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Spectrophotometric Determination of Manganese

↘︎ Nov 28, 2006 … 2′ … download⇠ | skip ⇢

Introduction

A spectrophotometer measures the amount of light absorbed by a solution at different wavelengths of light emitted. Beer’s Law says that absorbance is equal to molar absorptivity times the thickness of the sample times the concentration of the sample. Beer’s law also states that conformity of a solution is able to be determined by plotting its absorbances versus its concentrations, and if a straight line results crossing through the origin, the solution has conformity. Using this information, it is possible to determine an unknown concentration of a solution by finding its absorbance, or if given its concentration, its absorbance can be found without the use of a spectrophotometer.

Experimental

First, a spectrophotometer was turned on, allowed to warm up for about 15 minutes, and was set at a wavelength 400 nm. A cuvette filled with deionized water was used for blanking the spectrophotometer. A second cuvette was filled with a solution of potassium permanganate which was provided. Each cuvette was wiped with a Kimwipe before being placed in the spectrophotometer in order to eliminate smudges which could affect the light passing through. The spectrophotometer was blanked at 400 nm and the cuvette with the potassium permanganate solution was placed in, and its absorbance was read and recorded. It was taken out, and the spectrophotometer was then blanked at 410 nm. The cuvette with the potassium permanganate solution was once against placed in the spectrophotometer. Its absorbance was read and recorded again. This process was repeated, increasing the wavelength of the spectrophotometer by 10 nm until it reached 640 nm when recording ceased. The wavelength with the highest absorbance was used for the rest of the experiment.

Four volumetric flasks were then used to make solutions of KMnO4. Flask 1 was a 100 mL volumetric flask that contained 10 mL of 3.170 x 10-4 M KMnO4, which was dispensed into the flask using a buret. Flasks 2 through 4 were all 50 mL volumetric flasks that contained 20 mL, 30 mL, and 40 mL respectively of 3.170 x 10-4 M KMnO4. All four volumetric flasks were filled to the line on the neck with deionized water. All the flasks were agitated, and cuvettes were filled with each sample. Each cuvette was placed in the spectrophotometer and their absorbances were all recorded.

Next the unknown was placed into a 250 mL beaker and 10 mL of concentrated nitric acid was added to it. Then 0.5 g of potassium periodate was dissolved in 40 mL of deionized water. This solution was heated with a hot plate in order to aid the dissolving process. The contents of the 250 mL beaker were emptied into this solution and were heated for about 10 minutes, but the solution was never brought to a boil. After heating, the solution was put on ice and brought back to room temperature. A cuvette was then filled with this solution and its absorbance was determined and recorded using the spectrophotometer.

Results

Absorption vs. Wavelength for Maximum Absorbance Determination:

Wavelength (nm) Absorbance
400 0.051
410 0.043
420 0.057
430 0.056
440 0.059
450 0.107
460 0.159
470 0.253
480 0.373
490 0.515
500 0.671
510 0.842
520 0.961
530 1.063
540 0.991
550 0.971
560 0.657
570 0.612
580 0.357
590 0.161
600 0.127
610 0.105
620 0.119
630 0.088
640 0.076

Standard Solutions:

Standard solution Initial buret reading Final buret reading Volume added (mL)
Standard #1 5.00 15.00 10.00
Standard #2 15.00 35.00 20.00
Standard #3 1.50 31.50 30.00
Standard #4 0.60 40.60 40.00

Unknown Number: 14

Wavelength: 530

Standard solution Concentration of KMnO4 (M) Absorbance
Standard #1 3.170 x 10-5 (10%) 0.038
Standard #2 1.268 x 10-4 (40%) 0.313
Standard #3 1.902 x 10-4 (60%) 0.453
Standard #4 2.536 x 10-4 (80%) 0.605
Standard #5 3.170 x 10-4 (100%) 0.834
Unknown Solution 1.370 x 10-4 0.322

Calculations

To find the concentration of the standards, I figured out how much the KMnO4­ was diluted in each volumetric flask. I did this by taking the amount of KMnO4­ added, then divided by the total volume on the volumetric flask. I then multiplied this percentage by the original concentration of KMnO4, which was 3.170 x 10-4. To find the concentration of the unknown solution, I first got the equation of the standard curve line, which was y = 2701.2x – 0.048. I then substituted the absorbance I found for the unknown, which was 0.322, for y. I could then find the value of x, which was the concentration.

