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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You're on a roll — Here are the answers to your questions: 1. a) Write the electron configuration of each of the elements X and Y. Step 1: Determine the number of electrons for element X. Ion X^3+ has an electron configuration of 2.8. This means it has lost 3 electrons. Original number of electrons for element X = 2 + 8 + 3 = 13. Electron configuration of element X: 2.8.3 Step 2: Determine the number of electrons for element Y. Ion Y^2- has an electron configuration of 2.8. This means it has gained 2 electrons. Original number of electrons for element Y = 2 + 8 - 2 = 8. Electron configuration of element Y: 2.6 b) Write the formula of the compound formed when X and Y react. Element X has 3 valence electrons (from 2.8.3), so it forms an ion with a charge of +3 (X^3+). Element Y has 6 valence electrons (from 2.6), so it forms an ion with a charge of -2 (Y^2-). To balance the charges, two X^3+ ions are needed for every three Y^2- ions (2 × +3 = +6 and 3 × -2 = -6). The formula of the compound is X_2Y_3. c) What type of bond would be formed in the compound named in b) above? Element X forms a positive ion (X^3+), indicating it is a metal. Element Y forms a negative ion (Y^2-), indicating it is a non-metal. The bond formed between a metal and a non-metal is an ionic bond. Type of bond: Ionic bond 2. Draw an apparatus that is used in the laboratory to measure: a) 250 cm^3 of liquids or solutions. A volumetric flask of 250 cm^3 capacity is used for accurate measurement of a specific volume of liquid or for preparing solutions of known concentration. It has a narrow neck with a single calibration mark. b) 26.5 cm^3 of hydrochloric acid. A burette is used for accurate measurement of variable volumes of liquids, especially in titrations, to a precision of 0.05 cm^3. It is a long, graduated glass tube with a stopcock at the bottom. 3. Starting with Zinc; describe how solid Zinc carbonate can be prepared. Step 1: React zinc metal with a dilute acid, such as nitric acid, to form a soluble zinc salt (zinc nitrate). Zn(s) + 2HNO_3(aq) → Zn(NO_3)_2(aq) + H_2(g) Step 2: Add a solution of a soluble carbonate, such as sodium carbonate, to the zinc nitrate solution. This will cause insoluble zinc carbonate to precipitate. Zn(NO_3)_2(aq) + Na_2CO_3(aq) → ZnCO_3(s) + 2NaNO_3(aq) Step 3: Filter the mixture to separate the solid zinc carbonate precipitate from the solution. Step 4: Wash the collected zinc carbonate precipitate with distilled water to remove any soluble impurities. Step 5: Dry the solid zinc carbonate, for example, by placing it in a desiccator or in a warm oven. Send me the next one 📸