6.1 lab notes and coursework PDF

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chem laboratory Determination of Ideal Gas Law Constant...

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2021

Determination of Ideal Gas Law Constant

Student: 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 H2O2)

Trial 6 (3 mL H 2 O 2)

Trial 7 (4 mL H 2 O 2)

Trial 8 (4 mL H2O2)

Air temperature (°C)

25°C

25°C

25°C

25°C

25°C

25°C

25°C

25°C

Volume H2O2 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)

9.8mL0.0mL = 9.8mL

11.5mL1.0mL = 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

2021

Activity 2 Data Table 2 Show work for determining Concentration of H2O2 in the box below. Concentration should be mol/L. Concentration H 2O 2 (mol/L)

30.0g H2O2/ 1000mL 30.0g H2O2 x 1.0mol H2O2 / 34.01g H2O2 = 0.88mols H2O2 / L

Trial 1

Trial 2

Trial 3

Trial 4

Trial 5

Trial 6

Trial 7

Trial 8

1 mL H2O2

1 mL H2O2

2 mL H2O2

2 mL H2O2

3 mL H2O2

3 mL H2O2

4 mL H2O2

4 mL H2O2

Mole s H2O2

0.0009mol H2O2

0.0009mol H2O2

0.0018mol H2O2

0.0018mol s H2O2

0.0026mol s H2O2

0.0026mol s H2O2

0.0035mols H2O2

Mole s O 2*

0.00045mol 0.00045mol s O2 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

0.0183L

0.0195L

0.031L

0.0332L

0.0424L

0.041L

ΔV (L)

0.0098L

0.0105L

*Hint: Use reaction stoichiometry to solve for moles of O2. Show work for determining moles of H2O2 for Trial 1 here: 1.0mL H2O2 x 1L/1000mL x 0.88mols H2O2/1L = 0.00088mols H2O2 Show work for determining moles of O2 for Trial 1 here: 0.0009mols H2O2 x 1.0mol O2/2.0mols H2O2 = 0.00045mols O2 Show work for determining moles of H2O2 for Trial 3 here: 2.0mL H2O2 x 1L/1000mL x 0.88mols H2O2/1L = 0.00176mols H2O2 Show work for determining moles of O2 for Trial 3 here: 0.0018mols H2O2 x 1.0mols O2/2.0mol H2O2 = 0.0009mols O2 Show work for determining moles of H2O2 for Trial 5 here: 3.0mL H2O2 x 1L/1000mL x 0.88mols H202/1L = 0.00264mols H2O2 © 2016 Carolina Biological Supply Company

0.0035mols H2O2

2021 Show work for determining moles of O2 for Trial 5 here: 0.0026mols H2O2 x 1.0mol O2/2.0mols H2O2 = 0.0013mols O2 Show work for determining moles of H2O2 for Trial 7 here: 4.0mL H2O2 x 1L/1000mL x 0.88mols H2O2/1L = 0.00352mols H2O2 Show work for determining moles of O2 for Trial 7 here: 0.0035mols H2O2 x 1.0mol O2/2.0mols H2O2 = 0.00175mols O2

Insert a copy of your graph. Your graph must include the equation of the line displayed, your name and date in the title. You cannot complete the rest of this lab without this graph and the equation of the line. See the Week 6 Announcement for an example. Graph:

Volume (L) O2 vs. Moles O2 Jacen Duplain 06/07/2021 0.05 0.04

f(x) = 23.64 x

0.04

Volume (L) O2

0.03 0.03 0.02 0.02 0.01 0.01 0

0

0

0

0

0

Moles O2

© 2016 Carolina Biological Supply Company

0

0

0

0

0

2021

Show work, including all units, for the calculations/conversions for Air Temperature, Gas Constant R and Percent Error in the boxes below. Calculations must include complete units to be correct.

Air Temperature (K) Show work in the space provided.

25 + 273.15 = 298.15K

Air Pressure (atm) Show work in the space provided.

29.85inHg *

Equation of the Line

= 0.9976atm

Y = 23.64x

Gas Constant R Show work in the space provided.

Percent Error Show work in the space provided. Use 0.0821 L*atm/mol*K in your calculation

25.40 mmHg ∗1 atm 1 inHg 760 mmHg

23.64

(

l ( 0.9976 atm ) mols 298.15 k

=

0.0791

L∗atm mols∗ K

)

L∗atm L∗atm −0.0791 |0.0821 mol∗K mols∗ K| ∗100=3.65 % error 0.0821

L∗atm mol∗ K

Question 2: Identify at least two potential sources of error in the experiment. Are any assumptions made that would add to the experimental error? Some potential sources of error would be the amount of gas that is lost between the time of adding H2O2 and securing the cap of the bottle. Another potential error source would be if any liquid is still in the container after rinsing the residual yeast from a previous trial.

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