reacts with oxygen in the air to form magnesium oxide. The reaction can be represented by the following equation: 2Mg(s) + O2(g) → 2MgO(s) The magnesium in the ribbon combines with oxygen in the air to form solid magnesium oxide. This reaction is exothermic, which means that it releases heat and light energy. So, when magnesium ribbon is burned in air, it consumes the oxygen in the air to form magnesium oxide. However, nitrogen, argon, and carbon dioxide are not chemically reactive with magnesium, and therefore are not removed from the air by the burning of magnesium ribbon. In summary, the correct option is (i) only - burning magnesium ribbon in air removes oxygen only.

Read lesson note on Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

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Question 21 Report

$ \mathrm{SO_3} $ is not directly dissolved in water in the industrial preparation of $ \mathrm{H_2SO_4} $ by the contact process because

SO

_{3}

is an acidic gas

Acid is usually added to water and not water to acid

The reaction between SO

_{3}

and water is violently exothermic

SO

_{3}

does not dissolve readily in water

Acids, Bases And Salts

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Question 22 Report

Which of the following reactions is an oxidation process?

2H

^{+}

+ 2e

^{-}

\to

H

_{2}

2O

^{2 -}

+ 4e

^{-}

\to

O

_{2}

OCl

^{-}

+ H

_{2}

O + 2e

^{-}

\to

2OH

^{-}

+ Cl

^{-}

Cl + e

^{-}

\to

Cl

^{-}

Oxidation And Reduction

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Question 23 Report

A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years

75,000 particles

35,000 particles

25,000 particles

50,000 particles

Answer Details

The half-life of a radioactive nucleus is the time it takes for half of its particles to decay. This means that after 20 years, 100,000 particles will become 50,000 particles. After 40 years, we can find the number of particles remaining by counting the number of half-lives that have passed. Since 40 years is double the half-life of 20 years, this means that two half-lives have passed, so the number of particles will be halved twice. Starting with 100,000 particles: - After 1 half-life (20 years), there will be 50,000 particles remaining. - After 2 half-lives (40 years), there will be 25,000 particles remaining. So, exactly at the end of 40 years, there will be 25,000 particles remaining.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 24 Report

The cost of discharging 6.0g of a divalent metal, X from its salt is ₦12.00. What is the cost of discharging 9.0g of a trivalent metal, Y from its salt under the same condition?

[X = 63, Y = 27, 1F = 96,500C]

N12.00

N27.00

N63.00

N95.00

Answer Details

For X: X

^{2 +}

+ 2e

^{-}

\to

X
2F = 63g
xF = 6g
x =

\frac{6 \times 2}{63} = \frac{4}{21} F

\frac{4}{21}

F = N12.00
1F =

\frac{12}{\frac{4}{21}}

= N63.00
1F is equivalent to N63.00.
For Y: Y

^{3 +}

+ 3e

^{-}

\to

Y
3F = 27g
xF = 9g
x =

\frac{3 \times 9}{27}

= 1F
1F = N63.00

Read lesson note on Chemical Combination (JAMB)

Chemical Combination

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Question 25 Report

Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are

Ca(OH)

_{2}

and NaOH

Na

_{2}

CO

_{3}

NaCl

NaOH

Answer Details

The chemical constituent of soda lime used to prepare methane in the laboratory is Ca(OH)2 (calcium hydroxide) and NaOH (sodium hydroxide). Soda lime is a mixture of these two compounds. When sodium ethanoate (NaC2H3O2) is heated with soda lime, it undergoes a reaction known as the Kolbe's reaction, which produces methane gas (CH4) as one of the products. The reaction can be represented as follows: 2NaC2H3O2 + 2Ca(OH)2 → 2CH4 + 2NaOH + 2CaCO3 In this reaction, the sodium ethanoate reacts with the calcium hydroxide to form calcium acetate (Ca(C2H3O2)2) and sodium hydroxide. The calcium acetate then decomposes to produce methane gas and calcium carbonate (CaCO3), which is a solid precipitate. Therefore, the chemical constituents of soda lime used to prepare methane in the laboratory are calcium hydroxide (Ca(OH)2) and sodium hydroxide (NaOH).

