Tuesday, December 3, 2019

Lab Report free essay sample

However, if we add just one more drop of NaOH solution from the buret, the solution will immediately turn pink because the solution is now basic. This slight excess of NaOH is not much beyond the end point. The volume of the base is recorded and used to determine the molarity of the acetic acid solution. Experimental Data Standardization of NaOH solution | Trial 1| Trial 2| Trial 3| Mass of KHP| 0. 297 g| 0. 325 g| 0. 309 g| Initial buret reading, NaOH| 0. 00 mL| 0. 50 mL| 7. 70 mL| Final buret reading, NaOH| 32. 0 mL| 34. 0 mL| 38. 7 mL| Volume used, NaOH| 32. 0 mL| 33. mL| 31. 0 mL| Molarity of NaOH solution| 0. 0454 M| 0. 0475 M| 0. 0488 M| Average molarity of NaOH| 0. 0472 M| Titration of unknown | Trial 1| Trial 2| Trial 3| Initial buret reading, NaOH| 2. 70 mL| 19. 9 mL| 0. 00 mL| Final buret reading, NaOH| 19. 9 mL| 36. 2 mL| 19. 8 mL| Volume used, NaOH| 17. 2 mL| 16. 3 mL| 19. 8 mL| Molarity of acetic acid solution| 0. 0780 M| 0. 0769 M| 0. 0935 M| Average molarity of acetic acid solution| 0. 0828 M| Sample Calculations The following calculations were used for each Trial, but only inputs for Trial 1 will be shown below. We will write a custom essay sample on Lab Report or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page Volume = Final buret reading – Initial buret reading i. Volume of NaOH = Final buret reading of NaOH – Initial buret reading of NaOH ii. Volume of NaOH = 32. 0 mL NaOH – 0. 00 mL NaOH iii. Volume of NaOH = 32. 0 mL Molarity = Moles/Liters i. Molarity of NaOH solution = (mass of KHP/molar mass of KHP) / Volume of NaOH ii. Molarity of NaOH solution = (0. 2966 g/204. 22 g)/0. 032 L iii. Molarity of NaOH solution = 0. 0454 M Molarity of acetic acid = (Molarity NaOH * Volume NaOH) / Volume Acetic Acid i. Molarity of acetic acid = (0. 0472 M * 0. 0172 L)/ 0. 1 L ii. Molarity of acetic acid = 0. 0780 M Percent Error = Experimenal value-Accepted valueAccepted value*100 i. Percent Error of Molarity of NaOH = 0. 0472 M-0. 05 M0. 05 M*100 ii. Percent Error of Molarity of NaOH = 5. 6% i. Percent Error of Molarity of acetic acid = 0. 078 M-0. 080 M0. 080 M*100 ii. Percent Error of Molaarity of acetic acid = 2. 5% Discussion The results obtained from the experiment proved to t he principle that using the indictor we can find the end point, which is very close to the equivalence point of an acidic solution. Then using that point we were able to calculate the unknown molarity which was one of the goals of the experiment. The calculations also verify Boyle’s Theory. When we calculated the molarity of the acetic solution, an average value of 0. 078 M was obtained. The true value of the molarity of the acetic acid solution was 0. 08 M. Although it isn’t right on, it is very close to the true value which leads me into discussing the percent error. We found the percent error of the molarity of NaOH to be 5. 6%, and the percent error of the molarity of acetic acid to be 2. 5%, which are both pretty small. The error may have occurred when adding NaOH solution. Occasionally slightly more pressure was put on tilts of the piece on the buret to allow the solution to flow through. This means that more of the solution may have been used than needed. Overall, experiment agrees with the formulated hypothesis. Pre-Lab and Post Lab Questions Pre-Lab 1. Molarity of NaOH solution = (mass of KHP/molar mass of KHP) / Volume of NaOH a. Molarity = (0. 2816 g/204. 22 g)/29. 68 mL Molarity = 4. 64*10-5 M 2. Molarity of acetic acid = (Molarity NaOH * Volume NaOH) / Volume Acetic Acid b. Molarity = ((4. 64*10-5 M)*20. 22 mL)/10. 06 mL Molarity = 9. 34*10-5 M Post Lab 1. A. TD B. TD 2. A graduated cylinder with calibration type TD could be used to deliver a certain amount of a liquid into another container. A graduated cylinder marked TC could be used to contain an accurate volume of a liquid that is to be mixed with another solution, where the experiment is to be done inside of that graduated cylinder. 3. 50g * 1mol /49. 997g = 1 mol 100g * 1mL / 1. 53g = 1L / 15. 3 1mol / (1L / 1. 53) = 1mol* 1. 53 / 1L = 15. 3 mol/L= 15. 3 M Lab report free essay sample The purpose of this lab was to perform and understand the procedures of conducting an ELISA test to determine whether a particular antibody is present in a patient’s blood sample through a virtual simulation. Hypothesis: If I successfully complete this lab, I will then understand how to perform an ELISA test, the purpose an ELISA test, and also how to interpret the results of this test. Materials and Procedures: Materials: Howard Huges medical Institute: Virtual Immunology Lab Human Anatomy Physiology Marieb 9th edition textbook Web access Procedures: Use the hyperlink Virtual Immunology Lab to visit the website where the virtual lab is located Click on the image to the left of the summary to launch the lab Read the diagnosis and background for the lab before you begin Follow the instructions provided throughout the lab BE