ESAT Chemistry Mock 5

Statement 1 is correct: $\displaystyle \ce{HCl}$ is a strong acid and fully dissociates in aqueous solution, giving $\displaystyle [\ce{H+}] = 0.100\text{ mol dm}^{-3}$ . $\displaystyle \ce{CH3COOH}$ is a weak acid and only partially dissociates, giving $\displaystyle [\ce{H+}] \ll 0.100\text{ mol dm}^{-3}$ . The initial rate of reaction with $\displaystyle \ce{CaCO3(s)}$ depends directly on $\displaystyle [\ce{H+}]$ , so the initial rate is greater for $\displaystyle \ce{HCl(aq)}$ .

Statement 2 is incorrect: Since both solutions have equal volumes and equal concentrations of monoprotic acids, they contain identical amounts (moles) of ionisable hydrogen. In the presence of excess $\displaystyle \ce{Mg}$ , the reaction drives the weak acid dissociation equilibrium to completion. The stoichiometric equation for both monoprotic acids is $\displaystyle \ce{Mg + 2HA -> MgA2 + H2}$ , meaning both produce the exact same total number of moles (and therefore volume) of $\displaystyle \ce{H2(g)}$ .

Statement 3 is correct: Complete neutralisation with $\displaystyle \ce{Ba(OH)2(aq)}$ follows the stoichiometry $\displaystyle \ce{2HA + Ba(OH)2 -> BaA2 + 2H2O}$ . Because both solutions contain the same total number of moles of acid, the number of moles of $\displaystyle \ce{Ba(OH)2}$ required for complete neutralisation is identical, so the required volume of $\displaystyle 0.0500\text{ mol dm}^{-3}\ \ce{Ba(OH)2(aq)}$ is the same for both.

Therefore, statements 1 and 3 only are correct.

We determine the subatomic composition for each ion:

- $\displaystyle \ce{^{32}S^{2+}}$ :
- Protons = $\displaystyle 16$ - Neutrons = $\displaystyle 32 - 16 = 16$ - Electrons = $\displaystyle 16 - 2 = 14$ - Mass-to-charge ratio ( $\displaystyle m/z$ ) = $\displaystyle \dfrac{32}{2} = 16$ - $\displaystyle \ce{^{36}Ar^{2+}}$ :
- Protons = $\displaystyle 18$ - Neutrons = $\displaystyle 36 - 18 = 18$ - Electrons = $\displaystyle 18 - 2 = 16$ (electron configuration: 2,8,6)
- Mass-to-charge ratio ( $\displaystyle m/z$ ) = $\displaystyle \dfrac{36}{2} = 18$ - $\displaystyle \ce{^{35}Cl+}$ :
- Protons = $\displaystyle 17$ - Neutrons = $\displaystyle 35 - 17 = 18$ - Electrons = $\displaystyle 17 - 1 = 16$ (electron configuration: 2,8,6)
- Mass-to-charge ratio ( $\displaystyle m/z$ ) = $\displaystyle \dfrac{35}{1} = 35$ Evaluating each statement:
- Statement 1 is correct: Both $\displaystyle \ce{^{36}Ar^{2+}}$ and $\displaystyle \ce{^{35}Cl+}$ possess $\displaystyle 16$ electrons, giving them the same electron configuration ( $\displaystyle 2,8,6$ ).
- Statement 2 is incorrect: Both $\displaystyle \ce{^{36}Ar^{2+}}$ and $\displaystyle \ce{^{35}Cl+}$ have exactly $\displaystyle 18$ neutrons ( $\displaystyle 36 - 18 = 18$ and $\displaystyle 35 - 17 = 18$ ).
- Statement 3 is correct: The $\displaystyle m/z$ values are $\displaystyle 16$ for $\displaystyle \ce{^{32}S^{2+}}$ , $\displaystyle 18$ for $\displaystyle \ce{^{36}Ar^{2+}}$ , and $\displaystyle 35$ for $\displaystyle \ce{^{35}Cl+}$ . Thus, $\displaystyle \ce{^{32}S^{2+}}$ has the lowest $\displaystyle m/z$ .

Hence, statements 1 and 3 only are correct.

W displaces X, so $\displaystyle W>X$ . X reacts with acid while Y does not, so $\displaystyle X>Y$ . Z requires electrolysis, placing it above carbon, while W can be reduced by carbon, placing W below carbon. Hence $\displaystyle Z>W>X>Y$ , option A.

1. Identify which mixtures form a precipitate:
- $\displaystyle \ce{AgNO3(aq) + MgCl2(aq)}$ forms insoluble $\displaystyle \ce{AgCl(s)}$ .
- $\displaystyle \ce{BaCl2(aq) + Al2(SO4)3(aq)}$ forms insoluble $\displaystyle \ce{BaSO4(s)}$ .
- $\displaystyle \ce{NaOH(aq) + CuSO4(aq)}$ forms insoluble $\displaystyle \ce{Cu(OH)2(s)}$ .
- $\displaystyle \ce{NaCl(aq) + KNO3(aq)}$ contains only spectator ions (all sodium, potassium, chloride, and nitrate salts are fully soluble in water), so no precipitate forms ( $\displaystyle 0\text{ g}$ ). The question specifies combinations that *form a precipitate*, so this combination cannot be selected.

2. Calculate the theoretical mass of precipitate for each reacting procedure ( $\displaystyle 1.0\text{ dm}^3$ of $\displaystyle 1.0\text{ mol dm}^{-3}$ solution contains $\displaystyle 1.0\text{ mol}$ of solute):
- $\displaystyle \ce{AgNO3}$ and $\displaystyle \ce{MgCl2}$ : $\displaystyle 1.0\text{ mol}$ of $\displaystyle \ce{MgCl2}$ provides $\displaystyle 2.0\text{ mol}$ of $\displaystyle \ce{Cl-}$ , forming $\displaystyle 2.0\text{ mol}$ of $\displaystyle \ce{AgCl}$ . $\displaystyle \text{Mass} = 2.0\text{ mol} \times 143.5\text{ g mol}^{-1} = 287\text{ g}$ - $\displaystyle \ce{BaCl2}$ and $\displaystyle \ce{Al2(SO4)3}$ : $\displaystyle 1.0\text{ mol}$ of $\displaystyle \ce{Al2(SO4)3}$ provides $\displaystyle 3.0\text{ mol}$ of $\displaystyle \ce{SO4^{2-}}$ , forming $\displaystyle 3.0\text{ mol}$ of $\displaystyle \ce{BaSO4}$ . $\displaystyle \text{Mass} = 3.0\text{ mol} \times 233\text{ g mol}^{-1} = 699\text{ g}$ - $\displaystyle \ce{NaOH}$ and $\displaystyle \ce{CuSO4}$ : $\displaystyle 1.0\text{ mol}$ of $\displaystyle \ce{CuSO4}$ provides $\displaystyle 1.0\text{ mol}$ of $\displaystyle \ce{Cu^{2+}}$ , forming $\displaystyle 1.0\text{ mol}$ of $\displaystyle \ce{Cu(OH)2}$ . $\displaystyle \text{Mass} = 1.0\text{ mol} \times 98\text{ g mol}^{-1} = 98\text{ g}$ 3. Among the procedures that form a precipitate:
- Largest mass: $\displaystyle \ce{BaCl2}$ and $\displaystyle \ce{Al2(SO4)3}$ ( $\displaystyle 699\text{ g}$ )
- Smallest mass: $\displaystyle \ce{NaOH}$ and $\displaystyle \ce{CuSO4}$ ( $\displaystyle 98\text{ g}$ )

Therefore, row D is correct.

Statements 1 and 2 are correct. Chlorination is for disinfection, not desalination, so statement 3 is incorrect.

To determine which reactions involve oxidation, we determine the changes in oxidation states or electron transfers in each reaction:

1. In $\displaystyle \ce{H2O2}$ , oxygen has an oxidation state of $\displaystyle -1$ . In $\displaystyle \ce{O2}$ , oxygen has an oxidation state of $\displaystyle 0$ (oxidation), while in $\displaystyle \ce{H2O}$ , it is $\displaystyle -2$ (reduction). Because this disproportionation produces $\displaystyle \ce{O2}$ with an increased oxidation state (loss of electrons), oxidation is involved.

