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Dr. Jose Cerda

schoolwork | Teacher … see also: Mr. Michael Settanni / Dr. Paul J. Angiolillo / Ms. Ringle / Dr. Peter M. Graham / Mrs. Marie H. Flocco / Dr. Joseph N. Bartlett

Electronic Spectra of a Dye Molecule

↘︎ Apr 20, 2009 … 2′ … download⇠ | skip ⇢

Introduction

The proposed particle in a box experiment uses theories from quantum mechanisms in order to determine and prove the behavior of a molecule. In the experiment performed, various dye molecules were observed using a Spectronic 21 spectrophotometer to determine their wavelengths of maximum absorption. The spectrophotometer emits light through the sample and uses a detector to measure the absorbance through a range of wavelengths. The “box” is considered to be the space between the nitrogens of the dye molecules, where the behavior of the molecules is observed. Upon measuring different concentrations of the dye molecules with the Spectronic 21, standard linear plots of absorbance versus concentration could then be graphed. These plots could then be used to establish that the dyes follow Beer’s law, A = εbc. The measured wavelengths of maximum absorption could then be compared to the theoretical values, which can be found using the following equations:

λmax = 8ml2c (p + 3)2 / [h(N+1)]

λmax = 8ml2c (p + 3 + α)2 / [h(p+4)]

where h = Planck’s constant, m = mass of an electron, c = speed of light, l = distance between the nitrogens, N = number of electrons in the entire molecular orbital π system, p = number of carbon atoms, and α = extra distance the conjugate electrons go beyond the terminal nitrogens. These equations are derived from the basics of quantum mechanics. By comparing the measured values to the calculated values, one can validate the theories of quantum mechanics and also determine α.

Procedure

To begin, 1 L of 1 x 10-6 M solutions were prepared using the following compounds: 1,1’-diethyl-4,4’-cyanine iodide, 1,1’-diethyl-4,4’-carbocyanine iodide, and 1,1’-diethyl-4,4’-dicarbocyanine iodide. Then, the following dilutions of each of those solutions were prepared: 100%, 50%, 25%, 10%, and 5%. Each of the dilutions was then analyzed using a Spectronic 21 spectrophotometer to determine their measured absorbance and wavelength of maximum absorption. These measured values were saved on the computer and later analyzed.

Results

Cyanine Iodide Carbocyanine Iodide Dicarbocyanine Iodide
λ max (measured) (nm)

595

707

814

λ max (calculated) (nm)

604

736

868

α

-0.0732

-0.236

-0.441

% error

1.49

3.94

6.22

c (m/s)

2.998 x 108

2.998 x 108

2.998 x 108

m (kg)

9.1 x 10-31

9.1 x 10-31

9.1 x 10-31

p

7

9

11

h (m2kg/s)

6.626 x 10-34

6.626 x 10-34

6.626 x 10-34

N

10

12

14

ε

80281

226691

127172

l (m)

1.42 x 10-10

1.42 x 10-10

1.42 x 10-10

Calculations

λ max (calculated) =

8 (9.1 x 10-31 kg) (1.42 x 10-10 m)2 (2.998 x 108 m/s) (7 + 3) 2 / [ (6.626 x 10-34 m2kg/s) (10 + 1) ]

λ max (calculated) = 6.04 x 10-7 m (1 nm / 10-9 m) = 604 nm

 

λ max (measured) =

8 (9.1 x 10-31 kg) (1.42 x 10-10 m)2 (2.998 x 108 m/s) (7 + 3 + α) 2 / [ (6.626 x 10-34 m2kg/s) (10 + 1) ]

α = 0.001628

 

Percent Error = | λ max (calculated) – λ max (measured) | / λ max (calculated) x 100%

| 604 -595 | / 604 x 100% = 1.49%

 

Conclusions

The calculated λ max values were extremely close to the measured λ max values. The percent error was 1.49% for cyanine iodide, 3.49% for carbocyanine iodide, and 6.22% for dicarbocyanine iodide. These results seem very good considering the laboratory conditions; they are nearly the best results we could have hoped for. This seems to show the equations derived from quantum mechanics are indeed valid. However, the measured λ max values are all less than the calculated λ max values, which should not have been the case. The measured λ max values should have been greater than the calculated λ max values in order to account for α, the extra length past the terminal nitrogens. This in turn made the values for α all negative. The dyes all seems to follow Beer’s law, as shown by the high R2 values calculated by plotting the absorbance versus concentration for each dye. It was odd that the absorbance reading for dicarbocyanine iodide exceeded a value of 1. Normally this should never happen; 1 should be the maximum absorbance value. This may be accounted for by the spectrophotometer not being correctly calibrated.

