ESAT Mock Chemistry: Questions & Worked Solutions

Question 1

1 mark

Consider a neutral atom of

^{24}\text{Mg}

and the magnesium ion

^{24}\text{Mg}^{2+}

. The atomic number of magnesium is 12.

Which of the following quantities is/are the same for both the neutral atom and the ion?

1. The number of protons in the nucleus
2. The number of neutrons in the nucleus
3. The number of electrons orbiting the nucleus

2.1, 2 and 3
A.1 only
B.2 only
C.1 and 2 only
D.1 and 3 only

Answer: C

Worked solution

The identity of an element is determined by its atomic number, which is the number of protons. For magnesium, the atomic number is 12, so both the neutral atom and the

^{24}\text{Mg}^{2+}

ion have 12 protons (Statement 1 is correct).

The mass number of this isotope is 24. The number of neutrons is calculated as:

\text{mass number} - \text{atomic number} = 24 - 12 = 12

. Since both are the same isotope (

^{24}\text{Mg}

), they both have 12 neutrons (Statement 2 is correct).

A neutral atom has the same number of electrons as protons (12). However, a

2+

ion is formed when an atom loses 2 electrons. Therefore, the

^{24}\text{Mg}^{2+}

ion has

12 - 2 = 10

electrons. Thus, the number of electrons is different (Statement 3 is incorrect). Only statements 1 and 2 are correct.

Question 2

1 mark

Phosphorus has an atomic number of 15. Which of the following correctly describes the composition of a neutral atom of the isotope

^{31}\text{P}

?

A.15 protons, 16 neutrons, and an electron configuration of 2, 8, 5
B.15 protons, 31 neutrons, and an electron configuration of 2, 8, 5
C.16 protons, 15 neutrons, and an electron configuration of 2, 8, 6
D.15 protons, 16 neutrons, and an electron configuration of 2, 13
E.31 protons, 15 neutrons, and an electron configuration of 2, 8, 5

Answer: A

Worked solution

The atomic number is 15, so there are 15 protons. Since it is a neutral atom, there are also 15 electrons. The number of neutrons is found by subtracting the atomic number from the mass number:

31 - 15 = 16

. The electron configuration for 15 electrons fills the shells in order: the first shell holds 2, the second holds 8, leaving 5 for the third shell (

15 - 2 - 8 = 5

). This gives a configuration of 2, 8, 5. Option A matches all these values.

Question 3

1 mark

An ion of a specific isotope is denoted as

Q^{2+}

. The ion has a nucleon number of 40 and contains 22 neutrons. What is the ratio of the total number of subatomic particles located in the nucleus to the total number of electrons in this ion?

A.2.00
B.2.22
C.2.50
D.1.82
E.2.11

Answer: C

Worked solution

1. Identify the number of nucleons (protons + neutrons): The nucleon number is given as 40. These are the particles in the nucleus.
2. Calculate the number of protons: Number of protons = Nucleon number

-

Number of neutrons

= 40 - 22 = 18

.
3. Calculate the number of electrons: The ion has a charge of

2+

, meaning it has lost 2 electrons relative to the neutral atom. Number of electrons

= 18 - 2 = 16

.
4. Calculate the ratio: Ratio

= \frac{\text{particles in nucleus}}{\text{electrons}} = \frac{40}{16} = 2.5

.

Question 4

1 mark

In a simplified model of an atom, the radius of the atom is approximately

10^5

times larger than the radius of its nucleus. Assuming the nucleus contains nearly all the mass of the atom, what is the approximate ratio of the average density of the nucleus to the average density of the whole atom?

A.10^{5}
B.10^{10}
C.10^{15}
D.10^{-15}
E.1

Answer: C

Worked solution

1. Let

r

be the radius of the nucleus and

R

be the radius of the atom. Given

R = 10^5 r

.
2. Volume

V

is proportional to the cube of the radius (

V = \frac{4}{3}\pi r^3

).
3. The volume of the atom

V_{\text{atom}} = \frac{4}{3}\pi (10^5 r)^3 = 10^{15} \cdot \frac{4}{3}\pi r^3 = 10^{15} V_{\text{nucleus}}

.
4. Density

\rho = \frac{\text{mass}}{\text{volume}}

. Since the mass

M

is approximately the same for both (as electrons have negligible mass and reside outside the nucleus),

\rho_{\text{nucleus}} = \frac{M}{V_{\text{nucleus}}}

and

\rho_{\text{atom}} = \frac{M}{V_{\text{atom}}}

.
5. The ratio

\frac{\rho_{\text{nucleus}}}{\rho_{\text{atom}}} = \frac{M/V_{\text{nucleus}}}{M/V_{\text{atom}}} = \frac{V_{\text{atom}}}{V_{\text{nucleus}}} = \frac{10^{15} V_{\text{nucleus}}}{V_{\text{nucleus}}} = 10^{15}

.

Question 5

1 mark

An ion of element

X

is represented by the symbol

^{A}_{Z}X^{n+}

. The relationship between the atomic number, mass number, and number of electrons (

e

) for this ion is given by the equations below:

A = 2Z + 3

e = Z - 3

Which of the following statements about this ion is/are correct?

1. The value of the charge

n

is equal to 3.
2. The number of neutrons in the ion is

Z + 3

.
3. The total number of subatomic particles (protons, neutrons, and electrons) in the ion is equal to

3Z

.

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

Answer: E

Worked solution

We evaluate each statement using the definitions of atomic number (

Z

), mass number (

A

), and charge.

