JAMB UTME - Chemistry - 2000 | Green Bridge CBT

Tambaya 1 Rahoto

The fundamental difference between the three states of matter is the

shape of their particles

number of particles in each state Amsar da ta dace

shape of the container they occupy

degree movement of their particles

Bayanin Amsa

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Karanta bayanin darasi akan States Of Matter (WAEC)

Kinetic Theory Of Matter And Gas Laws

States Of Matter

Duba Amsa

Tambaya 2 Rahoto

A mixture of iodine and sulphur crystals can be separated by treatment with

water to filter off sulphur

carbon (IV) suiphide to filter off iodine Amsar da ta dace

ethanoic acid to filter off sulphur

methanol to filter off iodine

Bayanin Amsa

A mixture of iodine and sulfur crystals can be separated by treatment with carbon (IV) sulfide to filter off iodine. This is because iodine dissolves in carbon (IV) sulfide, while sulfur does not. Therefore, adding carbon (IV) sulfide to the mixture will dissolve the iodine, allowing it to be separated from the sulfur by filtration. Water, ethanoic acid, and methanol would not be effective in separating iodine and sulfur because they do not have the same solvent properties as carbon (IV) sulfide towards iodine.

Karanta bayanin darasi akan Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

Duba Amsa

Tambaya 3 Rahoto

50cm of carbon (II) oxide exploded with 150cm\(^{3}\) of air containing 20% oxygen by volume. Which of the reactants was in excess?

carbon (II) oxide

carbon (IV) oxide

oxygen Amsar da ta dace

nitrogen

Bayanin Amsa

2CO + O

_{2}

\to

2CO

_{2}

50cm

^{3}

\to

25cm

^{3}

20%

\times

150cm

^{3}

30cm

^{3}

is the volume of oxygen available for the reactions.
But only 25cm

^{3}

is required, therefore O

_{2}

is in excess.

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Duba Amsa

Tambaya 4 Rahoto

0.46g of ethanol when burned raised the temperature of 50g of water by 14.3K. Calculate the heat of combustion of ethanol.

[C = 12, O = 16, H = 1, Specific heat capacity of water = 4.2 jg-1K-1]

+3000 KJ mol-1

+300 KJ mol-1

-300 KJ mol-1 Amsar da ta dace

-3000 KJ mol-1

Bayanin Amsa

Heat, = -M C Q
= -50 * 4.2 * 14.3
= -300 KJ mol-1

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

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Tambaya 5 Rahoto

HCL(aq) + H2O(l) H3O+(aq) + Cl-(aq)

In the reaction above, Cl-(aq) is the

conjugate acid

acid

conjugate base Amsar da ta dace

base

Bayanin Amsa

In the given reaction, HCl donates a proton (H+) to H2O to form H3O+ and Cl-. The species that accepts the proton is the base, which in this case is water (H2O). The species that donates the proton is the acid, which in this case is HCl. The conjugate base is the remaining species after the acid has donated the proton, which is Cl- in this reaction. Therefore, the correct answer is "conjugate base".

Karanta bayanin darasi akan Acids, Bases And Salts (Ghana Only) (WAEC)

Karanta bayanin darasi akan Periodic Chemistry (Ghana Only) (WAEC)

Acids, Bases And Salts (Ghana Only)

Periodic Chemistry (Ghana Only)

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Tambaya 6 Rahoto

N2(g) + O2(g) ↔ 2NO(g); ΔH = + 180.6 kJ mol-1

In the reaction above, increase in temperature will

increase the quantity of N2

increase in the yield of NO Amsar da ta dace

decrease the yield of NO

decrease the quantity of O2

Bayanin Amsa

In the given chemical reaction, the enthalpy change is positive (+180.6 kJ/mol) indicating that the reaction is endothermic, meaning it absorbs heat from the surroundings. According to Le Chatelier's principle, an increase in temperature will favor an endothermic reaction. Therefore, an increase in temperature will increase the yield of NO, which is the product of the reaction. Hence, the correct answer is: "increase in the yield of NO."

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Duba Amsa

Tambaya 7 Rahoto

In which order are the following salts sensitive to light?

