JAMB UTME - Chemistry - 2017

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Pregunta 4 Informe

Water for town supply is chlorinate to make it free from

bad colour

bacteria

temporary hardness

permanent hardness

Detalles de la respuesta

Water for town supply is chlorinated to make it free from bacteria. Chlorine is added to water in small amounts to kill harmful bacteria and other microorganisms that may be present in the water. These microorganisms can cause illnesses such as typhoid, cholera, and dysentery if ingested. Chlorination is an effective and inexpensive method to disinfect water and protect public health. Chlorine works by breaking down the cell walls of microorganisms, which leads to their death. Chlorine also helps to control the growth of algae and other microorganisms in the water supply, which can cause taste and odor problems. Chlorination does not remove other impurities from the water such as bad color, temporary hardness, or permanent hardness. These impurities may require additional treatment methods such as filtration or water softening to be removed.

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Pregunta 5 Informe

The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is – (Ag = 108 F = 96500(mol-1)

54.0g

27.0g

13.5g

108.0g

Detalles de la respuesta

To calculate the mass of silver deposited, we can use Faraday's laws of electrolysis, which states that the amount of substance produced at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The formula to calculate the mass of substance produced is: Mass (g) = (Current × Time × Atomic mass) / (Faraday's constant × Valency) Here, we have a current of 10A passing through a solution of silver salt for 4830s, and we know that the atomic mass of silver is 108 g/mol and the valency is 1 (since silver has a charge of +1). The value of Faraday's constant is 96500 C/mol. Substituting the values in the formula, we get: Mass (g) = (10A × 4830s × 108 g/mol) / (96500 C/mol × 1) Mass (g) = 54.0g Therefore, the mass of silver deposited is 54.0g, which is option (A) in the given options.

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Electrolysis

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Pregunta 6 Informe

A sample of orange juice is found to have a PH of 3.80. What is the concentration of the hydroxide ion in the juice?

1.6

\times

10-4

6.3

\times

10-11

6.3

\times

10-4

1.6

\times

10-11

Detalles de la respuesta

The pH scale measures the acidity or basicity of a solution. The pH is defined as the negative logarithm (base 10) of the concentration of the hydronium ion (H3O+), which is given by the expression: pH = -log[H3O+] To calculate the concentration of the hydroxide ion (OH-) in the juice, we need to use the equation for the ion product of water: Kw = [H3O+][OH-] = 1.0 × 10-14 where Kw is the ion product constant of water. At 25°C, Kw has a value of 1.0 × 10-14. To solve for [OH-], we need to rearrange the equation: [OH-] = Kw / [H3O+] Now we can substitute the pH of the orange juice to calculate the concentration of hydronium ion [H3O+]. pH = -log[H3O+] 3.80 = -log[H3O+] [H3O+] = 10^(-3.80) [H3O+] = 1.58 × 10^(-4) mol/L Substituting this value into the ion product equation: [OH-] = Kw / [H3O+] [OH-] = 1.0 × 10^-14 / (1.58 × 10^-4) [OH-] = 6.33 × 10^-11 mol/L Therefore, the concentration of the hydroxide ion in the orange juice is 6.33 × 10^-11 mol/L.

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Acids, Bases And Salts

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Pregunta 7 Informe

The reaction of halogens in the presence of sunlight is an example of

oxidation reaction

addition reaction

hydrogenation reaction

substitution reaction

Detalles de la respuesta

The reaction of halogens in the presence of sunlight is an example of a substitution reaction. In this type of chemical reaction, one atom or group of atoms in a molecule is replaced by another atom or group of atoms. In the case of halogens, such as chlorine or bromine, they can react with certain organic compounds by breaking the carbon-hydrogen bond and replacing the hydrogen atom with a halogen atom. This reaction is usually initiated by the energy from sunlight, which provides the necessary activation energy to break the bond. For example, in the presence of sunlight, chlorine gas can react with methane gas to form chloromethane and hydrogen chloride: CH4 + Cl2 -> CH3Cl + HCl In this reaction, one of the hydrogen atoms in methane is replaced by a chlorine atom, forming chloromethane, while hydrogen chloride is also produced as a byproduct. Overall, the reaction of halogens in the presence of sunlight is an example of a substitution reaction, where one atom or group of atoms is substituted for another.

