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Ftir Lab Report

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Fourier Transform Infrared (FTIR) Spectroscopy

Group 1 (Monday 1-5pm)

Reshma Reji

Objective of the Experiment
* FTIR Spectroscopy * FTIR Spectrophotometer * Samples

Physical properties of reagents used
Solution Preparation
Instrument settings

Data 1. IR spectrum of Chloroform and D-chloroform 2. Rotational Spectrum of CO2 (Standard Resolution) 3. Rotational Spectrum of CO2 (High Resolution) 4. Carbonyl stretch in 2-butanone solutions (wavenumber vs. % T) 5. Carbonyl stretch in 2-butanone solutions (wavenumber Vs Absorbance) 6. Calibration curve of 2-butanone solutions (concentration vs. absorbance) *
a. Preparation of solutions b. Concentration of the unknown c. Percent error of observed and theoretical ratios of CH, CD stretch frequencies

Results and Conclusion


Objective The goal of the first part of the experiment was to study the effects of isotopes on bond stretching. In the second part of the experiment, the influence of instrument resolution on the rotational spectrum of carbon dioxide was studied. The purpose of the third part of the experiment was to create a calibration curve to find the unknown concentration of 2-butanone.
Fourier Transform Infrared Spectroscopy (FTIR) is a very useful analytical technique used for qualitative and quantitative analysis of organic and inorganic compounds. It can be used for a variety of purposes such as; finding the concentrations of unknown solutions, determining the constituents of unknown solutions, forensic purposes or finding individual components in paints, drugs, fibers residues, contaminated chemicals etc. Solids, liquids and gases can be analyzed using this method. The most important benefit of FTIR spectroscopy is its ability to identify the different types of functional groups (CH, CD, C=O, CN, CF etc) in molecules. The chemical bonds in these functional groups correspond to the wavelengths (wavenumbers) of absorbed light and represent specific areas(peaks)of the obtained spectrum. The identification of different functional groups and their chemical bonds facilitate the characterization of molecules. The spectra obtained from pure compounds are highly distinctive and therefore helps to recognize molecular structures. FTIR spectra therefore serve as “molecular finger prints” for several compounds [6].
FTIR spectroscopy is used to record the infrared intensity at different wave numbers of light. Infrared light is classified into three regions; far infrared (FR), mid infrared (MI) and near infrared (NI). The wavenumbers range from 4 ~ 400cm-1 for FR, 400 ~ 4,000cm-1 for MI and 4,000 ~ 14,000cm-1 for NI. The infrared region in the electromagnetic spectrum can be seen in Figure 1. The Infrared spectrum is formed due to the transitions within the vibrational energy states that are quantized. FTIR spectroscopy detects the vibrations of the chemical bonds of the functional groups in the sample [8].

Figure 1: Electromagnetic Spectrum (IR region to the left of visible light region)

The number of degrees of freedom for a non linear molecule which consists of N atoms is 3N, of which three motions represent translational motions and another three correspond to rotational motions. The degrees of freedom remaining in the molecule is 3N-6, which represents the number of vibrational modes of the molecule. The number of ways that atoms can vibrate in the molecule is the number of its vibrational modes [1]. The frequencies at which chemical bonds vibrate depend on the elements and the nature of bonds. Vibrations of individual bonds can occur at different frequencies depending on the ground states and excited states of bonds. Molecular vibrations occur at lower frequencies for ground states and higher frequencies for excited states [6]. A ground state chemical bond can be brought to its vibrational excited state through the absorption of IR radiation. Chemical bonds can stretch, contract or bend when exposed to IR light. Absorptions due to stretching normally generate stronger peaks than bending absorptions. Each functional group in a molecule absorbs infrared light at a unique wavenumber range (or frequency). This does not depend on the structure of the remaining parts of the molecule. Different molecules may contain equivalent functional groups. However, the stretch of a particular functional group will always appear in the same wavenumber range (or frequency) in all molecules [1,4]. Some of the common functional groups and the frequencies at which they absorb can be seen in Table 1 and Figure 2.
Table 1: Common functional groups and their frequencies of absorption Functional Groups | Frequency (cm-1) | Wavelength (nm) | Alkanes, C-H | 2850 - 3000 | 3509 – 3333 | Alkynes, C≡C | 2100 - 2500 | 4762 – 4000 | Amines, NH, RNH | 3300 - 3500 | 3030 – 2857 | Ketones, C=O | 1690 – 1780 | 5917 – 5618 | Nitriles, C≡N | 2100 - 2270 | 4762 – 4405 |