Me

circa 2013 (25 y/o)

about adam

Jump…

  • 09 Apr 20: Electronic Spectra of a Dye Molecule #CHM 2422 (Physical Chemistry Lab II) #Dr. Jose Cerda #Saint Joseph's University
  • 09 Feb 17: The Kinetics of the Bromate-Bromide Reaction #CHM 2422 (Physical Chemistry Lab II) #Dr. Jose Cerda #Saint Joseph's University
  • 08 Dec 9: Using a Parr Bomb to Measure Enthalpy #CHM 2412 (Physical Chemistry Lab I) #Dr. Jose Cerda #Saint Joseph's University
  • 08 Oct 6: Analysis of Ethanol and Butanol Solutions via Gas Chromatography #CHM 2412 (Physical Chemistry Lab I) #Dr. Jose Cerda #Saint Joseph's University

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CHM 2422 (Physical Chemistry Lab II) (Class) / Saint Joseph’s University (School) / schoolwork (Post Type)

The Kinetics of the Bromate-Bromide Reaction

↘︎ Feb 17, 2009 … 4′ … download⇠ | skip ⇢

Introduction

This laboratory experiment placed an emphasis on the determination of the order of reaction and ultimately the rate constant and activation energy of the bromate-bromide reaction. The rate equation of this reaction is represented by:

-d[BrO3–] / dt = k [BrO3–]x [Br–]y [H+]z

Hence, by using varying and constant amounts of bromate, bromide, and acid in different trial runs, it is possible to determine the x, y, and z exponents of this equation, which are actually the order of the reactions.

In order to better understand the bromate-bromide reaction, it is important to take note of the stoichiometric equation, which is noted as:

BrO3– + 5Br– + 6H+ —> 3Br2 + 3H2O

This reaction couples with the reaction of Br­2 and methyl orange, which produces bleach. By combining the two reactions, the basis is built for determining the rate of the reaction. As bromate and bromide react, they produce Br2, which then reacts with methyl orange to dissipate the color of the solution and show completion of the reaction. This dissipation of color can be timed and used for calculation.

However, this reaction happens very quickly, and under normal circumstances would not be able to be timed. By adding phenol to the solution, the reaction of Br2 with methyl orange can be slowed down with the following side reaction:

C­6H5OH + Br2 —> BrC6H4OH + H+ + Br–

This reaction of Br2 with phenol slows down the reaction with methyl orange enough that it is able to be timed. Otherwise, the reaction would happen very quickly and it would be untimable. Nitrate is added to the solution to also improve the quality of the reaction, by increasing the ionic strength of the ion. In theory, the rate equation of the reaction should equal:

-d[BrO3–] / dt = k (a x 2a x (3a)2) = 18ka4

Where a is equal to the original bromate concentration, 2a is the initial bromide concentration, and 3a is the initial hydrogen ion concentration. The plot of -∆[BrO3–] / t versus time at different temperatures should yield linear plots, with the slopes equaling 18ka4. Knowing the value of a, the rate constant (k), can be evaluated at each temperature.

Finally, once the rate constant is found, the activation energy can be determined by using the following equation:

log k = (-E / 2.303R) x (1 / T) + constant

By plotting the log k versus 1 / T° K for each temperature, the slope yielded also will provide the activation energy in terms of calorie per mole.

Procedure

To begin, aqueous solutions of the following reagents were prepared: 0.333 M potassium bromate, 0.667 M potassium bromide, 0.500 M perchloric acid, 0.030 M phenol, 0.500 M sodium nitrate, and 40 mg/L of methyl orange solution. Solutions comprised of these were then prepared according to the given specifications. However, potassium bromate, potassium bromide, perchloric acid, and water were combined in one flask while phenol, sodium nitrate, and methyl orange solution were combined in another flask. The flasks were then placed in the bath and allowed to come to a constant temperature. The flask containing the bromate and bromide was transferred into the flask with methyl orange. As soon as the two solutions made contact, timing began using a stopwatch, and the solutions were mixed using a glass stirring rod to ensure conformity. As soon as the orange color had completely dissipated, timing stopped. This process was repeated for every trial according to the specifications of the following charts:

Table #1 (At 25° C)

BrO3– (mL) Br– (mL) NO3– (mL) H2O (mL) HClO4 (mL) C6H5OH (mL) MO (mL)
Run 1 5 5 10 5 10 10 5
Run 2 5 10 3.5 6.5 10 10 5
Run 3 10 5 7 3 10 10 5
Run 4 5 5 0 5 20 10 5

Table #2 (At 20° C)

BrO3– (mL) Br– (mL) NO3– (mL) H2O (mL) HClO4 (mL) C6H5OH (mL) MO (mL)
Run 1 5 5 10 12 10 3 5
Run 2 5 5 10 9 10 6 5
Run 3 5 5 10 6 10 9 5
Run 4 5 5 10 3 10 12 5
Run 5 5 5 10 0 10 15 5

Table #3 (At 25° C)

BrO3– (mL) Br– (mL) NO3– (mL) H2O (mL) HClO4 (mL) C6H5OH (mL) MO (mL)
Run 1 5 5 10 12 10 3 5
Run 2 5 5 10 9 10 6 5
Run 3 5 5 10 6 10 9 5
Run 4 5 5 10 3 10 12 5
Run 5 5 5 10 0 10 15 5

Table #4 (At 30° C)

BrO3– (mL) Br– (mL) NO3– (mL) H2O (mL) HClO4 (mL) C6H5OH (mL) MO (mL)
Run 1 5 5 10 12 10 3 5
Run 2 5 5 10 9 10 6 5
Run 3 5 5 10 6 10 9 5
Run 4 5 5 10 3 10 12 5
Run 5 5 5 10 0 10 15 5

Table #5 (At 35° C)

BrO3– (mL) Br– (mL) NO3– (mL) H2O (mL) HClO4 (mL) C6H5OH (mL) MO (mL)
Run 1 5 5 10 12 10 3 5
Run 2 5 5 10 9 10 6 5
Run 3 5 5 10 6 10 9 5
Run 4 5 5 10 3 10 12 5
Run 5 5 5 10 0 10 15 5

Results and Calculations

Determining the Order of Reaction:

-d[BrO3–] / dt = k [BrO3–]x [Br–]y [H+]z

Run 1 & Run 3

(0.002M) / (54.25secs) = k [0.333] x [0.0667] y [0.1] z
(0.002M) / (27.15secs) k [0.667] x [0.0667] y [0.1] z

0.5 = 0.499 x
x = 1

Run 1 & Run 2

(0.002M) / (54.25 secs) = k [0.0333] x [0.0667] y [0.1] z
(0.002M) / (30.34 secs) k [0.0333] x [0.1333] y [0.1] z

0.559 = 0.5 y
ln (0.559) = y ln (0.5)
y = 0.839

Run 1 & 4

(0.002M) / (54.25 secs) = k [0.0333] x [0.06667] y [0.1] z
(0.002M) / (24.47 secs) k [0.0333] x [0.06667] y [0.2] z

0.451 = 0.5z
ln (0.451) = z ln (0.5)
z = 1.15

Concentrations versus Recorded Times

25° C (Run 1)

25° C (Run 2)

30° C

35° C

[BrO3–]

t (s)

[BrO3–]

t (s)

[BrO3–]

t (s)

[BrO3–]

t (s)

0.6

21.78

0.6

21.47

0.6

12.57

0.6

6.63

1.2

37.18

1.2

50.11

1.2

23.21

1.2

15.29

1.8

54.57

1.8

82.18

1.8

34.3

1.8

22.22

2.4

71.8

2.4

109.72

2.4

43.67

2.4

31.81

3

97

3

143

3

55.88

3

44.75

Determining the rate constant for each temperature

-d[BrO3–] / dt = k (a x 2a x (3a)2) = 18ka4

20° C: k = 0.0198 = 894.57
(18)(0.0333)4

25° C: k = 0.0321 = 1450.29
(18)(0.0333)4

30° C: k = 0.056 = 2530.10
(18)(0.0333)4

35° C: k = 0.0638 = 2882.51
(18)(0.0333)4

Activation Energy:

T (° K)

