JAMB UTME - Chemistry - 2022

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Question 2 Report

How many neutrons are present in atom with mass number and atomic number 37 and 17 respectively?

18

20

37

17

Answer Details

The atomic number of an atom represents the number of protons in the nucleus of the atom. Since the atomic number given is 17, it means that there are 17 protons in the nucleus. The mass number of an atom represents the total number of protons and neutrons present in the nucleus. Therefore, if the mass number is given as 37, it means that the total number of protons and neutrons in the nucleus is 37. To determine the number of neutrons in the nucleus, we can subtract the atomic number (which represents the number of protons) from the mass number (which represents the total number of protons and neutrons). Thus, the number of neutrons in the atom with a mass number of 37 and an atomic number of 17 is: Number of neutrons = Mass number - Atomic number = 37 - 17 = 20 Therefore, the answer is 20.

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 3 Report

How many bonding pair are present in carbon(iv)oxide?

4

3

2

1

Answer Details

Carbon dioxide (CO2) has a linear molecular geometry, with two oxygen atoms bonded to the central carbon atom. Each bond between carbon and oxygen is a double bond, consisting of two pairs of electrons shared between the atoms. Therefore, there are two bonding pairs in each of the carbon-oxygen double bonds, giving a total of four bonding pairs in CO2. The answer is 4.

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 4 Report

Which of the following roles does sodium chloride play in preparation? It

reacts with glycerol

purifies the soap, accelerates the decomposition of fat and oil

separates soap from glycerol

converts the fat acid to its sodium salt

Answer Details

The role that sodium chloride (NaCl) plays in soap preparation is to separate soap from glycerol. When fats or oils are hydrolyzed with an alkali, such as sodium hydroxide (NaOH), the result is a mixture of soap and glycerol. Adding NaCl to this mixture helps to induce the precipitation of the soap, allowing it to be separated from the glycerol. This process is known as "salting out" and is used to purify the soap and remove impurities. Sodium chloride does not react with glycerol or accelerate the decomposition of fat and oil. Also, it does not convert the fatty acid to its sodium salt as this conversion is done by the alkali (such as NaOH) during the saponification process.

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

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Question 5 Report

Which of the following does NOT contain aluminum as a component?

Over-head cables

Duralumin

Container for caustic soda

Container for trioxonitrate (v) acid

Metals And Their Compounds

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Question 6 Report

The boiling point of water, ethanol, toulene and butan-2-ol are 373.0k, 351.3k, 383.6k and 372.5k respectively, which liquid has the highest vapour pressure at 323.0k

water

toluene

ethanol

butan-2-ol

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Question 7 Report

Electrons enter into orbitals in order of increasing energy as exemplified by?

1S

^{2}

2S

^{2}

2PX

^{2}

2py

^{2}

2pz

^{0}

3s

^{0}

1S

^{2}

2S

^{2}

2PX

^{2}

2py

^{1}

2pz

^{1}

3s

^{0}

1S

^{2}

2S

^{2}

2PX

^{1}

2py

^{1}

2pz

^{1}

3s

^{1}

1S

^{2}

2S

^{2}

2PX

^{2}

2py

^{2}

2pz

^{0}

3s

^{1}

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Question 8 Report

An organic compound with fishy smell is likely to have a general formula?

RCONHR

^{1}

RCONH

_{2}

RNH

_{2}

RCOR

^{1}

Answer Details

The organic compound with a fishy smell is most likely to have the general formula RNH2, which represents a primary amine. Amines are organic compounds that contain a nitrogen atom bonded to one or more carbon atoms. Primary amines have one alkyl or aryl group and two hydrogen atoms bonded to the nitrogen atom. Some primary amines have a fishy smell, which is caused by the presence of volatile amines. These amines are small molecules that can easily evaporate and have a strong odor, similar to that of fish. Examples of compounds that have a fishy smell include trimethylamine, which is found in fish, and butylamine, which is used in the production of rubber and pharmaceuticals. In summary, the organic compound with a fishy smell is likely to have the general formula RNH2, which represents a primary amine.