Figure 2: Various functional groups and their frequencies and wavelengths of absorption.
The Infrared instruments used in the earlier periods were dispersive types. In these types of instruments, specific frequencies of energy from the radiation were split using a prism or a grating. The detector measures the energy at each wavelength and yields a spectrum of intensity versus frequency. The main intricacy of the dispersive instruments was their slow paced scanning. This limitation was resolved by the development of an optical device called the interferometer. In a FTIR spectrophotometer, an interferometer is used instead of the regular monochromator [10]. Albert Michelson invented the Michelson Interferometer in 1880. However, sensitive detector arrays and computers for FTS were not available at that time and therefore the technique was not commonly used. Today, with the use of Fourier Transforming computers and highly sensitive detector arrays, FTS has now become a widely used technique for interferometric measurements [3].
Contrary to dispersive instruments, FTIR spectrophotometers can detect all IR frequencies at the same time. FTIR spectroscopy is preferred over the dispersive methods because of its high-speed, precision (does not need external calibration), high signal to noise performance, sensitivity and mechanical simplicity [10].
A schematic design of an interferometer is displayed in Figure 3. The main components of an FTIR spectrophotometer include an Infrared light source, interferometer, sample compartment, detector and the computer. Infrared light is obtained from a radiant blackbody source [10]. A small aperture regulates the amount of light that passes through the sample. The interferometer is an optical device that has the ability to detect all of the IR frequencies at the same time. The distinctive signal that the interferometer generates has all the frequencies of IR radiation “encoded” into it [9]. The interferometer splits IR radiation using a beam splitter into two beams. The radiation is reflected by the mirrors present inside the interferometer in such a way that the signal which passes out of the interferometer is a result of the interference of the two beams. The signal obtained is referred to as the “interferogram”. Every data point in the interferogram contains information about every beam that originated from the source. Therefore, when the interferogram signal is obtained, all the infrared frequencies are measured simultaneously [10]. However, the desired spectrum is not an interferogram, but a frequency spectrum. In a frequency spectrum, frequency is plotted against intensity. The interferogram is complex and cannot be analyzed the way it is. Therefore, the individual frequencies are “decoded” to obtain the desired spectrum. This process is accomplished by the computer using a famous technique called “Fourier Transformation”.

Figure 3: Schematic design of an FTIR spectrophotometer
Solid samples and liquid samples are prepared differently for FTIR analysis. For liquids, drops of liquid samples are injected in a dismountable liquid cell (Figure 4). The structure of the liquid cell prevents leakage and therefore the sample gets trapped between the salt plates present inside. The salt plates for liquid samples are generally made of sodium chloride (NaCl). The liquid cell is then placed in the sample holder of the spectrophotometer for analysis. For solids, the samples are initially crushed with potassium bromide (KBr). The powdery mixture is then made into a thin pellet and a great pressure is applied to dry it. This is placed in the sample holder and analyzed. An alternative method for solid analysis is to dissolve a small amount of solid in a solvent and place it in a salt plate. One example of a solvent that is normally used is methylene chloride (CH2Cl2). Evaporating the solvent leaves a thin film (also called as cast film) of the solid sample and is analyzed. This method is generally used to identify polymers [6,7].

Figure 5: A dismountable liquid cell

Salts are typically used in IR analysis because they are transparent to infra red rays and therefore do not absorb in the infra red region. In addition to NaCl, plates made of caesium iodide(CsI), silver chloride (AgCl) or Germanium(Ge) can also be used. However, the salts plates that are chosen depend on the nature of the sample that is analyzed. Salt plats made of NaCl, Kbr or CsI are commonly used for analyzing organic compounds. Plates made of non hydroscopic substances such as AgCl and Ge are preferred for samples containing water. A downside of using salt plates is the fact that they lose their transparency and appear to be fogged as they get exposed to moisture. The values obtained for transmittance in the IR spectrum rely highly on the condition of the salt plates. As the windows get exceedingly fogged, the signal strength decreases. Therefore, higher concentrations becomes necessary to augment the signal strength and offset the scattering [6, 7].