Agl > AgCl > AgBr

AgCl > Agl > AgBr

AgBr > AgCl > Agl

AgCl > AgBr > Agl Amsar da ta dace

Bayanin Amsa

The sensitivity of salts to light depends on the stability of their respective silver halide compounds. The stability of these compounds is determined by the size of the halide ion. The larger the halide ion, the less stable the corresponding silver halide compound is. In the given options, the salts are arranged in the order of increasing size of halide ions. Thus, the stability of the corresponding silver halide compounds decreases in the same order. The most stable compound is the one with the smallest halide ion, and the least stable is the one with the largest halide ion. Therefore, the correct order of sensitivity to light for the given salts is: AgCl > AgBr > AgI This is because AgI has the largest halide ion, making its corresponding silver halide compound the least stable and most sensitive to light. On the other hand, AgCl has the smallest halide ion, making its corresponding silver halide compound the most stable and least sensitive to light.

Karanta bayanin darasi akan Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

Duba Amsa

Tambaya 8 Rahoto

The reaction of carbide with water gives

ethyne Amsar da ta dace

ethene

ethane

ethanal

Bayanin Amsa

The reaction of carbide (calcium carbide) with water gives ethyne (also known as acetylene). This reaction is an example of an acid-base reaction, where the carbide acts as a base and water acts as an acid. The balanced chemical equation for this reaction is: CaC2 + 2H2O → C2H2 + Ca(OH)2 The ethyne gas produced in this reaction is widely used as a fuel for welding and cutting metals, as well as for other industrial applications.

Karanta bayanin darasi akan Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

Duba Amsa

Tambaya 9 Rahoto

The repeating unit in natural rubber is

alkyne

isoprene Amsar da ta dace

n-propene

peoprene

Bayanin Amsa

The repeating unit in natural rubber is isoprene. Natural rubber is a polymer made up of repeating units of the monomer isoprene. Isoprene has a molecular formula of C5H8 and a structure that contains a double bond, which allows it to form long chains through a process called polymerization. These chains then link together to form the rubbery material that we know as natural rubber.

Karanta bayanin darasi akan Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Karanta bayanin darasi akan Organic Compounds (JAMB)

Chemistry Of Carbon Compounds (Nigeria Only)

Organic Compounds

Duba Amsa

Tambaya 10 Rahoto

Hydration of ions in solution is associated with

absorption of heat

reduction of heat

conduction of heat

liberation of heat Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Duba Amsa

Tambaya 11 Rahoto

The partial pressure of oxygen in a sample of air is 452mmHg and the total pressure is 780mmHg. What is the mole fraction of oxygen?

0.203

0.579 Amsar da ta dace

2.060

5.790

Bayanin Amsa

The mole fraction of oxygen can be calculated using the formula: Mole fraction of oxygen = Partial pressure of oxygen / Total pressure Plugging in the given values, we get: Mole fraction of oxygen = 452mmHg / 780mmHg Simplifying this fraction gives us: Mole fraction of oxygen = 0.579 Therefore, the correct option is: 0.579 This means that about 57.9% of the total pressure of the air sample is due to oxygen molecules.

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Duba Amsa

Tambaya 12 Rahoto

Sucrose is made up of

glucose and glucose

glucose and fructose Amsar da ta dace

fructose and fructose

galactose and glucose

Bayanin Amsa

Sucrose is a type of sugar that is commonly found in many foods. It is made up of two simple sugar molecules, which are bonded together through a glycosidic bond. The options given are all possible simple sugar molecules, but only one option represents the correct combination of sugar molecules that make up sucrose. Sucrose is made up of one molecule of glucose and one molecule of fructose. Glucose and fructose are both monosaccharides, which are simple sugar molecules that cannot be broken down into simpler sugars. When glucose and fructose are combined through a glycosidic bond, they form a disaccharide called sucrose. Sucrose is commonly found in many foods, including fruits, vegetables, and processed foods like candy and baked goods.

Duba Amsa

Tambaya 13 Rahoto

Sieving is a technique used to separate mixtures containing solid particles of

small sizes

large sizes

different sizes Amsar da ta dace

the same size

Bayanin Amsa

Sieving is a method that is commonly used to separate mixtures that contain solid particles of different sizes. The mixture is poured through a sieve, which is a device with a mesh or a net. The smaller particles that can pass through the mesh or net will fall through, while the larger particles will be left on top. Therefore, sieving is used to separate solid particles that are of different sizes.