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Question 9 Report

Which of the following compound is NOT the correct formed compound when the parent metal is heated in air?

Calcium oxide

Sodium oxide ( Na2O )

Copper (II) oxide (CuO)

tri-iron tetraoxide (Fe2O)

Answer Details

The compound that is NOT correctly formed when the parent metal is heated in air is: tri-iron tetraoxide (Fe2O). This is because the correct compound formed from the heating of iron in air is iron (III) oxide or Fe2O3. The formula for tri-iron tetraoxide is incorrect, as it implies that there are only three iron atoms in the compound when there should be four.

Read lesson note on Metals And Their Compounds (JAMB)

Metals And Their Compounds

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Question 10 Report

When marble is heated to 1473K, another whiter solid is obtained which reacts vigoriously with water to give an alkaline solution. The solution contains

NaOH

KOH

Mg(OH)2

Ca(OH)2

Answer Details

The white solid obtained when marble (calcium carbonate, CaCO3) is heated to 1473K is calcium oxide (CaO), also known as quicklime. When quicklime reacts vigorously with water, it forms calcium hydroxide (Ca(OH)2), which is an alkaline solution. Therefore, the solution obtained from the reaction of quicklime with water contains calcium hydroxide (Ca(OH)2).

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

Periodic Chemistry (Ghana Only)

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Question 11 Report

Which of the following statement is TRUE of the complete hydrolysis of a glyceride by sodium hydroxide?

3 moles of NaOH are required for each mole of glyceride

3 moles of glycerol are produced

only one mole of soap is formed

concentrated H2SO4 is essential for the completion of the reaction.

Answer Details

The statement that is TRUE of the complete hydrolysis of a glyceride by sodium hydroxide is: - 3 moles of NaOH are required for each mole of glyceride. During the hydrolysis of a glyceride (a triglyceride), the ester bonds between the fatty acid chains and glycerol are broken by the action of a strong base like sodium hydroxide. This results in the formation of glycerol and the corresponding salts of fatty acids, which are commonly known as "soaps." The reaction can be represented by the following equation: Triglyceride + 3 NaOH → 3 soap + glycerol As per the equation, 3 moles of NaOH are required to hydrolyze one mole of glyceride, and 3 moles of soap and one mole of glycerol are produced. The use of concentrated sulfuric acid (H2SO4) is not essential for the completion of the reaction, but it can be used as a catalyst to speed up the reaction.

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

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Question 12 Report

Which of the following is stable to heat

NaHCO3

( NH4 )2 SO4

Ag NH3

K2 CO3

Answer Details

Out of the given options, K2CO3 is stable to heat.

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

Non-metals And Their Compounds

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Question 13 Report

30 cm3 of oxygen at 10 atmosphere pressure is placed in a 20 dm3 container. Calculate the new pressure if the temperature is kept constant.

6.7 atm

15.0 atm

6.0 atm

66.0 atm

Answer Details

Given:

Initial pressure, 𝑃1=10P1=10 atmospheres

Initial volume, 𝑉1=30 cm3V1=30cm3

Final volume, 𝑉2=20 dm3V2=20dm3

First, convert all volumes to the same units. Since 1 dm3dm3 is 1000 cm3cm3:

𝑉2=20 dm3=20×1000 cm3=20000 cm3V2=20dm3=20×1000cm3=20000cm3

Now, using Boyle's Law:

𝑃1𝑉1=𝑃2𝑉2P1V1=P2V2

Substitute the known values into the equation:

10×30=𝑃2×2000010×30=P2×20000

300=𝑃2×20000300=P2×20000

Solve for 𝑃2P2:

𝑃2=30020000P2=20000300

𝑃2=0.015 atmospheresP2=0.015atmospheres

Therefore, the new pressure if the temperature is kept constant is:

0.015 atmospheres0.015atmospheres

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

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Question 14 Report

Which of the following conducts electricity

Sulphur

Graphite

Diamond

Red phosphorous

Answer Details

Graphite is the option that conducts electricity.

