JAMB UTME - Chemistry - 2019

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Question 5 Report

The molecular shape and bond angle of water are respectively

linear, 180°

bent, 109.5°

tetrahedral, 109.5°

bent, 105°

Answer Details

The shape of water molecule = Bent/ V- shaped
The bond angle of water = 104.5°/ 105°

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 6 Report

The shapes of water, ammonia, carbon (iv) oxide and methane are respectively

linear, tetrahedral, angular and pyramidal

angular, pyramidal, linear and tetrahedral

pyramidal, linear, tetrahedral and angular

tetrahedral, angular, pyramidal and linear

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Question 7 Report

A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years

75,000 particles

35,000 particles

25,000 particles

50,000 particles

Answer Details

The half-life of a radioactive nucleus is the time it takes for half of its particles to decay. This means that after 20 years, 100,000 particles will become 50,000 particles. After 40 years, we can find the number of particles remaining by counting the number of half-lives that have passed. Since 40 years is double the half-life of 20 years, this means that two half-lives have passed, so the number of particles will be halved twice. Starting with 100,000 particles: - After 1 half-life (20 years), there will be 50,000 particles remaining. - After 2 half-lives (40 years), there will be 25,000 particles remaining. So, exactly at the end of 40 years, there will be 25,000 particles remaining.

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

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Question 8 Report

Consider the equation below:

\(\mathrm{Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O}\).

The oxidation number of chromium changes from

+5 to +3

+6 to +3

-2 to +3

+7 to +3

Answer Details

Cr

_{2}

O

_{7}^{2 -}

+ 6Fe

^{2 +}

+ 14H

^{+}

\to

2Cr

^{3 +}

+ 6Fe

^{3 +}

+ 7H

_{2}

O
The oxidation of Cr in Cr

_{2}

O

_{7}^{2 -}

:
Let the oxidation of Cr = x;
2x + (-2 x 7) = -2

⟹

2x - 14 = -2
2x = 12 ; x = +6
Hence, the change in oxidation of Cr = +6 to +3

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 9 Report

How many electrons will be found in the nucleus of an atom with mass number 23 and 17 neutrons?

none

6

7

8

Answer Details

Electrons are not found in the nucleus of an atom. The nucleus of an atom only contains protons and neutrons, while electrons are located outside the nucleus in the electron cloud. The mass number of an atom is equal to the sum of the number of protons and the number of neutrons in the nucleus. Therefore, if an atom has a mass number of 23 and 17 neutrons, then the number of protons in the nucleus can be calculated as: Protons = Mass number - Neutrons Protons = 23 - 17 Protons = 6 This means that the nucleus of the atom contains 6 protons. The number of electrons in a neutral atom is equal to the number of protons, so the atom also contains 6 electrons in the electron cloud surrounding the nucleus. In summary, the answer is that there are 6 protons and 6 electrons in the atom.

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 10 Report

The oxidation state(s) of nitrogen in ammonium nitrite is/are

-3, +3

+5

-3, +5

-3

Answer Details

Ammonium nitrite = NH

_{4}

NO

_{2}

NH

_{4}^{+}

: Let the oxidation number of Nitrogen = x
x + 4 = 1

⟹

x = 1 - 4
x = -3
NO

_{2}^{-}

: x - 4 = -1
x = -1 + 4

⟹

x = +3.
The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3.

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 11 Report

A solution X, on mixing with \( \mathrm{AgNO_3} \) solution gives a white precipitate soluble in aqueous \( \mathrm{NH_3} \), a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain

Al

^{3 +}

, Cl

^{-}

Hg

^{+}

, SO

_{4}^{2 -}

Pb

^{2 +}

, Cl

^{-}

Ag

^{+}

, SO

_{4}^{2 -}

Chemistry Of Carbon Compounds (Ghana Only)

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Question 12 Report

The IUPAC name for \( \mathrm{CH_3CH_2COOCH_2CH_3} \) is

methylpropanoate

ethylethanoate

ethylpropanoate

methylethanoate

Answer Details

The IUPAC name for the given molecule is ethyl propanoate. To arrive at the IUPAC name, we first identify the longest continuous chain of carbon atoms, which in this case is a 4-carbon chain (propane). We then identify and name the substituent groups attached to this chain, which are a methyl group (CH3) attached to the second carbon atom and an ethoxy group (OC2H5) attached to the third carbon atom. The ethoxy group is named as an ethyl group, and the entire molecule is named as ethyl propanoate, following the standard IUPAC naming conventions for esters.