**Statement 1:** The charge

n+

is determined by the difference between the number of protons and electrons. In any ion,

\text{charge} = \text{protons} - \text{electrons}

. Here, protons

= Z

and electrons

e = Z - 3

.

\text{Charge} = Z - (Z - 3) = +3

. Thus,

n = 3

. Statement 1 is correct.

**Statement 2:** The number of neutrons (

N

) is the mass number minus the atomic number.

N = A - Z = (2Z + 3) - Z = Z + 3

. Statement 2 is correct.

**Statement 3:** The total number of subatomic particles is the sum of protons (

P

), neutrons (

N

), and electrons (

e

).

\text{Total} = P + N + e = Z + (Z + 3) + (Z - 3) = 3Z

. Statement 3 is correct.

Since 1, 2, and 3 are correct, the answer is E.

Question 6

1 mark

An atom has the atomic number 19. Which of the following options correctly identifies the electron configuration, the Period, and the Group of this element in the Periodic Table?

0.Configuration: 2,8,7,2; Period: 4; Group: 2
A.Configuration: 2,8,9; Period: 3; Group: 9
B.Configuration: 2,8,8,1; Period: 3; Group: 1
C.Configuration: 2,8,8,1; Period: 4; Group: 1
D.Configuration: 2,8,8,1; Period: 4; Group: 19
E.Configuration: 2,8,7,2; Period: 4; Group: 2

Answer: C

Worked solution

Element 19 is potassium (

K

). To find its configuration, we fill shells in order: the 1st shell holds 2, the 2nd holds 8, the 3rd holds 8, and the 19th electron goes into the 4th shell, giving 2,8,8,1. The number of shells (4) corresponds to the Period number, so it is in Period 4. The number of valence electrons (1) corresponds to the Group number (Group 1). Option C correctly matches all these details.

Question 7

1 mark

The relative formula mass of an anhydrous metal carbonate

M_2\text{CO}_3

is

138.2

. The metal ion

M^{+}

present in this compound is isoelectronic with a neutral atom of the noble gas Argon (

Z = 18

). Element

M

consists of two naturally occurring isotopes with mass numbers

39

and

41

. Given that the relative formula mass of the carbonate ion

\text{CO}_3^{2-}

is

60.0

, what is the percentage abundance of the heavier isotope of

M

in this sample?

A.5.0%
B.10.0%
C.50.0%
D.90.0%
E.95.0%

Answer: A

Worked solution

1. Determine the relative atomic mass (

A_r

) of element

M

: The formula for the compound is

M_2\text{CO}_3

. Therefore:

2 \times A_r(M) + M_r(\text{CO}_3^{2-}) = 138.2

. Substituting the known value:

2 \times A_r(M) + 60.0 = 138.2 \implies 2 \times A_r(M) = 78.2 \implies A_r(M) = 39.1

.

2. Verify the identity of

M

: The ion

M^{+}

has 18 electrons (isoelectronic with Ar). Thus, the neutral atom

M

has

18 + 1 = 19

protons. Element 19 is Potassium (

K

), which has an

A_r

of approximately

39.1

, consistent with our calculation.

3. Calculate isotopic abundance: Let

x

be the fractional abundance of the heavier isotope (

^{41}M

). The abundance of the lighter isotope (

^{39}M

) is therefore

(1 - x)

.

A_r = (39 \times (1 - x)) + (41 \times x) = 39.1

.

39 - 39x + 41x = 39.1

.

2x = 0.1 \implies x = 0.05

.

Converting to percentage:

0.05 \times 100\% = 5.0\%

.

Question 8

1 mark

A metal

X

has only one stable isotope,

^{121}X

. A halogen

Y

has two naturally occurring isotopes:

^{35}Y

(with an abundance of 75%) and

^{37}Y

(with an abundance of 25%). A sample of the compound

XY_2

is analyzed in a mass spectrometer. Focusing only on the molecular ion

XY_2^{+}

, how many peaks are observed in this region of the spectrum, and what is the ratio of the abundance of the peak with the lowest

m/z

value to the peak with the highest

m/z

value?

A.2 peaks; ratio 3:1
B.3 peaks; ratio 1:1
C.3 peaks; ratio 3:1
D.3 peaks; ratio 9:1
E.4 peaks; ratio 9:1

Answer: D

Worked solution

1. Identify the possible isotopic combinations for

XY_2^{+}

: Since

X

is monoisotopic (

^{121}X

), the variations in mass come only from the two

Y

isotopes (

^{35}Y

and

^{37}Y

). The possible combinations for the two

Y

atoms are:
-

^{35}Y^{35}Y

: total mass

= 121 + 35 + 35 = 191

.
-

^{35}Y^{37}Y

(or

^{37}Y^{35}Y

): total mass

= 121 + 35 + 37 = 193

.
-

^{37}Y^{37}Y

: total mass

= 121 + 37 + 37 = 195

.
There are 3 distinct mass peaks.

2. Calculate relative abundances: Let

p = 0.75

(abundance of

^{35}Y

) and

q = 0.25

(abundance of

^{37}Y

).
- Abundance of lowest

m/z

peak (

191

)

= p^2 = (0.75)^2 = 0.5625

.
- Abundance of middle

m/z

peak (

193

)

= 2pq = 2(0.75)(0.25) = 0.375

.
- Abundance of highest

m/z

peak (

195

)

= q^2 = (0.25)^2 = 0.0625

.

3. Determine the ratio: Lowest (

191

) to Highest (

195

) is

0.5625 : 0.0625

.
Dividing both by

0.0625

:

0.5625 / 0.0625 = 9

. The ratio is

9:1

.