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Oxidation And Reduction

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Pregunta 8 Informe

The shape of the S-orbital is

elliptical

spiral

circular

spherical

Detalles de la respuesta

The shape of the S-orbital is spherical. This means that it has a symmetrical, 3-dimensional shape that resembles a ball or a globe. It is not flat or elongated, but rather is evenly rounded in all directions. This is the most basic type of orbital and is used to describe the electrons in the outermost energy level of an atom.

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Pregunta 9 Informe

Tin is unaffected by air at ordinary temperature due to its

Low melting point

Weak electropositive character

High boiling point

White lustrous appearance

Detalles de la respuesta

Tin is unaffected by air at ordinary temperature because of its high reactivity, which means it doesn't easily react or corrode with other elements such as oxygen, the main component of air. Tin has a high reactivity, which helps it resist corrosion and oxidation. This makes tin a great material for products that need to be resistant to corrosion, such as food containers and metal pipes.

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Chemical Combination

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Pregunta 10 Informe

The shape of ammonia molecules is

trigonal planar

octahedral

square planar

tetrahedral

Detalles de la respuesta

The shape of ammonia (NH3) molecules is tetrahedral. The ammonia molecule has one nitrogen atom and three hydrogen atoms. The nitrogen atom is at the center of the molecule, with the three hydrogen atoms arranged around it. The shape of the molecule is determined by the arrangement of the electrons in the outermost energy level of the nitrogen atom, which is called the valence shell. In the valence shell of the nitrogen atom, there are four pairs of electrons, one of which is a lone pair, while the other three are bonded to hydrogen atoms. These four pairs of electrons repel each other, and try to get as far away from each other as possible. This results in a tetrahedral arrangement of the atoms around the nitrogen atom. A tetrahedron is a three-dimensional shape with four triangular faces, and it looks like a pyramid with a triangular base. In the case of ammonia, the three hydrogen atoms form a triangular base, and the lone pair of electrons sits at the top of the pyramid, creating a tetrahedral shape. Therefore, the shape of ammonia molecules is tetrahedral.

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Pregunta 11 Informe

The constituent common to duralumin and alnico is

Co

Mn

Al

Mg

Detalles de la respuesta

The constituent common to duralumin and alnico is aluminum (Al). Duralumin is a type of aluminum alloy that typically contains around 4% copper and small amounts of other metals such as magnesium and manganese. It is known for its high strength-to-weight ratio and was commonly used in aircraft construction in the early 20th century. Alnico, on the other hand, is a type of permanent magnet alloy that contains aluminum, nickel, and cobalt (hence the name "alnico"). It also often contains small amounts of other metals such as copper and iron. Alnico magnets are known for their high magnetic strength and resistance to demagnetization. Although duralumin and alnico are very different in their properties and applications, they both contain aluminum as a common constituent. Aluminum is a versatile metal that is widely used in many different alloys due to its light weight, strength, and corrosion resistance.

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Pregunta 12 Informe

Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol-1]

0.300g

0.250g

0.2242g

0.448g

Detalles de la respuesta

M =

\frac{Molar mass × Quantity of Electricity}{96500 × no of charge}

= $\frac{M m I T}{96500 n}$

Copper II Chloride = CuCl2

CuCl2 → Cu2+ + 2Cl2

Mass of compound deposited = $\frac{Molar mass × Quantity of Electricity}{96500 × no of charge}$

Q = IT

I = 0.5A

T = 45 × 60

T = 2700s

Q = 0.5 × 2700

= 1350c

Molarmass = 64gmol-1

no of charge = + 2

Mass = $\frac{64 \times 1350}{96500 \times 2}$

Mass = 0.448g

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Electrolysis

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Pregunta 13 Informe

The oxidation number of iodine in KIO3 is

+7

+3

+5

+6

Detalles de la respuesta

The oxidation number of an atom is a measure of the number of electrons that it has gained or lost in a chemical reaction. To determine the oxidation number of iodine in KIO3, we need to use the fact that the sum of the oxidation numbers of all the atoms in a compound is equal to the overall charge of the compound. In KIO3, potassium (K) has an oxidation number of +1 because it belongs to the alkali metal group and it always has a +1 oxidation state. Oxygen (O) usually has an oxidation number of -2, unless it is combined with a more electronegative atom or a peroxide, which is not the case in KIO3. Let's assume that iodine (I) has an oxidation number of x. Since KIO3 is a neutral compound, the sum of the oxidation numbers of all its atoms must be zero. Therefore, we can write the following equation: (+1) + x + 3(-2) = 0 Simplifying the equation, we get: x = +5 Therefore, the oxidation number of iodine in KIO3 is +5.

