**Part III: Enthalpy of Precipitation - Sodium Acetate Trihydrate, NaC2H3O2·3H2O**- **Mass of NaC2H3O2·3H2O in test tube:** 17.5 grams- **NaC2H3O2·3H2O molecular weight:** 136.08 g/mol- **Mass of H2O in test tube:** 5.00 grams**Temperature Measurements:**- **Initial Temperature (T_initial):** 23.4°C- **Final Temperature (T_final):** 34.1°C- **Change in Temperature (ΔT):** 10.7°C*(Data Calculations on Reverse side)* In part III, you precipitated NaC2H3O2·3H2O from a saturated solution. The specific heat capacity, C, for the solution/solid mixture is 1.86 J/g°C. In reality, only about 0.8 grams (out of 1.75 grams) of sodium acetate in the test tube actually precipitated from the solution. Use this information to calculate ΔH for the precipitation of NaC2H3O2·3H2O from a saturated solution. Do you think this value overestimates or underestimates the actual value? Think about heat loss.

The ∆H for the precipitation of NaC2H3O2.3H2O from a saturated solution is to be calculated from the…

Part IlI: Enthalpy of Precipitation Sodium Acetate Trihydrate, NaC2H3O2 3H20 Mass of NaC2H3O2 3H2O in test tube: 17.5 grams NaC2H3O2 3H2O molecular weight: 136.08 g/mol Mass of H20 in test tube: 5.00 grams Tinitial:23.4°C Timal: 34.1°C AT 10.7°( (Data Calculations on Reverse Side)

Part IlI: Enthalpy of Precipitation Sodium Acetate Trihydrate, NaC2H3O2 3H20 Mass of NaC2H3O2 3H2O in test tube: 17.5 grams NaC2H3O2 3H2O molecular weight: 136.08 g/mol Mass of H20 in test tube: 5.00 grams Tinitial:23.4°C Timal: 34.1°C AT 10.7°( (Data Calculations on Reverse Side)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

**Part III: Enthalpy of Precipitation - Sodium Acetate Trihydrate, NaC2H3O2·3H2O**

- **Mass of NaC2H3O2·3H2O in test tube:** 17.5 grams
- **NaC2H3O2·3H2O molecular weight:** 136.08 g/mol
- **Mass of H2O in test tube:** 5.00 grams

**Temperature Measurements:**

- **Initial Temperature (T_initial):** 23.4°C
- **Final Temperature (T_final):** 34.1°C
- **Change in Temperature (ΔT):** 10.7°C

*(Data Calculations on Reverse side)*

In part III, you precipitated NaC2H3O2·3H2O from a saturated solution. The specific heat capacity, C, for the solution/solid mixture is 1.86 J/g°C. In reality, only about 0.8 grams (out of 1.75 grams) of sodium acetate in the test tube actually precipitated from the solution. Use this information to calculate ΔH for the precipitation of NaC2H3O2·3H2O from a saturated solution. Do you think this value overestimates or underestimates the actual value? Think about heat loss.

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