Question



Show your work for all calculations, you MUST justify your answer to get credir. Answers should reflect the appropriate numbe
Part I. Determination of the Heat Capacity of a Calorimeter 1. Considering you mixed equal volumes of hot and cold water insi
u s 4. Calculate the heat of the calorimeter, al Should be rature da Trial Tun colorin id. Base, because the heat of the Cal
0 0
Add a comment Improve this question Transcribed image text
Answer #1

The enthalpy change when one equivalent of an acid is neutralized by one equivalent of base in their dilute solutions is known as enthalpy of neutralization. It is constant for strong acids and strong bases and is equal to - 57.1 kJ/equi.

As NaOH (aq) + HCl(aq) ==> NaCl(aq) + H2O

ionic symbols : Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) ==> Na+(aq) + Cl-(aq) + H2O

cancel the ions which are present on both side of the equation: we get:

H+ + OH-==> H2O, i.e the overall reaction is just the combination of H+ and OH- to form H2O, so the enthalpy of neutralization of strong base and strong acid is approx -57.1 kJ, the value can be calculated on the basis of bomb calorimeter

Add a comment
Know the answer?
Add Answer to:
Show your work for all calculations, you MUST justify your answer to get credir. Answers should reflect the appr...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 2. Answer the following questions about the neutralization reaction of nitric acid using sodium hydroxide (see...

    2. Answer the following questions about the neutralization reaction of nitric acid using sodium hydroxide (see Equation 1 from the Introduction): a. Write the ionic and net ionic equations for the reaction. b. How does the molar enthalpy of reaction for the neutralization of nitric acid compare to the value you obtained for the neutralization of hydrochloric acid? What would you expect it to be when you compare the net ionic equations for the two reactions? Explain. livier, using Eos...

  • 1. A volume of water was heated to 81.76 °C and immediately added to 48.80 mL of water at 20.55 °C contained within a co...

    1. A volume of water was heated to 81.76 °C and immediately added to 48.80 mL of water at 20.55 °C contained within a coffee cup calorimeter. The final temperature of the mixture was 37.73 °C. The final volume of water inside the calorimeter was 92.03 mL. Assuming that them heat capacity of the solution is 4.18 J/g/°C, calculate the following: a. The volume of hot water added ml b. The mass of hot water (dwater = 1.00 g/mL) g...

  • 1. A volume of water was heated to 83.90 °C and immediately added to 48.11 mL...

    1. A volume of water was heated to 83.90 °C and immediately added to 48.11 mL of water at 23.98 °C contained within a coffee cup calorimeter. The final temperature of the mixture was 40.86 °C. The final volume of water inside the calorimeter was 90.45 mL. Assuming that them heat capacity of the solution is 4.18 J/g/°C, calculate the following: a. The volume of hot water added ml b. The mass of hot water (dwater = 1.00 g/mL) c....

  • A volume of water was heated to 82.86C and immediately added to 50.31 mL of water...

    A volume of water was heated to 82.86C and immediately added to 50.31 mL of water at 22:45 °C contained within a coffee cup calorimeter. The final temperature of the mixture was 37,08°C. The final volume of water inside the calorimeter was 94.33 mL. Calculate the following: Note: Heat capacity a. volume of hot water added is 4.18 *Consider sig fig b. masses of hot and cold water (density of water = 1,00 g/mL) c. changes in temperature (AT) of...

  • 2. A 49.45 mL volume of 1.00 M HCl was mixed with 49.14 mL of 2.00...

    2. A 49.45 mL volume of 1.00 M HCl was mixed with 49.14 mL of 2.00 M NaOH in a coffee cup calorimeter (with calorimeter constant = 25.1 J/°C) at 21.34 °C. The final temperature of the aqueous solution after the reaction was 29.37 °C. Assuming that them heat capacity of the solution is 4.18 J/g/°C, calculate the following: e. The enthalpy change (∆H) for the neutralization in kJ/mol HCl ( this should be a negative number) e. The enthalpy...

  • DATE NAME INSTRUCTOR SECTION/GROUP: DATA ANALYSIS Acel Table 1: Heat Capacity of Calorimeter Show calculations separately...

    DATE NAME INSTRUCTOR SECTION/GROUP: DATA ANALYSIS Acel Table 1: Heat Capacity of Calorimeter Show calculations separately Temp of calorimeter and cold water before mixing Temp of hot water before mixing Temp of water after mixing 16 water 24-1910 Thoe water24-47 Heat gained by cold water, qt Heat lost by hot water, q Heat gained by calorimeter, Heat capacity of calorimeter,o Table 2: Heat of Neutralization (HCI as acid) Show calculations separately Temp of calorimeter and NaOH before mixing ATut (after...

  • A 48.53 mL volume of 1.00 M HCl was mixed with 47.70 mL of 2.00 M...

    A 48.53 mL volume of 1.00 M HCl was mixed with 47.70 mL of 2.00 M NaOH in a coffee cup calorimeter (with calorimeter constant = 26.0 J/°C) at 21.43 °C. The final temperature of the aqueous solution after the reaction was 29.71 °C. Assuming that them heat capacity of the solution is 4.18 J/g/°C, calculate the following: a. The total mass of aqueous solution inside the calorimeter (dsoln = 1.00 g/mL) g b. The change in temperature (∆T) of...

  • Table 4.2: Enthalpy of Neutralization for an Acid-Base Reaction Trial 1 Concentration of NaOH 1. i...

    Table 4.2: Enthalpy of Neutralization for an Acid-Base Reaction Trial 1 Concentration of NaOH 1. i mol/L Initial temperature of NaOH 22.1°C Volume of NaOH Concentration of HCI 45, I mL 1.2 mol/L 22.3°C Initial temperature of HCI Trial 2 lol mol/L 21.9°C 63 mL 45. Iml 1,2 mol/L 22.5°C 45.2 mL 27.5° 22.2°C 5.3°C 90.3 mL 4.18 Jg! °C! Volume of HCI 45.3 ml Final temperature of mixture Average initial temperature (HCl and NaOH) Ti Temperature change, (47) 28.6°C...

  • Enthalpy of Neutralization Reaction: A 25.0 mL sample of 0.200 M NaOH is mixed with a...

    Enthalpy of Neutralization Reaction: A 25.0 mL sample of 0.200 M NaOH is mixed with a 25.0 mL sample of 0.200 M HNO3 in a coffee cup calorimeter.  NaOH and HNO3 will undergo Neutralization Reaction according to the following balanced equation: NaOH(aq) +   HNO3(aq) --> NaCl (aq) +  H2O (l)   Both solutions were initially at 35.00°C and Tmax of the resulting solution was recorded as 37.00°C (from the graph). Assume 1) that no heat is lost to the calorimeter or the surroundings, and...

  • calorimeter and HCI Part III: Determination of (Hº Trial One 1. Mass of 14.5467 2. Mass...

    calorimeter and HCI Part III: Determination of (Hº Trial One 1. Mass of 14.5467 2. Mass of calorimeter 59.5684 3. Mass of 0.9547 4. Initial Tof calorimeter and solution °C) 22.5 5. Final Tof calorimeter and solution °C) 42.4 Mgo ( Tsoln, (Tsoln, Heat 11. J) absorbed by the son show calculation: by the calorimeter 12 Heat absorbed J) show calculation: 13. Heat released by th reaction (..) show calculation: Heat of Reaction OH kJ/mol show calculation: 15. Heat of...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT