Exp. 7 - Get Charged Up Part 2

docx

School

University of Nebraska, Lincoln *

*We aren’t endorsed by this school

Course

109

Subject

Chemistry

Date

Feb 20, 2024

Type

docx

Pages

Uploaded by JudgeMosquitoPerson47

Experiment 7 Get Charged Up – Part II Total Points: 78 Student’s Name: Rachel Melton Lab Section: 510 REPORT Your report should include the following:  All data tables (7-1 to 7-4, including all calculations under each table)  Graphs that you constructed in Excel for Part I and Part II o The Excel work is worth 10 points  Answers to questions 1 – 6 NOTE: You must show your work for all calculations; no work, no credit. Data Table 7-1. Pressure and Volume Measures (5 pts) Volume (mL) Trial 1 Pressure Units: atm Trial 2 Pressure Units: atm Trial 3 Pressure Units: atm Ave. Pressure Units: atm Predicted Pressure 10 (Reference) 0.9720 0.9720 0.9715 0.9718 5 1.5700 1.3015 1.6015 1.491 1.9436 7 1.0745 0.9228 1.3328 1.110 1.3880 15 0.6047 0.4909 0.6653 0.5869 0.6479 20 0.4765 0.3993 0.5163 0.4640 0.4859 Air Temp (°C): 22.4 Example calculations must include calculations for:  the number of moles from the 10 mL (reference) using Ideal Gas Law (3 pts) PV/RT = n ( 0.9718 atm ) ( 0.010 L ) ( 0.08206 ) ( 295.55 K ) = 4.00 x 10 − 4 mol  the predicted pressure for 5 mL using the simplified form of the following equation: P 1 V 1 n 1 T 1 = P 2 V 2 n 2 T 2 . (3pts)

Experiment 7 Get Charged Up – Part II Total Points: 78 ( 0.9720 atm ) ( 10 mL ) ( 4.00 x 10 − 4 mol ) ( 295.55 K ) ( 4.00 x 10 − 4 mol ) ( 295.55 K ) ( 5 mL ) = 1.9436 atm Data Table 7-2. Volume of Flask (2 pts) Mass of Dry Flask (g) Temp of Water (°C) Mass of Flask and Water (g) Mass of Water (g) Volume of Water (or Flask) (mL) 81.748 21.0 214.533 132.785 133.052 Example calculations must include the calculation of volume of water based on mass: (3 pts) V = m/d = 132.785 g / 0.997992 g/mL = 133.052 mL Data Table 7-3. Temperature and Pressure Changes (5 pts) Temperature (°C) Pressure Units: atm Moles of gas in flask Calculated Pressure using PV = nRT Units: atm Calculated Pressure using P 1 V 1 /n 1 T 1 = P 2 V 2 /n 2 T 2 Units: atm 22.4 *Reference Data (Room temperature and pressure) 0.97907 5.319 x 10^-3 mol 76.3 0.9713 1.1475 1.1475 67.7 0.9694 1.1193 1.1193 65.4 0.9697 1.1118 1.1118 62.0 0.9701 1.1006 1.1006 57.0 0.9694 1.0842 1.0842 52.6 0.9688 1.0698 1.0698 48.8 0.9701 1.0573 1.0573 45.6 0.9694 1.0468 1.0468 42.8 0.9694 1.0376 1.0376 40.3 *LabQuest Data Table Reading 0.9707 *LabQuest Data Table Reading *Found using data in Data Table 7-2 and Reference Data 1.0294 1.0294

Experiment 7 Get Charged Up – Part II Total Points: 78 Example calculations must include calculations for:  moles of gas in flask using Ideal Gas Law equation (3 pts) n = PV/RT = ( 0.9707 atm ) ( 0.133052 L ) ( 0.08206 ) ( 295.55 K ) = 5.319 x 10 − 3 mol  the second-row data for calculated pressure using o PV = nRT (3 pts) and P = nRT/V = ( 5.319 x 10 − 3 mol ) ( 0.08206 ) ( 349.8 K ) 0.133052 L = 1.1475 atm o P 1 V 1 n 1 T 1 = P 2 V 2 n 2 T 2 (1 = reference row) (3 pts) P2 = P1T2/T1 = ( 0.9707 atm ) ( 349.8 K ) 295.55 K = 1.1475 atm Data Table 7-4. Hydrogen Gas Produced (5 pts) Mass of Mg (g) Moles of Mg (mol) Initial Temp (°C) Initial Pressure Units: atm Initial Moles in Flask (n 1 ) Final Temp (°C) Final Pressure Units: atm Final Moles in Flask (n 2 ) Moles of H 2 Produced (mol) Limiting Reagent 0.029 0.00119 23.8 0.9701 5.29 x 10^-3 24.1 0.9715 0.00530 2.97 x 10^-5 Mg 0.051 0.00210 23.4 0.9699 5.29 x 26.1 0.9704 0.00526 4.512 x Mg