Question 9

1 mark

Element

\text{X}

is located in Period 3 and Group 16 of the Periodic Table.

A specific sample of element

\text{X}

contains only two isotopes:

^{32}\text{X}

and

^{34}\text{X}

. In this sample, the ratio of the abundance of atoms of

^{32}\text{X}

to atoms of

^{34}\text{X}

is

9:1

.

What is the relative atomic mass (

A_r

) of

\text{X}

in this sample, and how many neutrons are present in the nucleus of the heavier isotope?

A. $A_r = 32.2$ ; neutrons = 18
B. $A_r = 32.2$ ; neutrons = 16
C. $A_r = 33.0$ ; neutrons = 18
D. $A_r = 33.8$ ; neutrons = 18
E. $A_r = 33.8$ ; neutrons = 16

Answer: A

Worked solution

Step 1: Determine the atomic number of

\text{X}

.
Element

\text{X}

is in Period 3, Group 16. Period 3 elements have three occupied electron shells. Group 16 elements have 6 valence electrons. The electron configuration is

2, 8, 6

. Total electrons = 16, so the atomic number (

Z

) is 16.

Step 2: Calculate the number of neutrons in the heavier isotope (

^{34}\text{X}

).

\text{Neutrons} = \text{Mass number} - \text{Atomic number} = 34 - 16 = 18

.

Step 3: Calculate the relative atomic mass (

A_r

) using the abundance ratio

9:1

.
The total parts in the ratio are

9 + 1 = 10

.

A_r = \frac{(9 \times 32) + (1 \times 34)}{10} = \frac{288 + 34}{10} = \frac{322}{10} = 32.2

.

Comparing these values to the options,

A_r = 32.2

and 18 neutrons matches option A.

Question 10

1 mark

The element

M

is located in Period 3, Group 2 of the Periodic Table. The element

X

is located in Period 3, Group 17. Which of the following is correct regarding these elements and the compound they form?

A.Element $M$ is a non-metal and element $X$ is a metal.
B.They react to form an ionic compound with the formula $MX_2$ .
C.They react to form a covalent compound with the formula $MX_2$ .
D.Element $X$ has a larger atomic radius than element $M$ .
E.Element $M$ has a higher first ionisation energy than element $X$ .

Answer: B

Worked solution

First, identify the elements:

M

(Period 3, Group 2) is Magnesium (

Mg

) and

X

(Period 3, Group 17) is Chlorine (

Cl

).

Mg

is a metal and

Cl

is a non-metal; therefore, they form an ionic compound.

Mg

typically loses two electrons to form

Mg^{2+}

, while

Cl

gains one electron to form

Cl^-

. To balance the charges, the formula must be

MgCl_2

, which corresponds to

MX_2

.

Regarding the other options: A is incorrect as the types are swapped. C is incorrect because the bonding is ionic, not covalent. D is incorrect because atomic radius decreases across a period, so

Mg

is larger than

Cl

. E is incorrect because first ionisation energy increases across a period, making

Cl

have a higher ionisation energy than

Mg

.

Question 11

1 mark

Consider the layout of the Periodic Table according to IUPAC conventions. What is the atomic number (

Z

) of the element located in Period 4 and Group 15?

A.15
B.31
C.33
D.35
E.51

Answer: C

Worked solution

To find the atomic number without a periodic table, we can sum the number of elements in the preceding periods:
- Period 1 contains 2 elements (

Z = 1

to

2

).
- Period 2 contains 8 elements (

Z = 3

to

10

).
- Period 3 contains 8 elements (

Z = 11

to

18

).

Period 4 begins at

Z = 19

. Following the groups across Period 4:
- Group 1:

Z = 19

- Group 2:

Z = 20

- Groups 3 through 12 contain the 10 transition metals:

Z = 21

to

30

.
- Group 13:

Z = 31

- Group 14:

Z = 32

- Group 15:

Z = 33

Alternatively, knowing the element in Period 3, Group 15 is Phosphorus (

Z = 15

), we add 18 (the number of elements in Period 4 before reaching the same group) to get

15 + 18 = 33

.

Question 12

1 mark

Elements

P, Q, R,

and

S

have consecutive atomic numbers

n, n+1, n+2,

and

n+3

respectively. Element

R

is a noble gas in Period 3. Which of the following statements about these elements is correct?

A.Element $P$ is a metal in Group 16.
B.Element $Q$ forms a stable ion with a $1+$ charge.
C.Element $S$ is an alkali metal in Group 1.
D.Element $P$ is in Period 2.
E.The oxide of element $S$ has an acidic character.

Answer: C

Worked solution

First, identify the elements based on the information provided. Element

R

is a noble gas in Period 3, which is Argon (

Z = 18

).

Since the atomic numbers are consecutive:
-

R: n+2 = 18 \rightarrow n = 16

.
-

P: n = 16

, which is Sulfur (Group 16, Period 3).
-

Q: n+1 = 17

, which is Chlorine (Group 17, Period 3).
-

R: n+2 = 18

, which is Argon (Group 18, Period 3).
-

S: n+3 = 19

, which is Potassium (Group 1, Period 4).

Now evaluate the options:
- A is incorrect:

P

(Sulfur) is a non-metal.
- B is incorrect:

Q

(Chlorine) forms a

1-

ion (

Cl^-

).
- C is correct:

S

(Potassium) is an alkali metal in Group 1.
- D is incorrect:

P

is in Period 3.
- E is incorrect: The oxide of

S

(Potassium oxide,

K_2O

) is a basic metal oxide.

