Given my data, I\'m not sure how to approach these calculations. I did a couple

Question

Given my data, I'm not sure how to approach these calculations. I did a couple but I am not sure how to do the rest of them. If anyone could help me and check if I did the ones I already did correctly, that would be amazing. Thank you. Experiment 11 DATA Name: Instructor: MoaHE CSANTO Section: Date TABLE 11.1. Time-temperature data for an acid-base and a redox reaction. Run1 Run 2 Run3 A Volume of HCI solution (mL) Molarity of HCl solution (M) Volume of NaOH solution (mL) Molarity of NaOH solution (M) Initial temperature of HCl solution (C) 2223, Initial temperature of NaOH solution) 2. 23,o (1s-sec) intervals 2.042.06 52.0 52. 2.1 2, 11 Temperatures of solution (C) at 0.25-min 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00 IE Eleventh Edition, Updated

Explanation / Answer

Sample calculation for Run - 1:

A: The first two calculations are correct and 3rd is wrong

3rd : Here the moles of water produced will be 0.1 mol.

Moles of HCl takem = MxV = 2.061M x 0.050L = 0.103 mol

Moles of NaOH taken = MxV = 2.116M x 0.052 L = 0.110 mol

HCl + NaOH -- > NaCl + H2O

1 mol 1 mol 1 mol 1 mol

Since 1 mol of HCl produces 1 mol of water, hence 0.103 mol of HCl (limiting reactant) will produce 0.103 mol H2O

Hence moles of water formed in the reaction = 0.103 mol

4th: Given that 57320 J of heat is produced per mole of H2O formed.

Hence heat produced when 0.103 mol of H2O formed = 57320 J/mol x 0.103 mol = 5904 J

Hence Hence theoritical heat expected to be produced = 5904 J (answer)

5th: Heat absorbed by calorimeter = Heat expected - Heat actually produced

= 5904 J - 5000 J = 904 J (answer)

6th: Hence heat capacity of calorimeter = 904 J / 12 DegC = 75.3 J/DegC (answer)

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