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

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Question 13 Report

Sulphur exists in six forms in the solid state. This property is known as

Isomerism

Allotropy

Isotopy

Isomorphism

Answer Details

The property of sulfur existing in six different forms in the solid-state is known as allotropy. Allotropy is a phenomenon where an element can exist in multiple forms, called allotropes, that have different physical and chemical properties but are composed of the same atoms. These different forms arise due to differences in the arrangement of atoms or molecules within the substance. In the case of sulfur, it can exist in multiple solid-state allotropes, including rhombic, monoclinic, and plastic sulfur, among others. Each of these allotropes has a different crystal structure, melting point, and other physical and chemical properties, even though they are all composed of sulfur atoms. Allotropy is a common phenomenon observed in many elements, including carbon, oxygen, and phosphorus, among others.

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

Non-metals And Their Compounds

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Question 14 Report

Which of the following represents the kind of bonding present in ammonium chloride?

ionic only

covalent only

ionic and dative covalent

dative covalent

Answer Details

Ammonium chloride contains both ionic and covalent bonds. In ammonium chloride, the ammonium ion (NH4+) is positively charged and the chloride ion (Cl-) is negatively charged. These ions are held together by ionic bonds, which are formed between positively and negatively charged ions. However, the bond between the hydrogen atom in the ammonium ion and the nitrogen atom in the ammonium ion is also a covalent bond. This type of covalent bond is known as a dative covalent bond, or a coordinate covalent bond, because the electron pair being shared is supplied by one atom only (the nitrogen atom in this case). So, the kind of bonding present in ammonium chloride is both ionic and dative covalent. In simple terms, ammonium chloride contains both ionic bonds between its positive and negative ions, and a dative covalent bond between the hydrogen atom and nitrogen atom within the ammonium ion.

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Question 15 Report

Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?

Zinc

Manganese

Sodium

Iron

Answer Details

The metal that is most essential in the regulation of blood volume, blood pressure, and osmotic equilibrium is sodium. Sodium is a key electrolyte that helps maintain the balance of fluids in the body, including blood volume and blood pressure. Sodium ions are positively charged and are attracted to negatively charged ions, such as chloride (Cl-) and bicarbonate (HCO3-), which together help regulate the pH of the blood. Sodium is also essential for maintaining osmotic equilibrium, which refers to the balance of solutes between cells and the extracellular fluid. Osmotic equilibrium is critical for proper cellular function and is regulated by the movement of water and electrolytes, including sodium, in and out of cells. While the other metals listed (zinc, manganese, and iron) are important for various functions in the body, such as enzyme activity and oxygen transport, they are not directly involved in regulating blood volume, blood pressure, and osmotic equilibrium in the same way that sodium is. Therefore, the answer is not options 1, 2, or 4, and the correct answer is: sodium.

Read lesson note on Metals And Their Compounds (JAMB)

Metals And Their Compounds

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Question 16 Report

Burning magnesium ribbon in air removes which of the following

(i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(iv)oxide?

i and iv only

i, ii, iii and iv only

i only

i and ii only

Answer Details

Burning magnesium ribbon in air will remove oxygen (option i) from the air, but not nitrogen (option ii), argon (option iii), or carbon dioxide (option iv). When magnesium burns, it reacts with oxygen in the air to form magnesium oxide. The reaction can be represented by the following equation: 2Mg(s) + O2(g) → 2MgO(s) The magnesium in the ribbon combines with oxygen in the air to form solid magnesium oxide. This reaction is exothermic, which means that it releases heat and light energy. So, when magnesium ribbon is burned in air, it consumes the oxygen in the air to form magnesium oxide. However, nitrogen, argon, and carbon dioxide are not chemically reactive with magnesium, and therefore are not removed from the air by the burning of magnesium ribbon. In summary, the correct option is (i) only - burning magnesium ribbon in air removes oxygen only.