Karanta bayanin darasi akan Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

Duba Amsa

Tambaya 14 Rahoto

The distance between the nuclei of chlorine atoms in a chlorine molecule is 0.194nm. The atomic radius of chlorine atom is

0.097nm Amsar da ta dace

0.914nm

0.388nm

2.388nm

Bayanin Amsa

The atomic radius of an atom is defined as half the distance between the nuclei of two identical atoms when they are bonded together. In this case, we know that the distance between the nuclei of chlorine atoms in a chlorine molecule is 0.194nm. Since this is the distance between the two nuclei, we can find the atomic radius of chlorine by dividing this distance by 2. 0.194nm ÷ 2 = 0.097nm Therefore, the atomic radius of a chlorine atom is 0.097nm. Answer: 0.097nm

Karanta bayanin darasi akan Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

Duba Amsa

Tambaya 15 Rahoto

The substance often use for vulcanization of rubber is

chlorine

hydrogen peroxide

sulphur Amsar da ta dace

tetraoxosulphate (IV) acid

Bayanin Amsa

Vulcanization is a chemical process that is used to make rubber stronger and more durable. During vulcanization, the rubber is treated with a chemical agent that causes it to cross-link and form a stronger network. This process can improve the strength, durability, and elasticity of the rubber. The options given are all chemicals, but only one option is commonly used as the substance for vulcanization of rubber. The substance that is often used for vulcanization of rubber is sulfur. When sulfur is added to raw rubber and the mixture is heated, the sulfur molecules cross-link with the rubber molecules, creating a stronger and more durable material. This process is known as vulcanization and it has revolutionized the rubber industry, allowing for the creation of many new and useful rubber products.

Karanta bayanin darasi akan Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

Duba Amsa

Tambaya 16 Rahoto

Powdered marble reacts faster with hydrochloric acid solution than the granular form because the powdered form has

more molecules

more atoms

larger surface area Amsar da ta dace

relatively large mass

Bayanin Amsa

The correct answer is that powdered marble has a larger surface area. This is because the powdered form has smaller particle sizes and, therefore, a greater surface area per unit mass. This greater surface area allows for a greater number of collisions between the particles of marble and the hydrochloric acid solution, leading to a faster reaction.

Karanta bayanin darasi akan Chemical Combination (JAMB)

Chemical Combination

Duba Amsa

Tambaya 17 Rahoto

A gas that is not associated with global warming is

CO2 Amsar da ta dace

SO3

CH4

Bayanin Amsa

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Karanta bayanin darasi akan Environmental Pollution (JAMB)

Kinetic Theory Of Matter And Gas Laws

Environmental Pollution

Duba Amsa

Tambaya 18 Rahoto

The diagram above is the solubility curve of a solute, X. Find the amount of X deposited when 500cm3 of a solution of X is cooled from 60oC to 20oC.

0.750 mole Amsar da ta dace

0.950 mole

2.375 moles

4.750 moles

Bayanin Amsa

Karanta bayanin darasi akan Solubility (JAMB)

Solubility

Duba Amsa

Tambaya 19 Rahoto

Potassium vapour burns with a

blue flame

brick-red flame

violet flame Amsar da ta dace

golden-yellow flame

Bayanin Amsa

Potassium vapour burns with a violet flame.

Karanta bayanin darasi akan Oxidation And Reduction (JAMB)

Oxidation And Reduction

Duba Amsa

Tambaya 20 Rahoto

(1/2)Zn2+(aq) + e- → (1/2) Zn(s)

In the reaction above, calculate the quantity of electricity required to discharge zinc.

[F = 96 500 C mol-1]

0.965 * 104C

4.820 * 104C

9.650 * 104C Amsar da ta dace

48.200 * 104C

Bayanin Amsa

The quantity of electricity required to discharge zinc can be calculated using Faraday's law, which states that one mole of electrons carries a charge of 96,500 Coulombs. The given chemical equation shows that the reduction of 1 mole of Zn2+ requires the gain of 2 electrons (1/2 Zn2+ + e- → 1/2 Zn). Therefore, to discharge 1 mole of Zn2+, 2 moles of electrons are needed. Using Faraday's law, we can calculate the quantity of electricity required to discharge 1 mole of Zn2+: 1 mole of Zn2+ requires 2 * 96,500 C = 193,000 C Therefore, to calculate the quantity of electricity required to discharge zinc from the given half-reaction, we need to divide 193,000 C by 2 (since the reaction involves half-moles of Zn2+): Quantity of electricity required = 193,000 C / 2 = 96,500 C Comparing this value with the options given, we can see that the correct answer is: 9.650 * 104 C This matches the value of Faraday's constant, which was given in the question.

