JAMB UTME - Chemistry - 2017

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

Bayanin Amsa

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.

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

Bayanin Amsa

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.

The tincture of iodine means iodine dissolved in

Bayanin Amsa

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

Tin is unaffected by air at ordinary temperature due to its

Bayanin Amsa

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.

For a general equation of the nature xP + yQ ? mR + nS, the expression for the equilibrium constant is

Bayanin Amsa

Expression for equilibrium constant

k = $\frac{\text{concentration of product}}{\text{concentration of reactant}}$

The constituent common to duralumin and alnico is

Bayanin Amsa

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.

When few drops of concentrated trioxonitrate(V) acid is added to an unknown sample and wanned an intense yellow colouration is observed. The likely functional group present in the sample is

Bayanin Amsa

Xanthopreitic test for the presence of protein, when conc nitric acid is added to the drop, an intense yellow colouration is observed.

It contains all the functional group of protein which includes the amino, alkanol and the carboxylic group. Adding few drops of conc HNO3 to a protein, gives an intense yellow colouration.

It is called Xanthopreitic test.

The acid anhydride that will produce weak acid in water is

Bayanin Amsa

H2CO3 is an example of a weak acid while H2SO4 and HNO3 are examples of a strong acid.

CO2 combines with water to give a weak trioxocarbonate (IV) acid.

CO2 + H2O ? H2CO3

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]

Bayanin Amsa

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 = $\frac{RV}{RT}$

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

n= 2.536mol

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?

Bayanin Amsa

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.

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]

Bayanin Amsa

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.

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

Bayanin Amsa

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.

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

Bayanin Amsa

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

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

Bayanin Amsa

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.

The oxidation number of iodine in KIO3 is

Bayanin Amsa

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.

${}_{88}^{226}Ra$ → ${}_{86}^{x}Rn$ + alpha particle

Bayanin Amsa

The given equation represents the alpha decay of ^226Ra into an unknown nuclide ^86Rn and an alpha particle. During alpha decay, an atomic nucleus emits an alpha particle, which consists of two protons and two neutrons bound together, resulting in a decrease of two in the atomic number and a decrease of four in the mass number. To determine the atomic number and mass number of the unknown nuclide ^86Rn, we can subtract the atomic and mass numbers of the alpha particle from those of ^226Ra: Atomic number: 88 (Ra) - 2 (alpha particle) = 86 (Rn) Mass number: 226 (Ra) - 4 (alpha particle) = 222 (Rn) Therefore, the unknown nuclide produced in the alpha decay of ^226Ra is ^222Rn, and the correct option is (D) 222.

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

Bayanin Amsa

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

The shape of ammonia molecules is

Bayanin Amsa

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.

The carbon atoms on ethane are

Bayanin Amsa

The carbon atoms in ethane are sp3 hybridized. Hybridization is a concept in chemistry that describes how the orbitals of an atom combine to form new hybrid orbitals. In the case of carbon, which has four valence electrons, the four orbitals (s, px, py, and pz) combine to form four hybrid orbitals that have equal energy and are oriented in a tetrahedral arrangement around the carbon atom. In ethane, each carbon atom is bonded to three other atoms: two other carbon atoms and three hydrogen atoms. Each of the carbon-hydrogen bonds is formed by overlapping the sp3 hybrid orbitals on the carbon atom with the 1s orbitals on the hydrogen atoms. The carbon-carbon bond is formed by overlapping two of the sp3 hybrid orbitals on each carbon atom. Overall, the sp3 hybridization of the carbon atoms in ethane allows for the formation of strong and stable covalent bonds between the carbon and hydrogen atoms, as well as between the two carbon atoms.

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

Bayanin Amsa

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.

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

Bayanin Amsa

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.

The furring of kettles is caused by the presence in water of

Bayanin Amsa

Furring of kettles is caused by the temporary hardness in water. Temporary hardness in water is caused by calcium and magnesium trioxocarbonate (IV)
CaCO3 causes the furring of kettles

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)

Bayanin Amsa

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.