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Oxidation And Reduction

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Pregunta 14 Informe

In the diagram above, Y is

fused CaO

H2O

NaOH

Concentrated H2SO4

Detalles de la respuesta

In the preparation of sulphur dioxide by the action of dilute acids on sulphates and bisulphites. conc H2SO4 helps to release SO2 from the mixture.
The setup represents the production of sulphur dioxide

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Acids, Bases And Salts

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Pregunta 15 Informe

Calculate the amount in moles of silver deposited when 9650C of electricty is passed through a solution of silver salt [= 96500 Cmol-1]

0.05

10.80

10.00

0.10

Detalles de la respuesta

The amount of silver deposited can be calculated using Faraday's law of electrolysis, which states that the amount of a substance deposited at an electrode during electrolysis is proportional to the number of moles of electric charge passed through the solution. We can start by using the formula: moles of substance = charge passed / Faraday's constant Where Faraday's constant is equal to 96500 Cmol^-1. Plugging in the given values: moles of silver = 9650 C / 96500 Cmol^-1 moles of silver = 0.1 So, the amount of silver deposited is 0.1 moles.

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Electrolysis

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Pregunta 16 Informe

Due to the high reactivity of sodium, it is usually stored under

water

mercury

paraffin

phenol

Detalles de la respuesta

Na is kept under kerosene (paraffin) to avoid reactivity with air.
Paraffin is also known as Kerosene.
Na(sodium) is kept in kerosene to prevent it from coming in contact with oxygen and moisture. If this happens, it will react with the moisture present in air and form sodium hydroxide.

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Metals And Their Compounds

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Pregunta 17 Informe

In the diagram above. X is

SO3

SO2

S

H2S

Detalles de la respuesta

The setup represents the production of sulfur dioxide. And the cylinder marked X is SO2

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Separation Of Mixtures And Purification Of Chemical Substances

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Pregunta 18 Informe

An isomer of C5H12 is

2–ethyl butane

butane

2–methyl butane

2–methyl propane

Detalles de la respuesta

C5H12 is the molecular formula for pentane, which has five carbon atoms and 12 hydrogen atoms. An isomer is a molecule that has the same molecular formula as another molecule, but with a different arrangement of atoms. Therefore, we need to look for molecules with the molecular formula C5H12 that have a different arrangement of atoms than pentane. Let's consider the options: - Butane (C4H10) has a different molecular formula than pentane (C5H12), so it cannot be an isomer of pentane. - 2-Methyl propane (C4H10) also has a different molecular formula than pentane, so it cannot be an isomer of pentane. - 2-Methyl butane (C5H12) has the same molecular formula as pentane, but with a different arrangement of atoms. Specifically, the carbon atoms in 2-methyl butane are arranged in a branched chain, while the carbon atoms in pentane are arranged in a straight chain. Therefore, 2-methyl butane is an isomer of pentane. - 2-Ethyl butane (C6H14) has a different molecular formula than pentane, so it cannot be an isomer of pentane. Therefore, the isomer of C5H12 is 2-methyl butane.

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Pregunta 19 Informe

In the laboratory preparation of trioxonitrate (V) acid the nitrogen(iv) oxide formed as a by-product is removed by

further heating

adding concentrated H2SO4

cooling the acid solution with cold water

bubbling air through the acid solution

Detalles de la respuesta

In the laboratory preparation of trioxonitrate (V) acid, nitrogen(iv) oxide is formed as a by-product. Nitrogen(iv) oxide is a harmful gas that can cause respiratory problems, so it needs to be removed from the reaction mixture. One way to remove nitrogen(iv) oxide is by bubbling air through the acid solution. When air is bubbled through the solution, the nitrogen(iv) oxide reacts with the oxygen in the air to form nitrogen dioxide, which is a brown gas. Nitrogen dioxide is not as harmful as nitrogen(iv) oxide and can be safely vented out of the laboratory. Therefore, bubbling air through the acid solution is the method used to remove nitrogen(iv) oxide formed as a by-product in the laboratory preparation of trioxonitrate (V) acid.

