20 Must Know Year 11 Chemistry Questions with Solutions

The Aufbau principle states that electrons fill the lowest available energy levels first. Since, the \small 2p orbitals have a lower energy than the \small 3s orbital, the electrons will fill the \small 2p first. Therefore, the electron spin-orbital diagram in option (A) violates the Aufbau principle.

Back to question 1

Across a period, consecutive elements have one more proton but the same number of electron shells. Therefore, the nuclear charge increases but the effect of electron shielding is mostly unchanged. Hence the nucleus can exert a greater attraction to electrons. Thus, electronegativity increases across a period.

Down a group, successive elements have a greater number of protons and electron shells. However, the effect of shielding increases more than the attraction from the increased nuclear charge. Hence the nucleus can exert a smaller attraction to electrons. Thus, electronegativity decreases down a group.

Back to question 2

Back to question 3

\small SO_2 is a covalent molecular compound. The discrete \small SO_2 molecules are held together by relatively weak intermolecular forces.

\small SiO_2 is a covalent network substance that is held together in a lattice by extremely strong covalent bonds.

Covalent bonds in a lattice network are stronger than ionic bonds which are in turn stronger than intermolecular forces. Therefore, more thermal energy, and thus higher temperatures, are required to break the covalent bonds in \small SiO_2 than that which is required to break the ionic bonds in \small KBr and the weak intermolecular forces in \small SO_2 . As such, the boiling point (BP) and melting point (MP) of \small SiO_2 is much higher than the BP and MP of \small KBr which is again higher than the BP and MP of \small SO_2 .

Back to question 4

{90 \atop 38} Sr \rarr \boldsymbol{{90 \atop 39}Y} + {0 \atop -1}e 

Back to question 5

Step 1: Write a balanced equation.

  {HCl}_{(aq)} \rarr {H^+}_{(aq)} + {Cl^-}_{(aq)}

Step 2: Calculate the moles of the known substance from the known quantity.

\begin{aligned}  n(HCl) & = c \times V \\ \\ & = 2.0 \times 0.010 \\ \\ & = 0.020 \text{ mol} \end{aligned}

Step 3: Calculate the moles of the unknown substance using mole ratios.

\begin{aligned} n(Cl^-) &= n(HCl) \\\\ &=0.020 \text{ mol} \end{aligned}

Step 4: Convert the moles to the desired quantity.

\begin{aligned} \text{number of } Cl^- \text{ ions } & = n \times N_A \\ \\ & = 0.020 \times 6.022 \times 10^{23} \\ \\ & = 1.2 \times 10^{22} \text{ (2 s.f.)} \end{aligned}

Back to question 6

\begin{aligned} m(Mg) &= 30.34 - 29.80 \\ \\ &= 0.54 \text{ g} \end{aligned} 

Step 2: Calculate the moles of \small Mg.

\begin{aligned} n(Mg) &= \dfrac{0.54}{24.31} \\  \\ &= 0.0222 \text{ mol} \end{aligned}

Step 3: Calculate the mass of \small O.

\begin{aligned} m(O) &= 30.70-30.34 \\ \\ & = 0.35\ \text{ g} \end{aligned}

Step 4: Calculate the moles of \small O.

\begin{aligned} n(O) &= \dfrac{0.35}{16.00} \\ \\ &= 0.0225 \text{ mol} \end{aligned}

Step 5: Convert the mole ratio to the simplest whole number ratio

\begin{aligned} n(Mg) : n(O) &= 0.0222:0.0225 \\ & \approx 1:1 \end{aligned}

Therefore, the empirical formula is \small MgO.

Back to question 7

\begin{aligned} T_1 &= 25 \degree \text{C} \\\\ T_2 & = \ ? \\\\ V_2 & = (1-0.65) \times V_1 \\\\ & = 0.35 \times V_1 \end{aligned}

Step 2: Write the formula that relates the known and desired quantities.

\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}

Step 3: Calculate the desired quantity.

\begin{aligned} \frac{V_1}{T_1} &= \frac{V_2}{T_2} \\\\ \frac{\cancel{V_1}}{25+273.15} & = \frac{0.35 \times \cancel{V_1}}{T_2} \\\\ \frac{1}{25+273.15}&=\frac{0.35}{T_2}  \\\\ T_2 &= 0.35 \times (25+273.15) \\\\ & = 104.35 \text{ K} \\\\ \therefore T_2 & = -168.80 \degree \text{C} \end{aligned}

The temperature of the gases inside the balloon is 10 \small \degree \text{C} higher than the temperature of liquid nitrogen.

\begin{aligned} T_\text{ liquid nitrogen} & = -168.8 -10 \\\\ & = -178.8 \degree \text{C} \end{aligned}

Therefore, the temperature of liquid nitrogen is -178.8 \small \degree \text{C} .