Karanta bayanin darasi akan Electrolysis (JAMB)

Electrolysis

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Tambaya 21 Rahoto

The knowledge of half-life can be used to

create an element

detect an element Amsar da ta dace

spilt an element

irradiate an element

Bayanin Amsa

The knowledge of half-life can be used to detect an element. Half-life is the time taken for half of the radioactive atoms in a sample to decay. Since each radioactive element has a unique half-life, scientists can use the knowledge of half-life to determine which elements are present in a sample. By measuring the amount of radioactivity in a sample over time, they can determine the half-life of the element and identify it. Therefore, option "detect an element" is correct.

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Karanta bayanin darasi akan Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Karanta bayanin darasi akan Chemical Industry And Environment (WAEC)

Kinetic Theory Of Matter And Gas Laws

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Industry And Environment

Duba Amsa

Tambaya 22 Rahoto

The noble gas, argon, is used for

electrical arc welding Amsar da ta dace

welding brass

underwater welding

steel-welding

Bayanin Amsa

Argon is commonly used in welding processes, especially in the form of tungsten inert gas (TIG) welding. In TIG welding, the arc is maintained between a tungsten electrode and the workpiece while a stream of argon gas is used to shield the weld area from atmospheric gases like nitrogen and oxygen. This prevents oxidation and other contamination of the weld, resulting in a higher-quality, cleaner weld. Therefore, the correct option is "electrical arc welding".

Karanta bayanin darasi akan Chemical Combination (JAMB)

Chemical Combination

Duba Amsa

Tambaya 23 Rahoto

The least easily oxidized of the metals below is

Al Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Metals And Their Compounds (JAMB)

Metals And Their Compounds

Duba Amsa

Tambaya 24 Rahoto

A phenomenon where an element exist in different forms in the same physical state is known as

isomerism

amorphism

allotropy Amsar da ta dace

isotopy

Bayanin Amsa

The phenomenon where an element exists in different forms in the same physical state is known as "allotropy." Allotropy is a property of some chemical elements where they exist in two or more different physical forms, which are called allotropes. These allotropes can have different properties, such as different colors, densities, and crystal structures, but they consist of the same type of atoms. Isomerism refers to the existence of compounds with the same molecular formula but different structural arrangements. Amorphism refers to the lack of long-range order in a material, while isotopy refers to the existence of different isotopes of an element with different numbers of neutrons.

Karanta bayanin darasi akan Rates Of Chemical Reaction (JAMB)

Rates Of Chemical Reaction

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Tambaya 25 Rahoto

Use the option above to answer this question. The graph that describes a zero order reaction is

B Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Karanta bayanin darasi akan Chemical Kinetics And Equilibrium System (Nigeria Only) (WAEC)

Kinetic Theory Of Matter And Gas Laws

Chemical Kinetics And Equilibrium System (Nigeria Only)

Duba Amsa

Tambaya 26 Rahoto

The chlorinated alkane often used industrially to remove grease is

tetrachloromethane Amsar da ta dace

chloromethane

trichloromethane

dichlorometane

Bayanin Amsa

Karanta bayanin darasi akan Organic Compounds (JAMB)

Organic Compounds

Duba Amsa

Tambaya 27 Rahoto

The pOH of a solution of 0.25 mol dm-3 of hydrochloric acid is

12.40

13.40 Amsar da ta dace

14.40

14.60

Bayanin Amsa

pOH = -log [2.5 * 10-1]
= -log[-0.3979 + 1]
= 13.40

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Duba Amsa

Tambaya 28 Rahoto

Which of the following statements is correct about the periodic table?

elements in the same period have the same number of valence electrons Amsar da ta dace

the valence electrons of the elements in the same period increase progressively 'across the period

elements in the same group have the same number of electron shell

the non-metallic properties of the elements tend to decrease across each period

Bayanin Amsa

Karanta bayanin darasi akan Periodic Chemistry (Ghana Only) (WAEC)

Periodic Chemistry (Ghana Only)

Duba Amsa

Tambaya 29 Rahoto

1.0 dm3 of distilled water was used to wash 2.0g of a precipitate of AgCl. If the solubility product of AgCl is 2.0 * 10-10mol2dm-6, what quantity of silver was lost in the process?