Read lesson note on Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

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Question 17 Report

How many alkoxyalkanes can be obtained from the molecular formula \( \mathrm{C_4H_{10}O} \)?

4

2

1

3

Answer Details

Alkoxyalkanes have a general formula of R-O-R', where R and R' are alkyl groups. From the given molecular formula C4H10O, we can see that there are four carbon atoms, so the longest possible alkyl group is butyl (C4H9-). To form alkoxyalkanes, we need to attach an oxygen atom to the alkyl group. This can be done in three ways - by attaching the oxygen to one of the terminal carbon atoms (forming a primary alcohol), by attaching it to one of the central carbon atoms (forming a secondary alcohol), or by attaching it to the carbonyl carbon atom (forming an ester). So, we can obtain a maximum of three alkoxyalkanes from the given molecular formula. However, we need to take into account that there are different isomers possible for each type of alcohol or ester, depending on which carbon atom the oxygen is attached to. Therefore, the correct answer is (at least) 3.

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

Read lesson note on Acids, Bases And Salts (Nigeria Only) (WAEC)

Read lesson note on Chemical Equilibria (JAMB)

Acids, Bases And Salts (Nigeria Only)

Chemical Equilibria

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Question 18 Report

The electronic configuration of element Z is \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^1\). What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.

Z

_{3}

(SO

_{4}

)

_{2}

ZSO

_{4}

Z

_{2}

SO

_{4}

Z

_{2}

(SO

_{4}

)

_{3}

Answer Details

Z = 1s

^{2}

2s

^{2}

2p

^{6}

3s

^{2}

3p

^{1}

∴

We have Z

^{3 +}

and SO

_{4}^{2 -}

The reaction : Z

^{3 +}

+ SO

_{4}^{2 -}

\to

Z

_{2}

(SO

_{4}

)

_{3}

.

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 19 Report

When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is

propylmethanoate

ethylpropanoate

methyl ethanoate

methyl propanoate

Answer Details

The compound formed when the end alkyl groups of ethyl ethanoate are interchanged is ethyl propanoate. This is because ethyl ethanoate consists of two parts: the "ethyl" group and the "ethanoate" group. The ethyl group is a two-carbon chain, and the ethanoate group is a combination of a one-carbon chain and a carbonyl group (C=O) that is also attached to an oxygen atom. When the end alkyl groups are interchanged, the "ethyl" group is moved from the second carbon to the first carbon of the ethanoate group, and the "propanoate" group is formed. The "propanoate" group consists of a three-carbon chain and the carbonyl group. Therefore, the resulting compound is ethyl propanoate, which has a chemical formula of CH3CH2COOCH2CH3. This compound is commonly used as a flavoring agent and has a fruity odor reminiscent of pears.

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

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Question 20 Report

The hybridization in the compound \( \mathrm{CH_3-CH_2-C\equiv H} \) is

sp

^{3}

and sp

sp

^{2}

sp

^{3}

and sp

^{2}

Answer Details

The hybridization in a and b is sp

^{3}

hybridization while in c and d is sp hybridization.

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Question 21 Report

If acidified Potassium Dichromate(VI) (\(\mathrm{K_2Cr_2O_7}\)) acts as oxidizing agent, color changes from

yellow to green

orange to green

yellow to red

orange to red

Answer Details

Potassium Dichromate (VI), when it is acidified, acts as an oxidizing agent. When this happens, the color changes from orange to green. This is because the orange color of the potassium dichromate is due to the presence of Cr(VI) ions, which are oxidized to Cr(III) ions. The green color that is produced is due to the formation of chromium(III) ions. In this reaction, the dichromate ions are being oxidized, which means that they are losing electrons, and the chromium ions are being reduced, which means that they are gaining electrons. The transfer of electrons causes the color change from orange to green.

