This appears to be an essay question based on an experiment. Different oxidizing agents possess varying strengths. When compared to nitric acid (HNO3), sulfuric acid (H2SO4) serves as a weaker oxidizing agent. Therefore, using H2SO4 in place of HNO3 would not yield the same experimental results.
1) The ionic compound present in solution b is K₂CrO₄ (potassium chromate). This compound contains two potassiums (oxidation state +1), a single chromium (oxidation state +6), and four oxygen atoms. The oxidation state of oxygen is -2, resulting in a neutral compound: 2 · (+1) + 6 + x · (-2) = 0. Hence, x = 4, denoting the count of oxygen atoms. 2) The ionic compound in solution a is AgNO₃ (silver nitrate). ω(N) = 8.246% ÷ 100%. Thus, ω(N) = 0.08246, indicating the mass percentage of nitrogen. M(MNO₃) = M(N) ÷ ω(N). It follows that M(MNO₃) = 14 g/mol ÷ 0.08246, leading to M(MNO₃) = 169.8 g/mol; the molar mass of the metal nitrate. M(M) = M(MNO₃) - M(N) - 3 · M(O). Consequently, M(M) = 169.8 g/mol - 14 g/mol - 3 · 16 g/mol, resulting in M(M) = 107.8 g/mol which is the atomic mass of silver (Ag). 3) The balanced chemical equation is: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq). In ionic form: 2Ag⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq). The net ionic equation is: 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s). Thus, the red precipitate is identified as silver chromate (Ag₂CrO₄). 4) The mass of solid silver chromate created is m(Ag₂CrO₄) = 331.8 g. The amount is determined by n(Ag₂CrO₄) = m(Ag₂CrO₄) ÷ M(Ag₂CrO₄). Therefore, n(Ag₂CrO₄) = 331.8 g ÷ 331.8 g/mol yields n(Ag₂CrO₄) = 1 mol. From the balanced equation, n(Ag₂CrO₄): n(AgNO₃) = 1: 2, it follows n(AgNO₃) = 2 · 1 mol, which means n(AgNO₃) = 2 mol. Then, the mass of silver nitrate is computed as m(AgNO₃) = n(AgNO₃) · M(AgNO₃). Hence, m(AgNO₃) = 2 mol · 169.8 g/mol gives m(AgNO₃) = 339.6 g; thus, m(AgNO₄) equals m(K₂CrO₄). Therefore, m(K₂CrO₄) = 339.6 g; amount of potassium chromate is n(K₂CrO₄) = m(K₂CrO₄) ÷ M(K₂CrO₄). Thus, n(K₂CrO₄) = 339.6 g ÷ 194.2 g/mol thus arrives at n(K₂CrO₄) = 1.75 mol. 5) The dissociation of silver nitrate in water is expressed as: AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq). Volume of solution a = 500 mL ÷ 1000 mL/L results in V(solution a) = 0.5 L. Concentration equation c(AgNO₃) = n(AgNO₃) ÷ V(solution a), thus c(AgNO₃) = 2 mol ÷ 0.5 L, yielding c(AgNO₃) = 4 mol/L = 4 M. As a result: c(AgNO₃) = c(Ag⁺) = c(NO₃⁻). Thus, c(Ag⁺) = 4 M; the concentration of silver ions in the initial solution a. 6) The dissociation of potassium chromate in water is represented as: K₂CrO₄(aq) → 2K⁺(aq) + CrO₄²⁻(aq). Volume of solution b = 500 mL ÷ 1000 mL/L results in V(solution b) = 0.5 L. Following, c(K₂CrO₄) is calculated as n(K₂CrO₄) ÷ V(solution b). So c(AgNO₃) = 1.75 mol ÷ 0.5 L gives c(AgNO₃) = 3.5 mol/L = 3.5 M. Consequently: c(K⁺) = 7 M; the concentration of potassium ions in solution b. Therefore, c(CrO₄²⁻) = 3.5 M; the concentration of chromium ions in the same solution. 7) The total final volume is V(final solution) = V(solution a) + V(solution b). Thus, V(final solution) = 500.0 mL + 500.0 mL leads to V(final solution) = 1000 mL ÷ 1000 mL/L results in V(final solution) = 1 L. Then n(NO₃⁻) = 2 mol. Therefore, c(NO₃⁻) = n(NO₃⁻) ÷ V(final solution) finds c(NO₃⁻) = 2 mol ÷ 1 L and results in c(NO₃⁻) = 2 M; the concentration of nitrate anions in the final solution. 8) In solution b, there are 3.5 mol of potassium cations while part of that combines with 2 moles of nitrate anions: K⁺(aq) + NO₃⁻(aq) → KNO₃(aq). From the reaction: n(K⁺): n(NO₃⁻) = 1: 1. Thus, Δn(K⁺) = 3.5 mol - 2 mol results in Δn(K⁺) = 1.5 mol, signifying the remaining potassium anions in the final solution. Thus, c(K⁺) = Δn(K⁺) ÷ V(final solution) yields c(K⁺) = 1.5 mol ÷ 1 L, leading to c(K⁺) = 1.5 M; the final concentration of potassium cations.
<span>According to crystal field strength, the Cl ligand results in the longest d-d transition when coordinated with Ti(III) due to its classification as a weak field ligand that causes minimal d orbital splitting.</span>
To calculate the moles of MgSO4.7H2O, we find the molar mass equals 246, thus moles = 32 / 246 = 0.13 moles. Upon heating, all 7 H2O from one molecule will evaporate. The total moles of H2O present amount to 7 x 0.13 = 0.91, and the mass of that H2O is 0.91 x 18 = 16.38g. Therefore, the mass of the anhydrous MgSO4 that remains is 32 - 16.38 = 15.62 g.