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Acids, Bases And Salts

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Pregunta 20 Informe

The salt formed from a weak acid and a strong base hydrolyzes in water to form

A saturated solution

an acidic solution

a buffer solution

an alkaline solution

Detalles de la respuesta

This is a solution formed from the hydrolyses of an alkali in water.

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Pregunta 21 Informe

A given mass of gas occupies 2dm3 at 300k. At what temperature will its volume be doubled, keeping the pressure constant?

400k

480k

550k

600k

Detalles de la respuesta

To solve this problem, we can use the combined gas law, which states that the product of pressure and volume divided by temperature is constant, as long as the amount of gas and the pressure remain constant. In this case, the pressure is constant and the initial conditions of the gas are: - Volume (V1) = 2 dm3 - Temperature (T1) = 300 K Let's call the final temperature T2, and the final volume V2 = 2V1 = 4 dm3, since we want the volume to be doubled. Using the combined gas law, we can set up the following equation: P * V1 / T1 = P * V2 / T2 Simplifying the equation, we get: V1 / T1 = V2 / T2 Substituting the values we know, we get: 2 / 300 = 4 / T2 Cross-multiplying and solving for T2, we get: T2 = 4 * 300 / 2 = 600 K Therefore, the temperature at which the gas will double its volume, keeping the pressure constant, is 600 K. Option (d) is the correct answer.

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Pregunta 22 Informe

CH2S(g) + O2(g) → 2Cn + SO2(g)

What is the change in the oxidation number of copper in the reaction?

0 to + 2

0 to + 1

C + 1 to 0

+ 2 to + 1

Detalles de la respuesta

In the reactant;

Cu2S

2 Cu - 2(1) = 0

2 Cu = 2

Cu = $\frac{2}{2}$

Cu = +1

In the product, Cu

Cu = O

The oxidation number of Cu in Cu2S and Cu respectively is +1 and 0 respectively

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Oxidation And Reduction

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Pregunta 23 Informe

The tincture of iodine means iodine dissolved in

ethanol

bromine chloride

chlorine water

water

Detalles de la respuesta

It is also called weak iodine solution. Tincture solutions are characterized by the presence of alcohol.

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Pregunta 24 Informe

The enzyme used in the hydrolysis of starch to dextrin and maltose is

Zymase

Sucrase

Amylase

Lactase

Detalles de la respuesta

The enzyme used in the hydrolysis of starch to dextrin and maltose is called amylase. Starch is a complex carbohydrate found in many foods, including grains, potatoes, and vegetables. It consists of long chains of glucose molecules that are linked together. To digest starch, our bodies use the enzyme amylase, which breaks down the long chains into smaller molecules of dextrin and maltose. These smaller molecules are more easily absorbed into the bloodstream and used as energy by the body. Amylase is produced in the salivary glands and pancreas and is released into the digestive system when we eat. In the mouth, amylase begins breaking down starch as we chew our food, and then continues its work in the small intestine where it is released by the pancreas. In summary, amylase is the enzyme responsible for breaking down starch into smaller molecules of dextrin and maltose during digestion.

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Pregunta 25 Informe

The general formula of alkanones is

RCHO

R2CO

RCOOH

RCOOR

Detalles de la respuesta

The general formula of alkanones is R2CO, which represents a functional group that consists of a carbon atom double-bonded to an oxygen atom (C=O) and attached to two alkyl (or aryl) groups represented by the "R" symbol. In simpler terms, alkanones are a type of organic compound that contain a carbon atom double-bonded to an oxygen atom and two additional carbon-containing groups. The "R" groups can be any combination of alkyl or aryl groups, which are made up of carbon and hydrogen atoms arranged in a specific way. The general formula R2CO applies to all alkanones, regardless of the specific alkyl or aryl groups attached to the central carbon atom. By knowing this formula, we can easily recognize an alkanone when we see one and predict its chemical properties based on the functional group it contains.

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Pregunta 26 Informe

The constituent of air necessary in the rusting process are

O2 and H2O

Ar and CO2

CO2 and H2O

O2 and CO2

Detalles de la respuesta

The constituent of air includes O2, CO2, N2 and Noble gases. While for a rusting process to take place, the presence of O2, H2O and CO2 is important.

