61-lab-notes-and-coursework

.pdf

School

Fresno City College *

*We aren’t endorsed by this school

Course

Subject

Chemistry

Date

Feb 20, 2024

Type

pdf

Pages

Uploaded by yangster2005

6.1 lab notes and coursework Fundamentals of Chemistry Lab (Southern New Hampshire University) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university 6.1 lab notes and coursework Fundamentals of Chemistry Lab (Southern New Hampshire University) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by thao yang (tyang87@iclpoud.com) lOMoARcPSD|19832313

2021 Determination of Ideal Gas Law Constant S tudent: Jacen Duplain Date: 06/07/2021 Activity 1 If your concentration, moles, or R calculations are incorrect or your calculation work does not actually result in the desired final units or units are missing/incomplete or cancelled incorrectly, you will lose points. Be sure that every number you include has its associated unit. Email me if you are unsure or need assistance. Data Table 1 Trial 1 (1 mL H 2 O 2 ) Trial 2 (1 mL H 2 O 2 ) Trial 3 (2 mL H 2 O 2 ) Trial 4 (2 mL H 2 O 2 ) Trial 5 (3 mL H 2 O 2 ) Trial 6 (3 mL H 2 O 2 ) Trial 7 (4 mL H 2 O 2 ) Trial 8 (4 mL H 2 O 2 ) Air temperature (°C) 25°C 25°C 25°C 25°C 25°C 25°C 25°C 25°C Volume H 2 O 2 liquid (mL) 1.0mL 1.0mL 2.0mL 2.0mL 3.0mL 3.0mL 4.0mL 4.0mL Initial Volume Gas (mL) 0.0mL 1.0mL 0.0mL 0.0mL 0.0mL 0.0mL 0.0mL 0.0mL Final Volume Gas (mL) 9.8mL 11.5mL 18.3mL 19.5mL 31.0mL 33.2mL 42.4mL 41.0mL ΔV (mL) 9.8mL- 0.0mL = 9.8mL 11.5mL- 1.0mL = 10.5mL 18.3mL -o.omL = 19.5mL -0.0mL = 31.0mL -0.0mL = 33.2mL -0.0mL = 42.4mL –0.0mL =42.4m 41.0mL -0.0mL = Question 1: Would the volume of oxygen that is generated be affected if a smaller mass of yeast were used? Why or why not? There would be a smaller amount of catalase present to perform as a catalyst to the decomposition of H2O2 resulting in a lesser amount of O2 produced. © 2016 EW5 Carolina Biological Supply Company Downloaded by thao yang (tyang87@iclpoud.com) lOMoARcPSD|19832313

2021 Activity 2 Data Table 2 Show work for determining Concentration of H 2 O 2 in the box below. Concentration should be mol/L. Concentration H 2 O 2 (mol/L) 30.0g H2O2/ 1000mL 30.0g H2O2 x 1.0mol H2O2 / 34.01g H2O2 = 0.88mols H2O2 / L Trial 1 1 mL H 2 O 2 Trial 2 1 mL H 2 O 2 Trial 3 2 mL H 2 O 2 Trial 4 2 mL H 2 O 2 Trial 5 3 mL H 2 O 2 Trial 6 3 mL H 2 O 2 Trial 7 4 mL H 2 O 2 Trial 8 4 mL H 2 O 2 Mole s H 2 O 2 0.0009mol H2O2 0.0009mol H2O2 0.0018mol H2O2 0.0018mol s H2O2 0.0026mol s H2O2 0.0026mol s H2O2 0.0035mols H2O2 0.0035mols H2O2 Mole s O 2 * 0.00045mol s O2 0.00045mol s O2 0.0009mol s O2 0.0009mol s O2 0.0013mol s O2 0.0013mol s O2 0.00175mol s O2 0.00175mol s O2 ΔV (L) 0.0098L 0.0105L 0.0183L 0.0195L 0.031L 0.0332L 0.0424L 0.041L *Hint: Use reaction stoichiometry to solve for moles of O 2 . Show work for determining moles of H 2 O 2 for Trial 1 here: 1.0mL H2O2 x 1L/1000mL x 0.88mols H2O2/1L = 0.00088mols H2O2 Show work for determining moles of O 2 for Trial 1 here: 0.0009mols H2O2 x 1.0mol O2/2.0mols H2O2 = 0.00045mols O2 Show work for determining moles of H 2 O 2 for Trial 3 here: 2.0mL H2O2 x 1L/1000mL x 0.88mols H2O2/1L = 0.00176mols H2O2 Show work for determining moles of O 2 for Trial 3 here: 0.0018mols H2O2 x 1.0mols O2/2.0mol H2O2 = 0.0009mols O2 Show work for determining moles of H 2 O 2 for Trial 5 here: 3.0mL H2O2 x 1L/1000mL x 0.88mols H202/1L = 0.00264mols H2O2 © 2016 Carolina Biological Supply Company Downloaded by thao yang (tyang87@iclpoud.com) lOMoARcPSD|19832313

Your preview ends here

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