5070/11 Oct/Nov 2024 — Chemistry MCQ Answers

Q11MMedium-EasyStates of Matter

Which row shows both a property of a gas and the correct explanation for this property?

Options

	property	explanation for the property
A	Gases flow easily.	The bonds within the molecules are weak.
B	The pressure of a sample of gas increases when the volume is decreased.	The particles collide more frequently with the walls of the container.
C	The volume of a sample of gas increases when the temperature is increased.	The particles in a gas are far apart.
D	The spread of perfume particles is due to diffusion.	Diffusion is the movement of particles from an area of low concentration to one of high concentration.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

A is incorrect: Gases flow easily because the intermolecular forces between the molecules are very weak, not because the covalent bonds within the molecules are weak.
B is correct: Decreasing the volume of a gas decreases the space available for the particles, leading to more frequent collisions with the walls of the container per unit area, which increases the pressure.
C is incorrect: The increase in volume when a gas is heated at constant pressure is explained by particles gaining kinetic energy and moving faster, not simply because they are far apart.
D is incorrect: Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration, not low to high.

Answer

B

Final answer

B

Detailed explanation

Walkthrough

To determine the correct row, evaluate both the property and the scientific explanation in each option:

Option A: While it is true that gases flow easily (fluidity), this property is due to weak intermolecular attractions between molecules allowing them to move randomly and freely past one another. The strength of the internal covalent bonds within the molecules (intramolecular bonds) does not affect fluidity.
Option B: When the volume of a container is reduced at constant temperature, the gas particles are packed closer together into a smaller space. Consequently, they collide with the walls of the container more frequently. Since gas pressure is the result of forces exerted by these particle collisions per unit area of the wall, an increased collision frequency results in higher pressure. Both the property and explanation are correct.
Option C: Increasing the temperature increases the average kinetic energy and speed of gas particles. If the gas is allowed to expand (constant pressure), the particles move further apart to maintain equilibrium, increasing the volume. Simply stating that "particles in a gas are far apart" is a static property of gases, not the kinetic reason why volume expands upon heating.
Option D: While perfume spreading is an example of diffusion, the definition provided in the table is backwards. Diffusion is the net movement of particles down a concentration gradient—from a region of higher concentration to a region of lower concentration.

Therefore, row B is the only correct statement.

Key Takeaways

Gas pressure is caused by the collisions of gas particles with the inside walls of the container.
Decreasing container volume increases the number of collisions per unit time per unit surface area (collision frequency), thus increasing gas pressure.
Distinguish between forces between molecules (intermolecular forces) and bonds within molecules (intramolecular covalent bonds).
Diffusion is the random movement of particles resulting in a net movement from high concentration to low concentration.

Common Mistakes

Confusing intramolecular bonds (covalent bonds inside a molecule) with intermolecular forces when explaining physical properties such as boiling point or fluidity.
Reversing the concentration gradient in the definition of diffusion (i.e. thinking it goes from low to high concentration).

Things to Be Careful About

Ensure both columns (the property and the explanation) are scientifically accurate in multiple-choice pairing questions.

Techniques used

relate particle collision frequency to gas pressureevaluate kinetic particle theory explanations for gas propertiesidentify incorrect definitions and explanations

Q21MEasyStates of Matter

When measured under the same conditions of temperature and pressure, which gas diffuses at the same rate as nitrogen?

Options

A ammonia, NH₃
B carbon monoxide, CO
C ethane, C₂H₆
D oxygen, O₂

Pick an answer

DifficultyEasy

Worked solution

Working

Under the same conditions of temperature and pressure, gases with the same relative molecular mass ( $M_r$ ) diffuse at the same rate.

Calculate $M_r$ for nitrogen gas, $\text{N}_2$ :

M_r(\text{N}_2) = 2 \times 14 = 28

Calculate $M_r$ for each given option:

A $\text{NH}_3$ : $M_r = 14 + (3 \times 1) = 17$
B $\text{CO}$ : $M_r = 12 + 16 = 28$
C $\text{C}_2\text{H}_6$ : $M_r = (2 \times 12) + (6 \times 1) = 30$
D $\text{O}_2$ : $M_r = 2 \times 16 = 32$

Carbon monoxide ( $\text{CO}$ ) has the same relative molecular mass ( $M_r = 28$ ) as nitrogen gas ( $\text{N}_2$ ), so it diffuses at the same rate.

Answer

B

Final answer

B

Detailed explanation

Walkthrough

The rate of diffusion of a gas depends inversely on its relative molecular mass ( $M_r$ ) at a given temperature and pressure: gases with lower $M_r$ diffuse faster, while gases with higher $M_r$ diffuse slower. Two gases with identical $M_r$ values will diffuse at the same rate under the same conditions.

First, find the $M_r$ of nitrogen gas. Nitrogen is diatomic, $\text{N}_2$ , so $M_r = 2 \times 14 = 28$ .
Next, calculate the $M_r$ $M_{r}$ of each option:
- $\text{NH}_3$ : $14 + 3(1) = 17$
- $\text{CO}$ : $12 + 16 = 28$
- $\text{C}_2\text{H}_6$ : $2(12) + 6(1) = 30$
- $\text{O}_2$ : $2(16) = 32$
Since $\text{CO}$ has an $M_r$ of $28$ , identical to $\text{N}_2$ , it diffuses at the same rate.

Key Takeaways

Diffusion is the net movement of particles from an area of higher concentration to an area of lower concentration due to random particle motion.
The rate of diffusion of a gas is determined by its relative molecular mass ( $M_r$ ) — lighter molecules move and diffuse faster.
Gases with the same $M_r$ diffuse at the same rate under identical temperature and pressure conditions.

Common Mistakes

Forgetting that nitrogen gas is diatomic ( $\text{N}_2$ , $M_r = 28$ ) and mistakenly taking its mass as atomic nitrogen ( $\text{N}$ , $A_r = 14$ ).

Things to Be Careful About

Always check the formula of elemental gases — oxygen and nitrogen exist as diatomic molecules ( $\text{O}_2$ and $\text{N}_2$ ).

Techniques used

calculate relative molecular massrelate rate of diffusion to relative molecular mass

Q31MMedium-EasyAtoms, Elements and CompoundsStoichiometry

X and Y are both elements, one of which is a non-metal.

1 mole of X is added to 1 mole of Y and they are heated together.

X and Y react completely to form substance Z.

Substance Z cannot be easily separated to form X and Y.

Three statements are given.

Y must be the non-metal.
X and Y formed a compound on heating.
Substance Z has the empirical formula XY.

Which statements are correct?

Options

A 1 and 3
B 1 only
C 2 and 3
D 2 only

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Statement 1 is incorrect: the question states that one of the elements is a non-metal, but does not specify whether $\text{X}$ or $\text{Y}$ is the non-metal. Either could be the non-metal.
Statement 2 is correct: elements $\text{X}$ and $\text{Y}$ react chemically upon heating to form substance $\text{Z}$ , which cannot be easily separated back into its elements by physical means. Therefore, $\text{Z}$ is a compound.
Statement 3 is correct: $1\text{ mole}$ of $\text{X}$ reacts completely with $1\text{ mole}$ of $\text{Y}$ without any leftover reactants. The simplest whole-number molar ratio of $\text{X}$ to $\text{Y}$ is $1:1$ , so the empirical formula of $\text{Z}$ is $\text{XY}$ .

Thus, statements 2 and 3 are correct.

Answer

C

Final answer

C

Detailed explanation

Walkthrough

Let us evaluate each of the three statements:

Statement 1: "Y must be the non-metal."
The question informs us that one of the two elements is a non-metal and the other is a metal, but it gives no additional clues to assign identities specifically. $\text{X}$ could be the non-metal and $\text{Y}$ the metal, or vice versa. Therefore, statement 1 is not necessarily true.
Statement 2: "X and Y formed a compound on heating."
A compound is a substance formed when two or more elements chemically combine in a fixed ratio. Unlike mixtures, compounds cannot be separated into their constituent elements by simple physical separation techniques. Because $\text{X}$ and $\text{Y}$ react together completely upon heating to produce substance $\text{Z}$ , which cannot be easily separated into $\text{X}$ and $\text{Y}$ , substance $\text{Z}$ must be a chemical compound. Statement 2 is correct.
Statement 3: "Substance Z has the empirical formula XY."
The empirical formula represents the simplest whole-number ratio of the atoms of each element present in a compound. Since $1\text{ mole}$ of $\text{X}$ reacts completely with $1\text{ mole}$ of $\text{Y}$ with no excess reactant, the reacting mole ratio is:
$\text{moles of X} : \text{moles of Y} = 1 : 1$
This gives an empirical formula of $\text{XY}$ . Statement 3 is correct.

Combining statements 2 and 3 gives option C.

Key Takeaways

A compound consists of two or more different elements chemically joined together in fixed proportions; it cannot be separated by physical means.
The empirical formula is the simplest whole-number ratio of atoms/moles of each element in a compound.
When elements react in a $1:1$ molar ratio without any excess, their empirical formula has subscripts of $1$ for each element (e.g., $\text{XY}$ ).

Common Mistakes

Assuming the second named element is always the non-metal just because non-metals are written second in ionic formulae (e.g., $\text{NaCl}$ ).
Confusing the properties of a mixture (can be separated physically) with a compound (cannot be separated by physical methods).

Things to Be Careful About

Ensure you check whether each statement must be strictly true based only on the information provided in the stem.

Techniques used

distinguish between elements, mixtures, and compoundsdetermine empirical formula from mole reacting ratio

Q41MMedium-EasyAtoms, Elements and Compounds

An ion has an electronic configuration of 2,8,8.

It has a 2– charge and a nucleon number of 36.

How many neutrons are present in the nucleus of this ion?

Options

A 16
B 18
C 19
D 20

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Find the total number of electrons in the ion from its electronic configuration:

\text{Number of electrons} = 2 + 8 + 8 = 18

The ion carries a $2-$ charge, which means it has $2$ more electrons than protons:

\text{Number of protons} = 18 - 2 = 16

Calculate the number of neutrons using the nucleon number ( $A = 36$ ):

\text{Number of neutrons} = \text{nucleon number} - \text{number of protons} = 36 - 16 = 20

Therefore, the correct option is D.

Answer

D

Final answer

D

Detailed explanation

Walkthrough

To find the number of neutrons in the nucleus of an atom or ion, we use the fundamental relationship:

\text{Number of neutrons} = \text{nucleon number (mass number)} - \text{proton number (atomic number)}

Find the number of electrons:
The electronic configuration is given as $2,8,8$ . Summing the electrons in all shells gives:
$2 + 8 + 8 = 18\text{ electrons}$
Determine the number of protons:
A negatively charged ion (anion) has gained extra electrons. A $2-$ charge indicates that the ion has $2$ extra electrons compared to the neutral atom. Therefore:
$\text{Number of protons} = 18 - 2 = 16$
(This corresponds to the element sulfur, $^{36}_{16}\text{S}^{2-}$ ).
Calculate the number of neutrons:
The question states that the nucleon number is $36$ . Subtracting the proton number from the nucleon number:
$\text{Number of neutrons} = 36 - 16 = 20$

Matching with the given options, D is the correct answer.

Key Takeaways

The total number of electrons is found by summing the numbers in the electronic configuration.
For a negative ion of charge $n-$ , $\text{number of protons} = \text{number of electrons} - n$ .
For a positive ion of charge $n+$ , $\text{number of protons} = \text{number of electrons} + n$ .
The nucleon number (mass number) equals $\text{protons} + \text{neutrons}$ .

Common Mistakes

Option A (16): Calculating the number of protons correctly but forgetting to subtract from the nucleon number to find the number of neutrons.
Option B (18): Incorrectly assuming the number of protons is equal to the number of electrons ( $18$ ) and calculating $36 - 18 = 18$ .
Adding instead of subtracting charge: Adding $2$ to $18$ to get $20$ protons, leading to $36 - 20 = 16$ neutrons.

Things to Be Careful About

Pay close attention to the sign of the ionic charge: negative charge ( $2-$ ) means more electrons than protons, so subtract the magnitude of the charge from the total electron count to get the proton number.
Always distinguish clearly between what the question asks for (neutrons) and intermediate calculated quantities (protons or electrons).

Techniques used

calculate number of electrons from electronic configurationdeduce proton number from ionic charge and electron countcalculate number of neutrons from nucleon number and proton number

Q51MMedium-EasyAtoms, Elements and Compounds

The formulae for two ions are given.

^{72}_{34}\text{X}^{2-} \qquad ^{74}_{34}\text{X}^{2-}

Which statement is correct?

Options

A The chemical properties of both ions are the same.
B The number of electrons in each ion is different.
C The number of neutrons in each ion is the same.
D The number of protons in each ion is different.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

For $^{72}_{34}\text{X}^{2-}$ :

Number of protons $= 34$
Number of neutrons $= 72 - 34 = 38$
Number of electrons $= 34 + 2 = 36$

For $^{74}_{34}\text{X}^{2-}$ :

Number of protons $= 34$
Number of neutrons $= 74 - 34 = 40$
Number of electrons $= 34 + 2 = 36$

Evaluating the statements:

A is correct: Both species have the same number of protons ( $34$ ) and the same electron configuration ( $36$ electrons), so their chemical properties are identical.
B is incorrect: Both ions have $36$ electrons.
C is incorrect: The first ion has $38$ neutrons and the second has $40$ neutrons.
D is incorrect: Both ions have $34$ protons.

Answer

A

Final answer

A

Detailed explanation

Walkthrough

Identify the subatomic particle counts from the nuclide symbol $^{A}_{Z}\text{X}^{q}$ $_{Z}^{A} X^{q}$ :
- $Z$ is the proton number (atomic number). Both species have $Z = 34$ , meaning they both have $34$ protons.
- $A$ is the nucleon number (mass number). The first species has $A = 72$ , giving $72 - 34 = 38$ neutrons. The second has $A = 74$ , giving $74 - 34 = 40$ neutrons.
- The charge is $2-$ , meaning each ion has gained $2$ extra electrons: $\text{electrons} = 34 + 2 = 36$ .
Compare the two ions:
- They have the same number of protons and electrons, but a different number of neutrons (they are isotopic ions).
- Chemical properties depend entirely on the number and arrangement of electrons (and nuclear charge), which are identical in both ions. Therefore, their chemical properties are the same.

Key Takeaways

Isotopes of the same element have the same number of protons and electrons, but different numbers of neutrons.
Because they possess identical electronic configurations, isotopes (and their corresponding ions) exhibit the same chemical properties.

Common Mistakes

Forgetting that a $2-$ negative charge means adding electrons to the neutral atom, not subtracting.
Confusing mass number ( $A$ ) with neutron number ( $A - Z$ ).

Things to Be Careful About

Do not confuse chemical properties (which are determined by electrons) with physical properties like mass and density (which depend on the nucleus/neutrons).

Techniques used

deduce subatomic particle numbers from nuclide notationrelate electron configuration and proton number to chemical propertiesidentify isotope characteristics

Q61MMedium-EasyAtoms, Elements and Compounds

Which pair of molecules have the same number of electrons in covalent bonds?

Options

A CH₃OH and C₂H₄
B CO₂ and H₂O
C NH₃ and N₂
D O₂ and Cl₂

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Each single covalent bond consists of $2$ shared electrons, a double bond has $4$ shared electrons, and a triple bond has $6$ shared electrons.

Let us calculate the total number of electrons involved in covalent bonds for each pair:

A: In $\text{CH}_3\text{OH}$ , there are $5$ single bonds ( $3 \times \text{C-H}$ , $1 \times \text{C-O}$ , $1 \times \text{O-H}$ ), giving $5 \times 2 = 10$ bonding electrons. In $\text{C}_2\text{H}_4$ , there are $4$ single $\text{C-H}$ bonds and $1$ double $\text{C=C}$ bond, giving $(4 \times 2) + 4 = 12$ bonding electrons. ( $10 \neq 12$ )
B: In $\text{CO}_2$ , there are $2$ double $\text{C=O}$ bonds, giving $2 \times 4 = 8$ bonding electrons. In $\text{H}_2\text{O}$ , there are $2$ single $\text{O-H}$ bonds, giving $2 \times 2 = 4$ bonding electrons. ( $8 \neq 4$ )
C: In $\text{NH}_3$ , there are $3$ single $\text{N-H}$ bonds, giving $3 \times 2 = 6$ bonding electrons. In $\text{N}_2$ , there is $1$ triple $\text{N}\equiv\text{N}$ bond, giving $1 \times 6 = 6$ bonding electrons. Both molecules have $6$ bonding electrons.
D: In $\text{O}_2$ , there is $1$ double $\text{O=O}$ bond, giving $4$ bonding electrons. In $\text{Cl}_2$ , there is $1$ single $\text{Cl-Cl}$ bond, giving $2$ bonding electrons. ( $4 \neq 2$ )

Therefore, pair C has the same number of bonding electrons.