Incomplete oxidation of ethanol yields

Bayanin Amsa

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.

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

Bayanin Amsa

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.

Aluminium hydroxide is used in the dyeing industry as a

Bayanin Amsa

Aluminium hydroxide is used in the dyeing industry as a mordant. It combines with a dye and thereby fixes it in a material.

Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O2(l) + X. In the reaction above X is

Bayanin Amsa

In the given chemical equation, the reactants are solid calcium hydroxide (Ca(OH)2) and gaseous ammonium chloride (NH4Cl). When they react, they form solid calcium chloride (CaCl2), liquid water (H2O), and X, an unknown product. To determine the identity of X, we need to balance the chemical equation first. The balanced chemical equation is: Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O(l) + 2NH3(g) As per the balanced chemical equation, the product X is 2 molecules of ammonia (NH3). Therefore, the answer is (b) NH3.

The constituent of air necessary in the rusting process are

Bayanin Amsa

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

The reddish–brown rust on ion roofing sheets consists of

Bayanin Amsa

The reddish-brown rust on iron roofing sheets is mainly composed of Fe2O3.3H2O, which is a compound of iron and oxygen with water molecules. Iron roofing sheets, when exposed to air and moisture, undergo a chemical reaction. Oxygen in the air reacts with the iron to form a new compound called iron oxide (Fe2O3). As moisture (water) is also present, the iron oxide reacts with the water to form hydrated iron oxide (Fe2O3.3H2O), which is commonly known as rust. So, the reddish-brown rust on iron roofing sheets is mainly composed of Fe2O3.3H2O, which is a combination of iron, oxygen, and water molecules.

Ethene is prepared industrially by

Bayanin Amsa

Ethene, also known as ethylene, is prepared industrially by a process called cracking. Cracking involves breaking down larger hydrocarbon molecules, typically obtained from crude oil or natural gas, into smaller ones. This is usually done by heating the hydrocarbons to very high temperatures (up to 850°C) in the presence of a catalyst. The cracking process breaks the long chains of hydrocarbons into shorter ones, including the production of ethene. This is because the high temperature and catalyst cause the chemical bonds in the hydrocarbon molecules to break apart, leading to the formation of smaller molecules. These smaller molecules, including ethene, can then be separated from the other products and used in various applications such as the production of plastics. Therefore, out of the given options, the industrial preparation of ethene is accomplished by the process of cracking.

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

Bayanin Amsa

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.

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

Bayanin Amsa

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

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]

Bayanin Amsa

M = $\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}$

= $\frac{MmIT}{96500n}$

Copper II Chloride = CuCl2

CuCl2 → Cu2+ + 2Cl2

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

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

The IUPAC nomenclature of the structure above is

Bayanin Amsa

CH2CH2 - CCH3CH2
Start the numbering from the terminal carbon.

In the diagram above. X is

Bayanin Amsa

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

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

Bayanin Amsa

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.

The general formula of alkanones is

Bayanin Amsa

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.

In the diagram above, Y is

Bayanin Amsa

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

CH2S(g) + O2(g) → 2Cn + SO2(g)

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

Bayanin Amsa

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

The densities of two gases, X and Y are 0.5gdm-3 and 2.0gdm-3 respectively. What is the rate of diffusion of X relative to Y?

Bayanin Amsa

The rate of dimension of a gas inversely proportional to the square root of its molecular mass or its density, which is Graham's Law of diffusion of gas.

R ∝ $\frac{1}{\sqrt{Mm}}$ or R ∝ $\frac{1}{\sqrt{D}}$

Dx = 0.5gdm-3, Dy = 2gdm-3

R= $\frac{K}{\sqrt{D}}$

R$\sqrt{D}$ = k

R1$\sqrt{D_1}$ = R1$\sqrt{D_2}$

Rx$\sqrt{D_x}$ = Ry$\sqrt{D_y}$

$\frac{R_x}{R_y}$ = $\frac{\sqrt{D_y}}{\sqrt{D_x}}$

= $\frac{\sqrt{2}}{\sqrt{0.5}}$

= 2.0