Read lesson note on Chemical Combination (JAMB)

Chemical Combination

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Question 26 Report

The two ions responsible for hardness in water are

Ca

^{2 +}

and / or Mg

^{2 +}

K

^{+}

and / or Mg

^{2 +}

Ca

^{2 +}

and / or Li

^{+}

Ca

^{2 +}

and / or Na

^{+}

Answer Details

The ions responsible for hardness in water are Ca2+ and/or Mg2+. Hardness in water refers to the presence of calcium and magnesium ions, which are commonly found in natural water sources such as rivers, lakes, and groundwater. These ions can react with soap to form insoluble compounds, reducing the effectiveness of soap and causing scaling in pipes and appliances. The hardness of water is often measured in terms of the concentration of calcium and magnesium ions, expressed as calcium carbonate equivalents (CaCO3).

Read lesson note on Solubility (JAMB)

Read lesson note on Water (JAMB)

Solubility

Water

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Question 27 Report

The following are isoelectronic ions except

Na

^{+}

Mg

^{2 +}

Si

^{2 +}

O

^{2 +}

Answer Details

Two or more ions are said to be isoelectronic if they have the same electronic structure and the same number of valence electrons.
Na

^{+}

= 10 electrons = 2, 8
Mg

^{2 +}

= 10 electrons = 2,8
O

^{2 -}

= 10 electrons = 2,8
Si

^{2 +}

= 12 electrons = 2,8,2

⟹

Si

^{2 +}

is not isoelectronic with the rest.

Read lesson note on Acids, Bases And Salts (Nigeria Only) (WAEC)

Read lesson note on Atomic Structure And Bonding (JAMB)

Acids, Bases And Salts (Nigeria Only)

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Question 28 Report

The part of the total energy of a system that accounts for the useful work done in a system is known as

Enthalpy change

None of the above

Gibb's free energy

Entropy

Answer Details

The part of the total energy of a system that accounts for the useful work done in a system is known as "Gibbs free energy". Gibbs free energy is a thermodynamic property that represents the amount of energy that can be converted into useful work in a system. It takes into account both the energy of the system and the entropy, or disorder, of the system. In other words, Gibbs free energy is a measure of the energy available to do work, taking into account the energy that is unavailable due to entropy. In simple terms, if a system has a high Gibbs free energy, it has a lot of energy available to do work, and if a system has a low Gibbs free energy, it has little energy available to do work.

Read lesson note on Energy Changes (JAMB)

Energy Changes

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Question 29 Report

A secondary alkanol can be oxidized to give an

alkanone

alkyne

alkanoic acid

alkanal

Answer Details

A secondary alkanol is an alcohol with two carbon atoms attached to the carbon bearing the hydroxyl group (-OH). Secondary alkanols can be oxidized by a strong oxidizing agent, such as potassium dichromate (K2Cr2O7), to give an alkanone. During the oxidation process, the oxygen atom from the oxidizing agent replaces the hydroxyl group of the secondary alkanol to form a carbonyl group (C=O) in the alkanone. Since alkanones contain a carbonyl group, they are also known as ketones. Therefore, the answer to the question is alkanone, as secondary alkanols can be oxidized to form ketones.

Read lesson note on Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

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Question 30 Report

$200\text{cm}^3$ of $0.50\text{mol}/\text{dm}^3$ solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

