Experiment HCI+ NaOH Part D Volume and molarity of acid H₂C₂O + NaOH 50 mL and 2.0 M 50 mL and 2.0 M HCI+NH₂ H₂C₂O+NH, 5

[answerhappygod]

Experiment Hci Naoh Part D Volume And Molarity Of Acid H C O Naoh 50 Ml And 2 0 M 50 Ml And 2 0 M Hci Nh H C O Nh 5 1 (77.8 KiB) Viewed 59 times

Experiment Hci Naoh Part D Volume And Molarity Of Acid H C O Naoh 50 Ml And 2 0 M 50 Ml And 2 0 M Hci Nh H C O Nh 5 2 (77.8 KiB) Viewed 59 times

Experiment HCI+ NaOH Part D Volume and molarity of acid H₂C₂O + NaOH 50 mL and 2.0 M 50 mL and 2.0 M HCI+NH₂ H₂C₂O+NH, 50 ml, and 2.0 M 50 mL and 2.0 M Volume and molarity of base 186] ΑΣΦ 4 AH = -71.2 O... Type bases 50 ml, and 2.0 M 50 ml and 20 M 50 mL and 2.0 M 50 mL and 2.0 M Final temperature (C) 33.62 33.22 32.48 Change in enthlapy, AH, and heat of reaction, gras To calculate the total amount of heat (energy) being transferred in the calorimeter, the following equation for the heat of solution () is used 32.05 gmCAT where m is the mass, C is the specific heat, AT is the change in temperature of the solution, and g is equal to the energy change of the solution (e, it can be a positive or negative value) When the mass of solution is unknown, the density and volume of solution can be used to calculate it, where a density of 1.00 g/ml. (water) is used to approximate m if the exact density is unknown. The specific heat of water (4.184 3/(g-C)) can then be used to calculate quel kJ/mol solution compared to neutralizing HCI with NaOH 0J/g 1.67 J/g 4.77 J/g The calorimeter is insulated (preventing energy exchanges with the environment outside of the calorimeter), so the energy change for the reaction () is equal and opposite to the energy change taking place in the surroundings (g): 1.0 qris the enthalpy change, AH, is the amount of energy being released or absorbed by the reaction per mole (n) at constant pressure. The relationship between the - because the net flow of energy goes in only one direction. Lastly, enthalpy and heat of reaction is ΔΗ = x = = 6.57 J/g For this exercise, you can simulate the described conditions by changing the values in the Run Experiment tool, under the Experiment tab, in the simulation The hints run through each step in the calculation, where the equations are given before this part. The second hint calculates the temperature change, which is the only step that requires the simulation Use the simulator to react 76 mL of 1.6 MHCI with 76 mL of 1.6 M NaOH at an initial temperature of 20.00 "C, note the final temperature. Use the data from the simulation to calculate the change in enthalpy per mole of product (n) Estimate the density of the liquid to be 1.00 g/ml. Express your answer in kilojoules per mole to three significant figures. View Available Hint(s)