- The Ccal 140 J C 1 (41.41 KiB) Viewed 49 times
The Ccal=140 J/C
-
answerhappygod
- Site Admin
- Posts: 899603
- Joined: Mon Aug 02, 2021 8:13 am
The Ccal=140 J/C
The Ccal=140 J/C
Carefully work through the sample calculations for this lab under sub-heading "Sample Calculations for the Heat of Neutralization." When you are finished, return to this question. A 50.00 mL sample of 0.4998 M KOH at 18.2 °C was added to a calorimeter containing 50.00 mL of 0.5023 M HNO3 at 18.6 °C. The maximum temperature, as displayed on the resulting Temp C, degrees C vs. Time, s plot, was 20.9 °C. Using the density table for water (because the acid and base are both very dilute and their densities will approximate that of water) and your CCAL value from the previous question, determine for this experiment the following values: MKOH 49.93 g MHNO3 = 49.92 g q=1.3 x 10^3 J moles of limiting reagent = 0.7922 moles AH -52. kJ/mol
Carefully work through the sample calculations for this lab under sub-heading "Sample Calculations for the Heat of Neutralization." When you are finished, return to this question. A 50.00 mL sample of 0.4998 M KOH at 18.2 °C was added to a calorimeter containing 50.00 mL of 0.5023 M HNO3 at 18.6 °C. The maximum temperature, as displayed on the resulting Temp C, degrees C vs. Time, s plot, was 20.9 °C. Using the density table for water (because the acid and base are both very dilute and their densities will approximate that of water) and your CCAL value from the previous question, determine for this experiment the following values: MKOH 49.93 g MHNO3 = 49.92 g q=1.3 x 10^3 J moles of limiting reagent = 0.7922 moles AH -52. kJ/mol