2. $\displaystyle \ce{MnO4^-(aq) + 8H+(aq) + 5e^- -> Mn^{2+}(aq) + 4H2O(l)}$ is a reduction half-equation. Manganese is reduced from $\displaystyle +7$ to $\displaystyle +2$ by gaining electrons. No oxidised species is present in this half-equation, so it does not involve oxidation.

3. In $\displaystyle \ce{CrO4^{2-}}$ , chromium is in the $\displaystyle +6$ oxidation state. In $\displaystyle \ce{Cr2O7^{2-}}$ , chromium is also in the $\displaystyle +6$ oxidation state. Oxygen remains $\displaystyle -2$ and hydrogen remains $\displaystyle +1$ . This is an acid-base condensation reaction, not a redox reaction; no oxidation occurs.

4. In $\displaystyle \ce{SO2}$ , sulfur has an oxidation state of $\displaystyle +4$ . In $\displaystyle \ce{H2S}$ , sulfur has an oxidation state of $\displaystyle -2$ . In elemental $\displaystyle \ce{S}$ , sulfur has an oxidation state of $\displaystyle 0$ . The sulfur from $\displaystyle \ce{H2S}$ is oxidised from $\displaystyle -2$ to $\displaystyle 0$ , while the sulfur from $\displaystyle \ce{SO2}$ is reduced from $\displaystyle +4$ to $\displaystyle 0$ . Therefore, oxidation is involved.

Thus, reactions 1 and 4 involve oxidation.

Let us analyse the structure and bonding of each substance based on the data:

- Substance W: High melting point ( $\displaystyle 772\text{ }^\circ\text{C}$ ), does not conduct when solid, but conducts when molten and in aqueous solution. This is characteristic of a giant ionic lattice. In the molten state, electrical conduction occurs due to the movement of mobile ions, not delocalised electrons. Thus, statement 1 is incorrect.

- Substance X: Very high melting point ( $\displaystyle 1650\text{ }^\circ\text{C}$ ), non-conductive as a solid and molten, and insoluble in water. This is characteristic of a giant covalent lattice (such as $\displaystyle \ce{SiO2}$ ). Melting it requires breaking many strong covalent bonds throughout the network. Thus, statement 2 is correct.

- Substance Y: Low melting point ( $\displaystyle 122\text{ }^\circ\text{C}$ ) and non-conductive when molten, showing it consists of neutral molecules (simple molecular structure). However, its aqueous solution conducts electricity well. For a solution to conduct electricity, mobile charge carriers must be present; therefore, Y must ionise/react upon dissolving in water to produce mobile ions (e.g., a solid carboxylic acid such as benzoic acid). Thus, statement 3 is correct.

- Substance Z: Low melting point ( $\displaystyle 146\text{ }^\circ\text{C}$ ) and poor conductivity in all states, characteristic of a neutral simple molecular substance (such as glucose or sucrose) that does not ionise in water.

Therefore, only statements 2 and 3 are correct.

The test confirms the presence of water, but it does not establish purity, pH or boiling point.

Let $\displaystyle V$ be the volume and $\displaystyle c$ be the concentration in $\displaystyle \text{mol dm}^{-3}$ of each solution mixed.

- In Experiment 1: $\displaystyle \ce{MgSO4(aq) + Ba(OH)2(aq) -> BaSO4(s) + Mg(OH)2(s)}$ . Both barium sulfate and magnesium hydroxide are insoluble precipitates. Therefore, $\displaystyle cV$ moles of $\displaystyle \ce{BaSO4(s)}$ and $\displaystyle cV$ moles of $\displaystyle \ce{Mg(OH)2(s)}$ are formed. The total mass of dry solid is $\displaystyle cV \times [M_r(\ce{BaSO4}) + M_r(\ce{Mg(OH)2})]$ .

- In Experiment 2: $\displaystyle \ce{Na2SO4(aq) + Ba(OH)2(aq) -> BaSO4(s) + 2NaOH(aq)}$ . Only $\displaystyle \ce{BaSO4(s)}$ is insoluble; sodium hydroxide is soluble. The mass of dry solid formed is $\displaystyle cV \times M_r(\ce{BaSO4})$ .

- In Experiment 3: $\displaystyle \ce{K2SO4(aq) + Ba(OH)2(aq) -> BaSO4(s) + 2KOH(aq)}$ . Only $\displaystyle \ce{BaSO4(s)}$ precipitates; potassium hydroxide is soluble. The mass of dry solid formed is $\displaystyle cV \times M_r(\ce{BaSO4})$ .