Experimental Table 2: Chemicals used Reagent | Formula | Structure | Molecular Weight(g/mol) | Density(g/ml) | Melting Point(°C) | Boiling point(°C) | CAS No: | Chloroform | CHCl3 | | 119.38 | 1.474 | -63.5 | 61.2 | 67-66-3 | Chloroform-D | CDCl3 | | 120.38 | 1.500 | -64.0 | 60.9 | 8665-49-6 | 2-butanone | C4H8O | | 72.11 | 0.804 | -86 | 79.64 | 78-93-3 |

Table 3: Preparation of Solutions Solutions | Weight % of 2-Butanone | 2-Butanone (ml) | 2-Butanone (l) | Chloroform (ml) | 1 | 0.05 | 0.0230ml | 23.0l | 25ml | 2 | 0.075 | 0.0343ml | 34.3l | 25ml | 3 | 0.10 | 0.0460ml | 46.0l | 25ml | 4 | 0.125 | 0.5720ml | 57.2l | 25ml | 5 | 0.150 | 0.0680ml | 68.0l | 25ml | The volume of 2-Butanone required to prepare each solutions were calculated using Equation 2. To derive Equation 2, the following ratio and Equation 1 were used (see below)
Mass of ButanoneMass of Chloroform=Weight% of Butanone100

Mass of 2-Butanone = (Density of 2 Butanone*Volume of 2-Butanone) Mass of Chloroform = (Density of Chloroform*Volume of Chloroform) Wt% of each 2-Butanone solution = 0.10%, 0.050%, 0.125%, 0.075%,0.30%

Equation 1:
Volume of 2-Butanone= mass of 2-butanone(g)Density of 2-butanone(gml)

Equation 2:
Using Equation 1 and the ratio,
Volume of 2-Butanone(ml)= Density of CHCl3*Volume of CHCl3*(wt% 2butanone/100)Density of 2-butanone

Wt% of each 2-Butanone solution = 0.10%, 0.050%, 0.125%, 0.075% and 0.30% Volume of CHCl3 used to prepare each solution = 25ml Density of CHCl3=1.474g/ml Density of 2-Butanone=0.804g/ml

Part 1
In the first part of the experiment, two isotopes; Chloroform (CHCl3) and Deuterated Chloroform (CDCl3), were each scanned using a Fourier Transform Infrared (FTIR) spectrophotometer under standard resolution. To analyze CHCl3, several drops of CHCl3 were injected into salt plates made of sodium chloride. After obtaining the spectrum of CHCl3, the salt plate was rinsed with CDCl3. The spectrum of CDCl3 under standard resolution was also taken. Using the IR spectra of CHCl3 and CHCl3, the frequencies of the CH stretch and the CD stretch was determined. In the spectra, the CH stretch was observed at 3007.46 cm-1and the CD stretch was seen in 2250.53cm-1.The experimental ratio of these stretch frequencies was calculated.

Part 2
Background scans were performed to obtain the spectrum of carbon dioxide in air. The scans were taken under standard resolution and high resolution. The two spectra were compared.
Part 3
In the third part of the experiment, five solutions of 2-Butanone with varying concentrations (by weight %) were prepared in chloroform using the amounts listed in Table 3. The volumes of 2-butanone needed to prepare these solutions were calculated using Equation 2. All solutions were prepared in 25ml of chloroform. Initially a back ground spectrum was obtained. Using a syringe, the sample solutions were injected into a dismountable liquid cell (Figure 5) and placed into the sample holder of the spectrophotometer. Infrared spectrum of each solution was obtained under standard resolution.