Answer

C

Final answer

C

Detailed explanation

Walkthrough

A covalent bond is formed when atoms share a pair of electrons.

A single bond involves $1$ shared pair = $2$ electrons.
A double bond involves $2$ shared pairs = $4$ electrons.
A triple bond involves $3$ shared pairs = $6$ electrons.

To find the pair of molecules with the same number of electrons in covalent bonds, we determine the number of bonding electrons in each molecule:

$\text{CH}_3\text{OH}$ : Contains $3$ $\text{C-H}$ single bonds, $1$ $\text{C-O}$ single bond, and $1$ $\text{O-H}$ single bond, totalling $5$ single bonds = $10$ electrons.
$\text{C}_2\text{H}_4$ : Contains $4$ $\text{C-H}$ single bonds and $1$ $\text{C=C}$ double bond, totalling $4 \times 2 + 4 = 12$ electrons.
$\text{CO}_2$ : Contains $2$ $\text{C=O}$ double bonds = $2 \times 4 = 8$ electrons.
$\text{H}_2\text{O}$ : Contains $2$ $\text{O-H}$ single bonds = $2 \times 2 = 4$ electrons.
$\text{NH}_3$ : Contains $3$ $\text{N-H}$ single bonds = $3 \times 2 = 6$ electrons.
$\text{N}_2$ : Contains $1$ $\text{N}\equiv\text{N}$ triple bond = $6$ electrons.
$\text{O}_2$ : Contains $1$ $\text{O=O}$ double bond = $4$ electrons.
$\text{Cl}_2$ : Contains $1$ $\text{Cl-Cl}$ single bond = $2$ electrons.

Comparing the pairs:

Option A: $10$ vs $12$ electrons
Option B: $8$ vs $4$ electrons
Option C: $6$ vs $6$ electrons (identical)
Option D: $4$ vs $2$ electrons

Thus, option C is the correct answer.

Key Takeaways

Covalent bonding involves shared pairs of electrons between non-metal atoms.
Single, double, and triple covalent bonds contain $2$ , $4$ , and $6$ electrons respectively.
Make sure to count only the bonding electrons (shared electrons) and not lone pairs or inner-shell electrons when asked specifically about electrons in covalent bonds.

Common Mistakes

Confusing total valence electrons or total atomic electrons with bonding electrons.
Forgetting that double and triple bonds consist of $4$ and $6$ electrons respectively, rather than $2$ electrons per bond line.

Things to Be Careful About

Ensure all bonds in polyatomic molecules (such as the $\text{O-H}$ and $\text{C-O}$ bonds in methanol) are accounted for.
Read carefully whether the question asks for the number of bonds, the number of pairs of bonding electrons, or the total number of bonding electrons.

Techniques used

count shared pairs of electrons in moleculesdetermine the number of covalent bonding electrons

Q71MEasyStoichiometry

Chrome alum is a salt that contains two different cations and one anion.

The ions present in chrome alum are K⁺, Cr³⁺ and SO₄²⁻.

What is the formula of the salt?

Options

A KCrSO₄
B KCr(SO₄)₂
C KCr(SO₄)₃
D K₂Cr(SO₄)₃

Pick an answer

DifficultyEasy

Worked solution

Working

An ionic compound must be electrically neutral, meaning the total positive charge must equal the total negative charge.

The ions present are $\text{K}^+$ , $\text{Cr}^{3+}$ , and $\text{SO}_4^{2-}$ .
Total positive charge from one $\text{K}^+$ and one $\text{Cr}^{3+}$ :

(+1) + (+3) = +4

To balance a charge of $+4$ , two sulfate ions ( $\text{SO}_4^{2-}$ ) are required:

2 \times (-2) = -4

Combining one $\text{K}^+$ , one $\text{Cr}^{3+}$ , and two $\text{SO}_4^{2-}$ gives the formula $\text{KCr}(\text{SO}_4)_2$ .

Evaluating the options:

A $\text{KCrSO}_4$ : Total charge $= (+1) + (+3) + (-2) = +2 \neq 0$
B $\text{KCr}(\text{SO}_4)_2$ : Total charge $= (+1) + (+3) + 2(-2) = 0$
C $\text{KCr}(\text{SO}_4)_3$ : Total charge $= (+1) + (+3) + 3(-2) = -2 \neq 0$
D $\text{K}_2\text{Cr}(\text{SO}_4)_3$ : Total charge $= 2(+1) + (+3) + 3(-2) = -1 \neq 0$

Answer

B

Final answer

B

Detailed explanation

Walkthrough

To find the correct chemical formula for any ionic compound or salt (including double salts containing two different cations), the overall compound must have zero net electrical charge.

Identify the charges of the constituent ions:
- Potassium ion: $\text{K}^+$ has a charge of $+1$ .
- Chromium(III) ion: $\text{Cr}^{3+}$ has a charge of $+3$ .
- Sulfate ion: $\text{SO}_4^{2-}$ has a charge of $-2$ .
Add the positive charges of the cations:
If the formula contains one potassium ion and one chromium ion, the total positive charge is $(+1) + (+3) = +4$ .
Determine the number of sulfate ions needed to neutralise the $+4$ charge:
Each sulfate ion carries a $-2$ charge. To obtain a total negative charge of $-4$ , two sulfate ions are needed since $2 \times (-2) = -4$ .
Write the formula using brackets around the polyatomic sulfate ion:
$\text{KCr}(\text{SO}_4)_2$

Therefore, option B is the correct answer.

Key Takeaways

In any neutral compound, the sum of positive oxidation states/charges must equal the sum of negative oxidation states/charges (net charge = 0).
When polyatomic ions such as sulfate ( $\text{SO}_4^{2-}$ ) appear more than once in a formula, they must be enclosed in brackets with the subscript outside, e.g., $(\text{SO}_4)_2$ .

Common Mistakes

Forgetting that sulfate has a $-2$ charge and mistakenly treating it as $-1$ .
Incorrectly summing multiple cations with different charges.

Things to Be Careful About

Ensure all subscripts are properly accounted for when calculating net charge for each option.

Techniques used

balance ionic charges for electrical neutralitydeduce formula of double salt from constituent ions

Q81MMedium-EasyAtoms, Elements and CompoundsStoichiometry

The table gives some information about barium and chlorine.

	relative atomic mass	group in Periodic Table
barium	137	II
chlorine	35.5	VII

Using this information only, a student makes three statements.

There is more than one isotope of chlorine.
There may be more than one isotope of barium.
The relative formula mass of barium chloride is 208.

Which statements are correct?

Options

A 1, 2 and 3
B 1 and 3 only
C 2 and 3 only
D 3 only

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Statement 1: The relative atomic mass ( $A_r$ ) of chlorine is given as $35.5$ . Since mass numbers of individual isotopes must be whole numbers (integers), a non-integer $A_r$ proves that chlorine exists as a mixture of more than one isotope. Statement 1 is correct.
Statement 2: The $A_r$ of barium is $137$ . An integer relative atomic mass can arise either because an element has only one stable isotope or because the weighted average of multiple isotopes rounds to/equals a whole number. Therefore, there may be more than one isotope of barium. Statement 2 is correct.
Statement 3: Barium is in Group II, so it forms $\text{Ba}^{2+}$ ions. Chlorine is in Group VII, so it forms $\text{Cl}^-$ ions. The formula of barium chloride is $\text{BaCl}_2$ . $M_r(\text{BaCl}_2) = 137 + (2 \times 35.5) = 137 + 71 = 208$ Statement 3 is correct.

Since statements 1, 2, and 3 are all correct, option A is the correct answer.

Answer

A

Final answer

A

Detailed explanation

Walkthrough

Evaluate Statement 1:
- The relative atomic mass ( $A_r$ ) of an element is the weighted average of the masses of its naturally occurring isotopes relative to $^{12}\text{C}$ .
- Because individual protons and neutrons have mass numbers very close to 1, any individual isotope has an integer mass number (e.g., $^{35}\text{Cl}$ and $^{37}\text{Cl}$ ).
- A fractional $A_r$ such as $35.5$ is only possible if chlorine consists of a mixture of isotopes with different masses. Thus, Statement 1 is definitely correct.
Evaluate Statement 2:
- Barium has an $A_r$ of $137$ . Having an integer relative atomic mass does not rule out the presence of isotopes, because a mixture of isotopes can still have a weighted average that is a whole number (or rounds closely to 137). Therefore, it is correct to say that there may be more than one isotope. Statement 2 is correct.
Evaluate Statement 3:
- From the group numbers in the table:
  - Barium is in Group II $\rightarrow$ loses 2 electrons to form $\text{Ba}^{2+}$ .
  - Chlorine is in Group VII $\rightarrow$ gains 1 electron to form $\text{Cl}^-$ .
- To balance the charges, the formula of barium chloride is $\text{BaCl}_2$ .
- Calculate the relative formula mass ( $M_r$ ): $M_r = 137 + (2 \times 35.5) = 208$ Thus, Statement 3 is correct.

Since all three statements (1, 2 and 3) are correct, the correct option is A.

Key Takeaways

A non-integer relative atomic mass ( $A_r$ ) is direct evidence that an element consists of more than one isotope.
A whole number $A_r$ does not prove an element has only one isotope; it could be a mono-isotopic element or a mixture whose weighted average is an integer.
Group numbers directly give the charges on simple ions (Group II gives $+2$ , Group VII gives $-1$ ), allowing deduction of the correct chemical formula and subsequent $M_r$ calculation.

Common Mistakes

Incorrectly assuming that an integer relative atomic mass means an element cannot have isotopes (which would lead to rejecting statement 2).
Writing the formula of barium chloride incorrectly as $\text{BaCl}$ , leading to $M_r = 137 + 35.5 = 172.5$ instead of $208$ .

Things to Be Careful About

Note the wording in Statement 2: "There may be more than one isotope". "May" denotes possibility, which is scientifically accurate here.

Techniques used

relate non-integer relative atomic mass to the existence of isotopesdeduce ionic formula from periodic group numberscalculate relative formula mass

Q91MMedium-EasyStoichiometry

Which mass of carbon contains the same number of atoms as 16.0 g of sulfur?

Options

A 0.5 g
B 6.0 g
C 8.0 g
D 12.0 g

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Find the relative atomic masses from the Periodic Table:
- $A_r(\text{S}) = 32$
- $A_r(\text{C}) = 12$
Calculate the number of moles of sulfur atoms in $16.0\text{ g}$ :

\text{moles of S} = \frac{\text{mass}}{A_r} = \frac{16.0}{32} = 0.50\text{ mol}

Equal numbers of atoms correspond to equal numbers of moles. Therefore, $0.50\text{ mol}$ of carbon atoms is required:

\text{mass of C} = \text{moles} \times A_r = 0.50 \times 12 = 6.0\text{ g}

Answer

B

Final answer

B

Detailed explanation

Walkthrough

To have the same number of atoms, two samples of different elements must contain the same number of moles of atoms, because one mole of any substance contains Avogadro's number of particles ( $6.02 \times 10^{23}$ particles).

Step 1: Calculate the amount in moles of sulfur atoms present in $16.0\text{ g}$ of sulfur. The relative atomic mass ( $A_r$ ) of sulfur is $32$ .

\text{moles of S} = \frac{16.0\text{ g}}{32\text{ g / mol}} = 0.50\text{ mol}

Step 2: To contain the same number of atoms, we need $0.50\text{ mol}$ of carbon atoms. The relative atomic mass ( $A_r$ ) of carbon is $12$ .

\text{mass of C} = 0.50\text{ mol} \times 12\text{ g / mol} = 6.0\text{ g}

Therefore, option B is the correct answer.

Option A ( $0.5\text{ g}$ ) confuses the number of moles ( $0.50\text{ mol}$ ) with the mass in grams.
Option C ( $8.0\text{ g}$ ) is half the mass of sulfur, which does not take into account the ratio of their relative atomic masses.
Option D ( $12.0\text{ g}$ ) is the mass of $1.0\text{ mol}$ of carbon, which would contain twice as many atoms.

Key Takeaways

The number of particles (atoms, molecules, ions) is directly proportional to the amount in moles ( $n = \frac{\text{mass}}{A_r}$ ).
Equal amounts in moles of different elements contain exactly the same number of atoms.

Common Mistakes

Confusing the calculated number of moles ( $0.50\text{ mol}$ ) with the final mass, leading to selecting A.
Assuming equal masses contain equal numbers of atoms, ignoring the differences in relative atomic masses.

Things to Be Careful About

Always look up or verify the relative atomic masses from the Periodic Table ( $A_r(\text{S}) = 32$ , $A_r(\text{C}) = 12$ ).

Techniques used

convert mass to moles using Arequate number of moles to equal number of atomscalculate mass from moles and Ar

Q101MMedium-EasyStoichiometry

Compound X contains carbon, hydrogen and oxygen only.

It has an $M_r$ of 90.

100 g of compound X contains 40.0 g of carbon and 6.7 g of hydrogen.

How many oxygen atoms are there in each molecule of compound X?

Options

A 1
B 2
C 3
D 4

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Find the mass of oxygen in $100\text{ g}$ of compound X:

\text{mass of O} = 100 - (40.0 + 6.7) = 53.3\text{ g}

Determine the moles of each element in $100\text{ g}$ :

\begin{aligned} \text{moles of C} &= \frac{40.0}{12} = 3.33\text{ mol} \\ \text{moles of H} &= \frac{6.7}{1} = 6.70\text{ mol} \\ \text{moles of O} &= \frac{53.3}{16} = 3.33\text{ mol} \end{aligned}

Divide by the smallest value ( $3.33$ ) to find the simplest whole-number mole ratio:

\text{C} : \text{H} : \text{O} = \frac{3.33}{3.33} : \frac{6.70}{3.33} : \frac{53.3}{3.33} = 1 : 2 : 1

Empirical formula = $\text{CH}_2\text{O}$

Calculate the empirical formula mass ( $M_r$ of $\text{CH}_2\text{O}$ ):

M_r(\text{CH}_2\text{O}) = 12 + 2(1) + 16 = 30

Determine the molecular multiplier:

\text{multiplier} = \frac{90}{30} = 3

Molecular formula = $(\text{CH}_2\text{O})_3 = \text{C}_3\text{H}_6\text{O}_3$

Thus, there are $3$ oxygen atoms in each molecule of compound X.

Answer

C

Final answer

C

Detailed explanation

Walkthrough

Find the mass of oxygen:
The compound contains only carbon, hydrogen, and oxygen. In a $100\text{ g}$ sample:
$\text{mass of oxygen} = 100\text{ g} - 40.0\text{ g} - 6.7\text{ g} = 53.3\text{ g}$
Convert masses to moles:
Using the relative atomic masses ( $A_r(\text{C}) = 12$ , $A_r(\text{H}) = 1$ , $A_r(\text{O}) = 16$ ):
$\begin{aligned} n(\text{C}) &= \frac{40.0}{12} = 3.333\text{ mol} \\ n(\text{H}) &= \frac{6.7}{1} = 6.700\text{ mol} \\ n(\text{O}) &= \frac{53.3}{16} = 3.331\text{ mol} \end{aligned}$
Find the empirical formula:
Divide each mole value by the smallest number of moles ( $3.331$ ):
- $\text{C} = \frac{3.333}{3.331} = 1$
- $\text{H} = \frac{6.700}{3.331} \approx 2$
- $\text{O} = \frac{3.331}{3.331} = 1$
The empirical formula is $\text{CH}_2\text{O}$ .
Find the molecular formula:
The formula mass of the empirical unit $\text{CH}_2\text{O}$ is:
$12 + (2 \times 1) + 16 = 30$
The given relative molecular mass ( $M_r$ ) is $90$ .
$\text{Number of empirical units} = \frac{90}{30} = 3$
Therefore, the molecular formula is $(\text{CH}_2\text{O})_3 = \text{C}_3\text{H}_6\text{O}_3$ .
Each molecule contains $3$ oxygen atoms, corresponding to option C.

Key Takeaways

For a substance composed of three elements where percentages or masses are given for only two, find the third by subtracting from the total ( $100\text{ g}$ or $100\%$ ).
To find the empirical formula: convert mass $\rightarrow$ moles using $A_r$ $\rightarrow$ divide by the smallest value to get the simplest integer ratio.
The molecular formula is found by scaling the empirical formula by $\frac{M_r}{\text{empirical mass}}$ .

Common Mistakes

Forgetting to find the mass of oxygen and attempting to find a ratio using only carbon and hydrogen.
Dividing the mass by atomic number instead of relative atomic mass ( $A_r$ ).
Forgetting to convert the empirical formula to the molecular formula using the $M_r$ of 90 (which leads to choosing 1 oxygen atom, option A).