2.029 * 10-3mol dm-3

1.414 * 10-3mol dm-3

2.029 * 10-5mol dm-3

1.414 * 10-5mol dm-3 Amsar da ta dace

Bayanin Amsa

When solid AgCl is mixed with water, it partially dissolves and reaches equilibrium according to the following equation: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) The solubility product (Ksp) expression for this equilibrium is: Ksp = [Ag⁺][Cl⁻] Given that the solubility product (Ksp) of AgCl is 2.0 × 10⁻¹⁰ mol² dm⁻⁶, we can use this value and the initial volume of water (1.0 dm³) to calculate the maximum amount of Ag⁺ and Cl⁻ that can dissolve before reaching equilibrium. Let x be the molar solubility of AgCl, which is also equal to the concentration of Ag⁺ and Cl⁻ ions that dissolve. Then, at equilibrium: [Ag⁺] = [Cl⁻] = x Substituting these into the Ksp expression, we have: Ksp = x² Solving for x, we get: x = √Ksp = √(2.0 × 10⁻¹⁰ mol² dm⁻⁶) = 1.414 × 10⁻⁵ mol dm⁻³ This means that at equilibrium, the concentration of Ag⁺ and Cl⁻ ions in the solution is 1.414 × 10⁻⁵ mol dm⁻³. Now, we can use the volume of water and the concentration of Ag⁺ and Cl⁻ ions to calculate the total amount of AgCl that can dissolve: (1.0 dm³) × (1.414 × 10⁻⁵ mol dm⁻³) = 1.414 × 10⁻⁵ mol AgCl This is the maximum amount of AgCl that can dissolve in 1.0 dm³ of water. However, in the given problem, only 2.0 g of AgCl is used, which is equivalent to: (2.0 g) / (143.32 g/mol) = 0.01395 mol AgCl Since this amount is larger than the maximum amount that can dissolve in 1.0 dm³ of water, all of the AgCl will dissolve and none will be lost in the process. Therefore, the answer is option (D) 1.414 × 10⁻⁵ mol dm⁻³

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Duba Amsa

Tambaya 30 Rahoto

For a reaction in equilibrium, the species involved in the equilibrium constant expression are

gaseous and solid species

liquid and solid species

solid and dissolved species

gaseous and dissolved species Amsar da ta dace

Bayanin Amsa

The equilibrium constant expression is used to calculate the equilibrium constant of a chemical reaction. The species involved in the expression are those that are in a state of equilibrium, which means they are in constant flux between reactants and products. In order to calculate the equilibrium constant, we need to know the concentrations of the reactants and products in the equilibrium state. Therefore, the species involved in the equilibrium constant expression depend on the state of the reactants and products. Among the options given, the correct answer is the last one: gaseous and dissolved species. This is because when a reaction involves gaseous and dissolved species, their concentrations can change during the reaction, and therefore they need to be included in the equilibrium constant expression. Solid and liquid species, on the other hand, have a fixed concentration during the reaction, and so they are not included in the equilibrium constant expression.

Karanta bayanin darasi akan Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Karanta bayanin darasi akan Chemical Equilibria (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Equilibria

Duba Amsa

Tambaya 31 Rahoto

Alkanones are generally obtained by the oxidation of

primary alkanols

secondary alkanols Amsar da ta dace

tertiary alkanols

alkanoic acid

Bayanin Amsa

Karanta bayanin darasi akan Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Chemistry Of Carbon Compounds (Nigeria Only)

Duba Amsa

Tambaya 32 Rahoto

Synthesis gas is a mixture of

CH4 and H2O

CH4 and H2

CO2 and H2

CO and H2 Amsar da ta dace

Bayanin Amsa

Synthesis gas is a mixture of carbon monoxide (CO) and hydrogen gas (H2). Therefore, the correct option is "CO and H2".