10g

25g

20g

15g

Answer Details

To solve this problem, we need to use the concept of stoichiometry and the solubility product constant (Ksp) of calcium hydrogen trioxocarbonate (IV). First, we need to write the balanced equation for the reaction that occurs when the solution of calcium hydrogen trioxocarbonate (IV) is heated: Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g) From the balanced equation, we can see that 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate. Therefore, we need to determine the number of moles of calcium hydrogen trioxocarbonate (IV) in the solution: Number of moles = concentration x volume Number of moles = 0.50 mol/dm³ x 0.2 dm³ Number of moles = 0.1 mol Since 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate, the number of moles of calcium carbonate produced will also be 0.1 mol. Next, we need to use the solubility product constant (Ksp) of calcium carbonate to determine the maximum amount of solid that can be precipitated: Ksp = [Ca²⁺][CO3²⁻] Ksp = 3.3 x 10⁻⁹ (at 25°C) At the maximum amount of solid precipitated, all the calcium carbonate formed will have precipitated, and the concentration of calcium ions and carbonate ions will be equal. Therefore, we can assume that the concentration of calcium ions and carbonate ions is both x. Substituting into the Ksp expression: Ksp = x² 3.3 x 10⁻⁹ = x² x = 5.74 x 10⁻⁵ mol/dm³ The mass of calcium carbonate precipitated can now be calculated: Mass = number of moles x molar mass Mass = 0.1 mol x 100.1 g/mol Mass = 10.01 g Therefore, the maximum weight of solid precipitated is approximately 10 g. Note that this calculation assumes that all the calcium carbonate precipitated as a solid, which may not always be the case in a real-world experiment. Additionally, this calculation does not take into account any losses due to filtration or other experimental errors.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 31 Report

Which of the following factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below?

$2HCl + CaCO_3 \rightarrow CaCl_2 + H_2O + CO_2$

Increasing the concentration of the acid

Increasing the surface area of CaCO

_{3}

A and B

Reducing the concentration of the acid

Answer Details

The following factors increase a reaction rate

- Increase in concentration of reactants

- Increase in temperature

- Addition of catalyst

- Increase in the surface area of reactant(s)

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 32 Report

Which of the following alkaline metals react more quickly spontaneously with water?

Calcium

Beryllium

Magnesium

Strontium

Answer Details

The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) are the second most reactive metals in the periodic table, and, like the Group 1 metals, have increasing reactivity in the higher periods. Beryllium (Be) is the only alkaline earth metal that does not react with water or steam, even if metal is heated to red heat. Additionally, beryllium has a resistant outer oxide layer that lowers its reactivity at lower temperatures.

Magnesium shows insignificant reaction with water, but burns vigorously with steam or water vapor to produce white magnesium oxide and hydrogen gas:

Mg(s) + 2 H2O(l) ⟶ Mg(OH)2(s) + H2(g)

A metal reacting with cold water will produce metal hydroxide. However, if a metal reacts with steam, like magnesium, metal oxide is produced as a result of metal hydroxides splitting upon heating.

The hydroxides of calcium, strontium and barium are only slightly water-soluble but produce sufficient hydroxide ions to make the environment basic, giving a general equation of:

M(s) + 2 H2O(l) ⟶ M(OH)2(aq) + H2(g)

Order of reactivity	Metal	Reactions with water or steam
most reactive	potassium (K)	very vigorous reaction with cold water
↑	sodium (Na)	vigorous reaction with cold water
↓	calcium (Ca)	less vigorous reaction with cold water
least reactive	magnesium (Mg)	slow reaction with cold water, vigorous with steam

If metals react with cold water, hydroxides are produced.
Metals that react with steam form oxides.
Hydrogen is always produced when a metal reacts with cold water or steam.

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Metals And Their Compounds

Organic Compounds

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Question 33 Report

X is a substance which liberates $ \mathrm{CO_2} $ on treatment with concentrated $ \mathrm{H_2SO_4} $. A warm solution of X can decolorize acidified $ \mathrm{KMnO_4} $. X is

HCOOH

NaHCO

_{3}

FeCO

_{3}

H

_{2}

C

_{2}

O

_{4}

Answer Details

It should be noted that for X to liberate CO

_{2}

, X must be a carbonate or an oxalate. Since X decolorizes KMnO

_{4}

, X must be an oxalate.
Therefore, X is H

_{2}

C

_{2}

O

_{4}

.