Discussion/Conclusions

Potassium permanganate does indeed seem to follow Beer’s Law. When I plotted the absorbances found against the concentrations, I was left with nearly a straight line that goes almost directly through the origin. It is only 0.048 absorbances away from going through the origin, and the best fit line is very close to hitting every point plotted. This is one way to prove conformity and Beer’s Law.

Sources of error in this experiment could occur many different ways. If the cuvettes are not wiped off before being placed in the spectrophotometer, there could be smudges or fingerprints that would cause error. The wavelength on the spectrophotometer had to be set by eye, so there is some room for error there, too. If the dilutions are made inaccurately, that would also cause error in absorption readings. Overall, if anything measured in this experiment was measured inaccurately, that would cause error. Also, if the solution with the unknown in it was boiled, that may cause it to form something different than we wanted to measure and that would cause error, too.

Me

circa 2013 (25 y/o)

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  • 06 Nov 28: Spectrophotometric Determination of Manganese #CHM 1112 (General Chemistry Lab I) #Dr. Joseph N. Bartlett #Saint Joseph's University
  • 06 Nov 20: The Effect Light Intensity Has on the Photosynthesis of Spinach Chloroplasts #BIO 1011 (Biology I: Cells) #Dr. Denise Marie Ratterman #Saint Joseph's University
  • 06 Nov 14: Enthalpy of Hydration Between MgSO4 and MgSO4 ∙ 7 H2O #CHM 1112 (General Chemistry Lab I) #Dr. Joseph N. Bartlett #Saint Joseph's University
  • 06 Nov 7: Determining the Heat Capacity of Unknown Metals #CHM 1112 (General Chemistry Lab I) #Dr. Joseph N. Bartlett #Saint Joseph's University
  • 06 Nov 6: The Effect of Temperature on the Metabolism of Fructose and Glucose by Baker’s Yeast #BIO 1011 (Biology I: Cells) #Dr. Denise Marie Ratterman #Saint Joseph's University
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The Effect Light Intensity Has on the Photosynthesis of Spinach Chloroplasts

↘︎ Nov 20, 2006 … 4′ … download⇠ | skip ⇢

Abstract

Photosynthesis in plants is affected by the intensity of the light the plant is exposed to. For this experiment, DCPIP was added to cuvettes with spinach chloroplasts, which were exposed to an incandescent light at different distances for different intervals of time. After each exposure, the cuvettes were placed in a spectrophotometer set at 621 nm and the absorbance of DCPIP was measured. At 37 mm away from the light, the absorbance decreased at 0.033 absorbances/minute. At 27 mm away from the light, the absorbance decreased at 0.043 absorbances/minute. At 5 mm away from the light, the absorbance decreased at 0.050 absorbances/minute. As the distance from the light increased, the absorbance readings on the spectrophotometer went down at a faster rate. This means that the amount of DCPIP in the chloroplasts decreased quicker as the intensity of light was higher, and that photosynthesis occurs more quickly or slowly depending on the intensity of the light.

Introduction

Photosynthesis is the process that allows plants to survive (Chiras, 1993). It provides ATP (which can be used for energy), starch, cellulose, fats, and nucleic acids among other large molecules, and it consumes CO2 to produce O2 (Chiras, 1993). The plant chloroplast cell consists of an inner and outer membrane (Thorpe, 1984). Inside of the cell consists of chlorophyll and stroma. The stroma is a fluid substance comparable to cytosol in animal cells. Chlorophyll are particles that absorb light and pass the energy obtained onto thylakoid disks (Thorpe, 1984). Situated in the membrane of the thylakoid membrane is the electron transport chain. Along the electron transport chain, two electrons from H2O ­are excited by light in Photosystems I and II to reach a higher energy level. This energy is used to convert NADP+ to NADPH and to drive electrons into to thylakoid space, which creates a gradient. This gradient fuels ATP Synthase, which converts ADP + Pi to ATP. ATP is used for energy in many processes in plant cells. In the experiments performed, DCPIP was added to chloroplast cells, which replaced the NADP+ along the electron transport chain. Using a spectrophotometer, the amount of DCPIP present was able to be determined and tests using variable intensities of light were performed on spinach chloroplast cells.