Duba Amsa

Tambaya 33 Rahoto

The shapes of CO\( _2 \), H\( _2 \)O and CH\( _4 \) respectively are

bent, linear and tetrahedral

bent, tetrahedral and linear

linear, bent and tetrahedral Amsar da ta dace

tetrahedral, linear and bent

Bayanin Amsa

The shape of a molecule is determined by the arrangement of its atoms in space. The arrangement of atoms is influenced by the number of bonded electron pairs and the number of non-bonded electron pairs around the central atom. Based on this information, the shapes of the molecules are: - CO2: linear - H2O: bent - CH4: tetrahedral Therefore, the correct answer is bent, tetrahedral, and linear, respectively.

Karanta bayanin darasi akan Chemistry Of Carbon Compounds (Ghana Only) (WAEC)

Chemistry Of Carbon Compounds (Ghana Only)

Duba Amsa

Tambaya 34 Rahoto

The electron configuration of 22X 2+ ion is

1s2 2s22p6 3s2 3p6 4s2 3d2

1s2 2s22p6 3s2 3p6 4s2 3d1

1s2 2s22p6 3s2 3p6

1s2 2s22p6 3s2 3p6 4s2 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

Duba Amsa

Tambaya 35 Rahoto

A common characteristics of copper and silver in their usage as coinage metals is that they

have high metallic lustre

are not easily oxidized Amsar da ta dace

are easily oxidized

are not easil reduced

Bayanin Amsa

The correct option is: "are not easily oxidized." Copper and silver are commonly used as coinage metals because they are both relatively unreactive and do not easily corrode or oxidize. This makes them a good choice for coins, as they will not deteriorate or lose their value due to exposure to the air or other elements over time.

Karanta bayanin darasi akan Metals And Their Compounds (JAMB)

Metals And Their Compounds

Duba Amsa

Tambaya 36 Rahoto

MnO4(aq) + 8H+(aq) + Y → Mn2+(aq) + 4H2O(l)

Y in the equation above represents

2e-

3e-

5e- Amsar da ta dace

7e-

Bayanin Amsa

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Duba Amsa

Tambaya 37 Rahoto

The compound above is an

ether

ester Amsar da ta dace

alkanal

alkanol

Bayanin Amsa

Karanta bayanin darasi akan Periodic Chemistry (Ghana Only) (WAEC)

Periodic Chemistry (Ghana Only)

Duba Amsa

Tambaya 38 Rahoto

Given that M is the mass of substance deposited in an electrolysis and Q the quantity of electricity consumed, then Faraday's law can be written as

M = Z/Q

M = Q/Z

M = Z/2Q

M = QZ Amsar da ta dace

Bayanin Amsa

Faraday's law states that the amount of substance deposited in an electrolysis is directly proportional to the quantity of electricity consumed (Q). Mathematically, this can be expressed as M = ZQ, where M is the mass of substance deposited and Z is the electrochemical equivalent, a constant that depends on the substance being electrolyzed. Therefore, the correct option is "M = ZQ".

Karanta bayanin darasi akan Electrolysis (JAMB)

Electrolysis

Duba Amsa

Tambaya 39 Rahoto

Unsaturated organic compounds are identified by decolourization of

silver bromine and potassium tetraoxomanganate (VII) solutions

bromine water and acidified potassium tetraoxomanganate (VII) solution Amsar da ta dace

silver bromide solution and bromine water

bromine water and alkaline potassium tetraoxomanganate (VII) solution

Bayanin Amsa

The decolourization of bromine water and acidified potassium tetraoxomanganate (VII) solution is used to identify unsaturated organic compounds. When bromine water is added to an unsaturated organic compound, the brown color of bromine is decolorized due to the addition of bromine across the C=C double bond in the unsaturated compound. Similarly, when acidified potassium tetraoxomanganate (VII) solution is added to an unsaturated organic compound, the purple color of the solution is also decolorized due to the addition of the potassium tetraoxomanganate (VII) across the C=C double bond in the unsaturated compound. The decolorization of both these solutions is due to the reaction with unsaturated organic compounds, which is a characteristic test used to identify them. Therefore, the correct option is "bromine water and acidified potassium tetraoxomanganate (VII) solution".