Back to question 8

\begin{aligned} n(NaCl) & = \frac{m}{MM} \\\\ & = \frac{7.6}{(22.99 + 35.45)} \\\\ & = 0.13 \\\\ [NaCl] &= \frac{n}{V} \\\\ & = \frac{0.13}{0.250} \\\\ \therefore [NaCl] & = 0.52 \text{ mol L}^{-1} \text{(2 s.f.)} \end{aligned}

Part b)

Step 1: Identify the known and the desired quantity.

c_1 = 0.52 \text{ mol L}^{-1}

V_1 = 10 \text{ mL}

V_2 = 10+40 = 50 \text{ mL}

Step 2: Write the formula that relates the known and desired quantities.

c_1V_1 = c_2V_2

Step 3: Calculate the desired quantity.

\begin{aligned} c_1V_1 & = c_2V_2 \\\\  c_2 & = \frac{c_1V_1}{V_2} \\\\ & = \frac{0.52 \times 10}{50} \\\\ & = 0.10 \text{ mol L}^{-1} \text{(2 s.f.)}  \end{aligned}

Back to question 9

{BaCl_2}_{(s)} + {Na_2SO_4}_{(aq)} \rarr {BaSO_4}_{(s)} + 2{NaCl}_{(aq)}

Part b)

Step 1: Calculate the moles of each reactant present.

\begin{aligned} n(BaCl_2) & = \frac{m}{MM} \\\\ & = \frac{2.5}{(137.3 + 2\times 35.45)} \\\\ & =0.0120 \text{ mol} \\\\ n(Na_2SO_4) & = c\times V \\\\ & = 1.0 \times 0.010 \\\\ & = 0.010 \text{ mol} \end{aligned}

Step 2: Determine the moles of a product formed from each reactant.

\begin{aligned} 0.0120 \text{ mol}\ BaCl_2 &\rarr 0.0120 \text{ mol}\ BaSO_4\ (1:1\ \text{mole ratio}) \\\\ 0.010 \text{ mol}\ Na_2SO_4 &\rarr 0.010 \text{ mol}\ BaSO_4\ (1:1\ \text{mole ratio}) \end{aligned}

Step 3: Identify the limiting reactant.

From the calculations, \small Na_2SO_4 produces the least \small  BaSO_4 . Therefore, \small Na_2SO_4 is the limiting reagent.

Part c)

\begin{aligned} n(BaSO_4) &= n(Na_2SO_4) \\\\ & = 0.010 \text{ mol} \\\\ m(BaSO_4) & = n \times MM \\\\ & = 0.010 \times (137.3 + 32.07 + 4 \times 16.00) \\\\ & = 2.3 \text{ g} \text{ (2 s.f.)} \end{aligned}

Part d) 

\begin{aligned} n(BaCl_2)_\text{remaining} & = n(BaCl_2)_\text{initial} - n(BaCl_2)_\text{reacted} \\\\ & = 0.0120 - 0.010 \\\\ & = 0.0020 \text{ mol} \\\\ m(BaCl_2)_\text{remaining} & = n \times MM \\\\ & = 0.0020 \times (137.3 + 2\times 35.45) \\\\ & = 0.418 \\\\ \therefore m(BaCl_2)_\text{remaining} & = 0.42 \text{ g} \text{ (2 s.f.)} \end{aligned}

Back to question 10

Back to question 11

2{Al}_{(s)} + 3{CuSO_4}_{(aq)} \rarr {Al_2(SO_4)_3}_{(aq)} + 3{Cu}_{(s)}

Full ionic equation:

2{Al}_{(s)} + 3{Cu^{2+}}_{(aq)} + 3{{SO_4}^{2-}}_{(aq)} \rarr 2{Al^{3+}}_{(aq)} + 3{{SO_4}^{2-}}_{(aq)} + 3{Cu}_{(s)}

Net ionic equation:

2{Al}_{(s)} + 3{Cu^{2+}}_{(aq)} \rarr 2{Al^{3+}}_{(aq)} + 3{Cu}_{(s)}

Part b)

The oxidation numbers of reach of the elements are shown below.