Read lesson note on Electrolysis (JAMB)

Electrolysis

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Question 15 Report

$2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$

From the equation above, calculate the volume of unreacted oxygen gas if a mixture of $50\mathrm{cm}^3$ of hydroden and $75\mathrm{cm}^3$ of oxygen are involved

85cm

^{3}

50cm

^{3}

125cm

^{3}

55cm

^{3}

Answer Details

The balanced chemical equation shows that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor. Therefore, the stoichiometric ratio of hydrogen to oxygen is 2:1. In this problem, there are 50cm3 of hydrogen gas and 75cm3 of oxygen gas. Since the gases are at the same temperature and pressure, their volumes are directly proportional to the number of moles of gas present. Using the stoichiometric ratio, we can calculate that the amount of oxygen gas required to react completely with 50cm3 of hydrogen gas is (1/2) * 50cm3 = 25cm3. Since there are 75cm3 of oxygen gas present, there must be (75cm3 - 25cm3) = 50cm3 of unreacted oxygen gas remaining. Therefore, the volume of unreacted oxygen gas is 50cm3. Answer: 50cm3

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Stoichiometry And Chemical Reactions

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Question 16 Report

A sample of gas exerts a pressure of 8.2 atm when confined in a 2.93 dm3 container at 20c. The number of moles of gas in the sample is

1. 00

2.00

3.00

4.00

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Question 17 Report

Which of the following will act as both oxidizing agents and reducing agents?

H2 S

NH3

Cl2

SO2

Answer Details

The oxidizing and reducing properties of a substance depend on its ability to gain or lose electrons. A substance that can gain electrons acts as an oxidizing agent, while a substance that can lose electrons acts as a reducing agent. Among the given options, both Cl2 (chlorine gas) and SO2 (sulfur dioxide) can act as both oxidizing and reducing agents depending on the reaction conditions. - Cl2 can act as an oxidizing agent when it gains electrons to form Cl- ions, and it can act as a reducing agent when it loses electrons to form Cl+ ions. For example, in the reaction Cl2 + 2KBr → 2KCl + Br2, chlorine gas is acting as an oxidizing agent since it is gaining electrons from bromide ions to form bromine gas. However, in the reaction 2Cl- + Cl2 → 2Cl2-, chlorine gas is acting as a reducing agent since it is losing electrons to form chloride ions. - SO2 can act as an oxidizing agent when it gains electrons to form sulfite ions (SO32-), and it can act as a reducing agent when it loses electrons to form sulfur trioxide (SO3). For example, in the reaction SO2 + 2H2S → 3S + 2H2O, sulfur dioxide is acting as a reducing agent since it is losing electrons to form elemental sulfur. However, in the reaction 2SO32- + O2 → 2SO42-, sulfur dioxide is acting as an oxidizing agent since it is gaining electrons to form sulfate ions. H2S (hydrogen sulfide) and NH3 (ammonia) are not likely to act as both oxidizing and reducing agents under normal conditions. H2S tends to act as a reducing agent by donating electrons to oxidizing agents, while NH3 tends to act as a reducing agent by donating electrons to oxidizing agents or as a base by accepting protons.

Read lesson note on Chemical Bonds (Ghana Only) (WAEC)

Chemical Bonds (Ghana Only)

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Question 18 Report

An organic compound contains 69% carbon, 15.3% hydrogen and 30.7% oxygen. Calculate the the empirical formula [C=12, H = 1, O = 16]

C

_{4}

H

_{1} 2

O

C

_{3}

H

_{9}

O

C

_{4}

H

_{9}

O

C

_{3}

H

_{8}

O

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Question 19 Report

Copper (II) tetraoxosulphate (IV) is widely used as

fertilizer

fungicide

Disinfectant

purifier

Answer Details

Copper (II) tetraoxosulphate (IV), also known as copper sulfate or CuSO4, is widely used as a fungicide and a disinfectant. As a fungicide, copper sulfate is effective in controlling fungal diseases in plants, including mildew, leaf spots, and blights. It is also used as a fungicide in swimming pools to prevent the growth of algae. As a disinfectant, copper sulfate is effective in killing bacteria and viruses. It is used in a variety of applications, including in the production of animal feed, as a preservative for wood, and in water treatment to kill bacteria and algae. While copper sulfate has been used as a fertilizer in the past, its use in this capacity has largely been replaced by other compounds. It is not commonly used as a purifier.