1/T

log (k)

293

3.41 x 10-3

2.95

298

3.36 x 10-3

3.16

303

3.30 x 10-3

3.40

308

3.25 x 10-3

3.46

Slope = – E / (2.303 * R ) , R= 8.3145 J/mol K
-3287.7 = – E / (2.303 * 8.3145)
E = 62953.8 cal/mol

Conclusions

The first objective of this experiment was to determine the orders of the reaction. The values obtained vary in their likeness to the theoretical values. It was found that -d[BrO3–] / dt = k [BrO3–]1 [Br–]0.839 [H+]1.15 when it really should have been = k [BrO3–]1 [Br–]1 [H+]2. The calculated value for BrO3– was dead on, while the value for Br– was slightly off and the value for H+ was significantly different. This error could be attributed to a multitude of sources. For example, the concentrations of solutions used may have been different than believed. If the solutions were old, they may have lost strength, contributing towards this. This would in turn affect the timing with the stopwatch, which would throw off the calculation of the orders of reaction. To go along with this, if the solutions were not made accurately, meaning the exact aliquots were not used, this would also affect the time and ultimately rate.

The next objective was to determine the rate constant for each temperature and then the activation energy. The rate constants calculated seem fairly large, meaning the slopes that resulted from plotting [BrO3–] versus time were of small value. If the slopes were larger numbers, then the rate constants would have been smaller. This in turn made the activation energy seem excessive. With smaller rate constant values, the plot would have resulted with a smaller number for the slope, and thus a smaller rate constant. This problem can again be justified due to the fact that the believed concentrations and temperatures of solutions may not have been actually what we were working with.

Me

circa 2009 (21 y/o)

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Using a Parr Bomb to Measure Enthalpy

↘︎ Dec 9, 2008 … 3′ … download⇠ | skip ⇢

I. Introduction

The term enthalpy is used to describe the thermodynamic potential of a system. Heat of reaction, one form of enthalpy, is the energy given off by a substance upon combustion. It can be measured using a Parr bomb, which is an adiabatic container, held at a constant volume that ignites a substance. In this case, the heat of reaction is equal to the enthalpy because the reaction takes place at a constant volume and the change in pressure is negligible. The heat of reaction is also equal to the internal energy because of the constant pressure.

The substance to be measured is placed in the bomb, and the bomb is filled with oxygen and then sealed. The bomb is placed in a vessel of circulating water. The substance is ignited through wires touching it which are connected to a fuse box. The energy given off from combustion can then be measured from the changes in temperature of the water surrounding the bomb using a thermistor. The thermistor can be linked to a computer and the temperate as a function of time can be monitored and recorded. A substance of known enthalpy, or heat of reaction, is first used to create a constant for mass times Cp, so that a substance of unknown enthalpy can then be calculated from that information. The change in mass from each experimental run is negligible, making this method valid. The thermistor itself must also first be calibrated, by measuring a constant temperature bath and then changes in the temperature bath as a function of resistance.

II. Method

First, water that was a slightly above room temperature was placed in a vessel. The vessel was placed into the calorimeter and allowed to cool to room temperature. This change in temperature was measured and recorded with a thermistor and graphed on the computer as a function of resistance. This part of the experiment was performed without the actual bomb, as it was used for calibration purposes.

Next, the enthalpy of benzoic acid was measured. First, about 2000 g of water at room temperature was poured into the vessel. The exact weight of the water and vessel were recorded. Then a sample of about 1 g of benzoic acid was made into a pellet, and placed into the combustion cup. The exact weights of the benzoic acid pellet and combustion cup were also recorded. Fuse wire was attached to the electrodes, and it was made sure that the fuse wire securely touched the benzoic acid pellet. The lid of the bomb was securely attached, the bomb was purged with oxygen, and about 25 atmospheres of oxygen was pumped into the bomb.

The vessel was then placed into the calorimeter and bomb was placed into the vessel. The fuse connections were plugged into the bomb right before it was submerged in water. The lid of the calorimeter was placed on as to make sure the thermistor was in place in the water and so that the propeller could spin properly. The propeller was turned on and data was recorded on the computer. Once a stable temperature was reached, the bomb was ignited and again data was collected until a stable temperature was reached. This process was repeated for naphthalene and a food source (rice cakes).