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Experiment 7 Get Charged Up – Part II Total Points: 78 10^-3 10^-5 0.077 0.00317 22.4 0.9280 5.085 x 10^-3 26.1 0.9831 0.00533 2.45 x 10^-4 HCl 0.101 0.00415 22.8 0.9755 5.348 x 10^-3 28.7 0.9850 0.00535 2.00x 10^-6 HCl Example calculations must include the following for your first trial:  Calculate n 1 (2 pts) PV/RT = n ( 0.9701 atm ) ( 0.133052 L ) ( 0.08206 ) ( 23.8 + 273.15 K ) = 0.00529 mol  Calculate n 2 (2 pts) PV/RT = n ( 0.9715 atm ) ( 0.133052 L ) ( 0.08206 ) ( 24.1 + 273.15 K ) = 0.00530 mol  Calculate the moles of H 2 produced (2 pts) N2 – N1 = 2.97 x 10^ -5 moles of H2 QUESTIONS 1. Does the Ideal Gas Law do a reasonable job at predicting the behavior of a gas undergoing changes in temperature, pressure, and volume? Justify your answer with supporting evidence (i.e., your comparison of predicted and actual results). (5 pts) Yes, it does because the predicted values are very similar to the actual results. The calculated value at 42.8 degrees was 0.9694 atm and the actual value was 1.0376 atm which is very close. 2. With regards to the actual procedure, what were the advantages and disadvantages to the gas law method? (4 pts) Some advantages of the gas law method are that it was easy to use and remember. We also used the LabQuest which made getting the temperature and pressure very simple. It recorded our data points for us for Table 7-3. Some disadvantages to the gas law method are that it requires a lot of data to be known including pressure, volume, temperature, and number of moles. If you’re solving for one of those variables, you must know all the others which in some cases may not be possible. In our case, it just required extra time to get all the data. The ideal gas law also leaves a

Experiment 7 Get Charged Up – Part II Total Points: 78 considerable amount of room for human error, especially with having to get the perfect seal on the flask. 3. Based on your results , which of the three methods gave more precise results? Explain how you determined which method was more precise. (5 pts) Of the 3 methods, the most precise was the titration method. This method had the highest R^2 value at 0.9979 while the crystallization method had an R^2 value at 0.9836 and the gas law method had an R^2 value at 0.0406. Since the titration R^2 value is closer to 1, it is the most precise. 4. Based on your results , which of the three methods gave more accurate results? Explain how you determined which method was more accurate. (5 pts) HINT: The slope of the 3 graphs is expected to match the stoichiometry of the reaction, thus the expected slope might not be the same for all the graphs. Of the 3 methods, the titration method was the most accurate results as well. The slope of the trendline was 1.9142, 3.152 for the crystallization method, and -0.0329 for the gas law method. The known stoichiometric relationship for the titration and crystallization methods is 1:2 and 1:1 for the gas law method. Since 1.9142 is closer to 2 than the other values to their respective stoichiometric relationships, the titration method is the most accurate. 5. How does the mole ratio of H 2 produced to Mg reacted compare to the charge on the Mg ion after reacting with HCl? (3 pts) The mole ratio is H2 produce to MG reacted is 1:1 as seen from the reaction Mg(s) + 2HCl (aq) > MgCl2(aq) + H2(g). The charge of the Mg ion after reacting with the HCl is 2+. There is one mole of H2 for every mole of Mg2+, which means their ratio is 1:1. 6. Based on your results, which of the three methods would be better for determining the charge on a metal ion reacting with HCl? Explain why. (5 pts) I believe that the titration method is the best because it is the most precise and the most accurate. The method also leaves little room for human error in comparison to the gas law method.