Things to Be Careful About

Always use the atomic mass of oxygen ( $16$ ), not molecular oxygen ( $ext{O}_2 = 32$ ), when calculating the mole ratio of atoms in a chemical formula.

Techniques used

calculate mass of an element by differencedetermine empirical formula from mass compositiondeduce molecular formula using relative molecular mass

Q111MMedium-EasyElectrochemistry

Aqueous copper(II) sulfate is electrolysed using copper electrodes.

Which row describes the changes that take place during the electrolysis?

Options

	mass of anode	mass of cathode	colour of solution
A	increases	decreases	becomes paler
B	increases	decreases	stays the same
C	decreases	increases	becomes paler
D	decreases	increases	stays the same

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

During the electrolysis of aqueous copper(II) sulfate, $\text{CuSO}_4(\text{aq})$ , using copper electrodes:

At the anode (positive electrode):
Copper atoms lose electrons and dissolve as copper(II) ions into solution:
$\text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^-$
Therefore, the mass of the anode decreases.
At the cathode (negative electrode):
Copper(II) ions from the solution gain electrons and are deposited as solid copper:
$\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})$
Therefore, the mass of the cathode increases.
Colour of the solution:
For every $\text{Cu}^{2+}$ ion discharged at the cathode, a $\text{Cu}^{2+}$ ion is formed at the anode. The concentration of $\text{Cu}^{2+}$ ions in the solution remains constant, so the blue colour of the solution stays the same.

Matching these deductions with the options:

Mass of anode: decreases
Mass of cathode: increases
Colour of solution: stays the same

This corresponds to row D.

Answer

D

Final answer

D

Detailed explanation

Walkthrough

When aqueous copper(II) sulfate is electrolysed, the nature of the electrodes determines what happens:

With inert electrodes (such as carbon/graphite or platinum):
- Cathode: $\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})$ (copper deposits; cathode mass increases).
- Anode: $4\text{OH}^-(\text{aq}) \rightarrow \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^-$ (oxygen gas bubbles off).
- In this case, $\text{Cu}^{2+}$ ions are removed from the electrolyte without being replaced, so the blue colour of the solution gradually becomes paler.
With active copper electrodes (as in this question):
- Anode: The copper metal of the anode is oxidised more easily than $\text{OH}^-$ or $\text{SO}_4^{2-}$ ions. Copper dissolves: $\text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^-$ . Thus, the mass of the anode decreases.
- Cathode: Copper(II) ions in the electrolyte are preferentially discharged over $\text{H}^+$ ions because copper is lower in the reactivity series: $\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})$ . Thus, the mass of the cathode increases.
- Electrolyte: The rate at which $\text{Cu}^{2+}$ ions enter the solution at the anode equals the rate at which $\text{Cu}^{2+}$ ions leave the solution at the cathode. As a result, the concentration of $\text{Cu}^{2+}(\text{aq})$ ions remains unchanged, and the blue colour of the solution stays the same.

Therefore, row D is the correct choice.

Key Takeaways

Electrolysis with active (copper) electrodes results in the transfer of copper from the anode to the cathode.
The anode decreases in mass (oxidation), while the cathode increases in mass (reduction).
The overall concentration of copper(II) ions in the electrolyte does not change, so the solution maintains its blue colour.

Common Mistakes

Confusing inert electrodes with copper electrodes: students often recall that the blue colour fades, which is only true when inert electrodes (e.g. graphite or platinum) are used.
Mixing up the anode and cathode mass changes: remember that the anode dissolves (loses mass) and the cathode receives the deposited metal (gains mass).

Things to Be Careful About

Always read the question carefully to identify whether the electrodes are inert or active (copper).
At the anode, oxidation occurs (loss of electrons); at the cathode, reduction occurs (gain of electrons).

Techniques used

identify electrode reactions with active copper electrodesdeduce mass changes at the anode and cathoderelate electrolyte concentration to solution colour

Q121MEasyElectrochemistry

Which row describes one advantage and one disadvantage of using a hydrogen-oxygen fuel cell to power a road vehicle?

Options

	advantage	disadvantage
A	The fuel cell obtains the oxygen from the air.	The hydrogen has to be stored in a very strong tank.
B	The fuel cell obtains the oxygen from the air.	The only chemical product causes acid rain.
C	The fuel cell obtains the oxygen from water.	The hydrogen has to be stored in a very strong tank.
D	The fuel cell obtains the oxygen from water.	The only chemical product causes acid rain.

Pick an answer

DifficultyEasy

Worked solution

Working

Advantage: In a hydrogen-oxygen fuel cell, oxygen gas ( $\text{O}_2$ ) is taken directly from the ambient air, so only hydrogen fuel needs to be carried on board.
Disadvantage: Hydrogen is a highly flammable gas with a low density, so it must be stored under high pressure in very strong, reinforced fuel tanks to ensure safety.
Evaluating distractors:
- Oxygen is not extracted from water during the cell's operation (eliminating C and D).
- The only chemical product formed by the overall reaction ( $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ ) is water, which is non-polluting and does not cause acid rain (eliminating B and D).

Therefore, row A is correct.

Answer

A

Final answer

A

Detailed explanation

Walkthrough

A hydrogen-oxygen fuel cell generates electricity by combining hydrogen and oxygen chemically:

2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{l})

Reactant source: The cell uses hydrogen gas from a storage tank and draws oxygen directly from the surrounding air. This is an advantage because oxygen does not need to be carried in a separate tank.
Storage considerations: Because hydrogen has a very low density and is a flammable gas, storing enough mass of hydrogen to give a vehicle a reasonable driving range requires compressing it to high pressures (typically 350 to 700 bar). Consequently, fuel tanks must be exceptionally strong and crash-resistant, which is a major engineering challenge and disadvantage.
Product safety: The only product released during operation is pure water ( $\text{H}_2\text{O}$ ), which is harmless and produces zero pollutant emissions (unlike fossil fuels which produce $\text{CO}_2$ , $\text{CO}$ , $\text{SO}_2$ , and $\text{NO}_x$ ).

Thus, option A correctly identifies a genuine advantage and disadvantage.

Key Takeaways

In hydrogen-oxygen fuel cells, oxygen is supplied from the air, and hydrogen is stored on board.
The only chemical product of the fuel cell reaction is water ( $\text{H}_2\text{O}$ ).
Main disadvantages of hydrogen as a vehicle fuel include high-pressure storage requirements, flammability, and the current lack of refueling infrastructure.

Common Mistakes

Confusing the fuel cell with the electrolysis of water: the fuel cell uses hydrogen and oxygen to produce water and electricity; it does not take oxygen from water.
Believing fuel cells produce acidic emissions: the only product is water.

Things to Be Careful About

Ensure you distinguish between the overall product of the fuel cell (harmless water) and emissions associated with industrial hydrogen production (which may involve fossil fuels). The fuel cell itself produces only water.

Techniques used

evaluate advantages and disadvantages of hydrogen-oxygen fuel cellsidentify the reactants and products of a fuel cell

Q131MEasyChemical Energetics

Which statement about exothermic and endothermic reactions is correct?

Options

A In an endothermic reaction, energy is used to break bonds but no energy is released when bonds form.
B In an endothermic reaction, energy is released when bonds form but more energy is used to break bonds.
C In an exothermic reaction, energy is released both by breaking and by forming bonds.
D In an exothermic reaction, energy is released when bonds form but no energy is needed to break bonds.

Pick an answer

DifficultyEasy

Worked solution

Working

Bond breaking is always an endothermic process (energy is absorbed/used to break bonds).
Bond forming is always an exothermic process (energy is released when bonds form).
In an endothermic reaction, the energy taken in to break existing bonds in the reactants is greater than the energy released when new bonds form in the products.
Evaluating the options:
- A is incorrect because energy is always released when new bonds form.
- B is correct because energy is released during bond formation, but the energy required to break the bonds is greater, resulting in a net absorption of energy.
- C is incorrect because bond breaking requires an input of energy rather than releasing it.
- D is incorrect because energy is always required to break initial bonds.

Answer

B

Final answer

B

Detailed explanation

Walkthrough

Chemical reactions involve two energetic stages:

Bond Breaking: Energy must be supplied to separate bonded atoms in the reactant molecules. Hence, bond breaking is always an endothermic process.
Bond Making: Energy is released when new chemical bonds are established to form the product molecules. Hence, bond forming is always an exothermic process.

The overall energy change of a reaction ( $\Delta H$ ) is given by:

\Delta H = \text{energy absorbed to break bonds} - \text{energy released to form bonds}

In an endothermic reaction, $\Delta H > 0$ . This occurs when the energy absorbed during bond breaking is greater than the energy released during bond forming.
In an exothermic reaction, $\Delta H < 0$ . This occurs when the energy released during bond forming is greater than the energy absorbed during bond breaking.

Option B accurately describes an endothermic reaction: energy is released when new bonds form, but more energy is absorbed to break the initial bonds, yielding a net intake of thermal energy from the surroundings.

Key Takeaways

Bond breaking is endothermic (takes in energy; "MEXO BENDO" / B-IN, M-OUT).
Bond making is exothermic (releases energy).
A reaction is endothermic when $\text{energy taken in to break bonds} > \text{energy given out to make bonds}$ .
A reaction is exothermic when $\text{energy given out to make bonds} > \text{energy taken in to break bonds}$ .

Common Mistakes

Believing that endothermic reactions only involve bond breaking, or that exothermic reactions only involve bond making.
Confusing which process (breaking vs forming) absorbs energy and which releases energy.

Things to Be Careful About

Every chemical reaction (with negligible exceptions like radical combinations) involves both breaking of existing bonds and forming of new bonds. The overall type (exothermic or endothermic) is determined solely by the balance between the two energy quantities.

Techniques used

relate bond breaking and bond forming to energy changesdeduce overall enthalpy change from relative energy absorbed and released

Q141MEasyChemical Reactions

Two gases react together to produce a single product. The rate of the reaction is affected by an increase in pressure. The reaction is catalysed by platinum.

Which row describes one effect on this reaction of increasing pressure and one effect on this reaction of adding platinum?

Options

	one effect of increasing pressure	one effect of adding platinum
A	The rate of the reaction decreases.	The activation energy, $E_a$ , decreases.
B	The rate of the reaction decreases.	The activation energy, $E_a$ , increases.
C	The rate of the reaction increases.	The activation energy, $E_a$ , decreases.
D	The rate of the reaction increases.	The activation energy, $E_a$ , increases.

Pick an answer

DifficultyEasy

Worked solution

Working

Effect of increasing pressure: Increasing the pressure of a gaseous reaction pushes the particles closer together, increasing the number of particles per unit volume. This increases the frequency of collisions between reactant particles, thereby increasing the rate of reaction.
Effect of adding a catalyst (platinum): A catalyst provides an alternative reaction pathway with a lower activation energy ( $E_a$ ). Therefore, adding platinum decreases the activation energy.

Matching these two effects gives row C.

Answer

C

Final answer

C

Detailed explanation

Walkthrough

To determine the correct row, consider each factor separately:

Increasing Pressure in a Gaseous Reaction:
- When pressure is increased on a system containing reacting gases, the gas particles are forced closer together (concentration of gas particles increases).
- As a result, the frequency of successful collisions between particles increases per unit time.
- Therefore, the overall rate of the reaction increases (eliminating options A and B).
Adding a Catalyst (Platinum):
- A catalyst speeds up a chemical reaction without being permanently consumed.
- It works by providing an alternative reaction pathway with a lower activation energy ( $E_a$ ).
- Therefore, adding platinum decreases the activation energy (eliminating options B and D).

Combining these two correct deductions points directly to row C.

Key Takeaways

Increasing the pressure of gaseous reactants increases collision frequency, which increases the rate of reaction.
A catalyst provides an alternative pathway that lowers the activation energy ( $E_a$ ), allowing a greater fraction of collisions to have energy greater than or equal to $E_a$ .

Common Mistakes

Confusing the effect of a catalyst on activation energy: catalysts lower the activation energy, they do not increase it.
Confusing rate with reaction time: an increase in rate means the reaction occurs faster (takes less time).

Things to Be Careful About

Ensure you identify the required direction of change for both columns before selecting the final option.

Techniques used

apply collision theory to the effect of pressure on rateexplain the effect of a catalyst on activation energy

Q151MMedium-EasyChemical Reactions

Calcium carbonate reacts with excess dilute hydrochloric acid to form carbon dioxide.

The reaction is investigated in two experiments.

The rate of the reactions is compared by measuring the volume of carbon dioxide formed over time in each experiment. The two rates are compared by plotting graphs.

Fig. 1

Which statement about experiments 1 and 2 is correct?

Options

A Experiments 1 and 2 both slow down as the reaction proceeds.
B Experiments 1 and 2 must both use acid of the same concentration.
C Experiments 1 and 2 must have been done at the same temperature.
D Experiment 2 uses larger lumps of calcium carbonate. All other conditions stay the same.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

The gradient (slope) of a volume-time graph represents the rate of reaction. For both experiments 1 and 2, the gradient is steepest at $t = 0$ and continuously decreases over time until the curve becomes horizontal (gradient $= 0$ ), showing that both reactions slow down as the reactants are consumed. Thus, statement A is correct.
Experiment 2 has a steeper initial gradient than experiment 1, meaning experiment 2 has a faster rate. A higher rate could be caused by higher acid concentration, higher temperature, or smaller particle size (powder rather than lumps) of $\text{CaCO}_3$ , so neither concentration nor temperature must be identical (B and C are incorrect).
Larger lumps have a smaller surface area per unit volume, which would decrease the rate of reaction and produce a less steep curve, whereas curve 2 is steeper (D is incorrect).

Answer

A

Final answer

A

Detailed explanation

Walkthrough

Understanding the graph:
- The graph plots the volume of carbon dioxide gas produced against time.
- The gradient (slope) of the tangent to the curve at any point represents the rate of reaction at that instant.
- At the beginning ( $t = 0$ ), the curves are at their steepest, meaning the reaction is fastest because the concentration of acid and the available surface area of calcium carbonate are at their highest.
- As the reaction proceeds, reactants are used up. This lowers the frequency of effective collisions between particles, causing the reaction to slow down (the gradient decreases).
- Eventually, when the limiting reactant is completely used up, no more gas is produced, and the curve levels off to a horizontal plateau (rate $= 0$ ).
- Because both curves get progressively less steep until they flatten, both experiments slow down as the reaction proceeds. Therefore, A is correct.
Eliminating the distractors:
- B and C: Experiment 2 has a steeper initial slope than experiment 1, which indicates a faster initial rate. This faster rate could be achieved by using a higher temperature, a higher concentration of hydrochloric acid, or smaller pieces (powder) of calcium carbonate. Hence, it is not mandatory for temperature or concentration to be the same.
- D: If larger lumps were used in experiment 2, the total surface area would decrease. A smaller surface area leads to fewer collisions per unit time and a slower rate (a less steep curve). Experiment 2 is faster, so it cannot be using larger lumps if all other variables remain unchanged.

Key Takeaways

On a volume-against-time graph for a gas-producing reaction, the gradient represents the rate of reaction.
A decreasing gradient over time shows that the reaction slows down as reactants are consumed.
A steeper initial curve indicates a faster reaction rate, which can be caused by higher temperature, higher concentration, or smaller particle size (greater surface area).
When the total amount of limiting reactant is unchanged, the curves will reach the same horizontal plateau (final volume of gas).

Common Mistakes

Confusing the rate of reaction (steepness of the curve) with the total amount of product formed (height of the plateau).
Thinking larger lumps increase the rate of reaction; larger pieces have a smaller surface area-to-volume ratio, which slows down the reaction.

Things to Be Careful About

Always check both the slope (rate) and the final horizontal line (yield/stoichiometry) when interpreting reaction rate curves.

Techniques used

interpret rate graphs using gradientrelate reactant consumption to reaction rateevaluate factors affecting the rate of reaction

Q161MEasyChemical Reactions

If calcium carbonate is heated in a closed container, it will decompose, forming calcium oxide and carbon dioxide. Calcium oxide and carbon dioxide can recombine to form calcium carbonate.

After some time, a position is reached where calcium carbonate is decomposing, and calcium oxide and carbon dioxide are recombining at the same rate.

What is this position called?

Options

A activation energy
B backward reaction
C equilibrium
D neutralisation

Pick an answer

Q171MEasyChemical Reactions

Which set of conditions is used in the Contact process?

Options

	temperature / $^\circ\text{C}$	pressure / $\text{atm}$	catalyst
A	100	200	V₂O₅
B	300	200	Fe
C	450	2	Fe
D	450	2	V₂O₅

Pick an answer

DifficultyEasy

Worked solution

Working

The Contact process is used for the manufacture of sulfuric acid. In the key stage, sulfur dioxide is oxidised to sulfur trioxide:

2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})

The optimum conditions used in industry are:

Temperature: $450\ ^\circ\text{C}$
Pressure: $1\text{ to }2\text{ atm}$ (a low/moderate pressure is sufficient as the equilibrium yield is already very high)
Catalyst: Vanadium(V) oxide, $\text{V}_2\text{O}_5$

Matching these conditions to the options gives row D.