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 22 Report

Hydrogen diffused through a porous plug

thrice as fast as oxygen

at the same speed as oxygen

four times as fast as oxygen

twice as fast as oxygen

Answer Details

Hydrogen gas (H2) diffuses faster than oxygen gas (O2) through a porous plug. This is because the rate of diffusion of a gas through a porous plug is inversely proportional to the square root of its molar mass. Since the molar mass of hydrogen (2 g/mol) is much smaller than that of oxygen (32 g/mol), the rate of diffusion of hydrogen through a porous plug is much faster than that of oxygen. To be more specific, the ratio of the diffusion rates of two gases through a porous plug is given by the equation: Rate of diffusion of gas A / Rate of diffusion of gas B = √(Molar mass of gas B / Molar mass of gas A) Using the molar masses of hydrogen and oxygen, we get: Rate of diffusion of hydrogen / Rate of diffusion of oxygen = √(32 g/mol / 2 g/mol) = √16 = 4 Therefore, hydrogen diffuses through a porous plug four times as fast as oxygen. Thus, the correct answer is: four times as fast as oxygen.

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

Read lesson note on Acids, Bases And Salts (Nigeria Only) (WAEC)

Acids, Bases And Salts (Nigeria Only)

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Question 23 Report

Which two gases can be used for the demonstration of the fountain experiment?

SO

_{3}

and NH

_{3}

SO

_{2}

and NH

_{3}

HCl and NH

_{3}

SO

_{2}

and HCl

Answer Details

Two gases that can be used in the study of fountain experiment is ammonia gas and hydrogen chloride gas. The experiment introduces concepts like solubility and the gas laws at the entry level.

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

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Question 24 Report

\(200\text{cm}^3\) of \(0.50\text{mol}/\text{dm}^3\) solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

10g

25g

20g

15g

Answer Details

To solve this problem, we need to use the concept of stoichiometry and the solubility product constant (Ksp) of calcium hydrogen trioxocarbonate (IV). First, we need to write the balanced equation for the reaction that occurs when the solution of calcium hydrogen trioxocarbonate (IV) is heated: Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g) From the balanced equation, we can see that 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate. Therefore, we need to determine the number of moles of calcium hydrogen trioxocarbonate (IV) in the solution: Number of moles = concentration x volume Number of moles = 0.50 mol/dm³ x 0.2 dm³ Number of moles = 0.1 mol Since 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate, the number of moles of calcium carbonate produced will also be 0.1 mol. Next, we need to use the solubility product constant (Ksp) of calcium carbonate to determine the maximum amount of solid that can be precipitated: Ksp = [Ca²⁺][CO3²⁻] Ksp = 3.3 x 10⁻⁹ (at 25°C) At the maximum amount of solid precipitated, all the calcium carbonate formed will have precipitated, and the concentration of calcium ions and carbonate ions will be equal. Therefore, we can assume that the concentration of calcium ions and carbonate ions is both x. Substituting into the Ksp expression: Ksp = x² 3.3 x 10⁻⁹ = x² x = 5.74 x 10⁻⁵ mol/dm³ The mass of calcium carbonate precipitated can now be calculated: Mass = number of moles x molar mass Mass = 0.1 mol x 100.1 g/mol Mass = 10.01 g Therefore, the maximum weight of solid precipitated is approximately 10 g. Note that this calculation assumes that all the calcium carbonate precipitated as a solid, which may not always be the case in a real-world experiment. Additionally, this calculation does not take into account any losses due to filtration or other experimental errors.

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

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Question 25 Report

If the cost of electricity required to discharge 10g of an ion \(X^{3+}\) is N20.00, how much would it cost to discharge 6g of ion \(Y^{2+}\)?