SURE to pay close attention and follow each step precisely When you have completed the entire lab, copy the data results into the data analysis section of your report Data Analysis: Conclusion: The hypothesis stated was correct. We will write a custom essay sample on Lab report or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page After completing this lab I now understand how to perform an ELISA test, the purpose, and also how to interpret the results. An ELISA or Enzyme-linked Immunosorbent Assay test uses components of the immune system and chemicals to detect immune responses in the body (for example, to infectious microbes) and involves an enzyme and antibody or antigen (immunologic molecules). The purpose of an ELISA test is to detect substances that have antigenic properties, primarily proteins, which include hormones, bacterial antigens, and antibodies (MedicineNet, 2005). If done correctly, an ELISA test can confirm whether or not a patient has a particular disease or if there is a possibility that they could develop it. It is very important that a test such as this is performed correctly and repeated a few times just to be sure that the results aren’t a false-positive or false-negative. Notice in the lab results above that the ELISA plate was not washed like it should have been, thus yielding a false-positive indicating that all three patient’s may possibly have SLE. If all steps of the experiment would have been done correctly, then the results would have been as follows: patient A is likely to have SLE, patient C is likely to NOT have SLE, and patient B may have SLE, but further testing needs to be done. ELISA testing is very helpful, but it is not the final confirmation due to a variety of limitations that it has. Further testing and other factors would need to be taken into account before confirming whether or not a patient has a particular disease. Lab report free essay sample A molecular diffusion experiment of acetone-air (redundant w/ last sentence. . concise)was conducted with the goal of determining the diffusion coefficient of acetone into air. For this experiment, acetone was placed in a test tub 3mm OD, 2mm ID . . (is that correct? ) NMR tube? e and was allowed to diffuse into non-diffusing air that was passed over the test tube. The air that passed over the tube was from natural circulation in the room and no air was forced over the top of the test tube. The diffusion occurred over a period of approximately eight hours, with readings taken each hour. After analyzing the data collected from the performance of this experiment, tThe diffusion coefficient was calculated to be 0. 098 + 0. 02 cm2/s at T = ?. After completing our calculations, oOur results were then compared using the Chapman-Enskog equation as well as the Fuller, Schettler, and Giddings method. The diffusion coefficient calculated by the Chapman-Enskog was 0. We will write a custom essay sample on Lab report or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page 990 + 0. 001 cm2/s and the result of the Fuller, Schettler, and Giddings method was 0. 104 + . 002 cm2/s. The literature value found in Perry’s Chemical Engineer’s Handbook was 0. 125 + 0. 00 cm2/s. (at T = ?. . . or extrapolated from? ) The agreement of our method with the other methods available for calculating the diffusion coefficient was very good (how good is â€Å"very† good. .. significant discrepencies or not? ), and also agreed well with the literature value found. This led to a conclusion that this method of determining the diffusion coefficient of acetone into air can be aconsidered a reasonably reliable method. BACKGROUND Molecular diffusion is the transfer or movement of individual molecules through a fluid by random molecular movements (Geankoplis 412, year of publication). In the diffusion process, the molecules of interest flow from regions of high concentration to low concentration. Molecular diffusion can occur in both directions with the system. In the case of the diffusion tube experiment, however, acetone diffuses through non-diffusing air, which is passed over the top of the test tube containing the acetone. The air is allowed into the test tube, but does not diffuse into the acetone. Molecular diffusion of gases has been studied for many years. Molecular diffusion is a mass transport process Motivation for its study comes from the fact that chemical separation processes such as distillation, drying, ion exchange systems as well as many other processes depend on molecular diffusion (Kirk-Othmer Vol 8, p 149(check format)). EXPERIMENTAL METHODS For the performance of this experiment, a small test tube was filled approximately a third full of acetoneBe specific. . how small, starting height, diam, etc. This test tube was then vertically placed in a 10mL graduated cylinder which contained small beads. The purpose of the beads was to ensure that the test tube remained vertical. This assembly was then placed on a digital scale. The amount of air movement provided by the ventilation system was assumed to be adequate so as to ensure that the concentration of the acetone at the top of the tube was zero. An initial acetone level in the test tube was taken, as well as the mass of the assembly and the temperature of the area surrounding the assembly. After this initial data was taken, the area temperature and mass of the assembly were taken approximately every hour for the next eight hours. The final level of the acetone in the test tube was taken when the final temperature and mass reading were taken. DISCUSSION OF RESULTS From the data collected from the experiment, the diffusion coefficient was calculated using equation 6. 