A is incorrect because the temperature is far too low and the pressure is unnecessarily high.
B and C are incorrect because iron ( $\text{Fe}$ ) is the catalyst for the Haber process, not the Contact process.

Answer

D

Final answer

D

Detailed explanation

Walkthrough

The Contact process involves the catalytic oxidation of sulfur dioxide ( $\text{SO}_2$ ) to sulfur trioxide ( $\text{SO}_3$ ):

2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \quad \Delta H = -197\text{ kJ / mol}

Temperature ( $450\ ^\circ\text{C}$ ): The forward reaction is exothermic. A lower temperature favors a higher equilibrium yield of $\text{SO}_3$ , but causes the reaction rate to be impractically slow. Therefore, a compromise temperature of $450\ ^\circ\text{C}$ is chosen to balance yield and rate.
Pressure ( $1\text{--}2\text{ atm}$ ): There are 3 moles of gas on the left and 2 moles on the right. Increasing pressure shifts equilibrium to the right, but because the conversion is already around $98\%$ at atmospheric/low pressure, high-pressure equipment is economically unnecessary. A low pressure of $1\text{--}2\text{ atm}$ is used mainly to push gases through the reactor.
Catalyst ( $\text{V}_2\text{O}_5$ ): Vanadium(V) oxide speeds up the reaction to achieve equilibrium rapidly at this moderate temperature without affecting the equilibrium position.

Therefore, the correct row is D.

Key Takeaways

Contact process: $450\ ^\circ\text{C}$ , $1\text{--}2\text{ atm}$ , $\text{V}_2\text{O}_5$ catalyst.
Haber process: $450\ ^\circ\text{C}$ , $200\text{ atm}$ , $\text{Fe}$ catalyst.

Common Mistakes

Confusing the conditions of the Contact process with the Haber process (e.g., using $\text{Fe}$ catalyst or $200\text{ atm}$ pressure).

Things to Be Careful About

Ensure you recognise $\text{V}_2\text{O}_5$ as vanadium(V) oxide and associate it specifically with the oxidation of sulfur dioxide.

Techniques used

recall industrial conditions for the Contact processidentify the catalyst and operating parameters

Q181MMedium-EasyChemical Reactions

Copper forms a red oxide, Cu₂O, and a black oxide, CuO.

In the presence of a catalyst, aqueous hydrogen peroxide, H₂O₂, decomposes to form water.

The black oxide has copper in a ......1...... oxidation state than in the red oxide.

In forming water, hydrogen peroxide is ......2...... .

Which words correctly complete gaps 1 and 2?

Options

	1	2
A	higher	reduced
B	higher	oxidised
C	lower	reduced
D	lower	oxidised

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Determine the oxidation state of copper in each oxide:
- In $\text{CuO}$ (black oxide), oxygen has an oxidation number of $-2$ , so the oxidation state of copper is $+2$ .
- In $\text{Cu}_2\text{O}$ (red oxide), oxygen has an oxidation number of $-2$ , so $2 \times (\text{oxidation state of Cu}) = +2$ , meaning copper has an oxidation state of $+1$ .
- Therefore, the black oxide has copper in a higher oxidation state than in the red oxide.
Determine whether $\text{H}_2\text{O}_2$ is oxidised or reduced when forming $\text{H}_2\text{O}$ :
- The decomposition reaction is:

2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})

In $\text{H}_2\text{O}_2$ , hydrogen is $+1$ and oxygen has an oxidation number of $-1$ .
In $\text{H}_2\text{O}$ , hydrogen is $+1$ and oxygen has an oxidation number of $-2$ .
The decrease in oxidation number of oxygen from $-1$ to $-2$ means hydrogen peroxide is reduced in forming water.

This matches row A (1: higher, 2: reduced).

Answer

A

Final answer

A

Detailed explanation

Walkthrough

Gap 1 (Oxidation state of copper):
- In metal oxides, oxygen consistently has an oxidation state of $-2$ .
- In $\text{CuO}$ , the neutral compound requires the copper ion to balance the $-2$ charge, giving $\text{Cu}^{2+}$ (oxidation state $+2$ ).
- In $\text{Cu}_2\text{O}$ , two copper ions balance the $-2$ charge, giving $\text{Cu}^+$ (oxidation state $+1$ ).
- Since $+2 > +1$ , the black oxide (copper(II) oxide) has copper in a higher oxidation state than the red oxide (copper(I) oxide).
Gap 2 (Redox change of $\text{H}_2\text{O}_2$ to $\text{H}_2\text{O}$ ):
- Hydrogen peroxide, $\text{H}_2\text{O}_2$ , contains a peroxide bond where oxygen has an unusual oxidation state of $-1$ .
- When $\text{H}_2\text{O}_2$ converts into $\text{H}_2\text{O}$ , the oxidation state of oxygen decreases from $-1$ to $-2$ .
- A decrease in oxidation state corresponds to a gain of electrons, which is reduction.
- (Alternatively, the conversion of $\text{H}_2\text{O}_2$ to $\text{H}_2\text{O}$ involves the loss of oxygen, which also defines reduction).

Combining both conclusions gives higher for gap 1 and reduced for gap 2, which corresponds to option A.

Key Takeaways

For neutral compounds, the sum of all oxidation numbers equals zero.
Group 16 elements (like oxygen) typically have an oxidation state of $-2$ , except in peroxides where oxygen is $-1$ .
Reduction is defined as a decrease in oxidation state, a gain of electrons, or a loss of oxygen.
Oxidation is defined as an increase in oxidation state, a loss of electrons, or a gain of oxygen.

Common Mistakes

Forgetting that oxygen in peroxides has an oxidation state of $-1$ rather than $-2$ .
Confusing the formulae of copper(I) oxide ( $\text{Cu}_2\text{O}$ ) and copper(II) oxide ( $\text{CuO}$ ).
Looking at the formation of $\text{O}_2$ (where oxygen goes from $-1$ to $0$ , which is oxidation) instead of the specified formation of $\text{H}_2\text{O}$ .

Things to Be Careful About

The question specifically asks about what happens to hydrogen peroxide in forming water, not the other product (oxygen gas). In the disproportionation of $\text{H}_2\text{O}_2$ , one oxygen atom is reduced to form $\text{H}_2\text{O}$ and another is oxidised to form $\text{O}_2$ .

Techniques used

determine oxidation states of transition metals in compoundsidentify reduction using changes in oxidation state

Q191MEasyOrganic Chemistry

An organic compound, X, has a molecular formula C₄H₈O₂ and turns damp blue litmus paper red.

What is the displayed formula of X?

Options

Pick an answer

DifficultyEasy

Worked solution

Working

Turning damp blue litmus paper red indicates that compound $\text{X}$ is acidic.
Among organic compounds with the general formula $\text{C}_n\text{H}_{2n}\text{O}_2$ , carboxylic acids are acidic, whereas esters are neutral.
Therefore, compound $\text{X}$ must be a carboxylic acid containing the $-\text{COOH}$ functional group.
Structures A, B, and C are all esters containing the ester linkage ( $-\text{COO}-$ $- COO -$ ):
- A is methyl propanoate.
- B is ethyl ethanoate.
- C is propyl methanoate.
Structure D contains the carboxylic acid group ( $-\text{COOH}$ ) and has 4 carbon atoms, representing butanoic acid ( $\text{C}_4\text{H}_8\text{O}_2$ ).

Answer

D

Final answer

D

Detailed explanation

Walkthrough

An organic compound with the molecular formula $\text{C}_4\text{H}_8\text{O}_2$ has two oxygen atoms and one degree of unsaturation (a double bond or a ring). In O Level Chemistry, compounds with the general formula $\text{C}_n\text{H}_{2n}\text{O}_2$ are either carboxylic acids or esters.

The key chemical clue is that compound $\text{X}$ turns damp blue litmus paper red. This test indicates an acidic substance containing hydrogen ions ( $\text{H}^+$ ) in aqueous solution:

Carboxylic acids contain the carboxyl functional group ( $-\text{COOH}$ ) and weakly ionise in water to release $\text{H}^+(\text{aq})$ , turning blue litmus paper red.
Esters contain the ester linkage ( $-\text{COO}-$ ) within a carbon chain and are neutral, so they have no effect on blue litmus paper.

Evaluating the displayed formulae:

A is methyl propanoate, an ester (neutral).
B is ethyl ethanoate, an ester (neutral).
C is propyl methanoate, an ester (neutral).
D is butanoic acid, a four-carbon carboxylic acid featuring a terminal $-\text{C}(=\text{O})\text{OH}$ group, which is acidic and turns damp blue litmus red.

Thus, option D is correct.

Key Takeaways

Carboxylic acids have the functional group $-\text{COOH}$ and show acidic properties, such as turning damp blue litmus paper red and reacting with carbonates to produce carbon dioxide.
Esters and carboxylic acids with the same number of carbon atoms are structural isomers (both share the general molecular formula $\text{C}_n\text{H}_{2n}\text{O}_2$ ), but they have distinctly different functional groups and chemical properties.

Common Mistakes

Confusing the displayed formula of an ester ( $-\text{C}(=\text{O})-\text{O}-\text{C}-$ ) with that of a carboxylic acid ( $-\text{C}(=\text{O})-\text{O}-\text{H}$ ).
Forgetting that litmus paper must be damp to test gases or non-aqueous liquids for acidity.

Things to Be Careful About

Always inspect where the oxygen atoms and the hydrogen atom are attached: a carboxylic acid must have an $-\text{OH}$ group directly bonded to a carbonyl carbon ( $\text{C}=\text{O}$ ).

Techniques used

deduce functional group from litmus testidentify carboxylic acid structure from displayed formulae

Q201MMedium-EasyAcids, Bases and Salts

Which statement about oxides is correct?

Options

A All acidic oxides are gases at room temperature.
B All basic oxides dissolve in water to give alkalis.
C Amphoteric oxides react with acids, alkalis and water.
D Potassium oxide is a basic oxide.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

A is incorrect: Some acidic oxides are solids at room temperature (e.g. silicon dioxide, $\text{SiO}_2$ , and phosphorus(V) oxide, $\text{P}_4\text{O}_{10}$ ).
B is incorrect: Most basic oxides are insoluble in water (e.g. copper(II) oxide, $\text{CuO}$ , and iron(III) oxide, $\text{Fe}_2\text{O}_3$ ). Only soluble metal oxides (such as Group I oxides and calcium oxide) dissolve in water to form alkalis.
C is incorrect: Amphoteric oxides (e.g. $\text{Al}_2\text{O}_3$ and $\text{ZnO}$ ) react with both acids and alkalis to form salts, but they are insoluble in water and do not react with water.
D is correct: Potassium is a reactive metal, and its oxide, $\text{K}_2\text{O}$ , is a metal oxide that reacts with acids to form a salt and water; hence, it is a basic oxide.

Answer

D

Final answer

D

Detailed explanation

Walkthrough

Oxides are classified based on their acid-base characteristics:

Basic oxides: Oxides of metals (e.g. $\text{K}_2\text{O}$ , $\text{Na}_2\text{O}$ , $\text{CaO}$ , $\text{CuO}$ , $\text{MgO}$ ). They react with acids to form a salt and water. Most basic oxides are insoluble in water; only those of Group I metals and some Group II metals dissolve in/react with water to form alkaline solutions (metal hydroxides). Since potassium is a Group I metal, potassium oxide ( $\text{K}_2\text{O}$ ) is a basic oxide. Thus, D is correct.
Acidic oxides: Oxides of non-metals (e.g. $\text{CO}_2$ , $\text{SO}_2$ , $\text{NO}_2$ , $\text{SiO}_2$ , $\text{P}_4\text{O}_{10}$ ). While many are gases at room temperature, some are solids, such as $\text{SiO}_2$ (sand/quartz) and $\text{P}_4\text{O}_{10}$ , so statement A is false.
Amphoteric oxides: Oxides of certain metals (notably $\text{Al}_2\text{O}_3$ , $\text{ZnO}$ , $\text{PbO}$ ) that show both basic and acidic properties. They react with acids and with bases/alkalis to form salts and water, but they are insoluble in water and do not react with water, making statement C false.
Neutral oxides: Non-metal oxides (e.g. $\text{CO}$ , $\text{NO}$ , $\text{N}_2\text{O}$ ) that react with neither acids nor bases.

Key Takeaways

Metal oxides are generally basic (or amphoteric).
Non-metal oxides are generally acidic (or neutral).
An alkali is specifically a soluble base; not all basic oxides are soluble in water.
Amphoteric oxides (e.g. $\text{Al}_2\text{O}_3$ , $\text{ZnO}$ ) react with both acids and alkalis, but not with water.

Common Mistakes

Confusing the terms base and alkali: thinking all bases or basic oxides dissolve in water to give alkalis. Most transition metal oxides are insoluble bases.
Assuming all acidic oxides are molecular gases, forgetting giant covalent oxides like $\text{SiO}_2$ or solid non-metal oxides like $\text{P}_4\text{O}_{10}$ .

Things to Be Careful About

Check absolute words such as "all" in options A and B, which are frequently false due to well-known exceptions in chemistry (e.g. insoluble metal oxides, solid acidic oxides).

Techniques used

classify metal and non-metal oxidesrecall solubility and reactions of basic and amphoteric oxides

Q211MMediumAcids, Bases and Salts

A student is provided with suitable apparatus, distilled water and the following reagents.

\begin{gathered} \text{solid magnesium hydroxide} \qquad \text{solid lead carbonate} \\ \text{dilute nitric acid} \qquad \text{aqueous sodium chloride} \qquad \text{aqueous sodium sulfate} \end{gathered}

Which salts can the student prepare as a pure dry sample?

Options

A lead nitrate, magnesium nitrate, lead chloride, lead sulfate and magnesium sulfate
B lead nitrate, magnesium nitrate, lead chloride and lead sulfate only
C lead nitrate, magnesium nitrate and lead chloride only
D lead nitrate and magnesium nitrate only

Pick an answer

DifficultyMedium

Worked solution

Working

Preparation of soluble salts using an insoluble base/carbonate and dilute acid:
- $\text{Mg(NO}_3)_2$ (magnesium nitrate): Add excess solid $\text{Mg(OH)}_2$ to dilute $\text{HNO}_3$ , filter off unreacted solid, and crystallise the filtrate to obtain pure, dry magnesium nitrate.
- $\text{Pb(NO}_3)_2$ (lead nitrate): Add excess solid $\text{PbCO}_3$ to dilute $\text{HNO}_3$ , filter off unreacted solid, and crystallise the filtrate to obtain pure, dry lead nitrate.
Preparation of insoluble salts by precipitation:
- Once $\text{Pb(NO}_3)_2\text{(aq)}$ $Pb(NO_{3})_{2} (aq)$ is prepared:
  - Mix $\text{Pb(NO}_3)_2\text{(aq)}$ with $\text{NaCl(aq)}$ to precipitate insoluble lead(II) chloride, $\text{PbCl}_2$ . Filter, wash the residue with distilled water, and dry.
  - Mix $\text{Pb(NO}_3)_2\text{(aq)}$ with $\text{Na}_2\text{SO}_4\text{(aq)}$ to precipitate insoluble lead(II) sulfate, $\text{PbSO}_4$ . Filter, wash the residue with distilled water, and dry.
Magnesium sulfate:
- $\text{MgSO}_4$ is a soluble salt. Mixing $\text{Mg(NO}_3)_2\text{(aq)}$ and $\text{Na}_2\text{SO}_4\text{(aq)}$ does not form a precipitate, and the resulting mixture contains spectator ions ( $ext{Na}^+$ , $ext{NO}_3^-$ ) that cannot be easily separated by simple filtration to give a pure sample of $\text{MgSO}_4$ . Without dilute sulfuric acid, pure $\text{MgSO}_4$ cannot be prepared.

Therefore, the salts that can be prepared as pure dry samples are lead nitrate, magnesium nitrate, lead chloride, and lead sulfate only.