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Read lesson note on Environmental Pollution (JAMB)

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

Stoichiometry And Chemical Reactions

Environmental Pollution

Periodic Chemistry (Ghana Only)

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Question 20 Report

What volume of (dm$^{3}$) of water will be added to 10dm$^{3}$ of 2.0 mol/dm$^{3}$ HCL acid solution to give a final solution of 0.5 mol/dm$^{3}$?

30

40

20

50

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Question 21 Report

A chemical widely used as a fertilizer is?

galena

bauxite

emerald

nitrochalk

Answer Details

The chemical widely used as a fertilizer is nitrochalk. Nitrochalk is a type of fertilizer that contains a mixture of ammonium nitrate and calcium carbonate. Ammonium nitrate provides the necessary nitrogen for plant growth, while calcium carbonate helps to balance the soil's pH level. This combination of nutrients helps to promote healthy plant growth and increase crop yields. Nitrochalk is commonly used in agriculture and gardening to fertilize crops such as corn, wheat, and soybeans, as well as fruits and vegetables.

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

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Question 22 Report

On the basis of the electrochemical series, which of these ions will show the greater tendency to be discharged at the cathode in an electrolytic cell

cu

^{2 +}

sn

^{2 +}

fe

^{2 +}

zn

^{2 +}

Answer Details

The electrochemical series is a list of metals and ions arranged in order of their decreasing tendency to lose or gain electrons, and thus, their ability to act as reducing or oxidizing agents. The higher the position of a metal or ion in the electrochemical series, the greater its tendency to lose electrons and undergo oxidation, while the lower its position, the greater its tendency to gain electrons and undergo reduction. In an electrolytic cell, the cathode is the electrode where reduction occurs, meaning that cations (positively charged ions) are attracted and gain electrons to form neutral atoms or molecules. Based on the electrochemical series, the ion with the higher position in the series will have a greater tendency to gain electrons and be discharged at the cathode, while the ion with the lower position will have a lower tendency and may not be discharged at all. Among the given options, the electrochemical series order is: Cu2+ > Sn2+ > Fe2+ > Zn2+ Therefore, Cu2+ has the highest tendency to be discharged at the cathode and undergo reduction, while Zn2+ has the lowest tendency. So, in an electrolytic cell, Cu2+ will be discharged at the cathode, while Zn2+ may not be discharged at all, depending on the conditions of the cell.

Read lesson note on Electrolysis (JAMB)

Read lesson note on Redox Reactions (WAEC)

Electrolysis

Redox Reactions

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Question 23 Report

In order to electroplate spoon with silver, the arrangement of the electrolytic cell is?

the anode is a silver rod and the cathode is the spoon

the anode is the spoon and the cathode is a silver rod

the electrolyte is silver trioxonitrate(v)( solution and the cathode is a silver rod.

the electrolyte is silver trioxonitrate(v) solution and the anode is the spoon

Electrolysis

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Question 24 Report

Crude petroleum is converted to useful products by the process of?

fractional crystallization

fractional distillation

filtration

chromatography

Answer Details

The process of converting crude petroleum into useful products is known as fractional distillation. Crude petroleum is a mixture of different hydrocarbons, and fractional distillation separates these hydrocarbons based on their boiling points. During the process of fractional distillation, crude petroleum is heated to a high temperature, and the resulting vapors are passed through a tower called a fractionating column. This column contains a series of trays, and each tray contains a specific temperature range. As the vapors rise up the column, they cool and condense into liquids on the tray with a temperature that matches their boiling point. The liquids are then collected and further refined into useful products like gasoline, diesel, jet fuel, and heating oil. Fractional distillation is an important process because it allows us to separate and purify the different components of crude petroleum, which have different properties and uses. For example, gasoline has a lower boiling point and is more volatile than diesel fuel, which makes it ideal for use in cars. By separating these components, we can create products that meet specific needs and requirements.

