This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
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 (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. CH_4(g) + 2O_2(g) CO_2(g) + 2H_2O(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 CH_4 reacts with 2 volumes of 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 O_2 required = 2 × Volume of CH_4 Volume of O_2 required = 2 × 45 cm^3 = 90 cm^3 This result matches option D. The final answer is D) 90.0cm^3. Last free one today — make it count tomorrow, or type /upgrade for unlimited.