III. Results

Water
Equation (x is resistance) y = -1E-08x3 + 4E-05x2 – 0.0811x + 348.34
Change in Temperature (K) 6.78183
Measurement Benzoic Acid Naphthalene Rice Cake
Weight of Water (g) 2005.60 1999.60 2030.00
Weight of Vessel (g) 782.96 784.70 —-
Weight of Combustion Cup (g) 13.8628 —- —-
Weight Pellet (g) 0.9488 0.9068 0.5097
Final Resistance 1041.37 1027.268 1190.49
Initial Resistance 1166.984 1200.926 1143.122
Final Temperature (K) 295.97 296.40 291.60
Initial Temperature (K) 292.28 291.31 292.96
Temperature Difference (K) 3.69 5.09 1.36
Theoretical q (kJ/g) 26.43 47.86 16.736
Theoretical q (kJ) 25.08 43.40 8.53
Actual q (kJ) —- 34.6 9.25
Percent Error —- 20.31% 8.44%
Rice Cake
Weight of Sample Rice Cake, hydrated (g) 3.1655
Weight of Sample Rice Cake, dry (g) 2.8013
Weight of Water in Hydrated Rice Cake (g) 0.3642

Calculations:

a. Heat capacity (mCp) from benzoic acid

qb.a. = 26.43 kJ/g (theoretical)

q = [mCp] ΔT
q = K ΔT

K = q / ΔT
K = (26.43 kJ/g) (0.9488 g) / (3.69 K)
K = 6.80 kJ / K

 

b. Heat of reaction of naphthalene and % error

qnap = 47.86 kJ/g (theoretical)
qnap = (47.86 kJ/g) x (0.9068 g)
qnap = 43.42 kJ

qnap = K ΔT
qnap = (6.80 kJ / K) x (5.09 K)
qnap = 34.6 kJ (actual)

Percent Error:
(34.6 kJ – 43.42 kJ) / 43.42 kJ x 100% = 20.31%

 

c. Heat of reaction of rice cake and % error.

16 Cal / 4 g (looked up) = x / 0.5097 g
x = 2.0388 Cal (for the 0.5097 g pellet)
2.0388 Cal x (4.184 kJ / Cal) = 8.53 kJ

qrice cake = 8.53 (theoretical)

qrice cake = (6.80 kJ / K) x (1.36 K)
qrice cake = 9.25 kJ (actual)

Percent Error:
(9.25 kJ – 8.53 kJ) / 8.53 kJ x 100% = 8.44%

 

IV. Conclusion

The difference in heat of reaction between the actual and theoretical values for naphthalene was 20.1%. The measured value was lower than the theoretical value. This means the change in water temperature was lower than it should have been. This error could have come from a few sources. The water used may have been warmer than room temperature, meaning that difference between the temperatures that the water was at and the temperature of room temperature water would have not been accounted for in the temperature difference. Another problem could have been that bomb was not flushed correctly with oxygen. This would cause an uneven ignition of the naphthalene, and the temperature might not have gone up as much as it should.

The difference in the heat of reaction between the actual and theoretical values for the rice cake was 8.44%. This seems like a fairly reasonable result. The actual value was high than that of the theoretical value. This could be accounted for in a few ways. The water used may have been colder than room temperature, causing a bigger change in temperature than there should have been. The bomb may also have been dirty, causing debris to ignite along with the rice cake, which would add to the temperature difference. The rice cake may also have gained some water vapor, causing water to ignite along with the rice cake, also causing a bigger change in temperature. Overall, the measured values as somewhat close to the theoretical values, showing there is some merit to the theories we have learned.

Me

circa 1996 (9 y/o)

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Analysis of Ethanol and Butanol Solutions via Gas Chromatography

↘︎ Oct 6, 2008 … 3′ … download⇠ | skip ⇢

Introduction

During this laboratory experiment, a gas chromatograph was used to analyze the ethanol and butanol content of solutions of known mass ratios, and then construct a calibration curve using calculated area ratios from the graphs produced by the gas chromatograph. A solution of unknown mass ratio was then analyzed with the gas chromatograph in order to determine its area ratio. This area ratio was compared to the calibration curve in order to find the mass ratio of the unknown. Seven solutions of ethanol and butanol were prepared using different mass ratios of the two components. The solutions were then one by one injected into the gas chromatograph using a syringe. As each solution went through the gas chromatograph, the apparatus heated the solution. Ethanol and butanol have different chemical compositions and boiling points, so they reached a detector at different times. As the components reached the detector, the voltage given off by the components was read and recorded as a graph on a computer of time versus voltage.