[1 faraday = 96,500C, atomic masses are X = 27, Y = 24]

N10.00

N6.00

N20.00

N9.00

Answer Details

X

^{3 +}

+ 3e

^{-}

\to

X
3F = 27g
xF = 10g

\frac{x}{3} = \frac{10}{27} ⟹ x = \frac{10}{9} F

\frac{10}{9}

F

\equiv

N20.00
1F is equivalent to x

\frac{1}{\frac{10}{9}} = \frac{x}{20}

\frac{9}{10} = \frac{x}{20} ⟹ x = N 18.00

1F is equivalent to N18.00.
Y

^{2 +}

+ 2e

^{-}

\to

Y
2F = 24g
xF = 6g
x =

\frac{6 \times 2}{24} = \frac{1}{2} F

1F = N18.00

\frac{1}{2}

F =

\frac{1}{2} \times N 18.00

= N9.00

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Question 26 Report

\( \mathrm{SO_3} \) is not directly dissolved in water in the industrial preparation of \( \mathrm{H_2SO_4} \) by the contact process because

SO

_{3}

is an acidic gas

Acid is usually added to water and not water to acid

The reaction between SO

_{3}

and water is violently exothermic

SO

_{3}

does not dissolve readily in water

Acids, Bases And Salts

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Question 27 Report

Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?

But-2-ene

But-1-ene

But-2-yne

But-1-yne

Rates Of Chemical Reaction

Chemical Bonds (Ghana Only)

Redox Reactions

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Question 28 Report

The two ions responsible for hardness in water are

Ca

^{2 +}

and / or Mg

^{2 +}

K

^{+}

and / or Mg

^{2 +}

Ca

^{2 +}

and / or Li

^{+}

Ca

^{2 +}

and / or Na

^{+}

Answer Details

The ions responsible for hardness in water are Ca2+ and/or Mg2+. Hardness in water refers to the presence of calcium and magnesium ions, which are commonly found in natural water sources such as rivers, lakes, and groundwater. These ions can react with soap to form insoluble compounds, reducing the effectiveness of soap and causing scaling in pipes and appliances. The hardness of water is often measured in terms of the concentration of calcium and magnesium ions, expressed as calcium carbonate equivalents (CaCO3).

Read lesson note on Solubility (JAMB)

Read lesson note on Water (JAMB)

Solubility

Water

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Question 29 Report

Elements in the periodic table are arranged in the order of their

atomic numbers

metallic character

mass number

relative atomic mass

Answer Details

Elements in the periodic table are arranged in the order of their atomic numbers. The atomic number of an element is the number of protons in the nucleus of an atom of that element. The elements are arranged in order of increasing atomic number from left to right and from top to bottom in the periodic table. The elements in each row, also known as a period, have the same number of electron shells, while the elements in each column, also known as a group or family, have the same number of valence electrons. This arrangement makes it possible to predict the chemical and physical properties of an element based on its position in the periodic table. Therefore, the correct answer is: - atomic numbers

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 30 Report

X is a substance which liberates \( \mathrm{CO_2} \) on treatment with concentrated \( \mathrm{H_2SO_4} \). A warm solution of X can decolorize acidified \( \mathrm{KMnO_4} \). X is

HCOOH

NaHCO

_{3}

FeCO

_{3}

H

_{2}

C

_{2}

O

_{4}

Answer Details

It should be noted that for X to liberate CO

_{2}

, X must be a carbonate or an oxalate. Since X decolorizes KMnO

_{4}

, X must be an oxalate.
Therefore, X is H

_{2}

C

_{2}

O

_{4}

.

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

Periodic Chemistry (Ghana Only)

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Question 31 Report

The cost of discharging 6.0g of a divalent metal, X from its salt is ₦12.00. What is the cost of discharging 9.0g of a trivalent metal, Y from its salt under the same condition?

