B. Heat of Reaction Trial I Trial 2 volume of HCl, mL volume of NaOH, mL volume of solution, mL mass of solution, g = V x d 51.0mL 49.8 mL 48.4mL 50.5 mL 99.4mL 100.3mL (devolution T = 1.00 g/mL) solution 99.4 mL 100.3mL = T i, solution ATal i, cal Tf, cal = Tf, solution AT 'solution' (from your plot), °C 9cal, J = 21.0 J/°C XAT cal (from your plot), °C 22.8 °C 23.2°6 29.6°C 29.9°C °C = T - -T f, solution i, solution 6.8° 6-7°C 142.8 J 140.7J 9solution' J = m solution X 4.184 J/g °C XAT solution greaction' J = − (9cal + a solution) greaction, kJ moles of HCl used, mol (V x M = moles) AHDD, kJ/mol rxn' greaction, kJ == moles of HCI Average AH kJ/mol rxn'
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- 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 mLCalculate the Enthalpy of change in Kj/mol and % difference in the neutralization reaction below: HNO2 + NH3 = N2 + 2 H2O Initial Temp= 20 Deg celsius Final Temp= 22.3 Deg celsius Conc= 0.4 M [HNO3] and 0.4 M [NH3] Vol= 55 mL [HNO2] and 60 mL [NH3]= Mass is 115 g Specific Heat= 4.184 J/g For the other value in % difference use: -348 Kj/molFor the reaction: CuS (s) + H₂(g) → H₂S (g) + Cu (s) Given: ΔS ⁰ (CuS) = 66.5 J/mole-K ΔHf ⁰ (CuS) = -53.1 kJ/mol ΔS ⁰ (H₂S) = 205 J/ mole-K ΔHf ⁰ (H₂S) = -20.6 kJ/mol ΔS ⁰ (Cu) = 33.3 J/ mole-K ΔH f ⁰ (Cu) = 0.0 kJ/mol ΔS ⁰ (H₂) = 131 J/ mole-K ΔH f ⁰(H₂) = 0.0 kJ/mol The standard free energy of the reaction ΔG ⁰ rxn is A) -37.9 kJ/mole B) 44.6 kJ/mole C) -20.3 kJ/mole D) +20.3kJ/mole and spontaneous E) +37.9 kJ/mole
- 1.1The Ksp of Ca3 (PO4 ) 2 is 1.3 × 10−26 . Estimate the solubility of this salt in units of g. L −1 . You must show any reaction equation(s) that you may think are necessary. 1.2 If a sample of solid Ca3(PO4)2 is stirred into exactly one litre of a 0.550M solution of Na3PO4, how will the solubility of the salt compare with the answer that you have obtained in question 1.1? Explain you answer in a short sentence.Given: MnS + H+ < -- > Mn2+ + HS- CuS + H+ < -- > Cu2+ + HS- [A] Classify each reaction as exothermic or endothermic based on the DGr0 (Gibbs energy in kJ/mol). [B] What is the Ksp of MnS and CuS? [C] Which is more soluble? Gibbs free energy given in kJ/mol: CuS = −86.2 MnS = 30Table 2. Gibbs Free Energies of formation (kJ), ∆G°f, for Ions in 1M Solution and Ionic Solids Cations Cl--131.228 I--51.57 NO3--108.74 SO4-2-744.53 Ca2+-553.58 -748.1 -528.9 -743.07 -1797.28 W2 ∆G°f of water = -237.129 kJ/mol Calculated values of ∆G°rxn and the ∆Grxn of each box, Predicted results (ppt or no ppt).Observations (Rxn or No Rxn). S or support and R for Refute Cations Cl- I- NO3- SO4-2 Ca+2 help
- 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 vapor pressure of the solution to 3 decimal places in atm.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 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)Thallium(I) iodate (TlIO3) is only slightly soluble in water. Its Ksp at 25°C is 3.07 × 10-6 . Estimate the solubility of thallium(I) iodate in water in units of grams per 100.0 mL of water.The solubility of ZnS(s) in water at a certain temperature is 1.4 × 10–11 mol/L, i.e., x or the [Zn+2]=[S-2]. The value of the Ksp of ZnS is