2\overset{\textcolor{ff5748}{0}}{{Al}}_{(s)} + 3\overset{\textcolor{326df5}{+2}}{{Cu}^{2+}}_{(aq)} \rarr 2\overset{\textcolor{ff5748}{+3}}{{Al^{3+}}}_{(aq)} + 3\overset{\textcolor{326df5}{0}}{{Cu}}_{(s)}

Therefore the change in the oxidation number aluminium is +3.

Part c)

The oxidation number of aluminium has increased, while the oxidation number of copper has decreased. Therefore, aluminium has been oxidised, acting as the reducing agent while copper has been reduced, acting as the oxidising agent.

Back to question 12

Using the standard reduction potentials from HSC Data Sheet for zinc and silver, we can determine the following two redox half-equations and their corresponding standard potentials as shown in the table below.

The standard reduction potential of zinc is more negative than silver. Therefore, zinc will undergo oxidation at the anode while silver will undergo reduction at the cathode.

\begin{aligned} {Zn}_{(s)} &\rarr Zn^{2+} + 2e^- \kern{1cm}  \text{ (oxidation)}\\\\ Ag^+ + e^- &\rarr {Ag}_{(s)} \kern{2.1cm} \text{ (reduction)} \\\\ 2{Ag}^+ + {Zn}_{(s)} &\rarr 2{Ag}_{(s)} + {Zn}^{2+} \kern{0.7cm} \text{ (overall equation)} \end{aligned} 

Part b)

The electrons flow from the zinc anode to the silver cathode.

Part c)

A sodium chloride salt bridge is not a suitable salt bridge for this galvanic cell as the chloride ions can form insoluble precipitates with the silver ions in the solution.

{Ag^+}_{(aq)} + {Cl^-}_{(aq)} \rarr {AgCl}_{(s)}

An example of a more suitable alternative is sodium nitrate.

Back to question 13

Step 1: Identify the half-equations from the table of standard reduction potentials.

Step 2: Determine the standard potentials of the anode and cathode.

The standard reduction potential of manganese is more negative than tin. Therefore, manganese will undergo oxidation at the anode while tin will undergo reduction at the cathode.

\begin{aligned} {Mn}_{(s)} \rarr {Mn}^{2+} + 2e^- \kern{1cm} E_\text{\textcolor{326df5}{ox}}^\circ &= 1.18 \\\\ {Sn}^{2+} + 2e^- \rarr {Sn}_{(s)} \kern{1cm} E_\text{\textcolor{ff5748}{red}}^\circ &= -0.14 \end{aligned} 

Step 3: Calculate the standard cell potential of the electrochemical cell.

\begin{aligned} E_\text{cell}^\circ & = E_\text{\textcolor{326df5}{ox}}^\circ + E_\text{\textcolor{ff5748}{red}}^\circ \\\\ & = (1.18) + (-0.14) \\\\ & = 1.04 \\\\ \therefore E_\text{cell}^\circ &= 1.04 \text{ V}  \end{aligned}

Part b)

Step 1: Write a balanced overall cell equation for the galvanic cell.

{Mn}_{(s)} + {Sn^{2+}}_{(aq)} \rarr {Mn^{2+}}_{(aq)} + {Sn}_{(s)}

Step 2: Calculate the moles of the known substance from the known quantity.

The question states that cathode gained 0.64 \small \text{g} . In part a) we found that tin was reduced and therefore is the cathode.

\begin{aligned} m(Sn) &=  0.64 \text{ g} \\\\ n(Sn) & = \frac{m}{MM} \\\\ \frac{0.64}{118.7} \\\\ & = 0.00539 \dots \text{mol} \end{aligned}

Step 3: Calculate the moles of the unknown substance using mole ratios.

\begin{aligned} n(Mn) & = n(Sn) \\\\ & = 0.00539 \dots \text{mol} \end{aligned}

Step 4: Convert the moles to the desired quantity.

\begin{aligned} m(Mn)_\text{reacted} & = n\times MM \\\\ & = 0.00539 \dots \times 54.94 \\\\ & = 0.2962 \text{ g} \\\\ m(\text{Anode}) &= m(Mn)_\text{inital} - m(Mn)_\text{reacted} \\\\ & = 4 - 0.2962 \\\\ & = 3.7038 \\\\ \therefore m(\text{Anode})& = 3.7 \text{ g (2 s.f.)} \end{aligned}

Back to question 14

• Increasing concentration of acid used. For example, use a 2.0 \small \text{M} solution of \small HCl.
• Using a more reactive metal such as magnesium.
• Using smaller pieces of metal to increase the surface area exposed to the acid.
• Conducting the experiment at a higher temperature, for example, by placing the reaction vessel in a warm water bath.