Materials and Methods

First, a room was insulated from light and the lights were turned off. A green light was used in order to see. Six cuvettes were then obtained and labeled with the numbers 1 through 6. Tube 1 consisted of 0.5 mL chloroplast, 3.0 mL cold buffer, 1.5 mL cold distilled water, and 0.0 mL of DCPIP, which were all dispensed into the tube using a micropipettor. This tube was used as a blank for the spectrophotometer which was set to 621 nm. Tube 2 was used as a control and was covered completely with tin foil in order to insulate it from light. Tube 2 and 3 were filled with 0.5 mL chloroplast, 3.0 mL cold buffer, 0.5 mL cold distilled water, and last with 1.0 mL DCPIP. Immediately after being filled with the DCPIP and agitated, the tubes were both placed in the spectrophotometer and their absorbances were recorded. Tube 2 was first removed from its foil before being put in the spectrophotometer, and it was put back on after the reading. They were than placed 11.5 mm away from an incandescent light for 3 minutes. Their absorbances were again recorded and this was repeated until the control tube came to a constant reading while tube 3 gradually went down. Tubes 4 through 6 were also filled with 0.5 mL chloroplast, 3.0 mL cold buffer, 0.5 mL cold distilled water, and 1.0 mL DCPIP right before being subject to light. Tube tubes were placed 37 mm, 27 mm, and 5 mm away, respectively. Their initial absorbances were recorded and following absorbances were recorded every 60 seconds for 6 minutes was exposed to the light.

Results

Table I:

Reading Distance (in mm) Time (in minutes) Absorbance
Tube 2 (Control) 11.5 0 0.44
Tube 2 (Control) 11.5 3 0.44
Tube 2 (Control) 11.5 6 0.44
Tube 2 (Control) 11.5 9 0.44
Tube 3 11.5 0 0.51
Tube 3 11.5 3 0.385
Tube 3 11.5 6 0.29
Tube 3 11.5 9 0.21
Tube 4 37 0 0.465
Tube 4 37 1 0.46
Tube 4 37 2 0.45
Tube 4 37 3 0.40
Tube 4 37 4 0.395
Tube 4 37 5 0.30
Tube 5 27 0 0.465
Tube 5 27 1 0.45
Tube 5 27 2 0.39
Tube 5 27 3 0.35
Tube 5 27 4 0.29
Tube 5 27 5 0.25
Tube 5 27 6 0.21
Tube 6 5 0 0.46
Tube 6 5 1 0.42
Tube 6 5 2 0.38
Tube 6 5 3 0.33
Tube 6 5 4 0.25
Tube 6 5 5 0.205
Tube 6 5 6 0.16

Table II:

Reading Distance (mm) Rate of Decrease (Absorbance/min)
Tube 4 37 0.033
Tube 5 27 0.043
Tube 6 5 0.050

Tubes 2 and 3 were used to prove that it was indeed the light causing the absorbance to go down, and not the heat from the light. Tube 2 was covered with foil to prevent it from being exposed to light and its absorbance stayed constant, while the absorbance from tube 3 which was uncovered went down. As the tubes were placed closer to the light, their absorbances went down quicker, which was expected. The absorbance of tube 6 went all the way down to 0.16, which seemed very low (Table I). Test tube 5’s absorbance went low also, going down to 0.21 (Table I). Test tube 4 had very sporadic readings (Table I). This may be attributed to it not being thoroughly mixed enough.

Discussion

The absorbances readings recorded measured the amount of DCPIPoxidized in the cells. As photosynthesis occurred, electrons were donated to the DCPIPoxidized, forming DCPIPH2 reduced. DCPIPoxidized is able to absorb light from the spectrophotometer at 621 nm, hence that it why it was calibrated at 621 nm. As photosynthesis took place, there was less and less DCPIPoxidized available for absorbance. That is why the absorbance readings went down over time.