Answer

B

Final answer

B

Detailed explanation

Walkthrough

To determine which salts can be prepared as a pure, dry sample, we look at the methods of salt preparation available from the given reagents:

Soluble salts via excess insoluble reactant:
- $\text{Mg(OH)}_2\text{(s)} + 2\text{HNO}_3\text{(aq)} \rightarrow \text{Mg(NO}_3)_2\text{(aq)} + 2\text{H}_2\text{O(l)}$
  Excess solid magnesium hydroxide is added to dilute nitric acid, the excess is filtered off, and the filtrate is heated to saturation and cooled to form pure crystals of magnesium nitrate, $\text{Mg(NO}_3)_2$ .
- $\text{PbCO}_3\text{(s)} + 2\text{HNO}_3\text{(aq)} \rightarrow \text{Pb(NO}_3)_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$
  Similarly, excess lead(II) carbonate reacts with nitric acid to produce aqueous lead(II) nitrate, $\text{Pb(NO}_3)_2$ , which can be crystallised.
Insoluble salts via precipitation:
- Insoluble salts are made by mixing two aqueous solutions containing the required ions:
  - $\text{Pb}^{2+}\text{(aq)} + 2\text{Cl}^-\text{(aq)} \rightarrow \text{PbCl}_2\text{(s)}$
    Adding aqueous sodium chloride to aqueous lead(II) nitrate gives a white precipitate of $\text{PbCl}_2$ . This is collected by filtration, washed with distilled water, and dried.
  - $\text{Pb}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{PbSO}_4\text{(s)}$
    Adding aqueous sodium sulfate to aqueous lead(II) nitrate produces insoluble $\text{PbSO}_4$ , which is also purified by filtering, washing, and drying.
Why magnesium sulfate cannot be prepared pure:
- $\text{MgSO}_4$ is soluble in water. Mixing aqueous magnesium nitrate with aqueous sodium sulfate results in a solution containing $\text{Mg}^{2+}$ , $\text{NO}_3^-$ , $\text{Na}^+$ , and $\text{SO}_4^{2-}$ ions with no precipitation. Evaporating this solution would yield a mixture of salts, not a pure sample of $\text{MgSO}_4$ .

Thus, only four salts can be prepared as pure dry samples: lead nitrate, magnesium nitrate, lead chloride, and lead sulfate.

Key Takeaways

Soluble salts (like nitrates) are prepared by reacting an acid with an excess of an insoluble base or carbonate, filtering, and crystallising.
Insoluble salts (like $\text{PbCl}_2$ and $\text{PbSO}_4$ ) are prepared by precipitation (mixing two soluble solutions), filtering, washing, and drying the residue.
Soluble salts cannot be made purely by mixing two salt solutions unless an insoluble product forms to separate out.

Common Mistakes

Forgetting that $\text{PbCl}_2$ and $\text{PbSO}_4$ are insoluble and can be made by precipitation once $\text{Pb(NO}_3)_2$ is formed.
Assuming $\text{MgSO}_4$ can be prepared pure by mixing $\text{Mg(NO}_3)_2\text{(aq)}$ and $\text{Na}_2\text{SO}_4\text{(aq)}$ , overlooking that $\text{MgSO}_4$ is soluble and will remain mixed with sodium and nitrate ions.

Things to Be Careful About

Remember solubility rules: all nitrates are soluble; most chlorides are soluble except silver and lead; most sulfates are soluble except barium, calcium, and lead.

Techniques used

apply solubility rules to identify soluble and insoluble saltsdetermine appropriate salt preparation methods (excess base method and precipitation)evaluate feasibility of preparing pure dry salt samples from given reagents

Q221MEasyAcids, Bases and SaltsChemistry of the Environment

When heated, copper(II) sulfate crystals, CuSO₄•5H₂O, react as shown.

\text{CuSO}_4\bullet5\text{H}_2\text{O(s)} \rightarrow \text{CuSO}_4\text{(s)} + 5\text{H}_2\text{O(g)}

When ......1...... copper(II) sulfate, which is ......2...... in appearance, is heated, it forms ......3...... copper(II) sulfate, which is ......4...... in appearance.

Which words correctly complete gaps 1, 2, 3 and 4?

Options

	1	2	3	4
A	anhydrous	blue and crystalline	hydrated	white and powdery
B	hydrated	colourless and crystalline	anhydrous	blue and powdery
C	hydrated	blue and powdery	anhydrous	colourless and crystalline
D	hydrated	blue and crystalline	anhydrous	white and powdery

Pick an answer

DifficultyEasy

Worked solution

Working

$\text{CuSO}_4\cdot5\text{H}_2\text{O}$ contains water of crystallisation, so it is hydrated copper(II) sulfate (gap 1).
Hydrated copper(II) sulfate forms blue and crystalline solids (gap 2).
When heated, it loses water to form $\text{CuSO}_4$ , which is anhydrous copper(II) sulfate (gap 3).
Anhydrous copper(II) sulfate is a white and powdery solid (gap 4).

Matching these to the options corresponds to row D.

Answer

D

Final answer

D

Detailed explanation

Walkthrough

Hydrated vs Anhydrous:
- A salt containing chemically combined water within its crystal structure is called hydrated. $\text{CuSO}_4\cdot5\text{H}_2\text{O}$ is hydrated copper(II) sulfate.
- A salt that has lost its water of crystallisation is called anhydrous. $\text{CuSO}_4$ is anhydrous copper(II) sulfate.
  Therefore, gap 1 is 'hydrated' and gap 3 is 'anhydrous'.
Physical Appearance:
- Hydrated copper(II) sulfate crystals ( $ext{CuSO}_4\cdot5 ext{H}_2 ext{O}$ ) are bright blue and crystalline (gap 2).
- Anhydrous copper(II) sulfate powder ( $ext{CuSO}_4$ ) is white and powdery (gap 4).
Matching Options:
- Gap 1: hydrated
- Gap 2: blue and crystalline
- Gap 3: anhydrous
- Gap 4: white and powdery
  This completely matches option D.

Key Takeaways

Hydrated salts contain water of crystallisation, usually giving them a characteristic crystal shape and distinct colour.
Heating drives off the water of crystallisation, turning the hydrated crystal into an anhydrous powder.
Anhydrous copper(II) sulfate turning from white to blue upon the addition of water is a standard chemical test for the presence of water.

Common Mistakes

Confusing the terms hydrated (with water) and anhydrous (without water).
Mixing up the colours: thinking anhydrous copper(II) sulfate is blue or colourless rather than white.

Techniques used

distinguish between hydrated and anhydrous saltsrecall the colour change of copper(II) sulfate upon dehydration

Q231MEasyAtoms, Elements and CompoundsThe Periodic Table

Which formula represents the oxide of element Z in Group II of the Periodic Table?

Options

A ZO
B Z₂O₂
C ZO₃
D Z₂O

Pick an answer

DifficultyEasy

Worked solution

Working

Element $\text{Z}$ is in Group II of the Periodic Table, so each atom loses $2$ electrons to form a $\text{Z}^{2+}$ ion.
Oxygen is in Group VI and gains $2$ electrons to form an oxide ion, $\text{O}^{2-}$ .
Since the charges are equal and opposite ( $+2$ and $-2$ ), the ions combine in a $1:1$ ratio to form the neutral compound $\text{ZO}$ .

Answer

A

Final answer

A

Detailed explanation

Walkthrough

Identify the charge on the metal ion from its position in the Periodic Table: elements in Group II have two outer-shell electrons, which they lose to form positive ions with a charge of $+2$ (i.e., $\text{Z}^{2+}$ ).
Identify the charge on the oxide ion: oxygen is in Group VI, with six outer-shell electrons, so it gains two electrons to achieve a full shell, forming an oxide ion with a charge of $-2$ (i.e., $\text{O}^{2-}$ ).
Balance the ionic charges to make a neutral compound: one $\text{Z}^{2+}$ ion balances one $\text{O}^{2-}$ ion ( $+2 + (-2) = 0$ ). Thus, the simplest formula is $\text{ZO}$ .

Checking the distractors:

B ( $ext{Z}_2 ext{O}_2$ ): Formulae of ionic compounds are expressed as the simplest whole-number ratio of ions, which simplifies to $\text{ZO}$ .
C ( $ext{ZO}_3$ ): This would correspond to an element with a valency/oxidation state of $+6$ (like sulfur in $\text{SO}_3$ ).
D ( $ext{Z}_2 ext{O}$ ): This represents the oxide of a Group I metal where the metal forms a $+1$ ion (e.g., $\text{Na}_2 ext{O}$ ).

Therefore, option A is the correct answer.

Key Takeaways

Group number in the Periodic Table gives the number of valence electrons and allows you to deduce the stable ionic charge:
- Group I forms $\text{M}^+$
- Group II forms $\text{M}^{2+}$
- Group III forms $\text{M}^{3+}$
- Group VI forms $\text{X}^{2-}$
- Group VII forms $\text{X}^-$
The formula of an ionic compound is written in the lowest whole-number ratio of cations to anions that gives an overall neutral charge.

Common Mistakes

Writing unreduced formulae such as $\text{Z}_2 ext{O}_2$ instead of simplifying to the empirical formula $\text{ZO}$ .
Confusing Group I and Group II charges, leading to the incorrect formula $\text{Z}_2 ext{O}$ .

Things to Be Careful About

Always write ionic formulae in their simplest empirical ratio unless specifically dealing with peroxides or molecular species.

Techniques used

deduce ionic charge from group numbercombine ions in an electrically neutral ratio to determine formula

Q241MMedium-EasyThe Periodic Table

Element X is in Group I.

Some statements about element X are given.

X is not the least dense element in Group I.
X is more reactive than potassium.
X has an $A_r$ value less than 100.

Which element is X?

Options

A lithium
B sodium
C rubidium
D caesium

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

In Group I, reactivity increases down the group:
$\text{Li} < \text{Na} < \text{K} < \text{Rb} < \text{Cs}$
Since element $\text{X}$ is more reactive than potassium ( $\text{K}$ ), $\text{X}$ must be below potassium, so it can only be rubidium ( $\text{Rb}$ ) or caesium ( $\text{Cs}$ ). This eliminates lithium ( $\text{A}$ ) and sodium ( $\text{B}$ ).
Looking at relative atomic masses ( $A_r$ ) from the Periodic Table:
- Rubidium ( $\text{Rb}$ ): $A_r = 85.5$
- Caesium ( $\text{Cs}$ ): $A_r = 133$
Since $\text{X}$ has an $A_r < 100$ , element $\text{X}$ must be rubidium ( $\text{Rb}$ ).

Answer

C

Final answer

C

Detailed explanation

Walkthrough

To identify element $\text{X}$ , we evaluate the given clues using our knowledge of Group I (the alkali metals) and the Periodic Table:

Reactivity trend: Reactivity increases as you go down Group I because the outer electron gets further from the nucleus, experiences more shielding, and is lost more easily. Therefore, any element more reactive than potassium ( $\text{K}$ ) must lie below potassium in Group I: either rubidium ( $\text{Rb}$ ), caesium ( $\text{Cs}$ ), or francium ( $\text{Fr}$ ).
Relative atomic mass ( $A_r$ ): Checking the Periodic Table:
- $A_r(\text{Rb}) = 85.5$
- $A_r(\text{Cs}) = 133$
  Since element $\text{X}$ has $A_r < 100$ , caesium is ruled out, leaving rubidium as the correct element.
Density check: Rubidium is denser than lithium, confirming the first statement that $\text{X}$ is not the least dense element.

Thus, the correct option is C.

Key Takeaways

In Group I (alkali metals), chemical reactivity increases down the group.
Density generally increases down Group I, with lithium being the least dense alkali metal.
Relative atomic mass ( $A_r$ ) increases down any group in the Periodic Table.

Common Mistakes

Confusing the reactivity trend of Group I (increases down) with Group VII (decreases down).
Selecting caesium ( $\text{D}$ ) by overlooking the requirement that $A_r < 100$ .
Selecting sodium ( $\text{B}$ ) or lithium ( $\text{A}$ ) by wrongly assuming elements higher up in Group I are more reactive.

Things to Be Careful About

Always refer to the provided Periodic Table to check relative atomic mass ( $A_r$ ) values accurately rather than guessing.

Techniques used

apply group trends in reactivity for alkali metalsuse relative atomic mass from the Periodic Table to identify an elementeliminate options based on multiple physical and chemical criteria

Q251MEasyMetals

A student makes three statements about metals and non-metals.

All alloys contain at least one metal.
All metals are good thermal and electrical conductors.
All solid non-metals are malleable.

Which statements are correct?

Options

A 1, 2 and 3
B 1 and 2 only
C 1 and 3 only
D 2 only

Pick an answer

DifficultyEasy

Worked solution

Working

Statement 1: An alloy is a mixture of a metal with other elements (which can be other metals or non-metals such as carbon). Therefore, all alloys must contain at least one metal. (Statement 1 is correct)
Statement 2: All metals possess a lattice of positive ions surrounded by delocalised electrons, allowing them to conduct heat and electricity well. (Statement 2 is correct)
Statement 3: Solid non-metals are typically brittle, not malleable; when hammered or subjected to stress, they shatter rather than bend. (Statement 3 is incorrect)

Therefore, statements 1 and 2 only are correct.

Answer

B

Final answer

B

Detailed explanation

Walkthrough

Let us evaluate each of the three statements individually:

Statement 1: "All alloys contain at least one metal."
- By definition, an alloy is a mixture of a metal with other elements (e.g., steel is iron and carbon, brass is copper and zinc, bronze is copper and tin). Because a metal is always the base component, every alloy contains at least one metal. Thus, statement 1 is correct.
Statement 2: "All metals are good thermal and electrical conductors."
- The metallic bonding model consists of a giant lattice of positive metal ions immersed in a 'sea' of delocalised, freely moving valence electrons. These delocalised electrons can move rapidly through the structure to transfer charge and thermal energy, making all metals good conductors of electricity and heat. Thus, statement 2 is correct.
Statement 3: "All solid non-metals are malleable."
- Malleability (the ability to be hammered or rolled into sheets without breaking) is a characteristic property of metals, where layers of ions can slide over one another without disrupting the metallic bonding. Solid non-metals (like sulfur, phosphorus, iodine, and carbon in the form of diamond or graphite) are brittle rather than malleable; applying force fractures their fixed covalent bonds or overcomes weak intermolecular forces, causing the solid to shatter. Thus, statement 3 is incorrect.

Combining these evaluations gives 1 and 2 only, which corresponds to option B.

Key Takeaways

An alloy is a mixture of a metal with other elements.
Metals are characteristically good electrical and thermal conductors due to their delocalised electrons, and they are malleable because layers of positive ions can slide over each other.
Solid non-metals are brittle and generally poor conductors of heat and electricity (with the notable exception of graphite for electrical conductivity).

Common Mistakes

Confusing the terms malleable (can be hammered into sheets) and ductile (can be drawn into wires) with brittle (breaks easily when stressed).
Thinking that some non-metals (like graphite) are malleable because graphite is soft and slippery; graphite layers flake off, but it cannot be hammered into shape without crumbling (it is brittle).

Things to Be Careful About

Watch out for universal quantifiers like "all". While exceptions exist for electrical conductivity in non-metals (graphite conducts), physical malleability is never a property of solid non-metals.

Techniques used

recall properties of metals and non-metalsapply definition of an alloyevaluate truth value of multiple statements

Q261MMedium-EasyThe Periodic TableMetals

Group I elements and transition elements are metals.

Student X suggests that the Group I elements are above hydrogen in the reactivity series but that not all transition elements are above hydrogen.

Student Y suggests that the densities of Group I elements are lower than those of the transition elements.

Which students are correct?

Options

A both X and Y
B X only
C Y only
D neither X nor Y

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Evaluate Student X's statement:
- All Group I alkali metals (e.g. $\text{Li}$ , $\text{Na}$ , $\text{K}$ ) are highly reactive and lie well above hydrogen in the reactivity series, reacting vigorously with water to release hydrogen gas.
- Transition elements vary in reactivity: while some like iron ( $\text{Fe}$ ) lie above hydrogen, unreactive transition elements such as copper ( $\text{Cu}$ ), silver ( $\text{Ag}$ ), and gold ( $\text{Au}$ ) lie below hydrogen and cannot displace hydrogen from dilute acids.
- Thus, Student X is correct.
Evaluate Student Y's statement:
- Group I elements are soft metals with unusually low densities (lithium, sodium, and potassium are less dense than water).
- Transition elements are dense, typical heavy metals with closely packed metallic lattices.
- Thus, Student Y is correct.

Both students are correct.

Answer

A

Final answer

A

Detailed explanation

Walkthrough

Reactivity of Group I vs. Transition Elements (Student X):
- In the reactivity series, the Group I alkali metals (potassium, sodium, lithium) are positioned near the top, far above hydrogen. They readily lose their single valence electron to form stable $+1$ cations, reacting vigorously with water and dilute acids.
- Transition elements exhibit a much wider range of chemical reactivity. Some (such as iron, zinc, and nickel) are located above hydrogen and will react with dilute acids to produce hydrogen gas. However, others (notably copper, silver, gold, and platinum) lie below hydrogen in the reactivity series and do not react with dilute acids to release hydrogen. Therefore, not all transition elements are above hydrogen.
- Student X is correct.
Density Comparison (Student Y):
- Group I metals have large atomic radii relative to their mass, leading to loosely packed crystal structures and low densities (the densities of $\text{Li}$ , $\text{Na}$ , and $\text{K}$ are all less than $1.0\text{ g / cm}^3$ ).
- In contrast, transition elements have smaller atomic radii and stronger metallic bonding due to the involvement of inner d-electrons, resulting in high densities (typically greater than $5\text{ g / cm}^3$ ).
- Student Y is correct.