Evaluating the statements:
1. Correct: Experiment 1 forms both $\displaystyle \ce{BaSO4(s)}$ and $\displaystyle \ce{Mg(OH)2(s)}$ , whereas Experiment 2 forms only $\displaystyle \ce{BaSO4(s)}$ . Thus, Experiment 1 yields a strictly greater mass of dry solid.
2. Incorrect: Both Experiment 2 and Experiment 3 produce exactly $\displaystyle cV$ moles of $\displaystyle \ce{BaSO4(s)}$ as the only precipitate, so they yield the same mass of dry solid.
3. Correct: In Experiment 1, essentially all $\displaystyle \ce{Mg^{2+}}$ , $\displaystyle \ce{SO4^{2-}}$ , $\displaystyle \ce{Ba^{2+}}$ , and $\displaystyle \ce{OH-}$ ions precipitate out as solid residue, leaving very few dissolved ions in the filtrate. In Experiment 2, $\displaystyle \ce{Na+}$ and $\displaystyle \ce{OH-}$ ions remain completely dissolved in the filtrate at high concentration.

Therefore, statements 1 and 3 only are correct.

Let $\displaystyle n$ be the moles of electrons passed through each electrolytic cell.

**Solution 1 (concentrated $\displaystyle \ce{NaCl(aq)}$ ):**
- At the cathode: $\displaystyle \ce{Na+}$ is more reactive than hydrogen, so water/ $\displaystyle \ce{H+}$ is reduced: $\displaystyle 2\ce{H2O(l)} + 2\ce{e-} \rightarrow \ce{H2(g)} + 2\ce{OH-(aq)}$ . Moles of $\displaystyle \ce{H2} = \dfrac{n}{2}$ .
- At the anode: $\displaystyle \ce{Cl-}$ is oxidised: $\displaystyle 2\ce{Cl-(aq)} \rightarrow \ce{Cl2(g)} + 2\ce{e-}$ .
- The remaining solution contains $\displaystyle \ce{Na+(aq)}$ and $\displaystyle \ce{OH-(aq)}$ (aqueous $\displaystyle \ce{NaOH}$ ), so $\displaystyle [\ce{OH-}]$ increases and the pH increases.

**Solution 2 (dilute $\displaystyle \ce{H2SO4(aq)}$ ):**
- At the cathode: $\displaystyle 2\ce{H+(aq)} + 2\ce{e-} \rightarrow \ce{H2(g)}$ . Moles of $\displaystyle \ce{H2} = \dfrac{n}{2}$ .
- At the anode: $\displaystyle 2\ce{H2O(l)} \rightarrow \ce{O2(g)} + 4\ce{H+(aq)} + 4\ce{e-}$ . Moles of $\displaystyle \ce{O2} = \dfrac{n}{4}$ .

**Solution 3 (concentrated $\displaystyle \ce{CuSO4(aq)}$ ):**
- At the cathode: $\displaystyle \ce{Cu^{2+}(aq)} + 2\ce{e-} \rightarrow \ce{Cu(s)}$ .
- At the anode: $\displaystyle 2\ce{H2O(l)} \rightarrow \ce{O2(g)} + 4\ce{H+(aq)} + 4\ce{e-}$ .
- The generation of $\displaystyle \ce{H+(aq)}$ ions at the anode (leaving $\displaystyle \ce{H2SO4(aq)}$ ) increases $\displaystyle [\ce{H+}]$ , so the pH decreases.

Evaluating the statements:
- Statement 1 is correct: In both Solution 1 and Solution 2, $\displaystyle \dfrac{n}{2}\text{ mol}$ of $\displaystyle \ce{H2(g)}$ is evolved at the cathode. At the same temperature and pressure, equal amounts of gas occupy equal volumes.
- Statement 2 is incorrect: In Solution 2, $\displaystyle n(\ce{H2}) = \dfrac{n}{2}$ at the cathode and $\displaystyle n(\ce{O2}) = \dfrac{n}{4}$ at the anode. The volume of gas evolved at the cathode is twice that at the anode, not half.
- Statement 3 is correct: Solution 1 accumulates $\displaystyle \ce{OH-}$ (pH increases) while Solution 3 accumulates $\displaystyle \ce{H+}$ (pH decreases).