Bidwell (1979) reported that light absorption is not really affected by temperature. The results gained from the experiment were consistent with his findings. When tube 2, which was covered in tin foil, was exposed to the light, its absorbance stayed constant. Though light was not affecting the tube, it could still be heated up. Because its absorbance did not move, this showed that the heat did not affect any of the absorbance readings for any of the test tubes.

Tubes 3 through 6 were all exposed to the light and their absorbances went down over time. The rate at which their absorbances went down increased as their distance from the light source decreased. The intensity of light directly affects the rate of photosynthesis (Bidwell, 1979). Graphs show that the higher the intensity, the higher rate of photosynthesis. The intensity in this experiment was increased by moving the tube closer to the light.

Lastly, the reason a green light was used in order to see was because chlorophyll absorbs all colors of light except for green and yellow (Chiras, 1993). Additional experiments using different colors of light or different light bulbs would be interesting, as the chlorophyll trap different kinds of lights at different rates. This could show what range of the color spectrum chlorophyll accept best.

Literature Cited

Bidwell, R. G. S. 1979. Plant Physiology. (MacMillian Publishing Co., NY, NY) 726 p.

Chiras, Daniel D. 1993. Biology: The Web of Life. (West Publishing Co., St. Paul, Mn) 896 p.

Thorpe, N. O. 1984. Cell Biology. (John Wiley & Sons, NY, NY) 719 p.

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circa 2017 (29 y/o)

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Enthalpy of Hydration Between MgSO4 and MgSO4 ∙ 7 H2O

↘︎ Nov 14, 2006 … 2′ … download⇠ | skip ⇢

Introduction

Enthalpy of hydration is the energy change for converting 1 mol of an anhydrous substance to 1 mol of the hydrated substance. In order to find this number, it is necessary to first calculate the enthalpy of dissolution for each substance separately, and then find the different between the two. The enthalpy of dissolution is the energy change of dissolving 1 mol of a substance in water. It is calculated using temperature changes in the water, heat capacity of the substance, and the weight of the mixture. For this experiment, MgSO4 and MgSO4 ∙ 7 H2O were used and the enthalpy of hydration between the two was calculated.

Experimental

A Styrofoam cup and stirring bar were first obtained and weighed together. This mass was recorded. 100.0 mL of deionized water was measured with a graduated cylinder and then put into the cup with the stirring bar. The cup was again weighed and this new mass was recorded. The cup was then placed on a mixing plate set on medium to high and its temperature was recorded every 30 second for 4.5 minutes. An unknown amount of MgSO4 salt was added to the cup. The cup kept on the mixing plate set on medium to high and its temperature was recorded every minute for 15 minutes. Finally, the cup was weighed and its final mass was recorded. This process was repeated placing the MgSO4 with MgSO4 ∙ 7 H2O.

Results

Measurement MgSO4 ∙ 7 H2O Trial MgSO4 Trial
Mass of cup and stirring bar (g) 7.85 7.41
Mass of cup, stirring bar, and water (g) 107.21 106.70
Mass of water (g) 99.36 99.29
Mass of cup, stirring bar, water, and salt (g) 119.50 113.06
Mass of Mg salt (g) 12.29 6.36
Molar mass of solute (g) 246.476 120.369
Moles of solute added (mol) 0.04986 0.0528
Mass of salt and water (g) 111.68 105.65
Initial temperature at time of mixing (ºC) 20.90 21.60
Extrapolated final temperature of reaction mixture (ºC) 19.27 32.65
ΔT = Tfinal – Tinitial (ºC) -1.63 12.05
Heat Capacity of reaction mixture (J/(gºC)) 3.84 3.84
Heat transferred during dissolution, Q (Joule) 699. -4890.
ΔHdissolution (J/mole) 14000. (14.0 kJ) -92600. (-92.6 kJ)

