VCAA Chemistry What are the current and future options for supplying energy?

This is the total energy absorbed to produce one full mole of glucose. Producing one mole of glucose requires 6 moles of $CO_2$, not just one.

Photosynthesis is an endothermic process that absorbs energy from sunlight. This option incorrectly states that energy is released.

This describes the energy released during the combustion of one mole of glucose, rather than the energy absorbed per mole of $CO_2$ during photosynthesis.

This incorrect answer is obtained by calculating $42.0 / (19.0 / 0.39)$, which incorrectly divides the energy density by the efficiency instead of multiplying them to find the usable energy per tonne.

This incorrect answer is obtained by calculating $19.0 / (42.0 \times 0.39)$, which incorrectly divides the energy density by the required energy instead of dividing the required energy by the usable energy per tonne.

This incorrect answer is obtained by calculating $42.0 / 19.0$, which represents the mass needed if the power station were 100% efficient, completely ignoring the 39% efficiency factor.

This option incorrectly ranks S as stronger than Q. In Cell 2, Q acts as the anode (oxidized) and S as the cathode (reduced), which demonstrates that Q is a stronger reducing agent than S.

This option incorrectly identifies R as the strongest reducing agent. R produces the largest voltage when reduced by Q, indicating it has the highest reduction potential and is therefore the weakest reducing agent.

This option incorrectly ranks Q as stronger than T. In Cell 3, T acts as the anode (oxidized) and Q as the cathode (reduced), which demonstrates that T is a stronger reducing agent than Q.

This is incorrect because biodiesel is derived from renewable biomass and generally produces fewer net carbon emissions and particulates compared to fossil-fuel-derived petrodiesel.

This is incorrect because while petrodiesel consists of non-polar hydrocarbons, biodiesel consists of fatty acid methyl esters which contain polar ester groups.

This is incorrect because biodiesel has a higher cloud point than petrodiesel, meaning it is more prone to gelling and restricting flow in very cold climates.

This is a thermal decomposition reaction where the oxidation states of Calcium (+2), Carbon (+4), and Oxygen (-2) remain unchanged.

This is a combination reaction where no elements change their oxidation states (Ca remains +2, O remains -2, H remains +1).

This is an acid-base neutralization reaction where ions exchange partners without any change in oxidation numbers.

Converting $49\,097 \text{ MJ tonne}^{-1}$ to $\text{kJ g}^{-1}$ gives $49.1 \text{ kJ g}^{-1}$, which yields $4910 \text{ kJ}$ per $100 \text{ g}$. This is slightly less than the energy provided by ethyne.

An energy release of $11.63 \text{ kJ g}^{-1}$ yields $1163 \text{ kJ}$ per $100 \text{ g}$, which is the lowest energy output among the given choices.

Dividing $2670 \text{ kJ mol}^{-1}$ by the molar mass of $74 \text{ g mol}^{-1}$ gives $36.1 \text{ kJ g}^{-1}$, resulting in $3610 \text{ kJ}$ per $100 \text{ g}$.

Hydrogen fuel cells do not produce carbon dioxide ($CO_2$) during operation; they generate electricity through an electrochemical reaction rather than combustion.

While this describes the efficient energy conversion mechanism of a fuel cell, it does not explain the composition of the emissions or why they are less polluting.

This statement describes the operational requirement of a fuel cell (continuous fuel supply) but is unrelated to the environmental impact or chemical composition of its emissions.

This incorrect answer assumes that only 1 mole of electrons is transferred per mole of $O_2$, rather than the actual 4 moles required by the half-equation.

This incorrect answer uses the molar mass of atomic oxygen ($16.0$ g/mol) instead of molecular oxygen ($O_2$, $32.0$ g/mol) to calculate the final mass.

This value represents the number of moles of electrons produced ($n = \frac{36000}{96500} = 0.373$ mol), not the mass of $O_2$ consumed in grams.

While catalysts do lower the activation energy required, they function to speed up the reaction, not slow it down.

Catalysts lower the activation energy required for the reaction to occur, rather than requiring more energy.

This is incorrect because catalysts lower the activation energy and increase the rate of reaction.

This result comes from incorrectly multiplying the theoretical volume by the efficiency ($15.7 \text{ L} \times 0.25$), rather than dividing by it to account for the extra fuel needed due to energy loss.

This value is obtained by incorrectly multiplying the theoretical volume by $0.75$ (or $100\% - 25\%$), which is an incorrect application of the efficiency percentage.

This is the volume of octane required if the engine were $100\%$ efficient ($528 \text{ MJ} / 33.67 \text{ MJ/L}$). It fails to account for the $25.0\%$ efficiency of the engine, which requires more fuel to be burned.

This is incorrect. If manganese had an oxidation state of +1, the total charge of the ion would be $+1 + 4(-2) = -7$, which does not match the actual charge of -1.

This is incorrect. While +2 is a very common oxidation state for manganese (e.g., in $\text{Mn}^{2+}$ salts), it is not the oxidation state found in the permanganate ion.

This is incorrect. Manganese is in Group 7 and has only 7 valence electrons, so its maximum possible oxidation state is +7. It cannot reach +8.

This value is obtained by incorrectly dividing the moles of $\text{CO}_2$ (4.62 mol) by its molar mass (44.0 g/mol) instead of multiplying.

This incorrect answer results from a series of calculation errors, likely involving dividing the mass of butane by the product of the molar masses of butane and carbon dioxide.

This mass is calculated by incorrectly assuming a 1:1 molar ratio between butane and carbon dioxide, rather than the correct 2:8 ratio given in the balanced equation.

While Fe is oxidised, the oxidising agent is the specific species that accepts electrons. In this reaction, the copper(II) ion ($\text{Cu}^{2+}$) accepts electrons, not neutral copper (Cu).

Oxidation involves the reactant losing electrons. Here, solid Fe is oxidised, not the product $\text{Fe}^{2+}$. Additionally, the oxidising agent is the ion $\text{Cu}^{2+}$, not neutral Cu.

The species being oxidised is the reactant Fe, which loses electrons to become $\text{Fe}^{2+}$. The ion $\text{Fe}^{2+}$ is the product of oxidation, not the substance being oxidised.