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Chart 2: Charles Law experment Helium: P = 1 atm 3.000 intercept: 1.352 L slope: 0.004948 L/°C 2.000 Helium: P = 0.845 atm %3D intercept: 1.600 L slope: 0.005853 L/°C V(L) 1.000 100 300 200 t °C 400 500 close Calculations: For mula V= O {tv=o=-intercept/5lope}

Chart 2: Charles Law experment Helium: P = 1 atm 3.000 intercept: 1.352 L slope: 0.004948 L/°C 2.000 Helium: P = 0.845 atm %3D intercept: 1.600 L slope: 0.005853 L/°C V(L) 1.000 100 300 200 t °C 400 500 close Calculations: For mula V= O {tv=o=-intercept/5lope}

Chemistry
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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Record the information about the lines you have generated.  Calculate the temperature for V = 0:
       {tV=0 = - intercept / slope} for each set of data.

     Discuss your observations.
 

Chart 2: Charles Law expe riment Helium: P = 1 atm 3.000 intercept: 1.352 L slope: 0.004948 L/°C . Helium: P 0.845 atm intercept: 1.600 L slope: 0.005853 L/°C 2.000 V(L) 1,000 100 200 300 400 500 t °C close Calculations: For mula V= O {tv=o=- intercept/slope}
Transcribed Image Text:Chart 2: Charles Law expe riment Helium: P = 1 atm 3.000 intercept: 1.352 L slope: 0.004948 L/°C . Helium: P 0.845 atm intercept: 1.600 L slope: 0.005853 L/°C 2.000 V(L) 1,000 100 200 300 400 500 t °C close Calculations: For mula V= O {tv=o=- intercept/slope}
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