The constituent of air includes O2, CO2, N2 and Noble gases and for rusting process to take place.

The presence of O2 and CO2 as a constituent of air is indispensable

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Air

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Pregunta 27 Informe

In the laboratory preparation of ethyl ethanoate, the water present in the mixture is removed using a solution of

an hydrous CaCl4

concentrated NaCO4

dilute NaOH

concentrated H4SO4

Detalles de la respuesta

This is used in the preparation of ethyl ethanoate to remove the water present since it serves as a dehydrated agent.

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Pregunta 28 Informe

If 100cm3 of oxygen pass through a porous plug is 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is? [O = 16, H = 1]

10.0s

12.5s

17.7s

32.0s

Detalles de la respuesta

The rate at which a gas diffuses through a porous plug is inversely proportional to the square root of its molecular weight. The molecular weight of oxygen (O2) is 32 (16 x 2), while the molecular weight of hydrogen (H2) is 2 (1 x 2). Therefore, the square root of the ratio of the molecular weights of hydrogen and oxygen is: sqrt(2/32) = 0.25 This means that hydrogen diffuses through a porous plug 0.25 times as fast as oxygen. If it takes 50 seconds for 100cm3 of oxygen to pass through the porous plug, then it will take 0.25 times longer for the same volume of hydrogen to pass through the same porous plug. Thus, the time taken for the same volume of hydrogen to pass through the same porous plug is: 50 seconds x 0.25 = 12.5 seconds Therefore, the correct answer is 12.5s.

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Pregunta 29 Informe

An oxide XO2 has a vapour density of 32. What is the atomic mass of X?

20

32

14

12

Detalles de la respuesta

Molecular mass = vapour density X2

Mm of XO2 = x + 16(2) = x + 32

vapour density = 32

∴ x + 32 = 32 × 2

x + 32 = 64

x = 64 - 32

x = 32

∴ the relative molecular mass of X is 32
Relative molecular mass = vapour density × 2

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Pregunta 30 Informe

The ideal gas laws and equations are true for all gases at

low pressures and lower temperatures

low temperatures and high pressures

high pressures and high temperatures

low pressure and high temperatures

Detalles de la respuesta

The ideal gas laws and equations are most accurate for gases under conditions of low pressure and high temperature. At low pressure, the gas molecules are far apart and experience fewer intermolecular forces, allowing them to behave more ideally. At high temperature, the gas molecules have more kinetic energy, which helps to overcome any intermolecular forces that may be present. Under these conditions, the volume of a gas is directly proportional to its temperature and inversely proportional to its pressure, as described by the ideal gas law. However, as pressure and/or temperature increase, the behavior of real gases becomes more complex and may deviate from ideal gas behavior.

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Pregunta 31 Informe

According to Charle's law, the volume of a gas becomes zero at

−100°c

−273°c

−373°c

0°c

Detalles de la respuesta

According to Charles's law, the volume of a gas would theoretically become zero at -273°C or -459.4°F or 0 Kelvin, which is known as absolute zero. Charles's law is a gas law that describes the relationship between the volume and temperature of a gas, while the pressure and amount of gas are kept constant. The law states that at a constant pressure, the volume of a gas is directly proportional to its absolute temperature. This means that as the temperature of a gas decreases, its volume will also decrease proportionally. However, according to the laws of thermodynamics, it is impossible to cool a gas down to absolute zero or to completely eliminate its volume. This is because at absolute zero, the gas molecules would have zero kinetic energy, and therefore they would not be moving. This is a state that cannot be reached due to the third law of thermodynamics. Therefore, Charles's law is only valid within a certain temperature range and it cannot be applied to gases at extremely low temperatures or high pressures.

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Pregunta 32 Informe

The process that occurs when two equivalent forms of a compound are in equilibrium is

Isotopy

Resonance

Isomerism

Reforming

Detalles de la respuesta

Resonance involves two forms of a compound.
Isomerism involves two or more forms of an element.
Reforming involves the rearrangement of molecule.

This deals with a two forms of a molecule where the chemical connectivity is the same but the electrons are distributed differently around the structure.