[X = 63, Y = 27, 1F = 96,500C]

N12.00

N27.00

N63.00

N95.00

Answer Details

For X: X

^{2 +}

+ 2e

^{-}

\to

X
2F = 63g
xF = 6g
x =

\frac{6 \times 2}{63} = \frac{4}{21} F

\frac{4}{21}

F = N12.00
1F =

\frac{12}{\frac{4}{21}}

= N63.00
1F is equivalent to N63.00.
For Y: Y

^{3 +}

+ 3e

^{-}

\to

Y
3F = 27g
xF = 9g
x =

\frac{3 \times 9}{27}

= 1F
1F = N63.00

Read lesson note on Chemical Combination (JAMB)

Chemical Combination

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Question 32 Report

Which quantum divides shells into orbitals?

Spin

Principal

Azimuthal

Magnetic

Answer Details

The quantum that divides shells into orbitals is the "Azimuthal" quantum number, also known as the "angular momentum" quantum number. The azimuthal quantum number determines the shape of an electron's orbital, which is a region in space where there is a high probability of finding an electron. It describes the angular momentum of an electron in an atom and the number of subshells within a given shell. Each subshell is associated with a specific shape, and can hold a certain number of electrons. The azimuthal quantum number is represented by the letter "l" and can have integer values ranging from 0 to (n-1), where "n" is the principal quantum number. Each value of "l" corresponds to a different subshell shape: - l = 0 corresponds to an "s" subshell, which is spherical in shape. - l = 1 corresponds to a "p" subshell, which has a dumbbell shape with two lobes. - l = 2 corresponds to a "d" subshell, which has a more complex shape with four lobes and a doughnut-like ring. - l = 3 corresponds to an "f" subshell, which has an even more complex shape with eight lobes. The number of orbitals within a subshell is equal to 2l+1. For example, a "p" subshell (l = 1) has three orbitals (2l+1 = 3), which are labeled as "px", "py", and "pz". In summary, the azimuthal quantum number determines the shape of the electron's orbital and the number of subshells within a given shell, and it is represented by the letter "l".

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 33 Report

For the general equation of the nature

XP + yQ ⇌ mR + nS, the expression for the equilibrium constant is

k[P]

^{x}

[Q]

^{y}

\frac{m [R] [S]}{x [P] y [Q]}

\frac{[P]^{x} [Q]^{y}}{[R]^{m} [S]^{n}}

\frac{[R]^{m} [S]^{n}}{[P]^{x} [Q]^{y}}

Answer Details

The expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is: Kc = [R]m[S]n / [P]x[Q]y where Kc is the equilibrium constant, [R] and [S] are the concentrations of the products, and [P] and [Q] are the concentrations of the reactants, all raised to the stoichiometric coefficients (m, n, x, y) in the balanced equation. This equation is known as the equilibrium constant expression and it represents the ratio of the concentrations of the products and reactants at equilibrium for a particular chemical reaction. The equilibrium constant is a measure of how far a reaction proceeds towards completion, with a larger value indicating a greater extent of reaction. The equilibrium constant expression is derived from the law of mass action, which states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants raised to their stoichiometric coefficients. At equilibrium, the rates of the forward and reverse reactions are equal, and the equilibrium constant expression represents the ratio of the rate constants for these two reactions. Therefore, the correct expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is Kc = [R]m[S]n / [P]x[Q]y.

Read lesson note on Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Read lesson note on Chemical Equilibria (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Equilibria

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Question 34 Report

Which of the following could not be alkane?