Enthalpy of Hydration: -106.6 kJ

Time (minutes) Temperature of MgSO4 ∙ 7 H2O solution (ºC) Temperature of MgSO4 solution (ºC)
0.0 n/a n/a
0.5 20.90 21.63
1.0 20.90 21.63
1.5 20.90 21.63
2.0 20.90 21.62
2.5 20.90 21.60
3.0 20.90 21.60
3.5 20.90 21.59
4.0 20.89 21.57
4.5 20.89 21.57
5.0 (salt added) n/a n/a
5.5 19.57 26.50
6.0 19.30 27.18
7.0 19.29 28.72
8.0 19.30 29.12
9.0 19.32 29.50
10.0 19.35 31.20
11.0 19.38 31.65
12.0 19.40 31.60
13.0 19.42 31.44
14.0 19.49 31.28
15.0 19.50 31.10
16.0 19.51 30.91
17.0 19.58 30.76
18.0 19.60 30.58
19.0 19.65 30.43
20.0 19.69 30.23

Calculations

To find the mass of water used, I subtracted the weight of the cup with just the stirring rod from the weight of the cup with the stirring rod and water. To find the weight of the salt used, I subtracted the weight of the cup, stirring rod, and water from the final weight of the cup. In order to find the moles of solute used, I divided the mass of the salt by its molar mass. To find the change in temperature, I subtracted the initial temperature from the final temperature. In order to find Q, the heat capacity of the reaction mixture, I used the equation Q = – (mass of mixture) * (heat capacity of mixture) * (ΔT). To find the ΔHdissolution, I used the equation ΔH = Q / (number of moles of solute). Lastly, to calculate the enthalpy of hydration, I subtracted the ΔHdissolution of the MgSO4 ∙ 7 H2O from the ΔHdissolution of the MgSO4.

Discussion/Conclusions

I was surprised that while the MgSO4 salt heated the water, the MgSO4 ∙ 7 H2O salt cooled the water down. It was interesting that two substances very close in chemical makeup could have such different reactions in water. My graph for the temperature change of water with MgSO4 seems to only gradually jump in temperature after adding the salt. I believe this is because my lab partner forgot to turn the mixer on, so the salt was not completely mixing at first. Other than that, the procedure went well. The enthalpy of hydration of -106.6 kJ seems fairly high. Water takes 4.184 kJ to be raised only 1 ºC, so 106.6 kJ seems like a lot of energy.

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Determining the Heat Capacity of Unknown Metals

↘︎ Nov 7, 2006 … 3′ … download⇠ | skip ⇢

Introduction

One gram of water takes 4.184 joules of energy to increase its temperature 1 ºC. This is the most energy any substance takes to raise its temperature 1 ºC. In contrast to taking the most energy to raise its temperature 1 ºC, this means that it also takes the longest to cool down. This means that water has the highest heat capacity. It must release 4.184 joules of energy in order to decrease its temperature by just 1 ºC. The heat water releases is absorbed by its environment. Knowing the heat capacity of water, it is possible to find how well its environment insulates it. Also using the heat capacity of water, one can figure out the heat capacity of an unknown substance by putting it in water and measure the temperature change of the water and the unknown substance. In this experiment, this is exactly what was performed.

Experimental

First, an empty Styrofoam cup and lid were weighed and its mass was recorded. 70 mL of room temperature water was then added to the Styrofoam cup and it was reweighed and recorded. The temperature of the water was also recorded. Next, 30 mL of water was heated until boiling and this temperature was also recorded. The boiling water was poured into the Styrofoam cup and the final temperature of the combined water was measured and recorded. The final mass of the cup was also recorded.

For the next part of the experiment, an empty Styrofoam cup and lid were again weighed and its mass was recorded. 100 mL of room temperature water was added to the cup and it was reweighed and recorded. The temperature of the water was also recorded. Next, an unknown metal was heated to about 100 ºC and then poured into the Styrofoam cup. The final temperature of the water was measured and recorded, as was the final mass of the cup.