Since both statements are true, option A is the correct choice.

Key Takeaways

Group I Metals: Characterised by high chemical reactivity (always above hydrogen) and low physical densities/melting points compared to other metals.
Transition Elements: Characterised by high densities, high melting points, and variable reactivity, spanning positions both above and below hydrogen in the reactivity series.

Common Mistakes

Assuming all metals are above hydrogen in the reactivity series. Unreactive transition metals like copper ( $\text{Cu}$ ), silver ( $\text{Ag}$ ), and gold ( $\text{Au}$ ) are below hydrogen.
Confusing the physical softness/low density of Group I metals with transition metals.

Things to Be Careful About

Remember that position in the reactivity series relative to hydrogen dictates whether a metal can displace hydrogen from dilute non-oxidising acids (such as $\text{HCl}$ or dilute $\text{H}_2\text{SO}_4$ ).

Techniques used

compare properties of Group I metals and transition elementsapply the reactivity series to classify metals relative to hydrogen

Q271MMedium-EasyMetals

The equations for three reactions are given.

$2\text{O}^{2-} \rightarrow \text{O}_2 + 4\text{e}^-$
$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$
$\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3$

These reactions take place in the extraction of metals.

Which row is correct?

Options

	reaction in extraction of aluminium only	reaction in extraction of iron only	reaction in extraction of aluminium and iron
A	1	2	3
B	1	3	2
C	2	1	3
D	2	3	1

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Reaction 1: $2\text{O}^{2-} \rightarrow \text{O}_2 + 4\text{e}^-$ is the oxidation of oxide ions at the positive anode during the electrolysis of molten aluminium oxide (extraction of aluminium only).
Reaction 2: $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ occurs in both processes: in the blast furnace (combustion of coke to provide heat and form reducing gas) and during the electrolysis of aluminium (the oxygen produced reacts with the carbon/graphite anodes).
Reaction 3: $\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3$ is the reaction between calcium oxide (basic oxide) and silicon dioxide (acidic impurity) to form slag, which occurs in the blast furnace only (extraction of iron only).

Matching with the table:

Extraction of aluminium only: 1
Extraction of iron only: 3
Extraction of both: 2

This matches row B.

Answer

B

Final answer

B

Detailed explanation

Walkthrough

To determine the correct row, evaluate where each chemical reaction occurs:

Reaction 1: $2\text{O}^{2-} \rightarrow \text{O}_2 + 4\text{e}^-$
- This is an oxidation half-equation where oxide ions lose electrons to form oxygen gas.
- It occurs at the anode during the extraction of aluminium by electrolysis of molten $\text{Al}_2\text{O}_3$ in cryolite.
- This electrolytic step is unique to aluminium extraction, as iron is extracted chemically by reduction using carbon monoxide in a blast furnace.
- Therefore, Reaction 1 is in the extraction of aluminium only.
Reaction 2: $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$
- In the blast furnace for iron extraction, coke (carbon) burns in a blast of hot air to produce carbon dioxide and release heat.
- In the Hall-Héroult cell for aluminium extraction, the oxygen gas liberated at the positive carbon/graphite anodes reacts with the carbon at high temperatures ( $900\text{--}1000\text{ }^\circ\text{C}$ ) to form carbon dioxide, causing the anodes to burn away over time.
- Therefore, Reaction 2 occurs in the extraction of both aluminium and iron.
Reaction 3: $\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3$
- In the blast furnace, limestone ( $\text{CaCO}_3$ ) decomposes to calcium oxide ( $\text{CaO}$ ), which then neutralises acidic silicon dioxide (sand/ $\text{SiO}_2$ ) impurities present in the iron ore to form molten calcium silicate (slag, $\text{CaSiO}_3$ ).
- In aluminium extraction, bauxite ore is purified to pure $\text{Al}_2\text{O}_3$ before electrolysis (via the Bayer process), so slag formation does not take place in the cell.
- Therefore, Reaction 3 is in the extraction of iron only.

Matching these gives row B.

Key Takeaways

Aluminium is high in the reactivity series and must be extracted by electrolysis of its molten oxide dissolved in cryolite.
At the anode in aluminium extraction, oxide ions are oxidised to oxygen gas ( $2\text{O}^{2-} \rightarrow \text{O}_2 + 4\text{e}^-$ ), which subsequently burns the carbon anodes to form $\text{CO}_2$ .
Iron is extracted in the blast furnace where coke burns to form $\text{CO}_2$ , and limestone provides $\text{CaO}$ to remove $\text{SiO}_2$ as slag ( $\text{CaSiO}_3$ ).

Common Mistakes

Forgetting that carbon anodes burn away in the extraction of aluminium, leading students to think $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ is unique to the blast furnace.
Confusing the role of slag formation (limestone reaction) and thinking it happens in the electrolytic cell.

Things to Be Careful About

Ensure the three columns in the question's table match the assigned reactions (aluminium only, iron only, and both).

Techniques used

identify reactions in the Hall-Héroult electrolytic extraction of aluminiumidentify reactions in the blast furnace extraction of ironcompare common reaction steps across both industrial processes

Q281MEasyChemistry of the Environment

The domestic water supply is treated to make it safe to drink.

Which row identifies the treatment and its effect?

Options

	chlorination	filtration	sedimentation
A	removes nitrates	removes solids	removes nitrates
B	kills microbes	removes soluble compounds	removes solids
C	removes solids	kills microbes	removes soluble compounds
D	kills microbes	removes solids	removes solids

Pick an answer

DifficultyEasy

Worked solution

Working

Chlorination: Chlorine is added to water to kill harmful microbes/bacteria and sterilise the water.
Filtration: Water is passed through sand and gravel filters to remove insoluble solids.
Sedimentation: Water is allowed to stand so that larger suspended insoluble solids settle to the bottom and are removed.

Matching these effects with the table gives:

Chlorination: kills microbes
Filtration: removes solids
Sedimentation: removes solids

This matches row D.

Answer

D

Final answer

D

Detailed explanation

Walkthrough

Domestic water treatment involves several key physical and chemical steps to make water potable (safe to drink):

Sedimentation: Water is kept in large tanks, allowing heavier suspended insoluble particles to settle down under gravity. This removes large solids.
Filtration: Water is passed through fine beds of sand and gravel to trap and remove any remaining small, insoluble solid particles.
Chlorination: A carefully controlled amount of chlorine gas (or chlorine compounds) is added as a disinfectant to kill harmful bacteria, viruses, and other microorganisms.

Evaluating the table:

A is incorrect because neither chlorination nor sedimentation removes dissolved nitrates.
B is incorrect because simple sand filtration does not remove soluble (dissolved) compounds.
C mixes up the functions completely (e.g., chlorination does not remove solids; sedimentation does not remove soluble compounds).
D correctly states that chlorination kills microbes, while both filtration and sedimentation remove insoluble solids.

Key Takeaways

Physical methods (sedimentation and filtration) remove insoluble suspended solids of varying sizes.
Chemical disinfection (chlorination) kills bacteria and microorganisms to ensure the water is biologically safe.
Standard domestic water treatment does not remove dissolved mineral ions such as nitrates.

Common Mistakes

Confusing physical separation of insoluble particles (filtration/sedimentation) with the removal of dissolved/soluble substances.
Thinking that filtration kills microbes or removes dissolved chemical pollutants like nitrates.

Things to Be Careful About

Ensure you distinguish clearly between dissolved (soluble) impurities and suspended (insoluble) impurities.

Techniques used

identify stages in domestic water treatmentlink water treatment processes to their functions

Q291MMedium-EasyChemistry of the EnvironmentStoichiometry

Sodium phosphate, Na₃PO₄, and ammonium nitrate, NH₄NO₃, are both used as fertilisers.

Which row shows the correct percentage by mass of the element in each compound that improves plant growth?

Options

	% of the element in Na₃PO₄ that improves plant growth	% of the element in NH₄NO₃ that improves plant growth
A	19	18
B	19	35
C	42	18
D	42	35

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Identify the elements in NPK fertilisers that improve plant growth:
- In $\text{Na}_3\text{PO}_4$ , the essential nutrient element is phosphorus ( $\text{P}$ ). Sodium ( $\text{Na}$ ) is not an essential major fertiliser element.
- In $\text{NH}_4\text{NO}_3$ , the essential nutrient element is nitrogen ( $\text{N}$ ).
Calculate the percentage by mass of $\text{P}$ in $\text{Na}_3\text{PO}_4$ :
- Relative atomic masses ( $A_r$ ): $\text{Na} = 23$ , $\text{P} = 31$ , $\text{O} = 16$
$M_r(\text{Na}_3\text{PO}_4) = (3 \times 23) + 31 + (4 \times 16) = 69 + 31 + 64 = 164$ $\% \text{ P} = \frac{31}{164} \times 100\% \approx 18.9\% \approx 19\%$
Calculate the percentage by mass of $\text{N}$ in $\text{NH}_4\text{NO}_3$ :
- Relative atomic masses ( $A_r$ ): $\text{N} = 14$ , $\text{H} = 1$ , $\text{O} = 16$
- Total mass of nitrogen atoms = $2 \times 14 = 28$
$M_r(\text{NH}_4\text{NO}_3) = (2 \times 14) + (4 \times 1) + (3 \times 16) = 28 + 4 + 48 = 80$ $\% \text{ N} = \frac{28}{80} \times 100\% = 35\%$

Therefore, the correct row is B ( $19\%$ and $35\%$ ).

Answer

B

Final answer

B

Detailed explanation

Walkthrough

Identify the essential element in each fertiliser:
Fertilisers provide plants with three main elements for healthy growth: Nitrogen ( $\text{N}$ ), Phosphorus ( $\text{P}$ ), and Potassium ( $\text{K}$ ) (often known as NPK fertilisers).
- For $\text{Na}_3\text{PO}_4$ , the essential element is phosphorus ( $\text{P}$ ). (Sodium, $\text{Na}$ , is not a plant fertiliser nutrient; calculating sodium would give $\frac{3 \times 23}{164} \times 100\% \approx 42\%$ , which corresponds to distractor rows C and D).
- For $\text{NH}_4\text{NO}_3$ , the essential element is nitrogen ( $\text{N}$ ).
Calculate the percentage by mass of $\text{P}$ in $\text{Na}_3\text{PO}_4$ :
- Formula mass of $\text{Na}_3\text{PO}_4$ :
  $M_r = (3 \times 23) + 31 + (4 \times 16) = 164$
- Percentage of $\text{P}$ :
  $\% \text{ P} = \frac{31}{164} \times 100\% = 18.9\% \approx 19\%$
Calculate the percentage by mass of $\text{N}$ in $\text{NH}_4\text{NO}_3$ :
- Note that there are $2$ nitrogen atoms in the formula unit $\text{NH}_4\text{NO}_3$ .
- Formula mass of $\text{NH}_4\text{NO}_3$ :
  $M_r = 14 + (4 \times 1) + 14 + (3 \times 16) = 80$
- Percentage of $\text{N}$ :
  $\% \text{ N} = \frac{2 \times 14}{80} \times 100\% = \frac{28}{80} \times 100\% = 35\%$

Matching these values to the table gives $19\%$ and $35\%$ , corresponding to option B.

Key Takeaways

Plants require three main macronutrients from artificial fertilisers: nitrogen (for protein synthesis and leaf/stem growth), phosphorus (for root development), and potassium (for flowers and fruits).
To find the percentage by mass of an element in a compound:
$\% \text{ element} = \frac{\text{number of atoms of that element} \times A_r}{M_r \text{ of compound}} \times 100\%$

Common Mistakes

Calculating the percentage of sodium ( $\text{Na}$ ) instead of phosphorus ( $\text{P}$ ) in $\text{Na}_3\text{PO}_4$ , leading to $42\%$ and incorrectly choosing option C or D.
Counting only $1$ nitrogen atom instead of $2$ in $\text{NH}_4\text{NO}_3$ , leading to $\frac{14}{80} \times 100\% = 17.5\% \approx 18\%$ (options A and C).

Things to Be Careful About

Always inspect the chemical formula carefully to count all occurrences of an element (e.g. $\text{NH}_4\text{NO}_3$ contains two nitrogen atoms in different parts of the formula).

Techniques used

identify essential plant nutrient elements (NPK)calculate relative formula mass (Mr)calculate percentage by mass of an element in a compound

Q301MMedium-EasyChemistry of the Environment

Three processes are shown.

the decomposition of vegetation
emissions from a car engine
photosynthesis

Which row shows a gas produced in each process?

Options

	process 1	process 2	process 3
A	carbon monoxide	hydrogen	oxygen
B	carbon monoxide	nitrogen monoxide	carbon dioxide
C	methane	nitrogen monoxide	oxygen
D	methane	hydrogen	carbon dioxide

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Process 1 (decomposition of vegetation): Anaerobic bacterial decay of plant matter produces methane ( $\text{CH}_4$ ).
Process 2 (emissions from a car engine): High temperatures inside car engines cause nitrogen and oxygen from the air to react, forming nitrogen monoxide ( $\text{NO}$ ), which is released in exhaust fumes.
Process 3 (photosynthesis): Plants react carbon dioxide and water in the presence of light and chlorophyll to produce glucose and release oxygen ( $\text{O}_2$ ) gas.

Matching these gases to the table gives C.

Answer

C

Final answer

C

Detailed explanation

Walkthrough

To determine the correct option, evaluate each process individually:

Decomposition of vegetation (Process 1):
The anaerobic decay or decomposition of organic/plant material by bacteria (e.g., in paddy fields, swamps, and landfill sites) is a major natural source of methane ( $\text{CH}_4$ ).
Emissions from a car engine (Process 2):
Inside an internal combustion engine, the spark/combustion creates high temperatures and pressures. Under these conditions, atmospheric nitrogen ( $\text{N}_2$ ) and oxygen ( $\text{O}_2$ ) react together:
$\text{N}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}(\text{g})$
Hence, nitrogen monoxide is a primary pollutant in exhaust gases. (Hydrogen is not produced in car exhaust emissions).
Photosynthesis (Process 3):
Green plants use light energy to convert carbon dioxide and water into glucose and oxygen:
$6\text{CO}_2(\text{g}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(\text{aq}) + 6\text{O}_2(\text{g})$
Therefore, oxygen is released as a product, while carbon dioxide is consumed.

Comparing with the options, row C correctly identifies methane, nitrogen monoxide, and oxygen for processes 1, 2, and 3, respectively.

Key Takeaways

Methane is produced by the anaerobic decay of organic vegetation.
Oxides of nitrogen (such as $\text{NO}$ and $\text{NO}_2$ ) are formed when high temperatures in car engines cause $\text{N}_2$ and $\text{O}_2$ in air to react.
Photosynthesis consumes $\text{CO}_2$ and produces $\text{O}_2$ , whereas respiration consumes $\text{O}_2$ and produces $\text{CO}_2$ .

Common Mistakes

Confusing photosynthesis with respiration and thinking that photosynthesis produces $\text{CO}_2$ instead of $\text{O}_2$ .
Mistakenly identifying hydrogen as a component of car exhaust emissions.

Things to Be Careful About

Ensure you differentiate between reactants and products for processes like photosynthesis (it uses $\text{CO}_2$ and water, but produces glucose and $\text{O}_2$ ).

Techniques used

recall sources of greenhouse gasesrecall exhaust emissions from internal combustion enginesrecall products of photosynthesis

Q311MEasyOrganic Chemistry

Which compound is not an alkane, alkene, alcohol or carboxylic acid?

Options

Pick an answer

DifficultyEasy

Worked solution

Working

Structure A contains a carbonyl group ( $\text{C=O}$ ) bonded between two carbon atoms (a ketone, propanone). It is not an alkane, alkene, alcohol, or carboxylic acid.
Structure B contains a hydroxyl group ( $-\text{OH}$ ) attached to a saturated carbon, which makes it an alcohol (propan-2-ol).
Structure C contains only carbon and hydrogen atoms joined by single bonds ( $\text{C-C}$ and $\text{C-H}$ ), so it is a branched alkane (methylpropane).
Structure D contains a carbon-carbon double bond ( $\text{C=C}$ ), so it is an alkene (2-methylpropene).

Therefore, compound A is the only one that does not belong to any of the four named homologous series.