Fourier transform Infrared Spectrometer
Manufacturer: Mattson Instruments
Model: 4020 Galaxy Series
Part 1: Standard resolution (4.0cm-1)
Part 2: Standard resolution(4.0nm) and High resolution(0.4cm-1)
Part 3: Standard resolution(4.0cm-1) Data
Figure 6: Infrared Spectrum of Chloroform and D-chloroform (Absorbance vs. Wave numbers)

Figure 7: FT-IR Spectrum of Carbon dioxide under standard resolution

Figure 8: Rotational spectrum of Carbon dioxide under high resolution

Figure 9: Wavenumber vs. Percent Transmittance spectrum displaying the carbonyl stretch ( at 1710.561cm-1) in varied concentrations(wt%) of 2-butanone solutions.

Figure 10: Wavenumber vs. Absorbance Spectrum displaying the carbonyl stretch (at 1710.561cm-1) in varied concentrations (wt%) of 2-butanone solutions

Figure 11: Standard calibration curve of varying concentrations (wt %) of 2-butanone solutions, displaying the absorbencies of each solution at 1710.561cm-1

Lineast of concentration vs absorbance for Figure 11. m | 3.255041 | 0.153407 | b | Sm | 0.255667 | 0.027118 | Sb | R2 | 0.981828 | 0.020212 | Sy |


1. Finding the volume needed to prepare 2-Butanone solutions:

Equation 2 was derived using the calculations shown in experimental section:

Equation 2:

Volume of 2-Butanone(ml)= Density of CHCl3*Volume of CHCl3*(wt% 2butanone/100)Density of 2-butanone

Wt% of each 2-Butanone solution = 0.10%, 0.050%, 0.125%, 0.075% and 0.30% Volume of CHCl3 used to prepare each solution = 25ml Density of CHCl3=1.474g/ml Density of 2-Butanone=0.804g/ml

Weight % | Volume of 2-Butanone(ml) | 0.050 | 0.0230ml | 0.075 | 0.0343ml | 0.10 | 0.0460ml | 0.125 | 0.5720ml | 0.150 | 0.0680ml | The calculated volumes of 2-butanone used to prepare solutions is shown in the table to the right.

2. a. Calculation of unknown concentration: y=mx+b y= absorbance of the unknown obtained from Figure 10 = 0.598637841 x= unknown concentration

Equation obtained from calibration curve(Figure 11),

y=3.255x+0.1534 0.598637841=3.255x+0.1534 x= 0.137 wt%
Concentration of unknown 2 butanone solution =x=0.137 ±0.0068 wt%

b. Uncertainty:
Sx = Sy|m| 1k + 1n + (y-mean y)²m²Σ(xi-mean x)² n=5 n=5 k=1, y= 0.598637841, mean y=0.478911 y-mean y2=0.598637841-0.478911²=0.0143345 m= 3.255
Σxi-mean x2=0
Sy= 0.020212
Sx = 0.020212|3.255| 11 + 15 + 0.01433453.255²*0
Sx = ± 0.0068 (%wt)

3. Percent error of observed ratio and theoretical ratio of CH, CD stretch frequencies

a. Observed Ratio of frequencies: νCHνCD = 3007.46 cm-12250.53cm-1= 1.336

b. Theoretical Ratio:
According to Hooke’s Law, frequency; ν=12πkμ Where, k= Hooke’s constant (equal for both CH and CD) µ = the reduced mass of the two molecules

Therefore theoretical ratio of frequencies is,
Equation A νCHνCD=12πkμCH12πkμCD Reduced mass of C-H and C-D, μCH=massCmassHmassC+massH=12.01(1.008)12.01+(1.008)=0.929949=0.9299 μCH=massCmassDmassC+massD=12.01(2*1.008)12.01+(2*1.008)=1.726234=1.726234
From Equation A, the ratio of Theoretical frequencies, νCHνCD=μCDμCH=1.7220.9299=1.36 Therefore Observed Ratio or Actual Ratio = 1.336 Theoretical Ratio= 1.36

Percent error = [ |observed - theoretical | / theoretical value ] x 100 = = [(1.336-1.36)/1.36]*100