2-26 from Geankoplis: (Equation 1) As the z value was only recorded at the beginning and the end of the experiment, the intermediate values of z had to be calculated. The following equation was used for the calculation of the intermediate z values: (Equation 2) Thus, all values but DAB were known and could be plotted versus time to obtain a linear plot. By rearranging equation 1, it can be seen that the slope of this plot will be equal to 1/ DAB : (Equation 1. 1) The initial plot of data which includes all points is shown below in Figure 1. This plot contains all points and has an R2 value of 0. 9478. From this plot the molecular diffusivity coefficient was determined to be 0. 108 + 0. 022 cm2/s. Figure 1: First plot of data in Equation 1 The second point in the data (t=2700s) showed no diffusion occurred in the first 45 minutes, which seems unlikely (yes, good- sensitivity of balance, etc). If this point is taken as erroneous, the R2 value goes up to 0. 9639 (more important here will be the confidence interval on the slop. . . get that from Tools- Data Analyis-Regression menu in Excel or else in Polymath or TableCureve, etc) and the molecular diffusivity calculates out to be 0. 098 + 0. 021 cm2/s. The plot of the experimental data excluding the second point is presented below in Figure 2. Figure 2: Second plot of data in Equation 1. . forcing through zero point is good. . . looks to me like first FOUR points would give a lower Dab then the last 4. Problems with next 3 that lie below line? Anytihing suspicious happening here? To determine the time it takes for the system to reach steady state, the following equation can be used to calculate the fraction of steady state the system is at: (Equation 3) By plotting the value of ((NA)t/(NA)t=? ) versus time, the curve in Figure 3 was generated which demonstrates the systems approach to steady state. Wow, great! Cite source.  (still wonder about SST conditions of 1st 4 pts though. . . Figure 3: Fraction of steady state versus time From this plot, it could be said that the system achieves steady state in 115 minutes; however, there is strong evidence this may not be accurate. As mentioned earlier, the second point may be erroneous. This would change the path of the curve. In addition, data was not collected at a high enough frequency for this curve to be highly accurate at predicting the time to steady state. If in fact the second point is erroneous, the system could have come to steady state well before 115 minutes. This time of 115 minutes at best, could be the upper bound (or lower bound according to Whitaker’s criteria in his article (handout). . . not sure!! for the time it takes for the system to come to steady state. The scatter in the data can be attributed to various factors in the experiment. The scatter could be attributed to the changes in temperature, as the temperature did fluctuate slightly through the duration of the experiment – Good!. At what time did it stabilize?. The change in temperature would cause a change in the partial pressure of the acetone leading to further deviations. In addition, there was no measure of airflow past the tube. Changes in the airflow could also have contributed to the scatter as it could effect the concentration of the acetone at the top of the test tube (Good! ). The diffusion coefficient was also calculated using the Chapman Enskog equation, (Equation 4) and the Fuller, Schettler and Giddings method. (Equation 5) A literature value was also found for acetone at K(check Perrys), which was corrected to our experimental temperature using the correlation (Equation 6) The values obtained with these methods as well as those from the experimental data are presented in Table 2. Table 1: Values of molecular diffusivity coefficients found. ** ** A very good way to show this graphically in Excel would be to use a bar graph showing the values of Dab as height of a bar by method used, and error bars to easily demonstrate any overlap of uncertainty, discrepancy, etc. Example: The Chapman Enskog method is accurate within 8% and the Fuller Schettler and Giddings value has a lower accuracy than the Chapman Enskog (Geankoplis 425). The Chapman Enskog value is less than 1% different than the experimental value and the Fuller Schettler and Giddings value only about 6% different. From this analysis, it seems these equations predicted the experimental value very well. These calculated values are about 20% lower than the literature value. This variance may come from the inconsistent temperature in the room or from pressure fluctuations in the room caused perhaps by the starting and stopping of the HVAC systems. For the derivation of Equation 1, several assumptions are made. Beginning with the general equation (Geankoplis 6.  