Answer

A

Final answer

A

Detailed explanation

Walkthrough

To determine which molecule is not an alkane, alkene, alcohol, or carboxylic acid, examine the characteristic bonding and functional groups present in each given displayed formula:

Alkane: Saturated hydrocarbon containing only single covalent bonds ( $\text{C-C}$ and $\text{C-H}$ ). Option C is an alkane (methylpropane / isobutane).
Alkene: Unsaturated hydrocarbon containing at least one carbon-carbon double bond ( $\text{C=C}$ ). Option D contains a $\text{C=C}$ bond and is an alkene (2-methylpropene).
Alcohol: Contains a hydroxyl functional group ( $\text{-OH}$ ) attached to a carbon atom. Option B contains an $-\text{OH}$ group on the central carbon, making it an alcohol (propan-2-ol).
Carboxylic acid: Contains a carboxyl group ( $-\text{COOH}$ , i.e., a $\text{C=O}$ and an $-\text{OH}$ on the same carbon atom).

Structure A contains a $\text{C=O}$ group bonded to two alkyl groups. This functional group is a ketone (specifically propanone), which is none of the four listed classes.

Hence, A is the correct answer.

Key Takeaways

Alkanes have only single $\text{C-C}$ and $\text{C-H}$ bonds.
Alkenes contain a $\text{C=C}$ double bond.
Alcohols contain an $-\text{OH}$ functional group.
Carboxylic acids contain a $-\text{COOH}$ group.

Common Mistakes

Mistaking the $\text{C=O}$ in a ketone (A) for a carboxylic acid, overlooking that a carboxylic acid must have an $-\text{OH}$ group attached to the same carbonyl carbon atom ( $-\text{COOH}$ ).
Confusing branched structures (such as C) as belonging to a different family rather than recognising them as alkanes.

Things to Be Careful About

Ensure you check every atom and bond in the displayed formula: a carbonyl group alone ( $\text{C=O}$ ) flanked by carbon atoms is a ketone, not an acid or alcohol.

Techniques used

identify functional groups from displayed formulaeclassify organic compounds into homologous series

Q321MMedium-EasyOrganic Chemistry

Alkenes can be produced by the cracking of alkanes, such as decane, C₁₀H₂₂.

Which equation shows the cracking of decane to produce two different alkenes and at least one other product?

Options

A C₁₀H₂₂ → 2C₂H₄ + C₃H₆ + C₄H₁₀
B C₁₀H₂₂ → H₂ + 2C₂H₄ + C₃H₆ + C₅H₁₀
C C₁₀H₂₂ → 2C₃H₆ + C₄H₁₀
D C₁₀H₂₂ → H₂ + 2C₂H₄ + 2C₃H₆

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

To find the correct equation, check the conditions stated:

The equation must be balanced for $\text{C}$ and $\text{H}$ atoms (starting from $\text{C}_{10}\text{H}_{22}$ ).
It must produce two different alkenes (general formula $\text{C}_n\text{H}_{2n}$ ).
It must produce at least one other product (e.g., an alkane or $\text{H}_2$ ).

A: $\text{C}_{10}\text{H}_{22} \rightarrow 2\text{C}_2\text{H}_4 + \text{C}_3\text{H}_6 + \text{C}_4\text{H}_{10}$
- Products: $\text{C}_2\text{H}_4$ (ethene, alkene), $\text{C}_3\text{H}_6$ (propene, alkene), $\text{C}_4\text{H}_{10}$ (butane, alkane).
- Check balance: $\text{C} = 2(2) + 3 + 4 = 11 \neq 10$ . Not balanced.
B: $\text{C}_{10}\text{H}_{22} \rightarrow \text{H}_2 + 2\text{C}_2\text{H}_4 + \text{C}_3\text{H}_6 + \text{C}_5\text{H}_{10}$
- Products: $\text{C}_2\text{H}_4$ (alkene), $\text{C}_3\text{H}_6$ (alkene), $\text{C}_5\text{H}_{10}$ (alkene) — this gives three different alkenes, not two.
- Also check balance: $\text{C} = 2(2) + 3 + 5 = 12 \neq 10$ . Not balanced.
C: $\text{C}_{10}\text{H}_{22} \rightarrow 2\text{C}_3\text{H}_6 + \text{C}_4\text{H}_{10}$
- Products: $\text{C}_3\text{H}_6$ (propene, alkene) and $\text{C}_4\text{H}_{10}$ (butane, alkane).
- Contains only one type of alkene (propene), not two different alkenes.
D: $\text{C}_{10}\text{H}_{22} \rightarrow \text{H}_2 + 2\text{C}_2\text{H}_4 + 2\text{C}_3\text{H}_6$
- Check balance:
  - Carbon: $10 = 2(2) + 2(3) = 4 + 6 = 10$
  - Hydrogen: $22 = 2 + 2(4) + 2(6) = 2 + 8 + 12 = 22$
- Products:
  - $\text{C}_2\text{H}_4$ (ethene, alkene)
  - $\text{C}_3\text{H}_6$ (propene, alkene) $\rightarrow$ two different alkenes
  - $\text{H}_2$ (hydrogen gas) $\rightarrow$ at least one other product

Therefore, option D satisfies all conditions.

Answer

D

Final answer

D

Detailed explanation

Walkthrough

Cracking is the thermal decomposition of longer-chain alkanes into shorter, more useful molecules, which typically include smaller alkanes, alkenes, and hydrogen gas.

Let's evaluate each condition required by the question:

Two different alkenes: An alkene has the general formula $\text{C}_n\text{H}_{2n}$ . Ethene ( $\text{C}_2\text{H}_4$ ) and propene ( $\text{C}_3\text{H}_6$ ) are two different alkenes.
At least one other product: A product that is not one of those two alkenes, such as hydrogen gas ( $\text{H}_2$ ) or an alkane ( $\text{C}_n\text{H}_{2n+2}$ ).
Conservation of mass: The total number of carbon and hydrogen atoms in the products must equal 10 carbons and 22 hydrogens (from $\text{C}_{10}\text{H}_{22}$ ).

Checking option D:

\text{C}_{10}\text{H}_{22} \rightarrow \text{H}_2 + 2\text{C}_2\text{H}_4 + 2\text{C}_3\text{H}_6

Total $\text{C}$ atoms on right-hand side: $(2 \times 2) + (2 \times 3) = 4 + 6 = 10$ .
Total $\text{H}$ atoms on right-hand side: $2 + (2 \times 4) + (2 \times 6) = 2 + 8 + 12 = 22$ .
The products are ethene (alkene), propene (alkene), and hydrogen gas (other product).
All criteria are met, so D is correct.

Key Takeaways

Alkenes contain a $\text{C}=\text{C}$ double bond and have the general formula $\text{C}_n\text{H}_{2n}$ .
Alkanes have the general formula $\text{C}_n\text{H}_{2n+2}$ .
Cracking always conserves the total number of carbon and hydrogen atoms.
Cracking can produce smaller alkanes, alkenes, and/or hydrogen gas ( $\text{H}_2$ ).

Common Mistakes

Forgetting to check that the equation is balanced in both carbon and hydrogen atoms (options A and B fail on balancing alone).
Misidentifying the number of different alkenes formed (option C only forms one type of alkene, propene).

Things to Be Careful About

Count the coefficients carefully when summing up atoms: $2\text{C}_3\text{H}_6$ contains $2 \times 3 = 6$ carbons and $2 \times 6 = 12$ hydrogens.

Techniques used

balance the symbol equationidentify homologous series by general formula

Q331MMedium-EasyOrganic Chemistry

Alkenes undergo addition reactions with bromine to form dibromoalkanes.

Which statement is correct?

Options

A Ethene and bromine react to produce 1,1-dibromoethane.
B Propene and bromine react to produce 1,3-dibromopropane.
C But-2-ene and bromine react to produce 2,2-dibromobutane.
D But-1-ene and bromine react to produce 1,2-dibromobutane.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

In an addition reaction of an alkene with bromine ( $\text{Br}_2$ ), the carbon-carbon double bond ( $\text{C}=\text{C}$ ) breaks and one bromine atom attaches to each of the two previously double-bonded carbon atoms:

A: Ethene ( $\text{CH}_2=\text{CH}_2$ ) reacts with $\text{Br}_2$ to form 1,2-dibromoethane ( $\text{CH}_2\text{Br}-\text{CH}_2\text{Br}$ ), not 1,1-dibromoethane.
B: Propene ( $\text{CH}_3-\text{CH}=\text{CH}_2$ ) reacts with $\text{Br}_2$ to form 1,2-dibromopropane ( $\text{CH}_3-\text{CHBr}-\text{CH}_2\text{Br}$ ), not 1,3-dibromopropane.
C: But-2-ene ( $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3$ ) reacts with $\text{Br}_2$ to form 2,3-dibromobutane ( $\text{CH}_3-\text{CHBr}-\text{CHBr}-\text{CH}_3$ ), not 2,2-dibromobutane.
D: But-1-ene ( $\text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}_2$ ) reacts with $\text{Br}_2$ across carbons 1 and 2 to form 1,2-dibromobutane ( $\text{CH}_3-\text{CH}_2-\text{CHBr}-\text{CH}_2\text{Br}$ ).

Therefore, statement D is correct.

Answer

D

Final answer

D

Detailed explanation

Walkthrough

An addition reaction occurs when atoms or groups of atoms add across the $\text{C}=\text{C}$ double bond of an unsaturated hydrocarbon (alkene), breaking the double bond to form a saturated compound (alkane derivative) containing only single bonds.

When aqueous or liquid bromine ( $\text{Br}_2$ ) reacts with an alkene:

The $\text{C}=\text{C}$ double bond opens up into a single bond.
One bromine atom bonds to each of the two adjacent carbon atoms that originally shared the double bond.

Let us analyse each option:

Ethene: $\text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{Br}-\text{CH}_2\text{Br}$ (1,2-dibromoethane). Both carbons (1 and 2) gain a bromine atom.
Propene: $\text{CH}_3-\text{CH}=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_3-\text{CHBr}-\text{CH}_2\text{Br}$ (1,2-dibromopropane). Carbons 1 and 2 gain bromine atoms.
But-2-ene: $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 + \text{Br}_2 \rightarrow \text{CH}_3-\text{CHBr}-\text{CHBr}-\text{CH}_3$ (2,3-dibromobutane). Carbons 2 and 3 gain bromine atoms.
But-1-ene: $\text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_3-\text{CH}_2-\text{CHBr}-\text{CH}_2\text{Br}$ (1,2-dibromobutane). Carbons 1 and 2 gain bromine atoms.

Thus, only option D gives the correct IUPAC name of the addition product.

Key Takeaways

In addition reactions of halogens to alkenes, the halogen atoms always add to adjacent carbon atoms (the two carbons of the original double bond).
The positional numbers in the name of the product directly reflect the positions of the two carbons involved in the original double bond.

Common Mistakes

Confusing addition with substitution (assuming both bromine atoms could attach to the same carbon atom, e.g., forming 1,1-dibromoethane or 2,2-dibromobutane).
Misnumbering the carbon chain and thinking bromine atoms add to the ends of the molecule rather than across the double bond (e.g., forming 1,3-dibromopropane from propene).

Things to Be Careful About

Always identify the exact position of the double bond from the name of the starting alkene (e.g., but-1-ene has the double bond between carbons 1 and 2; but-2-ene between carbons 2 and 3).
Ensure the product is numbered from the end of the chain that gives the substituent halogen atoms the lowest possible locant numbers.

Techniques used

apply addition reaction rules across double bondsname haloalkanes using positional numbering

Q341MMedium-EasyOrganic ChemistryStoichiometry

How many moles of oxygen are required for the complete combustion of 2 moles of ethanol?

Options

A 3
B 4
C 6
D 7

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Write the balanced chemical equation for the complete combustion of ethanol, $\text{C}_2\text{H}_5\text{OH}$ :

\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}

From the equation, $1\text{ mole of }\text{C}_2\text{H}_5\text{OH}$ reacts completely with $3\text{ moles of }\text{O}_2$ .

For $2\text{ moles of ethanol}$ :

\text{moles of }\text{O}_2 = 2 \times 3 = 6\text{ moles}

A (3): This is the moles of oxygen needed for only $1\text{ mole}$ of ethanol.
B (4): Incorrect balancing result.
D (7): Corresponds to assuming hydrocarbon combustion without accounting for the oxygen atom already present in ethanol ( $\text{C}_2\text{H}_6 + 3.5\text{O}_2$ , scaled to $7$ ).

Answer

C

Final answer

C

Detailed explanation

Walkthrough

To find the number of moles of oxygen gas required:

Identify the formula of the alcohol: Ethanol has the molecular formula $\text{C}_2\text{H}_5\text{OH}$ (or $\text{C}_2\text{H}_6\text{O}$ ).
Write and balance the complete combustion equation:
- Combustion produces carbon dioxide ( $\text{CO}_2$ ) and water ( $\text{H}_2\text{O}$ ).
- 2 carbon atoms give $2\text{CO}_2$ .
- 6 hydrogen atoms give $3\text{H}_2\text{O}$ .
- Total oxygen atoms on the right-hand side: $(2 \times 2) + (3 \times 1) = 4 + 3 = 7\text{ oxygen atoms}$ .
- The ethanol molecule already contains $1\text{ oxygen atom}$ , so the remaining $6\text{ oxygen atoms}$ must come from $\text{O}_2$ gas: $\frac{7 - 1}{2} = 3\text{ molecules of }\text{O}_2$
- The balanced equation is: $\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}$
Use the molar ratio: The mole ratio of ethanol to oxygen is $1 : 3$ $1 : 3$ .
- For $2\text{ moles}$ of ethanol, the oxygen required is $2 \times 3 = 6\text{ moles}$ .

This matches option C.

Key Takeaways

Complete combustion of any alcohol produces $\text{CO}_2$ and $\text{H}_2\text{O}$ .
Always remember to subtract the oxygen atom already present inside the alcohol molecule when balancing the $\text{O}_2$ term.
Multiply the stoichiometric coefficient by the given amount in moles to obtain the required quantity.

Common Mistakes

Forgetting the oxygen atom inside ethanol ( $ext{C}_2 ext{H}_5 ext{OH}$ ), leading to balancing as if it were ethane ( $ext{C}_2 ext{H}_6$ ), which would give $3.5\text{ moles of }\text{O}_2$ per mole of fuel and incorrectly lead to $7\text{ moles}$ (option D).
Finding the correct stoichiometric coefficient ( $3$ ) for $1\text{ mole}$ of ethanol but forgetting that the question specifies $2\text{ moles}$ of ethanol (option A).

Things to Be Careful About

Ensure you check whether the question specifies complete or incomplete combustion (complete combustion always gives $\text{CO}_2$ , not $\text{CO}$ or $\text{C}$ ).
Double-check that all coefficients are balanced for both reactants and products before taking stoichiometric ratios.

Techniques used

construct and balance the combustion equation for ethanolapply stoichiometric ratios to calculate required moles

Q351MMedium-EasyOrganic Chemistry

Two reactions are shown.

\begin{aligned} \text{reaction 1} &\qquad \text{CH}_2\text{CH=CH}_2\text{OH} + \text{H}_2 \xrightarrow{\text{Ni}} \text{compound X} \\ \text{reaction 2} &\qquad \text{compound X} \xrightarrow{\substack{\text{acidified}\\\text{KMnO}_4\text{(aq)}}} \text{compound Y} \end{aligned}

Which row is correct?

Options

	name of compound X	type of compound Y
A	propan-1-ol	carboxylic acid
B	propan-1-ol	ester
C	butan-1-ol	carboxylic acid
D	butan-1-ol	ester

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

Identify compound X:
- The starting material has a 3-carbon chain with a $\text{C=C}$ double bond and an $-\text{OH}$ group on carbon 1: $\text{CH}_2\text{=CH}-\text{CH}_2\text{OH}$ (prop-2-en-1-ol).
- Reaction 1 is the catalytic addition of hydrogen ( $ext{H}_2$ with a nickel catalyst) across the $\text{C=C}$ double bond:

\text{CH}_2\text{=CH}-\text{CH}_2\text{OH} + \text{H}_2 \xrightarrow{\text{Ni}} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}

$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ has 3 carbon atoms and is propan-1-ol.

Identify the type of compound Y:
- Reaction 2 is the oxidation of propan-1-ol using acidified potassium manganate(VII), $\text{KMnO}_4(\text{aq})$ , which is a strong oxidising agent.
- Oxidation of a primary alcohol produces a carboxylic acid (propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$ ).

Therefore, compound X is propan-1-ol and compound Y is a carboxylic acid, matching row A.

