---### Calculating Enthalpy Change for a Neutralization ReactionIn a coffee-cup calorimeter, 120.0 mL of 1.4 M NaOH and 120.0 mL of 1.4 M HCl are mixed. Both solutions were originally at 25.7°C. After the reaction, the final temperature is 31.0°C. Assuming that the solutions have a density of 1.06 g/mL and a specific heat capacity of 4.18 J/g∙°C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.\[ \Delta H = \frac{\text{kJ}}{\text{mol}} \]---**Steps to Calculate Enthalpy Change (∆H):**1. **Measure Initial and Final Temperatures**: - Initial Temperature (Ti) = 25.7°C - Final Temperature (Tf) = 31.0°C2. **Calculate the Total Volume of Solution**: - Volume of NaOH solution = 120.0 mL - Volume of HCl solution = 120.0 mL - Total Volume = 120.0 mL + 120.0 mL = 240.0 mL3. **Determine the Total Mass of the Solution**: - Density of solution = 1.06 g/mL - Total Mass (m) = Total Volume * Density = 240.0 mL * 1.06 g/mL = 254.4 g4. **Find the Temperature Change (∆T)**: - ∆T = Tf - Ti = 31.0°C - 25.7°C = 5.3°C5. **Calculate the Heat Absorbed or Released (q)**: - Specific Heat Capacity (c) = 4.18 J/g∙°C - q = m * c * ∆T = 254.4 g * 4.18 J/g∙°C * 5.3°C = 5633.94 J (or 5.634 kJ)6. **Determine the Moles of Reactants**: - Volume of NaOH solution = 120.0

Volume of 1.4 M NaOH taken, V = 120.0 mL Volume of 1.4 M HCl taken, V = 120 mL

In a coffee-cup calorimeter, 120.0 mL of 1.4 M NaOH and 120.0 mL of 1.4 M HCl are mixed. Both solutions were originally at 25.7°C. After the reaction, the final temperature is 35.1°C. Assuming that all the solutions have a density of 1.0 g/cm³ and a specific heat capacity of 4.18 J/°C-g, calculate the enthalpy change for the neutralization of HCI by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter. ΔΗ = kJ/mol Submit Answer Try Another Version item attempt remaining

In a coffee-cup calorimeter, 120.0 mL of 1.4 M NaOH and 120.0 mL of 1.4 M HCl are mixed. Both solutions were originally at 25.7°C. After the reaction, the final temperature is 35.1°C. Assuming that all the solutions have a density of 1.0 g/cm³ and a specific heat capacity of 4.18 J/°C-g, calculate the enthalpy change for the neutralization of HCI by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter. ΔΗ = kJ/mol Submit Answer Try Another Version item attempt remaining

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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