Results

Identification of Metal: 12

Determination of Calorimeter Constant, B:

Trial 1 Trial 2
Mass of empty Styrofoam cup 3.56 g 3.55 g
Mass of cup + 70 mL water 72.20 g 73.26 g
Mass of cup + 70 mL water + 30 mL hot water 101.57 g 102.88 g
Initial temperature of water in calorimeter 23.70 ºC 24.20 ºC
Temperature of the boiling water bath 99.5 ºC 99.5 ºC
Final temperature of calorimeter + added hot water 43.99 ºC 44.3 ºC
Mass of cool water in cup, mCW 68.64 g 69.71 g
Mass of added hot water, mHW 29.37 g 29.62 g
Temperature change of cool water in calorimeter, ΔTCW 20.29 ºC, 293.29 K 20.1 ºC, 293.1 K
Temperature change of added hot water, ΔTHW 55.5 ºC, 328.5 K 55.2 ºC, 328.2 K
Calorimeter constant, B -425. J/K -430. J/K

Determination of the Heat Capacity of a Metal:

Trial 1 Trial 2
Mass of empty Styrofoam cup 3.59 g 3.58 g
Mass of cup + 100 mL water 103.40 g 103.33 g
Mass of cup + 100 mL water + hot metal 180.97 g 180.90 g
Initial temperature of water in calorimeter 24.35 ºC 23.9 ºC
Temperature of boiling water bath 99.0 ºC 99.5 ºC
Final temperature of calorimeter + added hot metal 29.75 ºC 29.375 ºC
Mass of cool water in cup, mCW 99.81 g 99.75 g
Mass of added hot metal, mHM 77.57 g 77.57 g
Temperature change of cool water in the calorimeter, ΔTCW 5.40 ºC, 278.40 K 5.5 ºC, 278.5 K
Temperature change of added hot metal, ΔTHM 69.3 ºC, 342.3 K 70.1 ºC, 343.1 K
Heat capacity of metal, ­­CP, M 0.0775 J/gK 0.132 J/gK
Molar mass of metal 323. g/mole 189. g/mole

Calculations

For the determination of the calorimeter constant, to find the mass of cool water in the cup, I simply subtracted the mass of the empty cup from the mass of the cup with 70 mL of water. To find the mass of hot water added, I subtracted the mass the cup with 70 mL of water from the mass of the cup with the 70 mL of cool water and 30 mL of hot water. To find the temperature changes, I found the difference in temperatures between the final and the initial readings. I then converted those temperatures to Kelvin from Celsius by adding 273. To find the calorimeter constant, I used the equation B = -CP(mCW ΔTCW + mHW ΔTHW) / ΔTCW.

For the determination of the heat capacity of a metal, I performed the same operation as I did for the determination of the calorimeter constant, only replacing the mass of hot water with the mass of the hot metal. The equation for the heat capacity of the metal was also different. It was CP, M = – ΔTCW (B + mCW CP) / (mHM ΔTHM). Lastly, to find the molar mass of the metal, I then divided 25 J/mole K by the heat capacity of the metal, which is the Law of Dulong and Petit.

Discussion/Conclusions

My final results do not seem very accurate. The calorimeter constants seem fairly close, but that difference affected the result of the heat capacity of my metal greatly. If I had gotten the same calorimeter constant for both trials, then the heat capacity of the metals would have came out nearly equal. As a result of the heat capacity of the metals being different, their molar masses were also thrown off. In conclusion, I must have made a mistake in a reading while finding the calorimeter constant.

Something that surprised me was how low the heat capacity of the metal was. Normally I think of metals as being very hot and staying hot, but this experiment proved how metals actually cool very quickly. People usually make this generalization, but they are wrong. I think it is because metals heat up more quickly, so they typically think they stay hot. In actuality, hot water is more dangerous than hot metal.