Back to question 15

{CH_3OH}_{(l)} + \frac{3}{2} {O_2}_{(g)} \rarr 2{H_2O}_{(l)} + {CO_2}_{(g)}

Step 2: Calculate the heat lost or gained by the substance (water and glass beaker).

\small \begin{aligned} q_\text{ total} & = m_\text{ water }c_\text{ water }\Delta T + m_\text{ glass beaker } c_\text{ glass beaker }\Delta T \\ \\ & = 0.100 \times 4.18 \times 10^{3} \times (36.5-20.8) + 0.025 \times 0.840\times 10^{-3}  \times (36.5-20.8) \\ \\ &= 6892.3 \text{ J}  \end{aligned}

Step 3: Calculate the heat lost or gained by the combustion process.

\begin{aligned} q_\text{ combustion process} &= -q_\text{ substance} \\ \\ &=-6892.3 \text{ J} \\ \\ &=-6.8923 \text{ kJ} \end{aligned}

Step 4: Calculate the enthalpy change of the process.

\begin{aligned} n(CH_3OH) & = \dfrac{n}{MM} \\ \\ &=\dfrac{0.35}{12.01+4\times 1.008+16} \\ \\ &=0.0109 \dots \text{ mol} \end{aligned}

\begin{aligned} \Delta H_c&=\dfrac{q_\text{ combustion process}}{n_\text{ combustion process}} \\ \\ & = -\dfrac{6.8923}{0.0109 \dots} \\ \\ \therefore \Delta H_c & = - 631 \text{ kJ mol}^{-1} \text{(3 s.f.)} \end{aligned}

Back to question 16

Step 2: Calculate the enthalpy of reaction using the sum of the bond energies.

\begin{aligned} {\Delta H}_{r} &= \sum {{\Delta H}_{\text{ reactant bond energies}}} − \sum {{\Delta H}_{\text{ product bond energies}}} \\\\ & = 1998 - 2044 \\\\ & = -46 \text{ kJ mol}^{-1} \end{aligned}

Therefore, the change in enthalpy for this reaction is -46 \small \text{ kJ mol}^{-1} .

Back to question 17

{H_2}_{(g)} + C_{(s)} + \frac{1}{2}{O_2}_{g} \rarr {H_2CO}_{(g)} \kern{1cm}  \Delta H_f^\circ (H_2CO)

Step 2: Write the intermediate equations that can be used to represent the desired process.

\begin{aligned} {H_2CO}_{(g)} + {O_2}_{(g)} \rarr {CO_2} + {H_2O}_{(l)} \kern{1cm} \Delta H_1 &= -563.5 \text{kJ mol}^{-1} \\\\ {C}_{(s)} + {O_2}_{(g)} \rarr {CO_2}_{(g)} \kern{1cm} \Delta H_2& = -393.5 \text{kJ mol}^{-1} \\\\ {H_2}_{(g)} + \frac{1}{2}{O_2}_{(g)} \rarr {H_2O}_{(l)} \kern{1cm} \Delta H_3 & = -285.8 \text{kJ mol}^{-1} \end{aligned}

Step 3: Identify the reactants and products that only appear in one intermediate equation.

\begin{aligned} \textcolor{326df5}{{H_2CO}_{(g)}} + {O_2}_{(g)} \rarr {CO_2}_{(g)} + {H_2O}_{(l)} \kern{1cm} \Delta H_1 &= -563.5 \text{kJ mol}^{-1} \\\\ \textcolor{7d5ee4}{{C}_{(s)}} + {O_2}_{(g)} \rarr {CO_2}_{(g)} \kern{1cm} \Delta H_2& = -393.5 \text{kJ mol}^{-1} \\\\ \textcolor{ff5748}{{H_2}_{(g)}} + \small{\frac{1}{2}}{O_2}_{(g)} \rarr {H_2O}_{(l)} \kern{1cm} \Delta H_3 &= -285.8 \text{kJ mol}^{-1} \end{aligned}

Step 4: Adjust each intermediate equation and the corresponding enthalpy values if required.

In the desired equation, formaldehyde is on the products side.

{H_2O}_{(l)} + {CO_2}_{(g)} \rarr  {O_2}_{(g)} + \textcolor{326df5}{{H_2CO}_{(g)}} \kern{1cm} \Delta H_1 = \textcolor{326df5}{-}(-563.5) \text{kJ mol}^{-1}

The other two equations do not need to be adjusted.