Question 13

1 mark

The elements in the Periodic Table are arranged by increasing atomic number. Element

T

is located in Group 14 and Period 3. Which of the following statements is correct?

A.The element with atomic number $Z-1$ (where $Z$ is the atomic number of $T$ ) is a non-metal.
B.The element with atomic number $Z+1$ is in Group 15.
C.Element $T$ forms a stable ion with a $4+$ charge in aqueous solution.
D.The element directly above $T$ in the Periodic Table has an atomic number of 7.
E.Element $T$ is a gas at room temperature and pressure.

Answer: B

Worked solution

Element

T

is in Period 3 and Group 14.
- Period 1:

H, He

.
- Period 2:

Li

to

Ne

.
- Period 3:

Na

(11),

Mg

(12),

Al

(13),

Si

(14).
So,

T

is Silicon (atomic number

Z = 14

).

Evaluating options:
- A is incorrect:

Z-1 = 13

is Aluminum, which is a metal.
- B is correct:

Z+1 = 15

is Phosphorus, which is in Group 15.
- C is incorrect: Silicon does not readily form

Si^{4+}

ions in aqueous solution; it forms covalent bonds.
- D is incorrect: The element above

T

(Group 14, Period 2) is Carbon, which has an atomic number of 6. (Nitrogen is 7).
- E is incorrect: Silicon is a solid metalloid with a giant covalent structure.

Question 14

1 mark

The first five successive ionisation energies of an element

Y

are

738

,

1451

,

7733

,

10543

, and

13630\,\text{kJ}\,\text{mol}^{-1}

.

Y

is in Period 3 of the Periodic Table.

Which of the following statements about

Y

is/are correct?

1. It is an alkaline earth metal.
2. It reacts with oxygen to form a compound with the formula

Y_2O

.
3. It has a smaller atomic radius than the element in Period 3, Group 17.

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

Answer: A

Worked solution

Step 1: Identify the group of element

Y

. The successive ionisation energies show a significant jump between the second (

1451\,\text{kJ}\,\text{mol}^{-1}

) and the third (

7733\,\text{kJ}\,\text{mol}^{-1}

) values. This indicates that the third electron is being removed from a core shell, meaning

Y

has two valence electrons. Therefore,

Y

belongs to Group 2.

Step 2: Identify the element. The Period 3 element in Group 2 is Magnesium (

Mg

).

Step 3: Evaluate the statements:
1. **Correct.** Group 2 elements are known as the alkaline earth metals.
2. **Incorrect.** Group 2 metals form ions with a

2+

charge, while oxide ions have a

2-

charge. They react with oxygen to form compounds with the formula

YO

(e.g.,

MgO

), not

Y_2O

(which is characteristic of Group 1 alkali metals).
3. **Incorrect.** Atomic radius decreases across a period from left to right due to the increasing nuclear charge attracting the same number of electron shells. Thus, Magnesium (Group 2) has a larger atomic radius than the element in Group 17 (Chlorine) in the same period.

Only statement 1 is correct.

Question 15

1 mark

Three elements are identified by their positions in the Periodic Table:

- Element X: Period 4, Group 2
- Element Y: Period 3, Group 16
- Element Z: Period 2, Group 17

Which of the following lists these elements in order of increasing first ionisation energy?

A. $X < Y < Z$
B. $Y < X < Z$
C. $Z < Y < X$
D. $X < Z < Y$
E. $Y < Z < X$

Answer: A

Worked solution

To order the elements by first ionisation energy (

IE_1

), we apply the general periodic trends:

IE_1

increases across a period (left to right) and increases up a group (bottom to top).

1. Element

X

(Period 4, Group 2) is Calcium (

Ca

). It is the furthest down (highest period number) and the furthest left (lowest group number) of the three. This combination results in the lowest

IE_1

due to high shielding and a large atomic radius.
2. Element

Y

(Period 3, Group 16) is Sulfur (

S

). It is further right and higher up than

X

, so its

IE_1

is significantly higher than that of

X

.
3. Element

Z

(Period 2, Group 17) is Fluorine (

F

). It is in the top right of the table (excluding noble gases), having the smallest atomic radius and highest effective nuclear charge of the three. Therefore, it has the highest

IE_1

.

Comparing the positions:

X

is bottom-left,

Y

is middle-right, and

Z

is top-right. The order of increasing

IE_1

is

X < Y < Z

.

Question 16

1 mark

A student investigates the reactivity of three Group 1 metals by adding small, equal-sized pieces of lithium, sodium, and potassium to separate troughs of water. The following observations are recorded:
- Metal

M_1

: Fizzes steadily and moves slowly across the surface until it disappears.
- Metal

M_2

: Fizzes rapidly, melts into a shiny ball, and skims quickly across the surface.
- Metal

M_3

: Reacts violently, producing a lilac flame and a small explosion.
Which of the following correctly identifies the metals and provides the correct explanation for the trend in reactivity observed?

0. $M_1 = \text{Na}, M_2 = \text{Li}, M_3 = \text{K}$ ; reactivity increases down the group because atomic mass increases.
A. $M_1 = \text{Li}, M_2 = \text{Na}, M_3 = \text{K}$ ; reactivity increases down the group as the outermost electron is further from the nucleus.
B. $M_1 = \text{K}, M_2 = \text{Na}, M_3 = \text{Li}$ ; reactivity increases down the group as the outermost electron is further from the nucleus.
C. $M_1 = \text{Li}, M_2 = \text{Na}, M_3 = \text{K}$ ; reactivity decreases down the group as the nuclear charge increases.
E. $M_1 = \text{Li}, M_2 = \text{Na}, M_3 = \text{K}$ ; reactivity increases down the group because the first ionisation energy increases.

