The question "What is the volume of energy required to burn 45cm³ of mixture at S.T.P" contains a conceptual error, as energy is not measured in volume units (cm³). However, given the options are in cm³ and it's a chemistry problem involving "burning" a "mixture" at S.T.P., it is highly probable that the question intends to ask for the volume of oxygen required to combust a specific gaseous fuel. We will assume the "mixture" refers to a common gaseous fuel, and the question asks for the volume of oxygen required for its complete combustion. Among common gaseous fuels, methane ($\text{CH}_4$) is a frequent subject in such problems. Step 1: Write the balanced chemical equation for the combustion of methane. The combustion of methane with oxygen produces carbon dioxide and water. $$\text{CH}_4(g) + 2\text{O}_2(g) \to \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$$ Step 2: Apply Gay-Lussac's Law of Combining Volumes. At Standard Temperature and Pressure (S.T.P.), the volumes of gaseous reactants and products are in simple whole-number ratios, which correspond to their stoichiometric coefficients in the balanced chemical equation. From the balanced equation, the ratio of methane to oxygen is 1:2. This means 1 volume of $\text{CH}_4$ reacts with 2 volumes of $\text{O}_2$. Step 3: Calculate the volume of oxygen required. Given the volume of the "mixture" (assumed to be methane) is 45 cm³. Volume of $\text{O}_2$ required = $2 \times \text{Volume of CH}_4$ Volume of $\text{O}_2$ required = $2 \times 45 \text{ cm}^3 = 90 \text{ cm}^3$ This result matches option D. The final answer is $\boxed{\text{D) 90.0cm}^3}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.