Conclusion As mentioned in the objective, the goal of the first part of the experiment was to study the effects of isotopes on bond stretching. The two isotopes used in this experiment were chloroform (CHCl3) and D-chloroform (CDCl3). An FTIR spectrum of these two chemicals was obtained. In the spectra, the CH stretch was observed at 3007.46 cm-1and the CD stretch was seen in 2250.53cm-1.The observed ratio was calculated to be 1.336 and theoretical ratio obtained was 1. 36. The theoretical ratio is slightly higher than the observed ratio. However, the percent error between these rations is 1.76%. The very low percent error shows the relative accuracy of the FTIR analysis. An important aspect that affects the stretching absorption frequency of bonds is the identity of the atoms that are involved. The two atoms involved in CHCl3 and CDCl3 are the hydrogen atom (H) and the deuterium atom (D). Deuterium is a heavy isotope of hydrogen. Theoretically, the higher the masses of the atoms involved, the lower will be the stretching frequencies of their absorption [9]. This can be understood from the Hooks Law which states that frequency is inversely proportional to the reduced mass: ν=12πkμ
This analogy can be seen in IR spectra of CHCl3 and CDCl3 obtained from this experiment (Figure 6). Since deuterium has a higher mass than hydrogen, the CD bond absorbs IR radiation at a lower stretch frequency of 2250.53cm-1 compared to the CH stretch frequency of 3007.46 cm-1. Therefore, heavy atoms make their attached bonds to absorb IR radiation at lower frequencies [9]. Theoretically, the CH stretch in alkanes should occur at 2850-3000cm-1. The absence of the C-H signal around the 2850-3000cm-1 range in D-chloroform (Figure 6) makes the spectrum unique and illustrates the fact that the IR spectra can be used as a finger print for the identification of molecules.
In the second part of the experiment, advantages and disadvantages of using high and low resolutions is understood. The spectra of carbon dioxide in both standard (4cm-1) and high resolutions (0.4cm-1) were obtained. Spectral resolution is defined as the capacity of a sensor to define and identify signals at very small wavelength intervals [11]. The resolving power is also described as the capability of the spectrograph to “resolve” features in the spectrum [11]. The resolution plays a major role in the IR spectrum that is obtained. This is obvious from the CO2 spectra displayed in Figure 7 and 8. The low resolution (4cm-1) spectrum in Figure 7 is broad and shows a CO2 stretch at 2356.6cm-1. However, the high resolution (0.4cm-1 ) rotational spectrum (Figure 8) displays a regularly spaced spectrum and is a more detailed version of the maxima seen in the low resolution spectrum (Figure 7). The spacing of the data in high resolution is at every 0.4cm-1 compared to 4.0cm-1 in low resolution spectrum. Therefore high resolution provides with a very detailed rotational spectrum at very fine wavelengths. High resolution will always provide with quality data which will resolve all bands. However, for analytical analyses, a lowest acceptable resolution is picked to increase the signal to noise ratio. Low resolutions will give maximum precision and sensitivity in the quantification of gas species. For analytical purposes, a high resolution spectrum would be too noisy[12].
The peaks in the high resolution spectrum represent the rotational transitions of the carbon dioxide stretch. Using the rotational spectrum, the change in energy of rotation can be found. The change in energy of the low resolution spectrum is higher than the change in energy of the high resolution spectrum. Therefore the spectrograph cannot resolve the rotational bands at low resolutions.
In an FTIR spectrophotometer, a background scan should be performed before obtaining the spectra of the sample. To obtain the rotational spectrum of CO2 (Figure 7, 8), no back ground scans were performed. This was intentionally done to understand how the presence of carbon dioxide affects IR spectra of any analyte. Atmospheric components (CO2, H2O) can interfere with the analyte and their signals might appear in the IR spectrum. This effect can be understood from the spectrum of CO2 in Figure 10 and 11. Running a background scan avoids this problem. Background spectrum contains the characteristics of the instrument. When the background spectrum is removed, the data obtained wholly comes from the sample. The background spectrum is automatically subtracted from the spectrum of the analyte and helps generate accurate data. Before recording the IR spectra of the 2-butanone solutions, a background scan was performed [9, 10].
In the third part of the experiment, the IR spectra of five butanone solutions were recorded in standard resolution. High resolution was not used because the carbonyl peak obtained from the spectra of 2-butanone is very broad. Running the solutions at high resolutions will give a much detailed spectrum. However, because the carbonyl peak is broad, a detailed spectrum will not be useful and will lead to higher acquisition times.
In an FTIR spectrum, wavenumber (cm-1) of absorption is plotted against transmittance (%T). Transmittance (T) is a fraction of the amount of light at a unique wavelength that passes through a sample. It is a ratio of the intensity of incident light and the light that passes out of the sample. Absorbance is the negative log of transmittance; A= -log (T). Therefore the %T values obtained from the spectra of the 2-butanone solutions were converted to absorbance values to graph the calibration curve. As seen in Figures 9 and 10, he carbonyl stretch of the 2-butanone solutions were seen in 1710cm-1. This was the wavenumber of maximum absorption. Using the data from the IR spectra, a calibration curve for the 2-butanone solutions (Figure 11) with concentration versus absorbance was obtained. The linearity of the graph (Figure 11) proves the beers law which states that concentration is directly proportional to absorbance. As seen in Figure 10, the 0.150Wt% butanone solution with its highest concentration has the highest absorbance. The high R2 values found in Lineast function of Figure 11 illustrates the accuracy and linearity of the measurements. Through the equation obtained from the calibration curve (Figure 11), y=3.255x+0.1534. The concentration of the unknown 2-butanone solution was found to be 0.137±0.0068 wt %( see calculations). This value seems reasonable because according to Figure 10, the concentration of the unknown should be between 0.125 wt% and 0.150wt%. The unknown concentration of 0.137wt% falls between this range and proves that this experiment was successful.
Theoretically, the y-intercept; 0.1534 of the equation; y=3.255x+0.1534 should be equal to zero. This is because when there is no concentration, there should be no absorbance. However, as seen in the equation; the y intercept obtained from the experiment is not zero. The R2 value obtained from the lineast function is 0.981828. Theoretically, this value represents linearity and accuracy of the measurements and should be equal to one.
The discrepancies mentioned above might have resulted due to a number of reasons. One reason might be the salt plates. The salt plates lose their transparency and appear to be fogged as they get exposed to moisture. The values obtained for transmittance in the IR spectrum rely highly on the condition of the salt plates. As the windows get exceedingly fogged, the signal strength decreases. Another reason that might have caused this error is erroneous concentrations of the 2-butanone solutions. The solutions tend to evaporate quicker results in disparity between the actual and recorded concentrations.