Specifically, the gas chromatograph functions using an inert gas, in the case of this experiment, nitrogen, to function as a carrier gas for the analyte. The analyte is injected through a septum using a microsyringe. The carrier gas aids the analyte flow through the column. As the analyte moves through the column, the different components are separated and reach the detector at different rates, as the components have different chemical properties. In the case of this experiment, once the analyte reached the detector, the thermal conductivity of the components were measured as voltage. The voltage readings were graphed as a function of time on a computer.

Results

Desired Ratio of Masses of Ethanol to Butanol Mass of Ethanol (g) Mass of Butanol (g) True Ratio of Masses of Ethanol to Butanol Ratio of Areas of Ethanol to Butanol
1:7 1.0159 7.0125 0.14487 0.202332
2:6 1.9839 6.0561 0.327587 0.402598
3:5 3.0484 5.0139 0.60799 0.700423
4:4 4.0454 4.0669 0.994713 1.100032
5:3 5.0325 3.0001 1.677444 0.846482
6:2 6.2265 2.0196 3.083036 2.95128
7:1 7.0155 1.0104 6.94329 6.061587
Unknown 4.2861 2.7397 1.56441 1.559821*

*The graph of the original unknown was unreadable to determine the ratio of the areas, so

Anwar’s group’s graph of unknown was used to find a ratio of areas.

Equation of Linear Regression Line y = 1.134x – 0.0184
Mass Ratio of Unknown (Ethanol vs. Butanol) 1.750437
Substance Weight of Cut-out Piece of Paper for 4:4 Composition (g) Ratio of Areas
Ethanol 0.1121 1.094
Butanol 0.1025

Discussion and Conclusions

The calibration curve came out fairly well. It follows almost exactly a straight line except for the value where the ratio of masses was 5:3. The ratio of the areas is much lower than it should be. There is a decrease in the ratio of the areas from the 4:4 composition (1.100032), to the 5:3 composition (0.846482), and then the area ratio goes back up in the 6:2 composition (2.95128). The value for the ratio of the areas should have been somewhere between 1.100032 and 2.95128. Other than that one value, all the other values are very linear. It should have been expected that the values would linear, considering that the ratio of the masses was not increased exponentially; the ratios were only decreased and increased by a constant value. This also means the slope of the area ratio versus the mass ratio should be close to being in unity (a value of one), because of the constant increase and decrease in the mass ratio.

The ratio of the areas for the original unknown was unreadable on the graph, so another group’s unknown graph was used to find a ratio of areas. That group did not record their original masses used for their unknown sample, so it is not possible to find the percent error. However, using the equation for the linear regression line, the expected ratio of masses could be found by subbing the ratio of the areas in for “x”. The mass ratio found for their unknown was 1.750437, which is close to the mass ratio for the 5:3 composition (1.677444), so it can be estimated that their unknown was composed of about 5 grams of ethanol and 3 grams of butanol. A percent error can be determined by comparing one of the points on the calibration curve to the actual equation of the line. For example, using the 6:2 mass compisition, the actual mass ratio was 3.083036. However, using the calculated area, according to the calibration curve the theoretical mass ratio is 3.31701. This yields a percent error of 7.59%. This is a fairly low percent error, but it should be, as the linear regression line is calculated to make it as close to the points on the graph as possible.

Finally, using the alternative method of finding the ratio of areas by physically weighing the two curves, the value was extremely close to the calculated value. The calculated ratio for the 4:4 mass ratio was 1.100032, while the ratio from weighing the curves was 1.094. This shows that weighing all the curves would have been an acceptable way of determining the area ratios, rather than using the trapezoidal rule to calculate the area ratios. The trapezoidal rule has error to it, so weighing the curves may even be more accurate than using the trapezoidal rule. If it was not a requirement to use the trapezoidal rule, it would have been preferred to simply weigh all the curves, as this would have been a much quicker way of determining the area ratios.

Me

circa 2013 (25 y/o)

More from…
CHM 2412 (Physical Chemistry Lab I) (Class) / Saint Joseph’s University (School) / schoolwork (Post Type)

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