Answer

A

Final answer

A

Detailed explanation

Walkthrough

Reaction 1 (Hydrogenation):
The reactant molecule is an unsaturated alcohol containing 3 carbon atoms. When hydrogen gas ( $ext{H}_2$ ) reacts in the presence of a nickel catalyst, an addition reaction takes place across the carbon–carbon double bond ( $ext{C=C}$ ). The double bond becomes a single bond ( $ext{C}- ext{C}$ ), saturating the carbon skeleton while leaving the $- ext{OH}$ functional group unchanged. The resulting product is $ext{CH}_3 ext{CH}_2 ext{CH}_2 ext{OH}$ , which is propan-1-ol (a 3-carbon chain, ruling out options C and D).
Reaction 2 (Oxidation):
Acidified aqueous potassium manganate(VII), $ext{KMnO}_4 ext{(aq)}$ , is a powerful oxidising agent used in organic chemistry to oxidise alcohols. When a primary alcohol like propan-1-ol is heated with an oxidising agent, it is oxidised to form a carboxylic acid (propanoic acid, $ext{CH}_3 ext{CH}_2 ext{COOH}$ ).

Combining both conclusions:

Compound X is propan-1-ol.
Compound Y is a carboxylic acid.

This corresponds to option A.

Key Takeaways

Alkenes undergo catalytic hydrogenation (addition of $ext{H}_2$ over a $ext{Ni}$ catalyst) to form alkanes/saturated chains.
Primary alcohols are oxidised by acidified $ext{KMnO}_4 ext{(aq)}$ or acidified $ext{K}_2 ext{Cr}_2 ext{O}_7 ext{(aq)}$ to form carboxylic acids.
Count the number of carbon atoms carefully: prop- represents a 3-carbon chain, whereas but- represents a 4-carbon chain.

Common Mistakes

Miscounting the carbon atoms in the formula and assuming it is a 4-carbon compound (butan-1-ol).
Confusing the formation of an ester (which requires an alcohol reacting with a carboxylic acid in the presence of an acid catalyst) with oxidation to a carboxylic acid.

Things to Be Careful About

Remember that hydrogenation targets the $ext{C=C}$ bond and does not reduce or alter the alcohol ( $- ext{OH}$ ) functional group under these conditions.
Note that acidified potassium manganate(VII) turns from purple to colourless when it oxidises an alcohol, confirming the reaction has taken place.

Techniques used

identify addition of hydrogen across a carbon-carbon double bondname an alcohol from its structural formuladeduce the product of oxidation of a primary alcohol by acidified potassium manganate(VII)

Q361MEasyOrganic Chemistry

A section of the structure of a protein is shown.

Fig. 1

How many amino acid monomer molecules have been used to make this section of the structure?

Options

A 2
B 3
C 4
D 5

Pick an answer

DifficultyEasy

Worked solution

Working

Each amino acid monomer unit has the general structure $-\text{NH}-\text{CH(R)}-\text{CO}-$ .

Counting the units along the displayed section from left to right:

$-\text{NH}-\text{CH(R)}-\text{CO}-$ (Unit 1)
$-\text{NH}-\text{CH(R)}-\text{CO}-$ (Unit 2)
$-\text{NH}-\text{CH(R)}-\text{CO}-$ (Unit 3)
$-\text{NH}-\text{CH(R)}-\text{CO}-$ (Unit 4)

There are 4 distinct $-\text{CH(R)}-$ groups and 4 corresponding carbonyl ( $\text{C}=\text{O}$ ) groups, meaning 4 amino acid molecules were used to form this section.

Answer

C

Final answer

C

Detailed explanation

Walkthrough

Proteins are natural condensation polymers (polyamides) made by linking together amino acid monomers. Each amino acid contains an amine group ( $-\text{NH}_2$ ) and a carboxylic acid group ( $-\text{COOH}$ ) attached to a central carbon atom that also bears a hydrogen atom and a side chain ( $-\text{R}$ ).

When amino acids join during condensation polymerisation, a water molecule is eliminated between the carboxylic acid group of one amino acid and the amine group of the next, forming an amide (peptide) link ( $-\text{CO}-\text{NH}-$ ).

Therefore, each amino acid residue remaining in the polymer backbone consists of:

-\text{NH}-\text{CH(R)}-\text{CO}-

To find the number of amino acid monomers present in the given structure:

Look for the central carbon atoms bonded to the $\text{R}$ groups: there are 4 $\text{R}$ groups shown.
Look for the carbonyl groups ( $\text{C}=\text{O}$ ): there are 4 $\text{C}=\text{O}$ groups shown.
Look for the $-\text{NH}-$ groups: there are 4 $-\text{NH}-$ groups shown.

Since each complete monomer unit contributes one $-\text{CH(R)}-$ group and one carbonyl group, exactly 4 amino acid monomers were joined to create this section.

Key Takeaways

Proteins are polyamides formed from amino acid monomers by condensation polymerisation.
The repeat unit for an amino acid residue in a protein backbone is $-\text{NH}-\text{CH(R)}-\text{CO}-$ .
Counting either the number of central carbon atoms ( $-\text{CH(R)}-$ ), the side chains ( $\text{R}$ ), or the carbonyl groups ( $\text{C}=\text{O}$ ) gives the number of monomer units.

Common Mistakes

Counting the number of internal peptide bonds (which is 3) instead of counting the total number of amino acid monomer units (which is 4).

Things to Be Careful About

Do not confuse the number of peptide bonds formed with the number of monomer residues. In an open chain of $n$ monomers, there are $n-1$ peptide bonds between them, but the total number of monomer residues shown is $n = 4$ .

Techniques used

identify the repeating unit of a polyamide chaindeduce the number of monomer units from the backbone structure

Q371MEasyExperimental Techniques and Chemical Analysis

Which piece of apparatus is used to measure exactly 27.3 $\text{cm}^3$ of a liquid?

Options

A a burette
B a condenser
C a measuring cylinder
D a volumetric pipette

Pick an answer

DifficultyEasy

Worked solution

Working

A burette can measure variable volumes of liquid accurately to the nearest $0.05\text{ cm}^3$ (or to one decimal place, such as $27.3\text{ cm}^3$ ).
A condenser is used to cool and condense vapours back into liquids, not to measure volume.
A measuring cylinder is used to measure approximate volumes of liquid, not exact decimal volumes.
A volumetric pipette only measures fixed volumes accurately (typically $10.0\text{ cm}^3$ or $25.0\text{ cm}^3$ ).

Therefore, a burette (A) is the correct apparatus.

Answer

A

Final answer

A

Detailed explanation

Walkthrough

In chemistry laboratories, different apparatus are used depending on the precision and type of volume being measured:

Burette: Designed to deliver accurately measured, variable volumes of liquids up to $50.0\text{ cm}^3$ , with graduations typically allowing readings to the nearest $0.05\text{ cm}^3$ (giving measurements to one or two decimal places, such as $27.30\text{ cm}^3$ or $27.3\text{ cm}^3$ ).
Volumetric Pipette: Delivers a single fixed, exact volume accurately (most commonly $10.0\text{ cm}^3$ or $25.0\text{ cm}^3$ ). It cannot deliver variable non-standard volumes like $27.3\text{ cm}^3$ .
Measuring Cylinder: Used to measure variable volumes reasonably quickly, but is only suitable for approximate measurements rather than exact or precise titration quantities.
Condenser: Used in distillation and reflux apparatus to condense vapours back into liquid; it is not a measuring instrument.

Thus, measuring exactly $27.3\text{ cm}^3$ requires a burette (A).

Key Takeaways

Use a burette for measuring accurate variable volumes of liquids (e.g., in titrations).
Use a volumetric pipette for measuring accurate fixed volumes (e.g., $25.0\text{ cm}^3$ ).
Use a measuring cylinder for approximate volumes.
Use a gas syringe for measuring volumes of gases.

Common Mistakes

Confusing a volumetric pipette with a burette: pipettes only measure specific pre-calibrated fixed volumes.
Choosing a measuring cylinder: while measuring cylinders can measure variable volumes, they lack the high precision denoted by the word "exactly".

Things to Be Careful About

Pay attention to keywords like "exact" / "accurate" versus "approximate", and "fixed" versus "variable" volume.

Techniques used

identify suitable laboratory apparatus for measuring variable volumesdistinguish between precision of volumetric glassware

Q381MEasyExperimental Techniques and Chemical Analysis

A student titrates aqueous sodium hydroxide with 0.1 $\text{mol / dm}^3$ hydrochloric acid. The titration results are used to calculate the concentration of the aqueous sodium hydroxide.

Which row is correct?

Options

	apparatus to measure the volume of dilute hydrochloric acid	apparatus to measure the volume of aqueous sodium hydroxide
A	burette	measuring cylinder
B	burette	volumetric pipette
C	measuring cylinder	burette
D	volumetric pipette	volumetric pipette

Pick an answer

DifficultyEasy

Worked solution

Working

In an acid-base titration, a fixed, accurate volume of one solution (often the alkali, aqueous sodium hydroxide) is measured into the conical flask using a volumetric pipette (usually $25.0\text{ cm}^3$ ).
The other solution (the titrant, dilute hydrochloric acid) is added dropwise from a burette so that the variable volume required to reach the end-point can be measured accurately (to the nearest $0.05\text{ cm}^3$ or $0.1\text{ cm}^3$ ).
A measuring cylinder is not sufficiently accurate for quantitative titration work.

Matching the columns:

Apparatus for dilute hydrochloric acid: burette
Apparatus for aqueous sodium hydroxide: volumetric pipette

This corresponds to row B.

Answer

B

Final answer

B

Detailed explanation

Walkthrough

In quantitative chemical analysis, particularly acid-base titrations, precision is crucial:

Volumetric Pipette: Used to deliver an exact, fixed aliquot of solution (commonly $25.0\text{ cm}^3$ or $10.0\text{ cm}^3$ ) into a conical flask. In standard procedures, this is used to measure the alkali (aqueous $\text{NaOH}$ ).
Burette: A calibrated tube with a stopcock used to dispense and measure variable volumes of liquid accurately up to $50.0\text{ cm}^3$ . It is used to deliver the acid (dilute $\text{HCl}$ ) gradually until the indicator changes colour at the neutralisation end-point.
Measuring Cylinder: Measures approximate volumes (typically $\pm 0.5\text{ cm}^3$ to $\pm 1\text{ cm}^3$ ), which is not precise enough for titration calculations requiring high accuracy.

Therefore, the dilute hydrochloric acid is placed in and measured from a burette, while the sodium hydroxide is measured accurately using a volumetric pipette (row B).

Key Takeaways

A volumetric pipette is designed for measuring a single, precise, fixed volume of liquid.
A burette is designed to measure accurate variable delivered volumes during a titration.
A measuring cylinder is used for approximate volume measurements and is not suitable for accurate titration analysis.

Common Mistakes

Confusing the roles of the burette and pipette, or thinking a measuring cylinder provides adequate accuracy for titration calculations.
Choosing row D by failing to realise that the titrant volume varies until the end-point is reached, so it cannot be measured using a fixed-volume pipette.

Things to Be Careful About

Ensure you identify which substance is delivered variably (the acid in the burette) and which is measured as a fixed volume (the alkali via pipette).

Techniques used

select suitable volumetric apparatus for titrationdistinguish between fixed-volume and variable-volume measuring instruments

Q391MMedium-EasyExperimental Techniques and Chemical Analysis

A mixture of four coloured dyes is analysed by chromatography.

The result is shown.

Fig. 1

Which change allows the four coloured dyes to be seen separately?

Options

A Measure the $R_f$ values of the spots carefully.
B Run the chromatogram for a longer time.
C Run the chromatogram using a different solvent.
D Use a locating agent.

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

The mixture contains four coloured dyes, but only three distinct spots are visible on the chromatogram. This indicates that two of the dyes have very similar solubilities in the chosen solvent and have moved together as a single spot.
Changing the solvent (option C) alters the relative solubilities and affinities of the dyes, allowing all four components to separate into individual spots.
Option A is incorrect because calculating $R_f$ values cannot make an unresolved spot split into two.
Option B is incorrect because the solvent front has already almost reached the top of the paper; running it longer will cause the solvent to run off the edge and blur the spots.
Option D is incorrect because the dyes are already coloured and clearly visible without a locating agent.

Answer

C

Final answer

C

Detailed explanation

Walkthrough

In paper chromatography, substances separate based on differences in their relative solubilities in the mobile phase (the solvent) and their attraction to the stationary phase (the chromatography paper):

Identify the problem: The question states that the sample is a mixture of four coloured dyes, yet the chromatogram displays only three spots. This means that two of the dyes have identical or nearly identical partition between the mobile and stationary phases in this particular solvent, causing them to travel the same distance and overlap.
Evaluate the solution: To achieve complete separation (resolution) of all four dyes, the solvent must be changed. A different solvent will interact differently with each dye's molecular structure, changing their relative solubilities so that the overlapping dyes travel at different rates and separate into four distinct spots.
Eliminate incorrect options:
- A: Calculating $R_f$ values is a measurement tool; it does not change the physical separation of the dyes.
- B: The solvent front is already near the top. Allowing the solvent to run off the top of the paper causes the spots to diffuse and merge, ruining the separation.
- D: Locating agents are used to make colourless substances (such as amino acids) visible. Since the dyes are already coloured and visible, a locating agent is unnecessary.

Key Takeaways

When the number of spots on a chromatogram is fewer than the known number of components in a mixture, two or more substances have overlapped.
Overlapping components can be separated by repeating chromatography with a different solvent.
Locating agents are only needed for colourless substances.

Common Mistakes

Confusing the purpose of a locating agent (used to visualise colourless spots) with a method to improve separation.
Assuming running the experiment for longer will always improve separation; once the solvent front reaches the top, running it longer leads to over-elution and loss of data.

Things to Be Careful About

Ensure you check whether the substances are described as "coloured" or "colourless" in the stem before considering locating agents.
The solvent front must never be allowed to run off the top edge of the paper, as $R_f$ values can no longer be determined accurately.

Techniques used

interpret a chromatogram to identify overlapping componentsselect a suitable modification to improve chromatographic separation

Q401MMedium-EasyExperimental Techniques and Chemical Analysis

An aqueous solution contains cations of metal X.

A precipitate forms when a few drops of aqueous sodium hydroxide are added to the solution.

The precipitate dissolves in excess aqueous sodium hydroxide.

What is a possible identity of metal X?

Options

	aluminium	ammonium	zinc	key
A	✓	✗	✓	✓ = yes
B	✓	✓	✗	✗ = no
C	✗	✗	✓
D	✗	✓	✗

Pick an answer

DifficultyMedium-Easy

Worked solution

Working

With a few drops of aqueous sodium hydroxide:

$\text{Al}^{3+}$ forms a white precipitate of $\text{Al(OH)}_3$ , which dissolves in excess sodium hydroxide.
$\text{Zn}^{2+}$ forms a white precipitate of $\text{Zn(OH)}_2$ , which also dissolves in excess sodium hydroxide.
$\text{NH}_4^+$ does not form a precipitate with sodium hydroxide; it gives ammonia gas.

So metal X could be aluminium or zinc, but not ammonium.

Answer

A

Final answer

A

Detailed explanation

Walkthrough

The question describes a cation test using aqueous sodium hydroxide. When a few drops of sodium hydroxide are added to a solution containing $\text{Al}^{3+}$ or $\text{Zn}^{2+}$ , a white precipitate forms because the metal hydroxide is insoluble in water:

$\text{Al}^{3+} + 3\text{OH}^- \rightarrow \text{Al(OH)}_3$

$\text{Zn}^{2+} + 2\text{OH}^- \rightarrow \text{Zn(OH)}_2$

Both of these hydroxides are amphoteric, meaning they react with excess hydroxide ions to form a soluble complex. So in excess sodium hydroxide, both precipitates dissolve.

Ammonium ions do not form a precipitate with sodium hydroxide. Instead, warming the mixture gives ammonia gas:

$\text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O}$

Therefore the metal X could be aluminium or zinc. Looking at the options, only option A shows aluminium = yes, ammonium = no, and zinc = yes.

Key Takeaways

Aqueous sodium hydroxide is used to test for cations.
$\text{Al}^{3+}$ and $\text{Zn}^{2+}$ both give white precipitates that dissolve in excess sodium hydroxide.
Ammonium ions give no precipitate with sodium hydroxide; they produce ammonia gas.
Questions like this often test whether you know more than one cation can behave in the same way.

Common Mistakes

Thinking ammonium forms a precipitate with sodium hydroxide. It does not; it produces ammonia gas.
Thinking only one of aluminium or zinc dissolves in excess sodium hydroxide. With sodium hydroxide, both do.
Confusing this test with the test using aqueous ammonia, where aluminium hydroxide does not dissolve in excess ammonia but zinc hydroxide does.

Things to Be Careful About

The reagent is aqueous sodium hydroxide, not aqueous ammonia. This matters because the behaviour in excess is different.
The precipitate is white for both aluminium and zinc hydroxide; the colour alone does not distinguish them.
In the table, a tick means the cation gives the described result, and a cross means it does not. Read the key carefully.

Techniques used

recall the action of aqueous sodium hydroxide on metal cationsidentify which hydroxides dissolve in excess sodium hydroxideeliminate ammonium as it does not form a precipitate

Chemistry 5070/11 — October/November 2024

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