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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Question 8: a) Step 1: Analyze the electronic arrangement of S. The electronic arrangement for S is 2:8:8. This means it has 2+8+8 = 18 electrons. Step 2: Define an ion. An ion is an atom or molecule that has gained or lost one or more electrons, resulting in a net electrical charge (positive or negative). Step 3: Explain why S is not an ion. The species P^2-, Q^2+, and R^+ all have charges indicated, signifying they are ions. S has no charge indicated, meaning it is a neutral species. A neutral atom has an equal number of protons and electrons. Therefore, S represents a neutral atom, not an ion. The electronic configuration 2:8:8 corresponds to the noble gas Argon, which exists as a neutral atom. b) Step 1: Determine the number of electrons and identify the neutral atoms for P, Q, R, and S. • P^2- has 2:8:8 electrons. The neutral atom P would have 18 - 2 = 16 electrons, which is Sulfur (S). • Q^2+ has 2:8 electrons. The neutral atom Q would have 10 + 2 = 12 electrons, which is Magnesium (Mg). • R^+ has 2:8 electrons. The neutral atom R would have 10 + 1 = 11 electrons, which is Sodium (Na). • S has 2:8:8 electrons. The neutral atom S has 18 electrons, which is Argon (Ar). Step 2: Compare the atomic radii of the neutral atoms (Sulfur, Magnesium, Sodium, Argon). All these elements are in Period 3 of the periodic table. Atomic radius generally decreases across a period from left to right due to an increasing nuclear charge pulling the electrons closer to the nucleus. Step 3: Identify the element with the largest atomic radius. Among Sodium (Group 1), Magnesium (Group 2), Sulfur (Group 16), and Argon (Group 18), Sodium is furthest to the left in Period 3. Therefore, Sodium has the largest atomic radius. The element with the largest atomic radius is R. Reason: R is Sodium, which is in Group 1, Period 3. Atomic radius decreases across a period due to increasing nuclear charge. As Sodium is the furthest left among the elements in Period 3, it has the largest atomic radius. Question 9: Step 1: Calculate the mass of 1 L of the solution. Given density = 1.134 gm/cm^3 = 1.134 g/mL. Volume = 1 L = 1000 mL. Mass of solution = Density × Volume = 1.134 g/mL × 1000 mL = 1134 g Step 2: Calculate the mass of pure HCl in 1 L of the solution. Percentage purity = 37\%. Mass of pure HCl = 37\% of 1134 g = 0.37 × 1134 g = 419.58 g Step 3: Calculate the molar mass of HCl. Molar mass of H = 1.008 g/mol Molar mass of Cl = 35.453 g/mol Molar mass of HCl = 1.008 + 35.453 = 36.461 g/mol Step 4: Calculate the moles of HCl in 1 L of the solution. Moles of HCl = Mass of pure HClMolar mass of HCl = 419.58 g36.461 g/mol = 11.5076 mol Step 5: Determine the molarity of the solution. Molarity = Moles of HClVolume of solution in L = 11.5076 mol1 L = 11.5076 M Rounding to a reasonable number of significant figures (e.g., three, based on 37% purity): The molarity of the solution is 11.5 M. Question 10: The question is incomplete as it provides pH values and solution labels but does not ask for a specific task or calculation. That's 2 down. 3 left today — send the next one.