Question

1. Table 1 shows the temperature-time data were recorded for the reaction between 50.0 mL of 1.06 M HA (a weak acid) and 50.0
was determined to be 31.2 °C. Use this data to determine AT 2. Given that the specific heat of the final salt solution is 4.1


Table II. Densities at 298 K. density solution (mol/L) (g/ml) H20(1) 0.997 pure HCI(aq) HNOs(aq) 1.0 1.02 1.0 1.03 0.992 NH4O
1. Table 1 shows the temperature-time data were recorded for the reaction between 50.0 mL of 1.06 M HA (a weak acid) and 50.0 mL of 0.90 MNH OH, ammonium hydroxide (a weak base, also known as aqueous ammonia) The solutions were mixed after 60 s of approximately constant temperature readings of 24.20°C. Table 1: Temperature-time Data Time (s) Temperature (C) Time (s) Temperature C) 31 90 24.25 30.9 105 24.22 15 120 30.8 24.20 30 30.7 135 24.20 45 30.6 150 24.20 60 30.5 165 30.8 75
was determined to be 31.2 °C. Use this data to determine AT 2. Given that the specific heat of the final salt solution is 4.10 J/g C and the density of the weak acid is 1.028 gml (the density of the NH4OH is given in Table II of the experiment and the volumes are given in question 1), calculate the heat capacity of the system in J°C. [Note: the tabulated density is close enough to use even if the solutions molanty is not exactly 1 M.] Hint 401 Your answer is incorrect- Check your calculations 3. Using the answers to questions 1 & 2, calculate Qeat for the reaction in kJ Hnt 2814 Close. It looks like you made a minor calculation error 4 Given the volumes and concentrations in question 1, and that the reaction has a 1.1 stoichiometric ratio between the reactants, calculate the moles of limiting reactant in the reaction 0047 mol 5 Using the answers to questions 3 &4, calculate the molar heat of reaction (All) in kJ mol Remember geal snd ΔΗ have opposite signs. kJ mol 049 You are on the right track However, the sign is wrong
Table II. Densities at 298 K. density solution (mol/L) (g/ml) H20(1) 0.997 pure HCI(aq) HNOs(aq) 1.0 1.02 1.0 1.03 0.992 NH4OH(aq) 1.0
0 0
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