Use the data in Thermodynamic Properties to calculate the vapor pressure of water at 50°C. Express your answer in torr. 88 X torr (please answer to 2 sig figs)

For the reaction 2Fe(s) + 3Cl2(g) →2FeC13 (8) AG-686 kJ and AS-440.3 J/K at 257 K and 1 atm. This reaction is favored under standard conditions at 257 K. The standard enthalpy change for the reaction of 2.15 moles of Fe(s) at this temperature would be kJ.

Determine the vapor pressure for a solution at 45°C for a solution that contained 25.0 g acetone and 75.0 g benzene. The vapor pressure for pure benzene is 0.296 atm and pure acetone is 0.757 atm at 45°C. THE ANSWER IS 0.438 atm. Please show work.

The dissociation of N2O4 into NO2 is 16.7% complete at 298 K and 1.013 bar.N2O4(g) ⇋ 2 NO2(g)Determine K and ΔG°rxn.

For a certain chemical reaction, the standard Gibbs free energy of reaction at  15.0   °C  is  142.   kJ . Calculate the equilibrium constant  K  for this reaction.Round your answer to  2  significant digits.

Please help me with my he. I can't seem to solve this problem. I hope you can help me This is a guide to answer the activity. HESS LAW (Law of Heat Summation) ​“The enthalpy change of the overall process is the sum of the enthalpy changes of the individual component steps.” - It is the standard enthalpy of Reaction, ΔHᵒ. - It can be illustrated in the following example Consider the following reactions: ​(a) Fe(s) + O2(g) à FeO(s)​​​​ΔHᵒ = -272 kJ ​(b) 2 Fe(s) + O2 (g) à Fe2O3(s)​​​ΔHᵒ = -82.5 kJ 1. Make sure to write the balance equation. 2. The final equation requires 2 moles of FeO which means we will multiply equation (a) and its enthalpy change by 2. 3. Reverse equation (a), which means, the product goes to the reactant side, (2) (FeO(s) à Fe(s) + O2(g))​​​ΔHᵒ = (-272 kJ)(2) = 544 kJ ​2 FeO(s) à 2 Fe(s) + O2(g))​​​ΔHᵒ = -825.5 kJ 4. Combine the final equation (a) and the initial equation (b). And then cancel the same components with same number of particles. In this…

For which case would ΔHsoln be expected to be negative?(A) if solute-solute interactions are much greater than solvent-solvent and solute-solvent interactions(B) if solute-solvent interactions are much greater than solvent-solvent and solute-solute interactions(C) if solute-solvent interactions are the same as solvent-solvent and solute-solute interactions(D) if solvent-solvent interactions are much greater than solute-solvent and solute-solute interactions

A chemical engineer is studying the two reactions shown in the table below. In each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 37.0 °C and constant total pressure. Then, he measures the reaction enthalpy AH and reaction entropy AS of the first reaction, and the reaction enthalpy AH and reaction free energy AG of the second reaction. The results of his measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and AS for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium. 6C(s) + 6H₂(g) + 30₂(g) → CH₁₂O6 (s) CH₂OH(g) + CO(g) → HCH₂CO₂ (1) ΔΗ = -1237. kJ J K AS-3953. AG = KJ Which is spontaneous? O this reaction. O the reverse reaction O neither ΔΗ = −…

Classify each of the reactions according to one of the four reaction types summarized in the table below. Reaction Spontaneous Process? Туре ΔΗ' (system) AS (system) (Standard Conditions) 1 Exothermic, 0 Spontaneous at all temperatures. AS (universe) > 0. Exothermic, 0 Positive, > 0 Depends on relative magnitudes of AH and AS". Spontaneous at higher temperatures. 3 Endothermic, > 0 Negative, < o Not spontaneous at any temperature. AS° (universe) < 0. 4 C6 H12O6 (s) + 6 02 (g) → 6 CO2 (g) + 6 H2O(l) a. A,H° = -673 kJ/mol-rxn A,S° = 60.4 J/K · mol-rxn O Spontaneous at all temperatures. O Spontaneous at lower temperatures. O Spontaneous at higher temperatures. ONot spontaneous at any temperature. b. ВаО(s) + C(graphite) — Ba(s) + СО(g) A,H° = 437.6 kJ/mol-rxn A,S° = 184.9 J/K· mol-rxn OSpontaneous at all temperatures. O Spontaneous at lower temperatures. O Spontaneous at higher temperatures. ONot spontaneous at any temperature.

Determine kc and kp at 298 K