C

_{4}

H

_{10}

C

_{5}

H

_{12}

C

_{7}

H

_{14}

C

_{8}

H

_{18}

Answer Details

An alkane is a type of hydrocarbon with only single bonds between the carbon atoms. It follows the general formula CnH2n+2, where "n" is the number of carbon atoms in the molecule. To determine whether a molecule is an alkane or not, we can calculate its molecular formula and check if it fits the general formula of alkane. Out of the given options, the third one (C7H14) cannot be an alkane. To see why, let's use the general formula of alkane, which is CnH2n+2. For C7H14 to be an alkane, it should have 2n+2 = 2(7) + 2 = 16 hydrogen atoms. However, C7H14 has only 14 hydrogen atoms, which means it does not follow the general formula of alkane. Therefore, C7H14 cannot be an alkane. The other options are as follows: - C4H10: This is butane, which is an alkane with four carbon atoms. - C5H12: This is pentane, which is an alkane with five carbon atoms. - C8H18: This is octane, which is an alkane with eight carbon atoms. In summary, the molecule C7H14 cannot be an alkane because it does not follow the general formula of alkane, while the other options are all examples of alkanes.

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

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Question 35 Report

The following are isoelectronic ions except

Na

^{+}

Mg

^{2 +}

Si

^{2 +}

O

^{2 +}

Answer Details

Two or more ions are said to be isoelectronic if they have the same electronic structure and the same number of valence electrons.
Na

^{+}

= 10 electrons = 2, 8
Mg

^{2 +}

= 10 electrons = 2,8
O

^{2 -}

= 10 electrons = 2,8
Si

^{2 +}

= 12 electrons = 2,8,2

⟹

Si

^{2 +}

is not isoelectronic with the rest.

Read lesson note on Acids, Bases And Salts (Nigeria Only) (WAEC)

Read lesson note on Atomic Structure And Bonding (JAMB)

Acids, Bases And Salts (Nigeria Only)

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Question 36 Report

Which of the following statements does not show Rutherford's account of Nuclear Theory? An atom contains a region

which contains protons and neutrons

which is positively charged

which is massive and can cause deflection of a few projectiles

which is very large and in which close to 98% of projectiles pass undeflected

Answer Details

Rutherford's account of Nuclear theory does not include the fact that atoms contain a massive region and cause deflection of from projectiles.

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 37 Report

When chlorine water is exposed to bright sunlight, the following products are formed

HCl

_{(a q)}

and Cl

_{2}

HOCl

_{(a q)}

and Cl

_{2}

HCl

_{(a q)}

and O

_{2}

HOCl

_{(a q)}

and O

_{2}

Chemistry Of Carbon Compounds (Ghana Only)

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Question 38 Report

What volume of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.

[Na = 23, N = 14, O = 16]

1.7 litres

0.420 litres

0.588 litres

2.35 litres

Answer Details

To calculate the volume of a solution, we need to use the formula: moles of solute = concentration x volume First, let's find the number of moles of sodium trioxonitrate (V) in 5g of the solute. The molar mass of NaNO3 is: Na = 23 N = 14 3 x O = 3 x 16 = 48 Molar mass = 23 + 14 + 48 = 85 g/mol The number of moles of NaNO3 in 5g is: moles = mass / molar mass = 5 / 85 = 0.0588 moles Now, we can use the formula above to find the volume of the solution: moles of solute = concentration x volume volume = moles of solute / concentration volume = 0.0588 moles / 0.100 M volume = 0.588 litres Therefore, the correct answer is 0.588 litres of 0.100M sodium trioxonitrate (V) solution contains 5g of solute.

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Stoichiometry And Chemical Reactions

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Question 39 Report

Which important nitrogen-containing compound is produced in Haber's process?

NH

_{3}

NO

_{2}

NO

HNO

_{3}

Answer Details

The important nitrogen-containing compound that is produced in Haber's process is NH3, which is also known as ammonia. Haber's process is a chemical process used to produce ammonia by reacting nitrogen gas (N2) and hydrogen gas (H2) under high pressure and temperature in the presence of an iron catalyst. The reaction between nitrogen and hydrogen produces ammonia as the main product, along with some nitrogen and hydrogen gases that do not react. NH3 is an important compound that is widely used in industry for the production of fertilizers, plastics, and other chemical products. It is also used as a cleaning agent, a refrigerant, and a fuel for engines. In addition, NH3 is an essential compound for life, as it is a key component of amino acids, which are the building blocks of proteins.

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