Read lesson note on Solubility (JAMB)

Read lesson note on Introduction To Chemistry (WAEC)

Solubility

Introduction To Chemistry

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Question 25 Report

In which of the following will hydrogen form ionic compound?

HCL

NaH

NH

_{3}

CH

_{4}

Chemical Combination

Organic Compounds

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Question 26 Report

The sulphide that is commonly used in coating electric fluorescent tubes is?

iron(ii)Sulphide

tin(ii)sulphide

Zinc Sulphide

lead(iv) Sulphide

Answer Details

The sulphide commonly used in coating electric fluorescent tubes is Zinc Sulphide. Zinc Sulphide is a type of material that glows when it is exposed to ultraviolet light. When ultraviolet light is generated inside a fluorescent tube, it excites the Zinc Sulphide particles, causing them to emit visible light. This visible light is what we see as the bright light coming from the tube. So, Zinc Sulphide acts as a phosphor and helps in producing the bright light in fluorescent tubes.

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

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Question 27 Report

Mixing aqueos solution of barium hydroxide and sodium tetraoxocarbonate (iv) yields a white precipitate of

Barium oxide

Sodium tetraoxocarbonate

Sodium hydroxide

Barium tetraoxocarbonate

Answer Details

Mixing aqueous solutions of barium hydroxide and sodium tetraoxocarbonate (IV) would result in a chemical reaction that produces a white precipitate of barium tetraoxocarbonate (IV). The balanced chemical equation for this reaction is: Ba(OH)2(aq) + Na2CO3(aq) → BaCO3(s) + 2NaOH(aq) In the above equation, the barium hydroxide (Ba(OH)2) reacts with sodium tetraoxocarbonate (IV) (Na2CO3) to form barium tetraoxocarbonate (IV) (BaCO3), which is a white precipitate, and sodium hydroxide (NaOH). Therefore, the correct option is 4) Barium tetraoxocarbonate.

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

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Question 28 Report

A metal which can be used as sacrificial anode for preventing corrosion of length of iron pipe is

copper

magnesium

silver

lead

Metals And Their Compounds

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Question 29 Report

The IUPAC name for CICH2-CH2-CH2-OH is

1-chloropropan-3-ol

3-chloropropan-1-ol

1 cholorpropano

3-chloropropanol

Answer Details

The IUPAC name for CICH2-CH2-CH2-OH is 3-chloropropan-1-ol. To name the compound using the IUPAC nomenclature system, we start by identifying the longest continuous chain of carbon atoms that contains the functional group (-OH). In this case, the longest chain contains three carbon atoms, so the root name is propane. Next, we identify the position of the substituent (-Cl) on the chain. The substituent is attached to the third carbon atom in the chain, so the name of the compound becomes 3-chloropropane. Finally, we add the suffix -ol to indicate that the compound contains an alcohol functional group (-OH), so the complete name of the compound is 3-chloropropan-1-ol. Therefore, the correct answer is 3-chloropropan-1-ol.

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

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Question 30 Report

An organic compound decolourized acidified KMnO4 solution but failed to react with ammonical AgNO3 solution. The organic compound is likely?

a carboxylic acid

Alkane

Alkene

Alkyne

Answer Details

The given information suggests that the organic compound is an unsaturated compound (because it decolorized the acidified KMnO4 solution), but it does not contain a functional group that reacts with ammonical AgNO3 solution. Therefore, the likely organic compound is an alkene or an alkyne. Carboxylic acids can also react with acidified KMnO4 solution, but they would also react with ammonical AgNO3 solution to form a silver carboxylate salt. Alkanes are saturated compounds and do not react with either reagent, so they would not decolorize the acidified KMnO4 solution. Therefore, based on the given information, the most likely option is either an alkene or an alkyne.