Works Cited











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...C8057 (Research Methods in Psychology): Writing Laboratory Reports Writing Lab Reports & APA Format Structure and Content of a Laboratory Report This following provides a brief overview of the structure and content of various elements in a laboratory report, based on APA guidelines. The APA publication manual can be referred to for more detail. The purpose of a lab report is to communicate research in a clear, systematic and standardised way. Primarily, a lab report should communicate the following things: → Why did I bother? Why did I do this experiment and why is it an interesting contribution to science? → How did I do it? How did I carry out this research (this should be detailed enough so that the experiment could be replicated precisely)? → What did I find? What were the results of your study? → So What? What do your findings mean in terms of your hypotheses and what theoretical contribution do they make? IMPORTANT! The advice that we give you in this handout is not inflexible, it is a guide! For example, when we say that the title should be fewer than 15 words, the world will not end if your title is 16 words. So, please use your common sense. In case you don’t have any, the snail of sense will appear to give you some advice. Format There are several sections to a lab report: → Title → Abstract → Introduction → Method o o o o o → Results → Discussion → References → Appendix (Optional) Title Your title should ideally be fewer than 15 words. A good title alerts the......

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Lab Report

...Name: Gertrude Edwards Date: 12/1/2014 Instructor’s Name: Staci Lynn Assignment: SCI203 Phase 2 Lab Report TITLE: Speciation • Purpose o The purpose of this lab is to evaluate what would happen if a species within a population were suddenly split into 2 groups. • Introduction o If a population is divided indefinitely by a barrier members of the divided population will not have the opportunity to breed with each other, over years, the biotic and abiotic conditions on either side of the barrier will vary from one another. (M.U.S.E). • Hypothesis/Predicted Outcome o Based on what I’ve learned I expect that species will undergo changes if they were split into 2 different groups, some would adapt and some wouldn’t. • Methods o The methods I used in this lab came from M.U.S.E. The initial separation would consist of some species from the mainland reaching the isolated Island, then after that the isolated population would begin to diverge because of the genetic drift and natural selection, then after that overtime divergence may eventually become sufficient to cause reproduction isolation. (M.U.S.E). • Results/Outcome o As a result, Natural selection will cause different selective and adaptive pressures to occur between the two divided populations and they will evolve forever. Over time this will result in speciation which is the creation of two new species. (M.U.S.E). •......

