Calculate the Ke for 2NO(g) + Br2(g) 2NOBR(9) with Kp=2.4 at 373 K. Show your solution. What is the implication of this result? (5 pts) Formula: Kp = Ke (RT)An
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- State your basis and show calculations. (Kf,water = 1.86 K kg mol-1) NaCl, MgCl2, and CaCO3 are candidate salts in the homemade ice cream-making process. The salts are to be dissolved in the ice-water mixture to lower the temperature of the mixture. The mixture will be then used to freeze ice cream. Which salt will be much more effective in lowering the temperature of the ice-water mixture considering that equal masses of each salt are to be compared?Table 2. Gibbs Free Energies of formation (kJ), ∆G°f, for Ions in 1M Solution and Ionic Solids cation anions Cl- -131.228 I- -51.57 NO3- -108.74 SO4-2 -744.53 Ba+2 -1296.32 W2 -663.9 -796.59 -1362.2 ∆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 Anions Cl- I- NO3- SO4-2 Ba+2(∆G°rxn) Ba+2 (∆Grxn) Ba+2 (ppt or no ppt) Ba+2 (Rxn or No Rxn) NO RXN NO RXN NO RXN RXN Ba+2 (S or support and R for Refute)A 5.00 g sample of an unknown salt is dissolved in 50.0 mL of water. During dissolution, the temperature of the solution increases from 24.53 oC to 26.82 oC. Which statement(s) is(are) true? qsurroundings > 0 qsystem < 0 qsystem > 0 qsurroundings < 0 (You can choose more than one)
- A 1.00-m solution of acetic acid, CH3COOH, in benzene has a freezing point of 2.96°C. Use the data in the Table to calculate the value of i and suggest an explanation for the unusual result. (Hint: If i is less than 1.0, each formula unit that dissolves yields less than one solute particle, an outcome suggesting aggregation of solute particles.) Answer is: i = 0.50; formation of dimers of composition (CH3COOH)2, need steps shown to understand thoughTable 2. Gibbs Free Energies of formation (kJ), ∆G°f, for Ions in 1M Solution and Ionic Solids cation anions Cl- -131.228 I- -51.57 NO3- -108.74 SO4-2 -744.53 Na+ -261.905 -384.138 -286.06 -367.00 -3646.85 W10 ∆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 Anions Cl- I- NO3- SO4-2 Na+(∆G°rxn) Na+ (∆Grxn) Na+ (ppt or no ppt) Na+ (Rxn or No Rxn) NO RXN NO RXN NO RXN NO RXN Na+ (S or support and R for Refute)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.
- In 1.00 atm of pure oxygen, the solubility of O2 (g) in water is 1.26*10-3 M at 25.0oC. The mole fraction of oxygen in air is 0.210. If the atmospheric pressure is 0.979 atm, what is the solubility of oxygen in air at 25oC? Answer is suppose to be 2.59*10-4 M, please show step by step.What is the boiling point rise in oF of a 15% solution of nitric acid in water boiling at 200oF? Need asapA 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. 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)