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Question 31 Report

An organic compound which decolorizes bromine water is likely to be?

C

_{3}

H

_{8}

C

_{2}

H

_{6}

C

_{4}

H

_{1} 0

C $_{2}$ H $_{4}$ $_{2}$ $_{4}$

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Question 32 Report

$ \mathrm{SO_2 + O_2 \rightarrow 2SO_3} $

In the reaction above, the most suitable catalyst is?

chromium(vi)oxide

iron(iii)oxide

copper(i)oxide

vanadium(v)oxide

Answer Details

The most suitable catalyst for the given reaction is vanadium(V)oxide (V2O5). Vanadium(V)oxide is a commonly used catalyst for the oxidation of sulfur dioxide (SO2) to sulfur trioxide (SO3). The reaction is an exothermic reaction, and it occurs at high temperatures (around 450-500°C) in the presence of a catalyst. V2O5 is an effective catalyst for this reaction because it has a high surface area and can provide active sites for the reaction to occur. The vanadium ions in the V2O5 catalyst undergo redox reactions with the sulfur dioxide and oxygen molecules, which promotes the formation of sulfur trioxide. Chromium(VI)oxide and iron(III)oxide are not suitable catalysts for this reaction because they are not effective at promoting the oxidation of sulfur dioxide to sulfur trioxide. Copper(I)oxide can be used as a catalyst for the reaction, but it is not as effective as vanadium(V)oxide.

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Question 33 Report

Which of the following will precipitate in dil. HCl

ZnS

Na2 S

Fe S

CuS

Answer Details

Among the given options, only CuS will precipitate in dilute HCl. CuS is insoluble in dilute HCl, and hence it will precipitate when added to dilute HCl. However, the other options will dissolve in dilute HCl, and hence they will not precipitate. ZnS will dissolve in dilute HCl to form ZnCl2 and H2S. Na2S will react with dilute HCl to produce H2S and NaCl. FeS will dissolve in dilute HCl to form FeCl2 and H2S. Therefore, the correct answer is (4) CuS.

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

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Question 34 Report

Complete dehydrogenation of ethyne yields

benzene

methane

ethane

propane

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Question 35 Report

Which of the following increases as boiling water changes to steam?

Temperature of the system

degree of disorder of the system

number of molecule

Activation energy

Answer Details

The degree of disorder of the system increases as boiling water changes to steam. When water is boiled and changes to steam, the water molecules gain energy and become more disordered, which means that the molecules move more rapidly and the entropy of the system increases. The temperature of the system also increases during this process, but the degree of disorder is the factor that specifically increases as the water changes to steam. The number of molecules and activation energy remain constant during this phase transition.

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

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Question 36 Report

$ \mathrm{Zn + 2HCl \rightarrow ZnCl_2 + H_2} $

What happens to zinc in the above reaction?

oxidized

a reactant

reduced

a metal

Answer Details

In the above reaction, zinc (Zn) reacts with hydrochloric acid (HCl) to form zinc chloride (ZnCl2) and hydrogen gas (H2). The chemical equation for the reaction is: Zn + 2HCl → ZnCl2 + H2 During the reaction, zinc atoms lose two electrons each and get oxidized to form positively charged zinc ions (Zn2+), as they react with the hydrogen ions (H+) from the hydrochloric acid to form zinc chloride. The hydrogen ions, on the other hand, gain an electron each and get reduced to form hydrogen gas molecules (H2). Therefore, in the given reaction, zinc is getting oxidized, as it loses electrons and forms a positively charged ion. Hence, the correct option is "oxidized."

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Oxidation And Reduction

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Question 37 Report

Alkanes are used mainly?

in the textile industry

in the hydrogenation of oils

as domestic and industrial fuels

as fine chemicals

Chemistry, Industry And The Environment

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Question 38 Report

What is the shape of a molecule of CCl4?

Pyramid

tetrahedral

Trigonal planar

linear

Answer Details

The shape of a molecule of CCl4 is tetrahedral.

Read lesson note on Atomic Structure And Bonding (JAMB)