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...Apparatus, materials and methods "As the name implies, the materials and methods used in the experiments should be reported in this section. The difficulty in writing this section is to provide enough detail for the reader to understand the experiment without overwhelming him or her. When procedures from a lab book or another report are followed exactly, simply cite the work, noting that details can be found in that particular source. However, it is still necessary to describe special pieces of equipment and the general theory of the assays used. This can be usually be done in a short paragraph, possibly along with a drawing of the experiment apparatus. Generally, this section attempts to answer the following questions: 1, What materials were used? 2. How were they used? 3. Where and when was the work done? (This question is most important in field studies.)" Observations and/or results with discussion Results "The results section should summarize the data from the experiments without discussing their implications. The data should be organized into tables, figures, graphs, photographs, and so on. But data included in a table should not be duplicated in a figure or graph. All figures and tables should have descriptive titles and should include a legend explaining any symbols, abbreviations, or special methods used. Figures and tables should be numbered separately and should be referred to in the text by number, for example: -Figure 1 shows that the activity......

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...Lab Report Using Gravitational Force as a Measurement Tool Answer the following questions about the results of this activity. Record your answers in the boxes. Send your completed lab report to your instructor. Don’t forget to save your lab report to your computer! Activity 1 Record your data from Activity 1 in the boxes below. Enter the data for the sample you used in each trial (5000 rpm, 10000 rpm, etc…) in the appropriate columns and the corresponding g-force, number of layers, and position of layers position results. You will need to use the following formula to assist with your laboratory report: • G-force =0 00001118 x radius of centrifuge arm x (rpm)2 • The radius of the centrifuge arm for this instrument is 10 cm. | | | | | | |Speed |5000 rpm |10000 rpm |15000 rpm |20000 rpm | | |2795 |1180 |2655 |4720 | |G-force | | | | | | | |4 |5 |3 | |Number of......

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Lab Report

...To do Lab Report for Single and Double Replacement reactions : This is a formal lab report. It must be typed or written by hand with blue or black ink. Make sure that you include the following: 1. Title 2. List of Materials 3. Safety that includes MSDS risk assessment for all the materials used ( instructions were given to you already and they are in Moodle) 4. Pre-lab questions 5. Data Tables with the results obtained ( observations and predictions) 6. Post Lab: Part I : Single Replacement Reactions a. For every reaction that took place you must write the balanced chemical equation b. Which metal reacted the most? c. Rank your metals from more to least active Part II: Double replacement Reactions a. For every reaction where you observed precipitate, write the complete balanced molecular equation, the complete ionic equation and the net ionic equation; use the solubility rules to identify the precipitate and the states of matter of each substance participating in the reaction. b. Which cation produced the most number of precipitates? c. Write general rules of solubility that you observed. 7. Final Conclusion and error analysis To do Lab Report for Single and Double Replacement reactions : This is a formal lab report. It must be typed or written with blue or black ink. Make sure that you include the following: 1. Title 2. List of Materials 3. Safety that includes MSDS risk assessment for all the materials used (......

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Lab Report

...Lab Report Introduction: Natural selection relies upon the assumption that all organisms produce more offspring than can survive in an environment demonstration limited resources. The heritable traits that are best suited for a particular environment should be passed on at a greater frequency then those less adaptive traits. Natural selection is unique in that it leads to adaptive change, and these changes are likely to lead to increased survival and reproductive success. To sum up, natural selection is that organisms can survive only because they can suit to the changed environment which means those organisms also need to change with the environment all the time. Hypothesis: during the experiment, the forks out of all the “predators” should survive, and the peas out of all those “preys” will be survived. Material and Metnods: We will use 4 kind of beans (preys items) which are fava beans, white beans, red beans and peas. And we also have 4 mouthparts (predators) which are knives, forks, spoons and foreceps. First, we get 100 ml for each prey and spread them on the ground. And the students will be divided into 4 groups with those 4 different predators. Each group of predators include 6 student. And all of the predators will have a cup which will serve as the stomach into which the prey will be placed. We also have a timer and a security that will make sure the process is fair. Here we begin the game. The predators will be gathered, they will be instructed not......

