This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to the questions: 16) When a burning candle is put into a gas jar of sulphur (IV) oxide gas, it goes off but a burning magnesium ribbon continues to burn. Explain these observations. A burning candle goes off in sulphur (IV) oxide (SO_2) gas because SO_2 is a non-supporter of combustion*. It does not provide oxygen for the candle to burn. A burning magnesium ribbon continues to burn in SO_2 because magnesium is a very reactive metal* and is a strong reducing agent. It reduces the sulphur (IV) oxide to sulphur, while magnesium itself gets oxidized to magnesium oxide. The reaction is: 2Mg(s) + SO_2(g) → 2MgO(s) + S(s) The heat from the burning magnesium is sufficient to sustain this reaction. 17) Starting with copper metal, describe how pure copper (II) carbonate can be prepared in the laboratory. Step 1: Convert copper metal to a soluble copper salt. Heat copper metal in air to form copper (II) oxide: 2Cu(s) + O_2(g) heat 2CuO(s) Then, react copper (II) oxide with a dilute acid (e.g., nitric acid or sulfuric acid) to form a soluble copper salt: CuO(s) + 2HNO_3(aq) → Cu(NO_3)_2(aq) + H_2O(l) (or CuO(s) + H_2SO_4(aq) → CuSO_4(aq) + H_2O(l)) Step 2: Precipitate copper (II) carbonate. Add a solution of a soluble carbonate (e.g., sodium carbonate or potassium carbonate) to the copper (II) nitrate (or sulfate) solution. Copper (II) carbonate is insoluble and will precipitate out. Cu(NO_3)_2(aq) + Na_2CO_3(aq) → CuCO_3(s) + 2NaNO_3(aq) Step 3: Purify the copper (II) carbonate. Filter the precipitate to separate it from the solution. Wash the precipitate with distilled water to remove any soluble impurities (like sodium nitrate). Finally, dry the pure copper (II) carbonate in an oven or desiccator. 18) a) State the observations that would be made when chlorine gas is bubbled through potassium bromide. The colorless potassium bromide solution would turn orange-brown* (or yellow-brown) as bromine is displaced. b) Write the ionic equation for the reaction in (a) above. Cl_2(g) + 2Br^-(aq) → 2Cl^-(aq) + Br_2(aq) 19) An element M has two isotopes. And. The relative atomic mass of the naturally occurring M is 63.5. Calculate the percentage abundance of each isotope. This question is incomplete as the mass numbers of the two isotopes are not provided. To solve this, we need the mass numbers of the two isotopes. Assuming the two isotopes are M-63 and M-65 (common for elements with average atomic mass around 63.5, like Copper): Step 1: Set up equations for percentage abundance. Let x be the percentage abundance of M-63. Then, the percentage abundance of M-65 is (100 - x). Step 2: Use the formula for relative atomic mass. Relative atomic mass = (mass of isotope 1 × \% abundance 1) + (mass of isotope 2 × \% abundance 2)100 63.5 = ((63 × x) + (65 × (100 - x)))/(100) Step 3: Solve for x. 63.5 × 100 = 63x + 6500 - 65x 6350 = 6500 - 2x 2x = 6500 - 6350 2x = 150 x = (150)/(2) x = 75 Step 4: Calculate the percentage abundance of each isotope. Percentage abundance of M-63 = 75\%. Percentage abundance of M-65 = (100 - 75)\% = 25\%. 20) Distinguish between the terms deliquescent and efflorescent as used in chemistry. A substance is deliquescent* if it absorbs moisture from the atmosphere and dissolves in it to form a solution. This process occurs when the vapor pressure of the saturated solution formed is less than the partial pressure of water vapor in the air. A substance is efflorescent* if it loses its water of crystallization to the atmosphere when exposed to air, becoming a powder. This occurs when the vapor pressure of the hydrate is greater than the partial pressure of water vapor in the air. Last free one today — make it count tomorrow, or type /upgrade for unlimited.