Explain why a burning magnesium continues to burn in a gas jar full of carbon (IV) oxide while a burning candle would be extinguished.

Here are the solutions to both questions. Question 7: Explain why a burning magnesium continues to burn in a gas jar full of carbon (IV) oxide while a burning candle would be extinguished. A burning candle requires oxygen (O_2) from the air to sustain its combustion. Carbon (IV) oxide (CO_2) is a stable compound and does not provide oxygen to support the combustion of most substances, including a candle. Therefore, a candle would be extinguished in a gas jar full of CO_2. Magnesium is a highly reactive metal. When burning, it can react directly with carbon (IV) oxide. In this reaction, magnesium reduces the carbon dioxide to elemental carbon and forms magnesium oxide. This reaction is highly exothermic, releasing enough energy to sustain the burning of magnesium. 2Mg(s) + CO_2(g) 2MgO(s) + C(s) Question 8: Determine the empirical formula of the hydrocarbon. Given: Mass of CO_2 = 8.4 g Mass of H_2O = 3.42 g Atomic mass of H = 1.0 Atomic mass of C = 12.0 Atomic mass of O = 16.0 Step 1: Calculate the molar masses of CO_2 and H_2O. Molar mass of CO_2 = 12.0 + (2 × 16.0) = 12.0 + 32.0 = 44.0 g/mol Molar mass of H_2O = (2 × 1.0) + 16.0 = 2.0 + 16.0 = 18.0 g/mol Step 2: Calculate the moles of CO_2 and H_2O produced. Moles of CO_2 = Mass of CO_2Molar mass of CO_2 = 8.4 g44.0 g/mol ≈ 0.1909 mol Moles of H_2O = Mass of H_2OMolar mass of H_2O = 3.42 g18.0 g/mol = 0.19 mol Step 3: Determine the moles of carbon and hydrogen atoms in the hydrocarbon. Since all carbon in the hydrocarbon forms CO_2, the moles of carbon atoms are equal to the moles of CO_2. Moles of C atoms = Moles of CO_2 = 0.1909 mol Since all hydrogen in the hydrocarbon forms H_2O, and each molecule of H_2O contains two hydrogen atoms, the moles of hydrogen atoms are twice the moles of H_2O. Moles of H atoms = 2 × Moles of H_2O = 2 × 0.19 mol = 0.38 mol Step 4: Determine the simplest whole-number ratio of C to H atoms. Divide the moles of each element by the smallest number of moles (0.1909 mol). For Carbon: 0.1909 mol0.1909 mol ≈ 1 For Hydrogen: 0.38 mol0.1909 mol ≈ 1.99 ≈ 2 The ratio of C:H is approximately 1:2. The empirical formula of the hydrocarbon is CH_2. Last free one today — make it count tomorrow, or type /upgrade for unlimited.