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Chemical Equilibria

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Pregunta 33 Informe

The gas that can be collected by downward displacement of air is

chlorine

sulphur (IV) oxide

carbon (IV) oxide

ammonia

Detalles de la respuesta

Upward delivery works well for hydrogen and ammonia, which are both less densed than air. Sometimes, they are collected over water.

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Separation Of Mixtures And Purification Of Chemical Substances

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Pregunta 34 Informe

Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 082 atm dm3 K-1 mol-1]

2.536

1.623

4.736

0.394

Detalles de la respuesta

For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm3

Pressure P = 6atm

Temperature T = 30°C + 273 = 303k

R, Gas constant = 0.082 atmdm3k-1 mol

Recall from ideal gas equation

pv = nRT

n = $\frac{R V}{R T}$

n = $\frac{6 \times 105}{0.082 \times 303}$

n= 2.536mol

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Pregunta 35 Informe

The alkyl group is represented by the general formula

CnH2n

CnH2n − 2

CnH2n + 1

CnH2n + 2

Detalles de la respuesta

The alkyl group is represented by the general formula CnH2n+1, where "n" represents the number of carbon atoms in the group. An alkyl group is a type of organic molecule that consists of a chain of carbon atoms, each of which is bonded to two hydrogen atoms. The formula CnH2n+1 represents this structure, where "n" is the number of carbon atoms in the chain. The number of hydrogen atoms in the chain is always two more than the number of carbon atoms, hence the formula CnH2n+1.

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Pregunta 36 Informe

The IUPAC nomenclature of the structure above is

3-methybut-3-ene

2-methylbut-1-ene

2-ethylprop-1-ene

2-methylbut-2-ene

Detalles de la respuesta

CH2CH2 - CCH3CH2
Start the numbering from the terminal carbon.

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Pregunta 37 Informe

Incomplete oxidation of ethanol yields

CH3COOH

CH3COCH3

CH3CH2OCH2CH3

CH2CHO

Detalles de la respuesta

C2H5OH → CH3CHO → CH3COOH
Ethanol.....oxidation Ethanal........Ethanoic Acid
Primary alcohol oxidises to aldehyde and later to carboxylic acid.
Secondary alcohol oxidises to ketones.
Ethanol is an example of primary alcohol and primary alcohol can be oxidised to aldehyde and carboxylic acid. Wherein, incomplete oxidation of primary alcohol yields aldehyde also known as alkaline while complete oxidation of primary alcohol yields carboxylic acid.

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Pregunta 38 Informe

Temporary hard water is formed when rain water containing dissolved carbon(IV) oxide flows over deposits of

CaCO3

Na2CO3

Na2SO4

CaSO4

Detalles de la respuesta

Temporary hard water is formed when rainwater, which is naturally soft, comes into contact with deposits of calcium carbonate (CaCO3) in the ground. As the rainwater seeps through the soil, it dissolves carbon dioxide (CO2) to form carbonic acid (H2CO3). This weak acid reacts with the calcium carbonate to form calcium bicarbonate (Ca(HCO3)2), which is soluble in water. As a result, the water that flows over these deposits contains dissolved calcium bicarbonate, which gives it temporary hardness. This temporary hardness can be removed by boiling the water, which causes the calcium bicarbonate to decompose back into calcium carbonate and carbon dioxide. The calcium carbonate then forms a white precipitate, which can be removed by filtration or settling. So, the correct answer to the given question is CaCO3.

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Oxidation And Reduction

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Pregunta 39 Informe

X(g) + 3Y(g) ---- 2z(g) H = +ve. if the reaction above takes place at room temperature, the G will be

negative

zero

positive

indeterminate

Detalles de la respuesta

$\begin{array}{cc} Enthalpy Change [ΔH] & Entropy Change [ΔS] & Gibbs free Energy[ΔG] \\ Positive & Positive & depends on T, may be + or - \\ Negative & Positive & always negative \\ Negative & Negative & depends on T, may be + or - \\ Positive & Negative & always positive \end{array}$

?G= ?H ? T?S.
To determine whether ?G will be positive or negative, the value of ?H(change in enthalpy) and ?S (change in entropy) must be given. Likewise the temperature.

Lee la nota de la lección en Kinetic Theory Of Matter And Gas Laws (JAMB)

Lee la nota de la lección en Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)