Ideal Gas Law Constant Lab Report Dr Z Edits
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Southern New Hampshire University *
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101L
Subject
Chemistry
Date
Feb 20, 2024
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Uploaded by DeanMoon13043
Determination of Ideal Gas Law Constant
Date: 12/04/2023
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)
17 °C
17 °C
17 °C
15 °C
15 °C
15 °C
25 °C
25 °C
Volume H
2
O
2
liquid (mL)
1 mL
1 mL
2mL
2mL
3 mL
3mL
4 mL 4 mL
Initial Volume
Gas (mL)
8 mL
18 mL
23 mL 25 mL
12 mL
7.5 mL
9.5 mL
12.5
mL
Final Volume
Gas (mL)
14 mL
24 mL
38 mL
42 mL
37 mL
30 mL
44 mL
41 mL
ΔV (mL)
6 mL
6 mL
15 mL
19 mL
25 mL
22.5
mL
34.5
mL
28.5
mL
Activity 2
Data Table 2
Show work for determining Concentration of H
2
O
2
in the box below. Concentration should be mol/L.
You need to begin your calculation using the provided 3%m/v H
2
O
2
concentration
.
Concentration
H
2
O
2
(mol/L)
3g H2O2/ 100 mL * 1000 mL/1L sol. * 1 mol H2O2/34.01 g H2O2
= .8820935 H2O2 mol/L
©2016 2 Carolina Biological Supply Company
2021
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
Moles
H
2
O
2
.00088
mol
H2O2
.00088
mol
H202
.00176
mol
H2O2
.00176
mol
H2O2
.00264
mol
H2O2
.00264
mol
H2O2
.00352
mol
H2O2
.00352
mol
H2O2
Moles
O
2
*
.00044
mol O2
.00044
mol O2
.00088
mol O2
.00088
mol O2
.00132
mol O2
.00132
mol O2
.00176
mol O2
.00176
mol O2
ΔV (L)
.006 L
.006 L
.015 L
.019 L
.025 L
.0225 L
.0345 L
.0285 L
*Hint: Use reaction stoichiometry/molar ratio to solve for moles of O
2
.
Show work for determining moles of H
2
O
2
for Trial 1
here: 1 mL H2O2 * 1L/1000mL * .88 H2O2 mol/L = .00088 moles of H2O2
Show work for determining moles of O
2
for Trial 1
here: .00088 moles H2O2 * 1 mol O2/2 moles H2O2 = .00044 moles of O2
Show work for determining moles of H
2
O
2
for Trial 3
here: 2 mL H2O2 * 1 L/1,000 mL * .88 H2O2 mol/L = .00176 moles H2O2
Show work for determining moles of O
2
for Trial 3
here: .00176 moles H2O2 * 1 mol O2/2 moles H2O2 = .00088 moles O2
Show work for determining moles of H
2
O
2
for Trial 5
here: 3 mL H2O2 * 1 L/1000 mL * .88 H2O2 mol/L = .00264 moles H2O2 Show work for determining moles of O
2
for Trial 5
here: .00264 moles H2O2 * 1 mol O2/2 moles H2O2 = .00132 moles O2
Show work for determining moles of H
2
O
2
for Trial 7
here: 4 mL H2O2 * 1L/1000 mL * .88 H202 mol/L = .00352 moles of H2O2
Show work for determining moles of O
2
for Trial 7
here: .00352
moles H2O2 * 1 mol O2/2 moles H2O2 = .00176 moles O2
© 2016 Carolina Biological Supply Company
2021
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. Don’t forget to set the y-intercept to 0. Graph:
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. This includes the correct units for the slope. Air Temperature (K)
Show work in the space
provided.
17 °C + 273 = 290 K
Air Pressure (atm)
Show work in the space
provided.
29.81 inHg * (25.4 mm/1 inch) * (1.00 atm/760 mm) = .996 or 1
atm Equation of the Line
y = 17.973x
Gas Constant R
Show work in the space
provided.
= .06197586
© 2016 Carolina Biological Supply Company
Your preview ends here
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2021
Percent Error
Show work in the space
provided.
Use 0.0821 L*atm/mol*K in
your calculation.
= -24.51 %
Question 1:
Would the volume of oxygen that is generated be affected if a smaller mass of yeast were used? Why or why not?
The volume of oxygen that is generated would not be affected since yeast acts as a catalyst in this reaction; it has the ability to impact the rate of which the reaction occurs but would not have an impact on the volume of oxygen that was produced. Question 2:
Identify at least two potential sources of error in the experiment. Are any assumptions made that would add to the experimental error?
The first potential source of error in this experiment that I identified would be measuring accuracy, since the yeast in particular consisted of incredibly small particles, which made it difficult to measure with accuracy with the scale since the pieces were so tiny and lightweight. A second potential source of error in this experiment would be potential errors when reading/recording measurements (such as the volume in the graduated cylinder) since it was, at times, in-between clear measurement lines, making it difficult to record the precise measurement
of each trial with supreme accuracy. © 2016 Carolina Biological Supply Company
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