1. In the calorimeter experiment, you mixed 50 mL of 1.0 M NaOH with 50 mL of approximately 1.0 M HCl. Suppose that 50 mL of a more concentrated solution of each were mixed. a. Should any additional heat have been released? Explain. b. Should you have obtained the same AT? Explain. c. Would the true molar enthalpy change, AH, for this reaction be different? Explain.

15. A kilogram of water and a kilogram of HCl are at the same temperature. Do the two substances contain the same amount of heat energy? Explain.

2. a. Determine the standard heat of formation, AH for the following reaction using the Data from the Table. C3H12 ) + 802(g) → 5CO2 + 6H₂O b. State whether the reaction is exothermic or endothermic. Explain. c. Rewrite the equation as thermochemical equations in two different ways to include the heat calculated in (2a). d. Draw a detail Energy Diagram representing the thermochemical equation in Ⓒ illustrating the the reactants and products and their positions.

What effect would leaving the lid off the coffee-cup calorimeter have on the specific heat(higher, lower or the same)? Why?

Pleasee don't provide handwritten solution .....

Please help me answer the questions from a to f using the data provided

The enthalpy change for a chemical reaction is the sum of the energy consumed in breaking bonds and the energy released during bond formation. One way to determine the overall energy change for a chemical reaction is to apply Hess’s law to add together a group of reactions which can be arranged such that the chemical equations, when combined, give the overall equation we are trying to characterize.     a. Write a balanced chemical equation for the combustion of gaseous propane in gaseous oxygen to produce gaseous carbon dioxide and liquid water.      C3​H8​(g)+    O2​(g) ---->CO2​(g)+     H2​O(g)   b. Combine the following equations to determine the enthalpy change for the combustion of 1 mole of propane. Assume that solid carbon is graphite.  3C(s,graphite)+4H2​(g)--->C3​H8​(g) ΔHo = -103.8 kJ/mol   C(s,graphite)+O2​(g)--->CO2​(g) ΔHo= -393.5 kJ/mol   H2​(g)+1/2​ O2​(g)--->H2​O(g) ΔHo= -285.8 kJ/mol      kJ/mol…