Therefore, statements 1 and 3 only are correct.

To find the values of $\displaystyle w$ and $\displaystyle y$ , we set up conservation equations for each element:

1. Nitrogen (N): $\displaystyle w = x$ 2. Hydrogen (H): $\displaystyle w = 2 + 2y$ 3. Oxygen (O): $\displaystyle 3w = 2(3) + 2x + y = 6 + 2x + y$ Substitute $\displaystyle x = w$ into the oxygen equation: $\displaystyle 3w = 6 + 2w + y \implies w = 6 + y$ Now equate the two expressions for $\displaystyle w$ : $\displaystyle 2 + 2y = 6 + y \implies y = 4$ Substitute $\displaystyle y = 4$ back into the expression for $\displaystyle w$ : $\displaystyle w = 6 + 4 = 10$ (This also gives $\displaystyle x = 10$ . Checking oxygen: $\displaystyle \text{LHS} = 3(10) = 30$ , $\displaystyle \text{RHS} = 6 + 2(10) + 4 = 30$ , which is consistent.)

Therefore, $\displaystyle w = 10$ and $\displaystyle y = 4$ , corresponding to row C.

1. Identify the equivalence point from the graph:
The extrapolated intersection occurs at:
- $\displaystyle \text{Volume of } \ce{H2SO4} = 20\text{ cm}^3$ - $\displaystyle \text{Volume of } \ce{NaOH} = 40\text{ cm}^3$ - $\displaystyle \text{Maximum temperature } T_{\text{max}} = 28.0\text{ }^\circ\text{C}$ 2. **Calculate the temperature rise ( $\displaystyle \Delta T$ ):** $\displaystyle \Delta T = 28.0\text{ }^\circ\text{C} - 18.0\text{ }^\circ\text{C} = 10.0\text{ }^\circ\text{C}$ 3. **Calculate the heat released ( $\displaystyle q$ ):** $\displaystyle \text{Total volume} = 60\text{ cm}^3 \implies m = 60\text{ g}$ $\displaystyle q = m c \Delta T = 60\text{ g} \times 4.0\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 10.0\text{ }^\circ\text{C} = 2400\text{ J} = 2.4\text{ kJ}$ 4. Determine moles of water formed:
The balanced equation is: $\displaystyle \ce{H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l)}$ At the stoichiometric point: $\displaystyle n(\ce{H2SO4}) = 0.020\text{ dm}^3 \times 1.0\text{ mol dm}^{-3} = 0.020\text{ mol}$ $\displaystyle n(\ce{NaOH}) = 0.040\text{ dm}^3 \times 1.0\text{ mol dm}^{-3} = 0.040\text{ mol}$ Therefore, the moles of $\displaystyle \ce{H2O}$ formed is: $\displaystyle n(\ce{H2O}) = 0.040\text{ mol}$ 5. Calculate the molar enthalpy change of neutralisation: $\displaystyle \Delta H_{\text{neut}} = -\dfrac{q}{n(\ce{H2O})} = -\dfrac{2.4\text{ kJ}}{0.040\text{ mol}} = -60\text{ kJ mol}^{-1}$

1. From the displayed structure of butanedioic acid (succinic acid), the molecular formula is $\displaystyle \ce{C4H6O4}$ .

2. Calculate the relative molecular mass ( $\displaystyle M_\text{r}$ ): $\displaystyle M_\text{r} = (4 \times 12.0) + (6 \times 1.0) + (4 \times 16.0) = 48.0 + 6.0 + 64.0 = 118.0\text{ g mol}^{-1}$ 3. Calculate the number of moles of acid: $\displaystyle n = \dfrac{1.18\text{ g}}{118.0\text{ g mol}^{-1}} = 0.010\text{ mol}$ 4. Each molecule contains two carboxylic acid groups ( $\displaystyle -\ce{COOH}$ ), so each mole of acid reacts with $\displaystyle 2\text{ mol}$ of $\displaystyle \ce{NaHCO3}$ to produce $\displaystyle 2\text{ mol}$ of $\displaystyle \ce{CO2(g)}$ : $\displaystyle \ce{HOOC-CH2-CH2-COOH + 2NaHCO3 -> NaOOC-CH2-CH2-COONa + 2CO2 + 2H2O}$ 5. Amount of $\displaystyle \ce{CO2}$ gas produced: $\displaystyle n(\ce{CO2}) = 2 \times 0.010\text{ mol} = 0.020\text{ mol}$ 6. Volume of gas collected: $\displaystyle V = 0.020\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.480\text{ dm}^3 = 480\text{ cm}^3$

