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

bubbling air through the acid solution

cooling the acid solution with cold water

bubbling air through the acid solution

JAMB UTME - Chemistry - 2017 | Green Bridge CBT

Swali 1 Ripoti

The acid anhydride that will produce weak acid in water is

SO3

NO2

SO2

CO2

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Periodic Chemistry (Ghana Only) (WAEC)

Periodic Chemistry (Ghana Only)

Tazama Jibu

Swali 2 Ripoti

The constituent of air necessary in the rusting process are

O2 and H2O

Ar and CO2

CO2 and H2O

O2 and CO2

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Air (JAMB)

Air

Tazama Jibu

Swali 3 Ripoti

When ΔH is negative, a reaction is said to be

endothermic

exothermic

reversible

ionic

Maelezo ya Majibu

When ΔH is negative, a reaction is exothermic. In simpler terms, exothermic reactions release energy into their surroundings, which is often felt as heat. When ΔH is negative, it means that the enthalpy change of the reaction is negative, indicating that energy is being released. This can be seen as a decrease in temperature in the surrounding environment or a release of heat or light. Examples of exothermic reactions include combustion reactions, where a fuel combines with oxygen to release heat and light, or the reaction between an acid and a base to form a salt and water. In both cases, energy is released into the surrounding environment, and the enthalpy change is negative (ΔH < 0).

Soma maelezo ya somo kwenye Energy Changes (JAMB)

Energy Changes

Tazama Jibu

Swali 4 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Electrolysis (JAMB)

Electrolysis

Tazama Jibu

Swali 5 Ripoti

The reddish–brown rust on ion roofing sheets consists of

Fe3 + (H2O)6

FeO.H2O

Fe2O3.3H2O

Fe3O4.2H22O

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

Tazama Jibu

Swali 6 Ripoti

The alkyl group is represented by the general formula

CnH2n

CnH2n − 2

CnH2n + 1

CnH2n + 2

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Chemistry Of Carbon Compounds (Nigeria Only)

Tazama Jibu

Swali 7 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Electrolysis (JAMB)

Electrolysis

Tazama Jibu

Swali 8 Ripoti

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

Isotopy

Resonance

Isomerism

Reforming

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Chemical Equilibria (JAMB)

Chemical Equilibria

Tazama Jibu

Swali 9 Ripoti

The oxidation number of iodine in KIO3 is

+7

+3

+5

+6

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Oxidation And Reduction (JAMB)

Oxidation And Reduction

Tazama Jibu

Swali 10 Ripoti

Tin is unaffected by air at ordinary temperature due to its

Low melting point

Weak electropositive character

High boiling point

White lustrous appearance

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Chemical Combination (JAMB)

Chemical Combination

Tazama Jibu

Swali 11 Ripoti

$ {}^{226}_{88}\mathrm{Ra} \rightarrow {}^{x}_{86}\mathrm{Rn} $ + alpha particle

226

220

227

222

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Chemical Combination (JAMB)

Soma maelezo ya somo kwenye Atomic Structure And Bonding (JAMB)

Chemical Combination

Atomic Structure And Bonding

Tazama Jibu

Swali 12 Ripoti

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

Maelezo ya Majibu

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

?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.

Soma maelezo ya somo kwenye Kinetic Theory Of Matter And Gas Laws (JAMB)

Soma maelezo ya somo kwenye Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Kinetic Theory Of Matter And Gas Laws

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Tazama Jibu

Swali 13 Ripoti

An isomer of C5H12 is

2–ethyl butane

butane

2–methyl butane

2–methyl propane

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Chemistry Of Carbon Compounds (Nigeria Only)

Tazama Jibu

Swali 14 Ripoti

Ethene is prepared industrially by

Reforming

Polymerization

Distillation

Cracking

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Chemical Industry And Environment (WAEC)

Chemical Industry And Environment

Tazama Jibu

Swali 15 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Tazama Jibu

Swali 16 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Tazama Jibu

Swali 17 Ripoti

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

water

mercury

paraffin

phenol

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Metals And Their Compounds (JAMB)

Metals And Their Compounds

Tazama Jibu

Swali 18 Ripoti

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?