Answer: A

Worked solution

The reactivity of Group 1 metals increases down the group (

\text{Li} < \text{Na} < \text{K}

). The observations match this trend: Lithium (

M_1

) bubbles, Sodium (

M_2

) melts due to the heat of reaction and skims, and Potassium (

M_3

) is reactive enough to ignite the hydrogen gas produced with a lilac flame. The reason for this increase in reactivity is that down the group, the atomic radius increases and there is more shielding from inner electron shells. This makes the electrostatic attraction between the nucleus and the valence electron weaker, so the electron is lost more easily.

Question 17

1 mark

The reactivity of Group 17 elements (the halogens) can be compared using displacement reactions. A student mixes aqueous solutions of halogens with aqueous solutions of sodium halides. In which of the following mixtures will a displacement reaction occur?
1.

\text{Cl}_2(aq) + 2\text{NaI}(aq)

2.

\text{I}_2(aq) + 2\text{NaBr}(aq)

3.

\text{Br}_2(aq) + 2\text{NaCl}(aq)

4.

\text{Br}_2(aq) + 2\text{NaI}(aq)

A.1 and 2 only
B.1 and 4 only
C.2 and 3 only
D.3 and 4 only
E.1, 3, and 4 only

Answer: B

Worked solution

In Group 17, reactivity decreases down the group:

\text{F} > \text{Cl} > \text{Br} > \text{I}

. A displacement reaction occurs when a more reactive halogen replaces a less reactive halide ion from its salt.
1.

\text{Cl}_2

is more reactive than

\text{I}

, so it will displace

\text{I}^-

to form

\text{I}_2

. (Reaction occurs)
2.

\text{I}_2

is less reactive than

\text{Br}

, so it cannot displace

\text{Br}^-

. (No reaction)
3.

\text{Br}_2

is less reactive than

\text{Cl}

, so it cannot displace

\text{Cl}^-

. (No reaction)
4.

\text{Br}_2

is more reactive than

\text{I}

, so it will displace

\text{I}^-

to form

\text{I}_2

. (Reaction occurs)
Thus, reactions 1 and 4 occur.

Question 18

1 mark

A sample of

1.30\,\text{g}

of pure

^{13}\text{C}

is completely burned in an excess of

^{16}\text{O}_2

to form

^{13}\text{C}^{16}\text{O}_2

. Given that the Avogadro constant

N_A = 6.0 \times 10^{23}\,\text{mol}^{-1}

and the atomic numbers (

Z

) of carbon and oxygen are

6

and

8

respectively, what is the total number of neutrons in the nuclei of the product molecules?

A. $4.2 \times 10^{23}$
B. $1.32 \times 10^{24}$
C. $1.38 \times 10^{24}$
D. $1.50 \times 10^{24}$
E. $2.70 \times 10^{24}$

Answer: C

Worked solution

Step 1: Calculate the moles of

^{13}\text{C}

atoms. The molar mass of

^{13}\text{C}

is

13.0\,\text{g/mol}

. Thus,

n = \frac{1.30\,\text{g}}{13.0\,\text{g/mol}} = 0.10\,\text{mol}

.
Step 2: Determine the stoichiometry of the product. The reaction is

^{13}\text{C} + ^{16}\text{O}_2 \rightarrow ^{13}\text{C}^{16}\text{O}_2

. Therefore,

0.10\,\text{mol}

of

^{13}\text{C}

produces

0.10\,\text{mol}

of

^{13}\text{C}^{16}\text{O}_2

.
Step 3: Calculate the number of neutrons in one molecule of

^{13}\text{C}^{16}\text{O}_2

. For

^{13}\text{C}

, neutrons

N = A - Z = 13 - 6 = 7

. For

^{16}\text{O}

, neutrons

N = 16 - 8 = 8

. One molecule contains one

^{13}\text{C}

and two

^{16}\text{O}

nuclei, so total neutrons per molecule

= 7 + 2(8) = 23

.
Step 4: Calculate total neutrons. Total neutrons

= 0.10\,\text{mol} \times 23\,\text{neutrons/molecule} \times 6.0 \times 10^{23}\,\text{mol}^{-1} = 2.3 \times 6.0 \times 10^{23} = 13.8 \times 10^{23} = 1.38 \times 10^{24}

.

Question 19

1 mark

In the formation of the ionic compound magnesium fluoride from its elements (

2\text{Mg} + \text{F}_2 \rightarrow 2\text{MgF}_2

is incorrect, the stoichiometry is

\text{Mg} + \text{F}_2 \rightarrow \text{MgF}_2

), which of the following statements correctly describes the fundamental rearrangement of subatomic particles according to the principles of chemical reactions?

A.Each magnesium nucleus loses two valence electrons, which are then captured by the nuclei of two fluorine atoms.
B.The total number of protons and neutrons in the system remains constant, while the valence electrons of each magnesium atom are transferred to fluorine atoms.
C.The fluorine atoms undergo nuclear transmutation to achieve a stable noble-gas electron configuration.
D.The mass of the products is exactly equal to the mass of the reactants because the total number of atoms, electrons, and photons is conserved.
E.Electrons are transferred from the fluorine atoms to the magnesium atoms to establish a stable crystal lattice.

