Write a one or two paragraphs of -what was study in identification of investigating entropy -Summarize what happened in these results shown below -what results did this experiment end up with whether this lab was successful or not The Data and results collected for this experiment For solution of Sodium hydroxide (NaOH) Trial 1: ΔT=7.8∘C, qsurroundings=1.633 kJ Trial 2: ΔT=7.8∘C, qsurroundings=1.632 kJ Trial 3: ΔT=9.1∘C, qsurroundings=1.906 kJ The average enthalpy change (ΔH) for the dissolution process is: ΔH= −33.71 kJ/mol Determination of qsystem Trial 1: qsystem = -1631.4 J Trial 2: qsystem = 1630.8 J Trial 3: qsystem = -1902.4 J Determination of ΔH ΔΗ = qsystem / moles of NaOH Trial 1: ΔΗ = -32032.2 J/mol Trial 2: ΔΗ = -31452.3 J/mol Trial 3: ΔΗ = -37522.7 J/mol For the solution of Sodium nitrate (NaOH3) Step 1: Calculate triangle T for each trial. Trial 1: - 4.2 C Trial 2: - 4.2C Trial 3:- 4.1C Step 2: Calculate qsurroundings for each trial Trial 1: -967.2 J Trial 2: -961.9 J Trial 3: -935.2 J Step 3: Calculate the moles of NaNO3 Trial 1: 0.0504 mol Trial 2: 0.0504 mol Trial 3: 0.0502 mol Step 4: Calculate the enthalpy change (ΔH) for each trial Trial 1: - 19.19kJ / mol Trial 2: - 19.08KJ / mol Trial 3: - 18.63kJ / mol Step 5: Calculate the average enthalpy change Average ΔH = - 18.97KJ / mol Therefore, the average enthalpy change for the dissolution of NaNO3 is approximately - 18.97kJ/mol This negative value indicates that the dissolution process is exothermic.
Write a one or two paragraphs of
-what was study in identification of investigating entropy
-Summarize what happened in these results shown below
-what results did this experiment end up with whether this lab was successful or not
|
The Data and results collected for this experiment For solution of Sodium hydroxide (NaOH)
Trial 1: ΔT=7.8∘C, qsurroundings=1.633 kJ Trial 2: ΔT=7.8∘C, qsurroundings=1.632 kJ Trial 3: ΔT=9.1∘C, qsurroundings=1.906 kJ
The average enthalpy change (ΔH) for the dissolution process is: ΔH= −33.71 kJ/mol
Determination of qsystem Trial 1: qsystem = -1631.4 J Trial 2: qsystem = 1630.8 J Trial 3: qsystem = -1902.4 J
Determination of ΔH ΔΗ = qsystem / moles of NaOH Trial 1: ΔΗ = -32032.2 J/mol Trial 2: ΔΗ = -31452.3 J/mol Trial 3: ΔΗ = -37522.7 J/mol
For the solution of Sodium nitrate (NaOH3) Step 1: Calculate triangle T for each trial. Trial 1: - 4.2 C Trial 2: - 4.2C Trial 3:- 4.1C Step 2: Calculate qsurroundings for each trial Trial 1: -967.2 J Trial 2: -961.9 J Trial 3: -935.2 J Step 3: Calculate the moles of NaNO3 Trial 1: 0.0504 mol Trial 2: 0.0504 mol Trial 3: 0.0502 mol Step 4: Calculate the enthalpy change (ΔH) for each trial Trial 1: - 19.19kJ / mol Trial 2: - 19.08KJ / mol Trial 3: - 18.63kJ / mol Step 5: Calculate the average enthalpy change Average ΔH = - 18.97KJ / mol Therefore, the average enthalpy change for the dissolution of NaNO3 is approximately - 18.97kJ/mol This negative value indicates that the dissolution process is exothermic.
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