0.1

0.5

2.0

4.0

Maelezo ya Majibu

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 ∝ $\frac{1}{\sqrt{M m}}$ 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

Soma maelezo ya somo kwenye Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Tazama Jibu

Swali 19 Ripoti

The tincture of iodine means iodine dissolved in

ethanol

bromine chloride

chlorine water

water

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Acids, Bases And Salts (Ghana Only) (WAEC)

Acids, Bases And Salts (Ghana Only)

Tazama Jibu

Swali 20 Ripoti

The shape of ammonia molecules is

trigonal planar

octahedral

square planar

tetrahedral

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

Tazama Jibu

Swali 21 Ripoti

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

NH3—C—C=O

CHO

CO

CNH2

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

Tazama Jibu

Swali 22 Ripoti

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

−100°c

−273°c

−373°c

0°c

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Tazama Jibu

Swali 23 Ripoti

The general formula of alkanones is

RCHO

R2CO

RCOOH

RCOOR

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Chemistry Of Carbon Compounds (Nigeria Only)

Tazama Jibu

Swali 24 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Oxidation And Reduction (JAMB)

Oxidation And Reduction

Tazama Jibu

Swali 25 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Tazama Jibu

Swali 26 Ripoti

The carbon atoms on ethane are

sp2 hybridized

sp3 hybridized

sp4 hybridized

sp hybridized

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Chemistry Of Carbon Compounds (Nigeria Only)

Tazama Jibu

Swali 27 Ripoti

Aluminium hydroxide is used in the dyeing industry as a

dye

dispersant

salt

mordant

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Soma maelezo ya somo kwenye Chemical Bonds (Ghana Only) (WAEC)

Soma maelezo ya somo kwenye Periodic Chemistry (Ghana Only) (WAEC)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Bonds (Ghana Only)

Periodic Chemistry (Ghana Only)

Tazama Jibu

Swali 28 Ripoti

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

NO2

NH3

N2O

H2O

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Tazama Jibu

Swali 29 Ripoti

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

amylase

diastatse

invertase

zymase

Maelezo ya Majibu

This is any amylase or a mixture of amylases that converts starch to dextrin and maltose.

Soma maelezo ya somo kwenye Structure Of The Atom (Ghana Only) (WAEC)

Soma maelezo ya somo kwenye Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Structure Of The Atom (Ghana Only)

Chemistry Of Carbon Compounds (Nigeria Only)

Tazama Jibu

Swali 30 Ripoti

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

CaCO3

Na2CO3

Na2SO4

CaSO4

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Oxidation And Reduction (JAMB)

Oxidation And Reduction

Tazama Jibu

Swali 31 Ripoti

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

chlorine

sulphur (IV) oxide

carbon (IV) oxide

ammonia

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

Tazama Jibu

Swali 32 Ripoti

The shape of the S-orbital is

elliptical

spiral

circular

spherical

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

Tazama Jibu

Swali 33 Ripoti

In the diagram above. X is

SO3

SO2

S

H2S

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

Tazama Jibu

Swali 34 Ripoti

Water for town supply is chlorinate to make it free from

bad colour

bacteria

temporary hardness

permanent hardness

Maelezo ya Majibu

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.

Tazama Jibu

Swali 35 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Tazama Jibu

Swali 36 Ripoti

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

calcium hydrogentrioxocarbonate (IV)

calcium trioxocarbonate (IV)

calcium tetraoxosulphate (VI)

calcium hydroxide

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

Tazama Jibu

Swali 37 Ripoti

Incomplete oxidation of ethanol yields

CH3COOH

CH3COCH3

CH3CH2OCH2CH3

CH2CHO

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Chemistry Of Carbon Compounds (Nigeria Only)

Tazama Jibu

Swali 38 Ripoti

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

Zymase

Sucrase

Amylase

Lactase

Maelezo ya Majibu

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.

Tazama Jibu

Swali 39 Ripoti

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

20

32

14

12

Maelezo ya Majibu

Molecular mass = vapour density X2

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

vapour density = 32

∴ x + 32 = 32 × 2