Answer: B

Worked solution

In a chemical reaction, the identity and number of nuclei (protons and neutrons) are strictly conserved. This eliminates options C and D (which suggests mass-energy equivalence/photon conservation which is irrelevant to basic chemical stoichiometry). Chemical reactions involve the rearrangement of electrons between atoms. In the case of

\text{Mg} + \text{F}_2 \rightarrow \text{MgF}_2

, magnesium (a metal) is oxidized. Each magnesium atom loses 2 valence electrons (

3s^2

) to form

\text{Mg}^{2+}

. These electrons are transferred to the fluorine atoms (non-metals) to form

\text{F}^-

ions. Option A is scientifically incorrect because nuclei do not 'lose' or 'capture' electrons into their own structure; electrons reside in orbitals surrounding the nucleus. Therefore, B is the only accurate description.

Question 20

1 mark

Which one of the following rows correctly identifies the chemical formulae for the covalent molecules methane, ammonia, and sulfur dioxide?

A.methane: $CH_4$ ; ammonia: $NH_4$ ; sulfur dioxide: $SO_2$
B.methane: $CH_3$ ; ammonia: $NH_3$ ; sulfur dioxide: $SO_2$
C.methane: $CH_4$ ; ammonia: $NH_3$ ; sulfur dioxide: $SO_2$
D.methane: $CH_4$ ; ammonia: $NH_3$ ; sulfur dioxide: $SO_3$
E.methane: $C_2H_4$ ; ammonia: $NH_3$ ; sulfur dioxide: $SO_2$

Answer: C

Worked solution

To identify the correct row, we must recall the standard chemical formulae for these common covalent compounds:

1. **Methane**: A hydrocarbon consisting of one carbon atom and four hydrogen atoms. Its formula is

CH_4

.
2. **Ammonia**: A compound consisting of one nitrogen atom and three hydrogen atoms. Its formula is

NH_3

. Note that

NH_4^+

is the ammonium ion, not the neutral molecule ammonia.
3. **Sulfur dioxide**: A compound consisting of one sulfur atom and two oxygen atoms. Its formula is

SO_2

. Note that

SO_3

is sulfur trioxide.

Evaluating the options:
- A is incorrect because it gives the formula for ammonia as

NH_4

.
- B is incorrect because it gives the formula for methane as

CH_3

(the methyl radical).
- C is correct as all three formulae (

CH_4

,

NH_3

,

SO_2

) are correct.
- D is incorrect because it gives the formula for sulfur dioxide as

SO_3

.
- E is incorrect because it gives the formula for methane as

C_2H_4

(ethene).

Question 21

1 mark

A student needs to calculate the relative formula mass (

M_r

) of sodium carbonate for a titration experiment.

Using the relative atomic masses:

Ar(Na) = 23.0

,

Ar(C) = 12.0

,

Ar(O) = 16.0

, what is the relative formula mass of sodium carbonate?

A.83.0
B.90.0
C.106.0
D.74.0
E.143.0

Answer: C

Worked solution

First, determine the chemical formula for sodium carbonate. Sodium forms a

Na^+

ion and the carbonate ion is

CO_3^{2-}

. To balance the charges, two sodium ions are required for every one carbonate ion, giving the formula

Na_2CO_3

.

Next, calculate the relative formula mass (

M_r

) by summing the relative atomic masses of all atoms in the formula:

M_r(Na_2CO_3) = (2 imes Ar(Na)) + (1 imes Ar(C)) + (3 imes Ar(O))

M_r = (2 imes 23.0) + 12.0 + (3 imes 16.0)

M_r = 46.0 + 12.0 + 48.0

M_r = 106.0

Distractor analysis:
- A: 83.0 corresponds to

NaCO_3

(incorrect charge balancing).
- B: 90.0 corresponds to

Na_2CO_2

(incorrect number of oxygen atoms).
- D: 74.0 corresponds to

Na_2CO

.
- E: 143.0 corresponds to

Na(CO_3)_2

.

Question 22

1 mark

Which one of the following chemical formulae is correctly matched to its compound name?

A.ammonium nitrate, $NH_4(NO_3)_2$
B.copper(II) chloride, $Cu_2Cl$
C.iron(III) hydroxide, $Fe(OH)_3$
D.potassium sulfate, $KSO_4$
E.silver chloride, $AgCl_2$

Answer: C

Worked solution

To find the correct formula, the total positive charge from the cations must equal the total negative charge from the anions.

- **A: Ammonium nitrate.** Ions are

NH_4^+

and

NO_3^-

. Both have a magnitude of 1, so the formula is

NH_4NO_3

. The given formula is incorrect.
- **B: Copper(II) chloride.** Ions are

Cu^{2+}

and

Cl^-

. To balance, we need two

Cl^-

ions for every one

Cu^{2+}

ion. The formula is

CuCl_2

. The given formula is incorrect.
- **C: Iron(III) hydroxide.** Ions are

Fe^{3+}

and

OH^-

. To balance the

3+

charge of the iron ion, three

OH^-

ions (each

1-

) are required. The formula is

Fe(OH)_3

. This is correct.
- **D: Potassium sulfate.** Ions are

K^+

and

SO_4^{2-}

. To balance, two

K^+

ions are needed. The formula is

K_2SO_4

. The given formula is incorrect.
- **E: Silver chloride.** Silver ions are typically

Ag^+

and chloride ions are

Cl^-

. The formula is

AgCl

. The given formula is incorrect.

