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- The bisulfate (or hydrogen sulfate) anion, HSO4 , is a weak acid. The equilibrium constant for the aqueous acid reaction HSO4 (ag) = H* (ag) + SO,2- (ag) is 1.2 x 10-2 (a) Calculate AG° for this equilibrium. Assume a tempera- ture of 25.0°C. (b) At low concentrations, activity coefficients are approxi- mately 1 and the activity of a dissolved solute equals its molality. Determine the equilibrium molalities of a 0.010-molal solution of sodium hydrogen sulfate.What would (triangle) Hrxn be for this new situation with 40 ml of both 1.0 M NaOH and 1.0 M HCl, Ccal=166.8, Tempertre final was 28.9 C, the initial temperature was 22.1 C.Perfluorocarbons (PFCs), hydrocarbons with all the H re-placed by F, have very weak cohesive forces. One unusual appli-cation shows a mouse surviving submerged in O₂-saturated PFCs. (a) At 298 K, perfluorohexane (C₆F₁₄, M=338 g/mol,d=1.674 g/mL) in equilibrium with 101,325 Pa of O₂ has a mole fraction of O₂ of 4.28X10⁻³. What is the kH in mol/Latm?(b) According to one source, the kH for O₂ in water at 25°C is 756.7 L atm/mol. What is the solubility of O₂- in water at 25°C in ppm? (c) Rank the kH values in descending order for O₂- in water,ethanol, C₆F₁₄, and C₆H₁₄. Explain your ranking.
- Lactic acid (CH3 - CH(OH) - COOH) is a weak acid and therefore a weak electrolyte.found in "cut milk". The freezing point of an aqueous solution 0.01 m of lactic acid is 0.0206 ºC. Knowing that in aqueous solution the following equilibrium occurs:CH3 -CH(OH) -COOH(ac) = CH3 -CH(OH) -COO(-)(ac) + H(+)(ac)and that, therefore, there are three species (solutes) in solution, calculate their percentage of ionization.Lactic acid (CH3 - CH(OH) - COOH) is a weak acid and therefore a weak electrolyte. found in "cut milk". The freezing point of an aqueous solution 0.01 m of lactic acid is 0.0206 ºC. Knowing that in aqueous solution the following equilibrium occurs: CH3 -CH(OH) -COOH(ac) = CH3 -CH(OH) -COO(-)(ac) + H(+)(ac) and that, therefore, there are three species (solutes) in solution, calculate their percentage of ionization.A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Find the freezing point of the solution(in C to 2 decimal places)
- A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Find the osmotic pressure in atm to three decimal placesA solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Determine the boiling point of the solution(in C to 2 decimal places)A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Determine the following: Boiling point of solution (in °C to two decimal places) Freezing point of solution (in °C to two decimal places) Vapor pressure of the solution (in atm to three decimal places) Osmotic pressure (in atm to three decimal places)
- Consider an aqueous solution of pyridine, C5NH5(aq), with F = 5.0 mM, at 1 bar and at 298 K. Pyridine is a weak Brønsted-Lowry base. Structure of pyridine, List all the species in solution. If you like, you can abbreviate pyridine as “pyr”.Sucrose (C12H22O11), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula C6H12O6: C12H22O11 (aq) + H2O (l) → C6H12O6 (aq) + C6H12O6 (aq) Sucrose Glucose Fructose At 23 o C and in a 0.5 M HCl, the following data were obtained for the disappearance of sucrose Time (min) [C12H22O11] (M) 0 0.316 39 0.274 80 0.238 140 0.19 210 0.146 Is this reaction first order or second order? What is the value of the rate constant?The following evidence was obtained from an experiment to determine the solubility of calcium chloride at room temperature. A sample of saturated calcium chloride solution was evaporated to dryness, and the mass of solid residue was measured.EvidenceVolume of solution (mL) = 15.0Mass of empty beaker (g) = 90.54Mass of beaker and residue (g) = 101.36The solubility of calcium chloride is g/100 mL