Karanta bayanin darasi akan Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

Duba Amsa

Tambaya 40 Rahoto

The refreshing and characteristic taste of soda water and other soft drinks is as a result of the presence in them of

carbon (IV) oxide Amsar da ta dace

carbon (II) oxide

soda

glucose

Bayanin Amsa

The refreshing and characteristic taste of soda water and other soft drinks is due to the presence of carbon dioxide gas, which is also known as carbon (IV) oxide. When carbon dioxide gas dissolves in water, it forms carbonic acid, which gives the drinks their slightly acidic taste. The carbon dioxide gas also creates the fizz or bubbles in the drink, which adds to its refreshing quality. Therefore, carbon (IV) oxide, is the correct answer. Options 2, 3, and 4 are not responsible for the taste of soda water and soft drinks.

Karanta bayanin darasi akan Acids, Bases And Salts (Nigeria Only) (WAEC)

Acids, Bases And Salts (Nigeria Only)

Duba Amsa

Tambaya 41 Rahoto

A side effect of soft water is that

it gives offensive taste

excess calcium is precipitated

it attacks lead contained in pipes Amsar da ta dace

it encourage the growth of bacteria

Bayanin Amsa

Soft water is water that has a low concentration of dissolved minerals such as calcium and magnesium. A side effect of soft water is that it may attack lead contained in pipes, causing it to dissolve into the water. Lead in drinking water can be harmful to human health, especially in young children and pregnant women. Therefore, it is important to be aware of the quality of the water you consume and take necessary precautions if the water is too soft or has any other potential contaminants.

Duba Amsa

Tambaya 42 Rahoto

In which of the following is the oxidation number of sulphur equal to -2?

H2S Amsar da ta dace

SO 2

SO 32-

Bayanin Amsa

The oxidation number of sulfur (S) is the hypothetical charge that an atom would have if all its bonds to atoms of different elements were completely ionic in nature. To determine the oxidation number of sulfur in each compound given, we need to follow some rules: - The oxidation number of an atom in its free (uncombined) state is zero. - The oxidation number of a monatomic ion is equal to its charge. - The sum of the oxidation numbers of all atoms in a neutral molecule must be zero. - The sum of the oxidation numbers of all atoms in a polyatomic ion must equal the charge on the ion. Using these rules, we can find the oxidation number of sulfur in each compound: - In S8, sulfur is in its elemental form, and the oxidation number of each atom is zero. - In H2S, hydrogen (H) has an oxidation number of +1, and the sum of the oxidation numbers in the molecule is zero. Therefore, sulfur must have an oxidation number of -2 to balance out the charge. - In SO2, oxygen (O) has an oxidation number of -2, and the sum of the oxidation numbers in the molecule is zero. Therefore, sulfur must have an oxidation number of +4 to balance out the charge. - In SO32-, oxygen has an oxidation number of -2, and the sum of the oxidation numbers in the ion is -2. Therefore, sulfur must have an oxidation number of +4 to balance out the charge. Therefore, the compound in which the oxidation number of sulfur is equal to -2 is H2S.

Karanta bayanin darasi akan Oxidation And Reduction (JAMB)

Oxidation And Reduction

Duba Amsa

Tambaya 43 Rahoto

The ratio of the initial to the final pressure of a given mass of gas is 1:1.5. Calculate the final volume of the gas if the initial volume was \(3000\text{cm}^3\) at the same temperature

120cm

^{3}

200cm

^{3}

Amsar da ta dace

450cm

^{3}

750cm

^{3}

Bayanin Amsa

The problem involves using Boyle's law, which states that the pressure of a gas is inversely proportional to its volume at a constant temperature. Mathematically, P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, respectively, and P2 and V2 are the final pressure and volume, respectively. Given the ratio of the initial to final pressure as 1:1.5, we can assume that the final pressure is 1.5 times the initial pressure. Therefore, P2 = 1.5 * P1. Using the formula P1V1 = P2V2 and substituting the known values, we get: P1 * 3000 = (1.5 * P1) * V2 Simplifying the equation, we get: V2 = (P1 * V1) / (1.5 * P1) = 2000cm3 Therefore, the final volume of the gas is 2000cm3. So the answer is 200cm3.

Karanta bayanin darasi akan Solubility (JAMB)

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Solubility

Kinetic Theory Of Matter And Gas Laws

Duba Amsa

Tambaya 44 Rahoto

Which of the following compounds is composed of Al, Si, O and H?