Me

circa 2008 (20 y/o)

More from…
CHM 1112 (General Chemistry Lab I) (Class) / Dr. Joseph N. Bartlett (Teacher) / Saint Joseph’s University (School) / schoolwork (Post Type)

The Effect of Temperature on the Metabolism of Fructose and Glucose by Baker’s Yeast

↘︎ Nov 6, 2006 … 1′ … download⇠ | skip ⇢

This study determined whether or not temperature had an affect on CO2 production between yeast and different sugars. For this experiment, 2 mL of 10% fructose solution, 10% glucose solution, and distilled water were each combined with 2 mL of baker’s yeast. Each of these three solutions was measured for CO2 production at three different temperatures: 3 ºC, 22 ºC (room temperature), and 42 ºC. None of the solutions produced CO2 when they were at 3 ºC. At room temperature, the fructose and yeast solution produced 0.37 mL of CO2 in 60 minutes, while the glucose and yeast solution produced 0.45 mL of CO2 in 60 minutes. The control sample did not produce any CO2. At 42 ºC, the fructose and yeast solution produced 1.28 mL of CO2 in 45 minutes, while the glucose and yeast solution produced 1.33 mL of CO2 in 45 minutes. Again, the control group did not produce any CO2. Temperature had a great impact of the production of CO2. When chilled, the solutions did not even produce any CO2 at all, but when heated, the solutions produced CO2 almost four times as fast as normal.

Glucose
Room Temp. (22 degrees C) Cold Temp. (3 degrees C) Hot Temp. (32 degrees C)
10 min. 0.00 ml. 0.00 ml. 0.11 ml.
15 min. 0.00 ml. 0.00 ml. 0.40 ml.
20 min. 0.00 ml. 0.00 ml. 0.47 ml.
25 min. 0.00 ml. 0.00 ml. 0.77 ml.
30 min. 0.17 ml. 0.00 ml. 0.91 ml.
35 min. 0.26 ml. 0.00 ml. 1.07 ml.
40 min. 0.30 ml. 0.00 ml. 1.20 ml.
45 min. 0.31 ml. 0.00 ml. 1.33 ml.
50 min. 0.33 ml. 0.00 ml. 1.33 ml.
55 min. 0.35 ml. 0.00 ml. 1.33 ml.
60 min. 0.42 ml. 0.00 ml. 1.33 ml.
65 min. 0.45 ml. 0.00 ml. 1.33 ml.

1.33= pipet exhausted all of the yeast

Fructose
Room Temp. (22 degrees C) Cold Temp. (3 degrees C) Hot Temp. (32 degrees C)
5 min. 0.00 ml. 0.00 ml. 0.09 ml.
10 min. 0.00 ml. 0.00 ml. 0.24 ml.
15 min. 0.05 ml. 0.00 ml. 0.36 ml.
20 min. 0.11 ml. 0.00 ml. 0.59 ml.
25 min. 0.14 ml. 0.00 ml. 0.76 ml.
30 min. 0.17 ml. 0.00 ml. 0.92 ml.
35 min. 0.22 ml. 0.00 ml. 1.05 ml.
40 min. 0.30 ml. 0.00 ml. 1.18 ml.
45 min. 0.33 ml. 0.00 ml. 1.28 ml.
50 min. 0.34 ml. 0.00 ml. 1.33 ml.
55 min. 0.45 ml.* 0.00 ml. 1.33 ml.
60 min. 0.47 ml.* 0.00 ml. 1.33 ml.

*1 ml. bubble appeared in the pipet

Me

circa 2009 (21 y/o)

More from…
BIO 1011 (Biology I: Cells) (Class) / Dr. Denise Marie Ratterman (Teacher) / Saint Joseph’s University (School) / schoolwork (Post Type)

The Failure of Rogerian Argument in “Super Size Me”

↘︎ Nov 6, 2006 … 3′ … download⇠ | skip ⇢

Obesity, a common problem in America, is thought to be fueled partly by sugary, fatty foods sold at fast food restaurants. People generally know that fast food is bad for them, but they continue to eat it. Two extremely overweight teenage girls sued McDonald’s claiming that their food was the cause of their weight gain. Their case was ultimately dropped because they could not prove that McDonald’s food caused their obesity. This lawsuit intrigued Morgan Spurlock, who decided to see if McDonald’s food truly does cause obesity. In order to test this theory, Spurlock embarked on a diet consisting only of McDonald’s food for one month. He entered the experiment in very good physical shape. By the end of the experiment, Spurlock had gained almost twenty-five pounds, acquired heart problems, and his liver was badly damaged. He became very lethargic and had constant headaches. Besides the physical problems Spurlock endured, he also suffered mentally from depression.