1. Calculate the amount of $\displaystyle \ce{CO}$ converted: $\displaystyle \text{Moles of } \ce{CO} \text{ reacted} = 8.00\text{ mol} \times 0.750 = 6.00\text{ mol}$ 2. Determine the moles of each species present at equilibrium:
- Remaining $\displaystyle \ce{CO}$ : $\displaystyle 8.00\text{ mol} - 6.00\text{ mol} = 2.00\text{ mol}$ - Reacted $\displaystyle \ce{H2}$ : $\displaystyle 2 \times 6.00\text{ mol} = 12.0\text{ mol}$ - Remaining $\displaystyle \ce{H2}$ : $\displaystyle 20.0\text{ mol} - 12.0\text{ mol} = 8.00\text{ mol}$ - Formed $\displaystyle \ce{CH3OH}$ : $\displaystyle 6.00\text{ mol}$ 3. Calculate the total moles of gas at equilibrium: $\displaystyle n_{\text{total}} = 2.00\text{ mol} + 8.00\text{ mol} + 6.00\text{ mol} = 16.0\text{ mol}$ 4. Calculate the total volume of the equilibrium mixture: $\displaystyle V = 16.0\text{ mol} \times 45.0\text{ dm}^3\text{ mol}^{-1} = 720\text{ dm}^3$

To evaluate the statements, we first identify the type of structure and bonding present in each substance:

- Substance P has a low melting point ( $\displaystyle -78\,^{\circ}\text{C}$ ) and does not conduct electricity as a solid or a liquid. This indicates a simple molecular structure. Melting substance P involves overcoming relatively weak intermolecular forces; the strong intramolecular covalent bonds remain intact. Thus, statement 1 is incorrect.

- Substance R has a very high melting point ( $\displaystyle 1420\,^{\circ}\text{C}$ ) and does not conduct electricity in either solid or molten states. This is characteristic of a giant covalent (macromolecular) lattice (such as silicon or silicon dioxide). Melting requires breaking many strong covalent bonds throughout the network lattice. Thus, statement 2 is correct.

- Substance Q has a high melting point ( $\displaystyle 770\,^{\circ}\text{C}$ ), does not conduct when solid, but conducts well when molten. This is characteristic of a giant ionic lattice. In the liquid state, ions become mobile and carry the electric current.

- Substance S conducts electricity well in both solid and molten states with a high melting point ( $\displaystyle 1538\,^{\circ}\text{C}$ ), which is characteristic of a giant metallic lattice. In solid metals, electrical conductivity is due to the movement of delocalised electrons through the lattice of metal cations. Thus, statement 3 is correct.

Statements 2 and 3 are correct, so the correct option is G.

Element Q is in Group 17, Period 4 (bromine).

1. Let $\displaystyle x$ be the fractional abundance of the isotope with mass number 79. Then $\displaystyle (1 - x)$ is the fractional abundance of the isotope with mass number 81: $\displaystyle 79x + 81(1 - x) = 79.9$ $\displaystyle 81 - 2x = 79.9$ $\displaystyle 2x = 1.1 \implies x = 0.55 = 55\%$ Thus, statement 1 is correct.

2. Halogen reactivity and oxidising ability decrease down Group 17. Chlorine is above element Q (bromine), so chlorine is a stronger oxidising agent than Q. Therefore, aqueous Q cannot oxidise chloride ions to chlorine (chlorine oxidises bromide, not vice versa). Thus, statement 2 is incorrect.

3. Calcium is in Group 2 and forms $\displaystyle \ce{Ca^{2+}}$ ions, while element Q is in Group 17 and forms $\displaystyle \ce{Q-}$ ions. The resulting compound has the empirical formula $\displaystyle \ce{CaQ2}$ . It is an ionic giant lattice, so its ions become mobile when molten, allowing it to conduct electricity. Thus, statement 3 is correct.

Therefore, statements 1 and 3 only are correct (option F).