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...04.03 Periodic Table - Lab Report Please complete the following lab report for assignment 04.03 Introduction to the Periodic Table. Take notes on the properties of the three types of elements. Metals|Good conductors of heat and electricity. Metal are ductile, and malleable. Not only have that metals also had a distinct shiny look which is called metallic luster.| Nonmetals|Poor in conducting heat and electricity generally as insulators, only at room temperature. Non-metals can be either solids or gases; they are neither ductile nor malleable. They are mainly either opaque or transparent.| Metalloids|Moderate at conducting heat and electricity, only solid at room temperature, can be ductile and malleable only to varying degrees. How at metallic luster in varying degrees.| Fill in the following data table as you complete the activity. Material|Conductivity(strong, moderate, none)|Ductility(ductile or not ductile)|Appearance(bright shine, moderate shine, or no shine)| 1|The light is bright.|Can be drawn into wire.|Shines brightly.| 2|The light is dim.|Can be drawn into wire|Shines moderately.| 3|There is no light.|Cannot be drawn into wire|Does not shine.| 4|The light is bright.|Can be drawn into wire.|Shines brightly.| Identify each of the following as a metal, nonmetal, or metalloid. Please use complete sentences to describe the observations that led you to identify each. Material 1: Metal- Because it can be drawn into wire, shines brightly, and the......

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...Gustavo Duran Per#5 Making Dilutions of Concentrated Solutions Purpose: To make dilutions of concentrated solutions and report their concentration in different ways. Hypothesis: I predict that at least one of the tubes will not change color because of the little concentration it has. Materials: Balance, analytical Balance, tabletop milligram weigh paper, 7.6x7.6 cm weigh boat, 3.5”x3.5” Lab scoops cupric sulfate 5-hydrate Tubes, 15 mL, capped Tube racks for 15mL tubes Deionized water Permanent lab marker pens Tubes, glass, 13x100mm Peg racks for 13x100 mm tubes pipets, 5mL pipet pump, green Spectrophotometer, spectronic 20 D+ Procedure: ● In matrixes similar to those shown in table 3.21 and 3.22, record all your calculations and diagrams. Confirm your calculations with another person’s calculations before you begin. Draw a diagram to show how each sample is diluted. Use the equation to make the calculations. Be sure to make the columns te appropriate width for the material they will contain. ● Label all tubes with the name and concentration of the sample, your initials, and the date. Table 3.21 Dilutions of the 300x stock CuSO4 volume to be made (mL) concentration to be made (x) 5 150x 7 30x 5 15x 5 3x 4 Calculations Volume of stock to use Solution preparation diagram 1x 1. Prepare 25mL of 300-mg/mL CuSO4 solution. Mix well. Do not use previously made solutions. consider this a......

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...Why do we keep a lab notebook? In the ‘real world’, you will be expected to record what you do and how you do it, with the data clearly identifiable. There are 2 reasons for this: • Any scientific experiment is only considered valid if it can be repeated by someone else using your notebook. • If you develop a new product or process, your notebook becomes a legal document that can be scrutinized in a court of law, especially in cases of patent infringement. In order to meet these requirements, certain conditions must be met: • Notebooks shall have bound pages, loose-leaf and spiral bound are not acceptable. • All pages must be numbered prior to use. This will show if any pages have been removed that contained data that might compromise interpretation of your conclusions. • All data shall be entered chronologically. Do not leave empty pages to insert information later. If you finish with a large expanse of blank page, put a line through it to show that it will not be used. • All notes are to be made in indelible pen – pencil will be penalized. If you make a mistake, just draw a single line through it, not a childish scribble to hide all traces! • All data will be recorded in your lab notebook at the moment it is generated. No notes shall be written on scraps of paper or memorized for later. Other notes about the lab notebook: All pre-lab and experimental work is hand-written in your lab notebook. ......

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