Spurlock created a documentary, Super Size Me, showing his experience and how McDonald’s food affected him. The purpose of the documentary was to show the danger of McDonald’s food on a person’s health. The viewer could then decide for themselves whether or not they would continue eating McDonald’s food. Spurlock tried to portray the documentary in an unbiased mood, but for the most part, Spurlock did not do an adequate job of gathering and presenting the information he found in a fair manner. In a Rogerian argument, the person wanting change tells theirs views, but then also tells the views of the defendant, sympathizing with them. By showing knowledge of how the defendant feels, the prosecutor gains trust from the defendant, then suggests a common ground for agreement. By being respectful of the opponent’s ideas and thoughts, the opponent will feel less threatened and be more inclined to change their ways. Spurlock did not treat McDonald’s views and opinions with respect, and thus did not succeed in Rogerian argument.

A way the documentary failed in Rogerian argument was by containing a copious amount of sarcasm. Almost every time Spurlock showed McDonald’s side of an argument, he would present it in a sarcastic way, making McDonald’s look wrong and foolish. He sarcastically ordered and ate his food, which created humor and generally made McDonald’s seem bad. Cartoons and animations shown also added to the whole comical routine. This mood made the viewer take anything McDonald’s said unseriously. If Spurlock were to eliminate sarcasm, the documentary would be entirely different and McDonald’s would not seem nearly as bad as they are presented.

Another way Spurlock failed in creating Rogerian argument was by presenting McDonald’s stance on a subject, and then giving his opinion on the subject directly afterwards. He often showed a fact McDonald’s presented and then trumped it with an even better fact he found, or gave the McDonald’s fact and then asked an open ended question, which would make McDonald’s always look wrong. Spurlock almost never gave his side of the argument first, and then showed McDonald’s side last. If he did show McDonald’s side of an argument last, he would show their argument in a downgrading sarcastic way, not in the confident way he presented his arguments. The way he presented his arguments leaves the viewer thinking that Spurlock was right in every issue discussed.

There are also some specific parts of Spurlock’s experiment that he could have performed better. For example, under almost any diet, one is bound to become overweight and out of shape if they do not exercise. Spurlock seemed to drastically change his daily routine for the experiment. He was in very good physical shape before the experiment, so he should have kept doing whatever he did to stay in shape. If that entailed going to the gym and working out, he should have continued doing that during the month he ate only McDonald’s food. Changing his daily routine most likely skewed the results of the experiment.

Also, most people that eat McDonald’s do not eat it three times a day, and most people do not eat it every day. Spurlock could have shown what would happen if McDonald’s was eaten only once or twice a day, or it was eaten only every other day. It is unrealistic to think that people eat all three meals at McDonald’s every day of the week. By limiting the amount of McDonald’s eaten, Spurlock would not have gone under such a dramatic transformation and McDonald’s would not have looked so bad. If he had eaten McDonald’s food in combination with healthy food, he may have gotten much different results. He also could have gathered information on how much McDonald’s food the obese teenage girls that sued McDonald’s ate, and then went on diet similar to theirs. That would have been a more accurate representation of how much McDonald’s food an overweight customer eats.

Spurlock succeeds in showing how eating only McDonald’s for a month without exercise will affect a person, but he does not succeed in Rogerian argument. If he were to perform the experiment under more normal conditions and present information in a less biased way, then it would be fairer to McDonald’s. It seems that he tried his best to portray McDonald’s in a negative way. Spurlock delivers information to make McDonald’s look naïve and foolish much of the time. The humor he adds makes the documentary more enjoyable and captivating for the audience, but it detracts from its viability. Overall, Spurlock could have done a much better job of portraying McDonald’s in a respectable manner.

Me

circa 2018 (30 y/o)

More from…
ENG 1011 (Craft of Language) (Class) / Mrs. Marie H. Flocco (Teacher) / Saint Joseph’s University (School) / schoolwork (Post Type)

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ADAM CAP is an elastic waistband enthusiast, hammock admirer, and rare dingus collector hailing from Berwyn, Pennsylvania.

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