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Periodic Chemistry (Ghana Only)

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Question 34 Report

Which of the following is a physical change?

Exposing white phosphorus to air

Burning kerosene

Dissolving calcium in water

Freezing ice cream

Answer Details

A physical change refers to a change in a substance that does not result in a change in its chemical composition. Out of the options provided, freezing ice cream is a physical change. This is because when ice cream is frozen, it changes from a liquid state to a solid state without any chemical reaction occurring. Exposing white phosphorus to air is a chemical change, as it reacts with oxygen in the air to form a new substance, phosphorus oxide. Burning kerosene is also a chemical change, as it undergoes combustion to form new substances, such as carbon dioxide and water vapor. Dissolving calcium in water is a physical change, as it simply involves the physical mixing of two substances without any chemical reaction occurring. Therefore, the only option that is a physical change is freezing ice cream.

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Question 35 Report

For the general equation of the nature

XP + yQ ⇌ mR + nS, the expression for the equilibrium constant is

k[P]

^{x}

[Q]

^{y}

\frac{m [R] [S]}{x [P] y [Q]}

\frac{[P]^{x} [Q]^{y}}{[R]^{m} [S]^{n}}

\frac{[R]^{m} [S]^{n}}{[P]^{x} [Q]^{y}}

Answer Details

The expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is: Kc = [R]m[S]n / [P]x[Q]y where Kc is the equilibrium constant, [R] and [S] are the concentrations of the products, and [P] and [Q] are the concentrations of the reactants, all raised to the stoichiometric coefficients (m, n, x, y) in the balanced equation. This equation is known as the equilibrium constant expression and it represents the ratio of the concentrations of the products and reactants at equilibrium for a particular chemical reaction. The equilibrium constant is a measure of how far a reaction proceeds towards completion, with a larger value indicating a greater extent of reaction. The equilibrium constant expression is derived from the law of mass action, which states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants raised to their stoichiometric coefficients. At equilibrium, the rates of the forward and reverse reactions are equal, and the equilibrium constant expression represents the ratio of the rate constants for these two reactions. Therefore, the correct expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is Kc = [R]m[S]n / [P]x[Q]y.

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Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Equilibria

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Question 36 Report

Which two gases can be used for the demonstration of the fountain experiment?

SO

_{3}

and NH

_{3}

SO

_{2}

and NH

_{3}

HCl and NH

_{3}

SO

_{2}

and HCl

Answer Details

Two gases that can be used in the study of fountain experiment is ammonia gas and hydrogen chloride gas. The experiment introduces concepts like solubility and the gas laws at the entry level.

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Question 37 Report

How many electrons will be found in the nucleus of an atom with mass number 23 and 17 neutrons?

none

6

7

8

Answer Details

Electrons are not found in the nucleus of an atom. The nucleus of an atom only contains protons and neutrons, while electrons are located outside the nucleus in the electron cloud. The mass number of an atom is equal to the sum of the number of protons and the number of neutrons in the nucleus. Therefore, if an atom has a mass number of 23 and 17 neutrons, then the number of protons in the nucleus can be calculated as: Protons = Mass number - Neutrons Protons = 23 - 17 Protons = 6 This means that the nucleus of the atom contains 6 protons. The number of electrons in a neutral atom is equal to the number of protons, so the atom also contains 6 electrons in the electron cloud surrounding the nucleus. In summary, the answer is that there are 6 protons and 6 electrons in the atom.

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Question 38 Report

Hydrocarbons which will react with Tollen's reagent conform to the general formula

C

_{n}

H

_{2 n}

C

_{n}

H

_{2 n + 2}

C

_{n}

H

_{2 n - 2}

C

_{n}

H

_{2 n + 1}

OH

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Question 39 Report

The IUPAC nomenclature of the compound

$\mathrm{H_3C - CH(CH_3) - CH(CH_3) - CH_2 - CH_3}$

3,4- dimethylhexane

2,3- dimethyl pentane

2- ethyl hexane

2,3- dimethylhexane