Question 23

1 mark

An aqueous solution of barium hydroxide is reacted with an aqueous solution of sulfuric acid. This reaction produces a white precipitate of barium sulfate and liquid water. Which of the following is the correctly balanced chemical equation for this reaction, including all state symbols?

0. $\text{Ba(OH)}_2(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{BaSO}_4(\text{s}) + 2\text{H}_2\text{O}(\text{l})$
A. $\text{Ba(OH)}_2(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{BaSO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})$
B. $\text{Ba(OH)}_2(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{BaSO}_4(\text{s}) + 2\text{H}_2\text{O}(\text{aq})$
C. $\text{Ba(OH)}_2(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{BaSO}_4(\text{s}) + 2\text{H}_2\text{O}(\text{l})$
E. $\text{BaOH}(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{BaSO}_4(\text{s}) + \text{H}_2\text{O}(\text{l})$

Answer: C

Worked solution

To write the balanced chemical equation, we first determine the formulae of the reactants and products. Barium hydroxide contains

\text{Ba}^{2+}

and

\text{OH}^-

, so its formula is

\text{Ba(OH)}_2

. Sulfuric acid is

\text{H}_2\text{SO}_4

. The products are barium sulfate (

\text{BaSO}_4

) and water (

\text{H}_2\text{O}

).

Step 1: Balance the atoms.

\text{Ba(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{H}_2\text{O}

. This ensures there are 1 Ba, 1 S, 6 O, and 4 H on both sides.

Step 2: Assign state symbols. The problem states both reactants are aqueous solutions:

\text{Ba(OH)}_2(\text{aq})

and

\text{H}_2\text{SO}_4(\text{aq})

. Barium sulfate is a precipitate, meaning it is a solid:

\text{BaSO}_4(\text{s})

. Water is the solvent and is a pure liquid:

\text{H}_2\text{O}(\text{l})

. Note that water cannot be

(\text{aq})

because

(\text{aq})

means 'dissolved in water'.

The correct equation is:

\text{Ba(OH)}_2(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{BaSO}_4(\text{s}) + 2\text{H}_2\text{O}(\text{l})

. Option C matches this.

Question 24

1 mark

Octane (

\text{C}_8\text{H}_{18}

) is a liquid fuel. Which of the following equations correctly represents the complete combustion of octane in excess oxygen at

25^\circ\text{C}

and

100\,\text{kPa}

including state symbols?

A. $2\text{C}_8\text{H}_{18}(\text{l}) + 25\text{O}_2(\text{g}) \rightarrow 16\text{CO}_2(\text{g}) + 18\text{H}_2\text{O}(\text{l})$
B. $2\text{C}_8\text{H}_{18}(\text{g}) + 25\text{O}_2(\text{g}) \rightarrow 16\text{CO}_2(\text{g}) + 18\text{H}_2\text{O}(\text{g})$
C. $2\text{C}_8\text{H}_{18}(\text{l}) + 25\text{O}_2(\text{g}) \rightarrow 16\text{CO}_2(\text{g}) + 18\text{H}_2\text{O}(\text{g})$
D. $\text{C}_8\text{H}_{18}(\text{l}) + 12\text{O}_2(\text{g}) \rightarrow 8\text{CO}_2(\text{g}) + 9\text{H}_2\text{O}(\text{l})$
E. $2\text{C}_8\text{H}_{18}(\text{l}) + 17\text{O}_2(\text{g}) \rightarrow 16\text{CO}(\text{g}) + 18\text{H}_2\text{O}(\text{l})$

Answer: A

Worked solution

Complete combustion of a hydrocarbon involves reacting with oxygen to produce carbon dioxide and water.

Step 1: Write the skeleton equation:

\text{C}_8\text{H}_{18} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}

.

Step 2: Balance the carbon atoms: 8 C on the left, so

8\text{CO}_2

on the right.

Step 3: Balance the hydrogen atoms: 18 H on the left, so

9\text{H}_2\text{O}

on the right.

Step 4: Balance the oxygen atoms:

8 \times 2 + 9 = 25

oxygen atoms on the right. This requires

\frac{25}{2}\text{O}_2

on the left. Multiplying the whole equation by 2 to get integer coefficients:

2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O}

.

Step 5: Determine state symbols at

25^\circ\text{C}

(Standard Ambient Temperature). Octane is a liquid (

\text{l}

). Oxygen is a gas (

\text{g}

). Carbon dioxide is a gas (

\text{g}

). Water is a liquid (

\text{l}

) at

25^\circ\text{C}

.

The correct equation is

2\text{C}_8\text{H}_{18}(\text{l}) + 25\text{O}_2(\text{g}) \rightarrow 16\text{CO}_2(\text{g}) + 18\text{H}_2\text{O}(\text{l})

.

Question 25

1 mark

A piece of magnesium ribbon is placed into a beaker containing an aqueous solution of silver nitrate. A displacement reaction occurs, producing solid silver and aqueous magnesium nitrate. Which of the following represents the balanced net ionic equation for this reaction including state symbols?

A. $\text{Mg}(\text{s}) + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + 2\text{Ag}(\text{s})$
B. $\text{Mg}(\text{s}) + \text{Ag}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{Ag}(\text{s})$
C. $\text{Mg}(\text{s}) + 2\text{AgNO}_3(\text{aq}) \rightarrow \text{Mg(NO}_3)_2(\text{aq}) + 2\text{Ag}(\text{s})$
D. $\text{Mg}^{2+}(\text{aq}) + 2\text{Ag}(\text{s}) \rightarrow \text{Mg}(\text{s}) + 2\text{Ag}^+(\text{aq})$
E. $\text{Mg}(\text{aq}) + 2\text{Ag}^+(\text{s}) \rightarrow \text{Mg}^{2+}(\text{s}) + 2\text{Ag}(\text{aq})$

Answer: A

Worked solution

The reaction involves magnesium metal (

\text{Mg}

) reacting with silver nitrate (

\text{AgNO}_3

) in solution.