Epson salt

Limestone

Clay Amsar da ta dace

Urea

Bayanin Amsa

The compound composed of Al, Si, O and H is clay. Clay is a mineral that is primarily composed of hydrous aluminum silicates and various impurities such as magnesium, calcium, potassium, or iron. The chemical formula of clay is typically expressed as Al₂O₃•2SiO₂•2H₂O, indicating the presence of aluminum, silicon, oxygen, and water in its composition. Therefore, the correct option is "Clay".

Karanta bayanin darasi akan Chemical Combination (JAMB)

Chemical Combination

Duba Amsa

Tambaya 45 Rahoto

Water molecules can be liquids especially when they are bonded to

alkaline earth metals

alkali metals

transition metals Amsar da ta dace

group VII elements

Bayanin Amsa

Karanta bayanin darasi akan Water (JAMB)

Water

Duba Amsa

Tambaya 46 Rahoto

How many moles of HCl will be required to react with potassium heptaoxodichromate (VI) to produce 3 moles of chlorine?

14 Amsar da ta dace

Bayanin Amsa

_{2}

_{2}

_{7}

+ 14HCl

\to

2CrO

_{13}

+ 2KCl + 3C

_{2}

+ 7H

_{2}

Karanta bayanin darasi akan Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Karanta bayanin darasi akan Stoichiometry And Chemical Reactions (WAEC)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Stoichiometry And Chemical Reactions

Duba Amsa

Tambaya 47 Rahoto

The conditions necessary for the extraction of a water molecule from two molecules of ethanol are

less acid and lower temperature

excess acid and a lower temperature

excess acid and a higher temperature Amsar da ta dace

less acid and a higher temperature

Bayanin Amsa

The conditions necessary for the extraction of a water molecule from two molecules of ethanol are excess acid and a higher temperature. This process is known as dehydration, where water is removed from a molecule or compound. In the case of ethanol, excess acid (usually concentrated sulfuric acid) acts as a catalyst to remove a water molecule (H2O) from two ethanol molecules (C2H5OH), leaving behind ethene (C2H4). The excess acid helps to facilitate the reaction by providing an acidic medium, while the higher temperature helps to break the bond between the two molecules and make the reaction more efficient. Less acid and lower temperature would not be as effective in extracting the water molecule, while excess acid and a lower temperature could slow down or inhibit the reaction altogether.

Karanta bayanin darasi akan Kinetic Theory Of Matter And Gas Laws (JAMB)

Karanta bayanin darasi akan Chemistry, Industry And The Environment (WAEC)

Kinetic Theory Of Matter And Gas Laws

Chemistry, Industry And The Environment

Duba Amsa

Tambaya 48 Rahoto

The air pollutant unknown in nature is

HCHO Amsar da ta dace

DDT

Bayanin Amsa

Karanta bayanin darasi akan Environmental Pollution (JAMB)

Environmental Pollution

Duba Amsa

Tambaya 49 Rahoto

Which of the following types of bonding does not involve the formation of new substances?

Metallic Amsar da ta dace

covalent

co-ordinate

electrovalent

Bayanin Amsa

Of the types of bonding listed, metallic, covalent, co-ordinate, and electrovalent, metallic bonding is the type of bonding that does not involve the formation of new substances. Metallic bonding occurs between metal atoms, where the outer electrons of each metal atom are delocalized and can move freely throughout the entire metallic crystal. These delocalized electrons are responsible for the characteristic properties of metals, such as high electrical conductivity and luster. However, metallic bonding does not involve the formation of new substances, as the metal atoms are simply held together in a lattice structure by the electrostatic attraction between the positive metal ions and the negatively charged delocalized electrons. In contrast, covalent, co-ordinate, and electrovalent bonding all involve the sharing or transfer of electrons between atoms to form new chemical substances.

Duba Amsa

Tambaya 50 Rahoto

Haematite is an ore of

zinc

lead

iron Amsar da ta dace

copper

Bayanin Amsa

Haematite is an ore of iron. It is a mineral form of iron oxide, with a chemical formula of Fe2O3. Haematite is one of the most important sources of iron for manufacturing iron and steel products. When heated with coke (a form of coal), haematite releases oxygen and forms iron, which is then used to produce various steel and iron products.

Karanta bayanin darasi akan Metals And Their Compounds (JAMB)

Metals And Their Compounds

Duba Amsa






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