Step 5: Add the adjusted intermediate equations and the corresponding enthalpy values.

\begin{array}{ccc}  {H_2O}_{(l)} + {CO_2}_{(g)} \rarr  {O_2}_{(g)} + {{H_2CO}_{(g)}} &\Delta H_1 = +563.5 \text{kJ mol}^{-1} \\\\ {{C}_{(s)}} + {O_2}_{(g)} \rarr {CO_2}_{(g)}  &\Delta H_2 = -393.5 \text{kJ mol}^{-1} \\\\ {{H_2}_{(g)}} + \small{\frac{1}{2}}{O_2}_{(g)} \rarr {H_2O}_{(l)}  &\Delta H_3  = -285.8 \text{kJ mol}^{-1} \\\\ \hline \\ {H_2}_{(g)} + C_{(s)} + \frac{1}{2}{O_2}_{g} \rarr {H_2CO}_{(g)}   &\Delta H_f^\circ  = -115.8 \text{kJ mol}^{-1} \end{array} 

Therefore, the enthalpy of formation of formaldehyde is -115.8 \small \text{kJ mol}^{-1} \text{ (4 s.f.)}

Back to question 18

Back to question 19

{C_6H_{12}O_6}_{(s)} + {6O_2}_{(g)}  \rarr 6{CO_2}_{(g)} + 6{H_2O}_{(l)}

Step 2: Calculate the enthalpy and entropy changes of the system.

Enthalpy:

\begin{aligned} \Delta H_{r}^{\circ} &= \sum\Delta H^{\circ} _{f\text{ products}} − \sum\Delta H^{\circ}_{f\text{ reactants}} \\ \\ &= (6\times \Delta H^{\circ} _{f (CO_2)} + 6\times\Delta H^{\circ}_{f(H_2O)}) - ( \Delta H^{\circ}_{f(C_6H_{12}O_6)} + 6 \times \Delta H^{\circ}_{f(O_2)})\\ \\ &= (6 \times - 393.5 + 6 \times - 285.8) - (-1274 + 6 \times 0) \\ \\ &= -4075.8 + 1274 \\ \\ \therefore \Delta H_{r}^{\circ} &= -2801.8 \text{ kJ mol}^{-1} \\\\ \end{aligned}

Entropy:

\begin{aligned} \Delta S_{r}^{\circ} &= \sum S^{\circ} _{\text{ products}} − \sum S^{\circ}_{\text{ reactants}} \\ \\ &= (6 \times S^{\circ}_{CO_2} + 6\times S^{\circ}_{H_2O}) − (S^{\circ}_{C_6H_{12}O_6} + 6\times S^{\circ}_{O_2}) \\ \\ & = (6 \times 213.7 + 6 \times 69.91)  - (212.0 + 6 \times 205.1) \\ \\ &= 1701.66 - 1442.6 \\ \\ &= 259.06 \text{ J K}^{-1} \text{ mol}^{-1} \end{aligned}

Step 3: Convert the units of temperature, enthalpy and entropy changes if required.

\begin{aligned} \Delta S & = 259.06 \text{ J K}^{-1} \text{ mol}^{-1} \\\\ & = 0.25906 \text{ kJ K}^{-1}\text{mol}^{-1} \end{aligned}

Step 4: Substitute the terms into the equation for Gibbs free energy.

\begin{aligned} \Delta G_r^\circ & = \Delta H_r^\circ - T \Delta S_r^\circ \\\\  \Delta G_r^\circ & = -2801.8 \text{ kJ mol}^{-1} - T\times 0.25906 \text{ kJ K}^{-1}\text{mol}^{-1} \end{aligned}

Step 5: Calculate the temperature range for the reaction by setting Gibbs free energy to less than zero.

\small  \begin{aligned} \Delta G_r^\circ & = -2801.8 \text{ kJ mol}^{-1} - T\times 0.25906 \text{ kJ K}^{-1}\text{mol}^{-1} < 0 \\\\  0&> -2801.8 \text{ kJ mol}^{-1} - T\times 0.25906 \text{ kJ K}^{-1}\text{mol}^{-1} \\\\ T\times 0.25906 \text{ kJ K}^{-1}\text{mol}^{-1} &> -2801.8 \text{ kJ mol}^{-1} \\\\ T & > \frac{-2801.8 \text{ kJ mol}^{-1}}{0.25906 \text{ kJ K}^{-1}\text{mol}^{-1}} \\\\ & > -10815 \text{ K} \end{aligned}

Since the minimum possible temperature is absolute zero, (0 \small \text{ K} ). The reaction must be spontaneous for all temperatures.

Back to question 20