For many substances, the variation with temperature of the molar heat capacity at constant pressure of is given by the expression: Cp.m = a + bT + C/T² For copper, a = 22.64 J K-1 mol-1, b = 6.28 x 10-3 J K-2 mol-1 with the value of 'c' being negligible. Calculate the change in the molar enthalpy of copper when it is heated from 293 to 323 K.

Question 10 please

please answer the question while referring to the diagram attached

Use Table B.8 to calculate the energy change, in kJ, as 1 mole of N₂ goes from 800°C to 300°C. The answer must be positive if the substance is heated-up and negative if it is cooled-down.

The Heat of Solution (AHsol) for CaSO(s) in water is -18.0 kJ/mol. Which of the following is true? More than one of these answer choices are true. O The intermolecular forces between solvent-solute are greater than the sum of the solute-solute and solvent- solvent intermolecular forces. O Crystallization is exothermic. O Dissolution is exothermic.

If a food product is being frozen in a system capable of removing 6200 kJ of thermal energy per product, with the product having a specific heat of 4.1 kJ/kg°C  above the freezing temperature of -1.7°C , a latent heat of fusion of 284 kJ/kg, and the product has a specific heat of 2.2 kJ/kg°C below the freezing temperature. If 10.4 kg of product enters the system at 15°C, and exits it with a temperature of –21 °C, what percentage (to the nearest percent) of the full capability of the freezer has been used?

11.29. It is proposed to cool a stream of 75-wt-% sulfuric acid solution at 330 K by diluting it with chilled water at 280 K. Determine the amount of water that must be added to 1 kg of 75-wt-% acid before cooling below 330 K actually occurs.

At 150 °C, linalyl acetate exerts a vapor pressure of 93 tor. If a steam still were run using steam heated to this temperature in place of boiling water, how much water must co-distill to collect 1.0 grams of linalyl acetate? Include your calculations.

Show clean and complete process. Thank you

The figure below is the basic plot of pressure vs. volume for the Dual Standard Cycle. Constant pressure heat addition 3 4 = constant Нeat IN Adiabatic expansion Heat OUT (Constant volume heat rejection) Constant volume heat addition Adiabatic compression Volume (V) Using the Dual Air Standard Cycle and the following data: Assume constant properties of air. R= 287 J/kgK; Cp = 1005 J/kgK; Cv = 718 J/kgK;y = 1.4 Cycle compression ratio = 11 (v1/v2) Air conditions prior to compression: Pressure = 1.6 bar; Temperature = 298K Heat added at constant volume = 290000 J/kg Heat added at constant pressure = 579000 J/kg a) Calculate the pressure at the end of the compression process. P2 = bar Pressure (P) LO