Step 1: Write the full molecular equation:

\text{Mg}(\text{s}) + 2\text{AgNO}_3(\text{aq}) \rightarrow \text{Mg(NO}_3)_2(\text{aq}) + 2\text{Ag}(\text{s})

.

Step 2: Write the total ionic equation by dissociating aqueous ionic compounds:

\text{Mg}(\text{s}) + 2\text{Ag}^+(\text{aq}) + 2\text{NO}_3^-(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + 2\text{NO}_3^-(\text{aq}) + 2\text{Ag}(\text{s})

.

Step 3: Remove spectator ions. Nitrate (

\text{NO}_3^-

) is present as

(\text{aq})

on both sides and does not change.

Step 4: The net ionic equation is:

\text{Mg}(\text{s}) + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + 2\text{Ag}(\text{s})

.

Step 5: Verify charge and atom balance. Left side charge:

0 + 2(+1) = +2

. Right side charge:

+2 + 0 = +2

. Atoms are balanced (1 Mg, 2 Ag). Option A is the correct balanced net ionic equation.

Question 26

1 mark

The dichromate(VI) ion,

\text{Cr}_2\text{O}_7^{2-}

, reacts with iodide ions,

\text{I}^-

, in an acidic solution to produce chromium(III) ions,

\text{Cr}^{3+}

, iodine,

\text{I}_2

, and water.

A partially balanced equation for this reaction is:

\text{Cr}_2\text{O}_7^{2-} + a\text{I}^- + b\text{H}^+ \rightarrow 2\text{Cr}^{3+} + d\text{I}_2 + 7\text{H}_2\text{O}

What is the value of the sum

a + b + d

in the simplest balanced equation?

A.16
B.19
C.23
D.26
E.33

Answer: C

Worked solution

To balance the redox equation

\text{Cr}_2\text{O}_7^{2-} + a\text{I}^- + b\text{H}^+ \rightarrow 2\text{Cr}^{3+} + d\text{I}_2 + 7\text{H}_2\text{O}

, we can use oxidation states or half-equations.

1. **Oxidation State changes:**
- Chromium in

\text{Cr}_2\text{O}_7^{2-}

is

+6

. In

2\text{Cr}^{3+}

, it is

+3

. For two chromium atoms, the total decrease is

2 \times 3 = 6

electrons.
- Iodine in

\text{I}^-

is

-1

. In

\text{I}_2

, it is

0

. To balance the electrons, the 6 electrons lost by Cr must be gained by the iodine part. Since each iodine atom changes by 1, we need 6 iodide ions (

a = 6

).
- These 6 iodide ions form 3 iodine molecules (

d = 3

).

2. **Balancing Hydrogen and Oxygen:**
- There are 7 oxygen atoms in the dichromate ion, which are balanced by the 7 water molecules on the right.
- The 7 water molecules contain

7 \times 2 = 14

hydrogen atoms. Therefore, there must be 14 hydrogen ions on the left (

b = 14

).

3. **Verifying Charge:**
- Left side:

(-2) + 6(-1) + 14(+1) = -2 - 6 + 14 = +6

- Right side:

2(+3) + 3(0) + 7(0) = +6

The equation is balanced with

a=6, b=14, d=3

.

4. **Summing the coefficients:**

a + b + d = 6 + 14 + 3 = 23

.

Question 27

1 mark

Consider the following reversible gas-phase reaction at equilibrium in a closed container:

\text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \rightleftharpoons \text{PCl}_5(\text{g})

The forward reaction is exothermic (

\Delta H = -88\,\text{kJ}\,\text{mol}^{-1}

).

Which of the following changes, when applied independently and assuming all other conditions remain constant, will increase the equilibrium yield of

\text{PCl}_5(\text{g})

?

1. Increasing the temperature
2. Decreasing the volume of the container
3. Adding more

\text{Cl}_2(\text{g})

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

Answer: D

Worked solution

To determine which changes increase the equilibrium yield of

\text{PCl}_5(\text{g})

, we apply Le Châtelier's Principle to each change independently:

1. **Increasing the temperature**: The reaction is exothermic (

\Delta H

is negative). According to Le Châtelier's Principle, an increase in temperature favours the endothermic (reverse) reaction to absorb the added thermal energy. This shifts the equilibrium to the left, which decreases the yield of

\text{PCl}_5(\text{g})

. Action 1 is incorrect.

2. **Decreasing the volume**: Decreasing the volume of the container increases the total pressure and the concentration of all gaseous species. The system responds by shifting the equilibrium to the side with the fewer number of moles of gas. On the left side (reactants), there are

1 + 1 = 2

moles of gas. On the right side (product), there is only

1

mole of gas. Therefore, the equilibrium shifts to the right, increasing the yield of

\text{PCl}_5(\text{g})

. Action 2 is correct.

3. **Adding more

\text{Cl}_2(\text{g})

**:

\text{Cl}_2

is a reactant. Increasing the concentration of a reactant shifts the equilibrium to the right to counteract the change by consuming the added reactant. This increases the yield of

\text{PCl}_5(\text{g})

. Action 3 is correct.

Since actions 2 and 3 increase the yield, the correct answer is D.