2-14): (Equation 7) One assumption was that because the case examined was a diffusing A (acetone) into non-diffusing B (air), the diffusion flux of air into the acetone (NB) was equal to zero. Another assumption made was that since the total pressure was low, the acetone gas diffusing into air was an ideal gas. This allowed for the term c to be replaced with its ideal gas equivalent, P/RT. Add itionally, the air passing over the test tube was assumed to contain no water vapor. An average air velocity that was uniform was passing over the acetone containing test tube was also assumed. There are non-idealities that exist in the molecular diffusion of acetone into air. Some of these non-idealities are corrected for in the journal from Lee and Wilke. Acetone displays surface tension effects which, instead of having a perfectly horizontal liquid surface, give the liquid acetone a slightly downward curved liquid level. Because of this curvature, the actual diffusion path length that the acetone travels is smaller than what the diffusion length would appear to be based on center liquid level or calculated liquid volume (Lee 2384). Along with a non ideal liquid surface, the air passing over the open end of the tube may cause some turbulence to exist in the top portion of the tube. With its existence, the turbulent area of the tube will cause a length to exist inside the tube where the concentration of acetone is zero. With the presence of this acetone vapor-free region, the diffusion length is again shorter than it would appear to be. To account for the non-idealities in the diffusion process, Lee and Wilke do not use the apparent diffusion path. Instead, they use an effective average diffusion path which they give by: (Equation 8) Where x is the effective average diffusion path, ? xs is the length of the curvature of the non-ideal liquid to account for the surface tension forces, ? xe is the length of the tube where the acetone vapor-free region exists due to turbulence that exists from the passage of the air, and ? x ­ is the sum of ? xs and ? xe (Lee 2384). When this is substituted back into the diffusion equation, it becomes the following: (Equation 9) Where Da is the apparent diffusion coefficient and D is the true diffusion coefficient based on the true diffusion path (Lee 2384). The way our experiment was setup, the driving force for the air across the test tube was natural air flow and did not employ forced air flow. Because of this, the length of the tube where the turbulence existed in the Lee and Wilke journal would most likely not have been present in our experiment. Also, the initial liquid acetone level selected in our experiment was such that the length of the curvature due to the surface tension forces on the acetone would have been negligible when compared to the apparent diffusion length of the tube. The initial height of the liquid in the tube for this experiment was chosen wisely. The reason for this is that with the initial level that was chosen, a sufficiently long diffusion path existed such that the non-idealities that were accounted for in the Lee and Wilke journal entry would have had a very insignificant impact on the results of our experiment. CONCLUSIONS AND RECOMMENDATIONS From the data collected an analyzed, it has been determined that the experimental procedure used here can determine the molecular diffusivity coefficient with some level of accuracy. For future experiments, some form of air flow regulation should be investigated. Something as simple as a room fan could be placed next to the scale to ensure a more constant air flow. Another increase in accuracy could be achieved by regulating the temperature with more consistency. If the experiment could be performed in a large insulated room, the temperature may not vary as much. Good job on Discussion, Conclusions, etc. . . to improve maybe expand to relate what YOU think are the main ‘uncertaintys’ that caused problems in your particular case and show evidence to support. Lab Report free essay sample LAB REPORT NUMBER TWO DATE: 3/25/2010 inal attachment Lab Experiment number 11 PURPOSE: To learn the Gram stain technique, the reason for the stain, and how to identify the results of the organisms stained. MATERIALS: Bunsen burner, inoculating loop, staining tray, glass slides, bibulous paper, lens paper, oil, and microscope   METHODS: Apply Crystal Violet (Primary stain) for 1 minute. Rinse with D-water Apply Iodine (Mordant) for 1 minute. Rinse with D-water. Apply Alcohol (Decolorize) for 30 seconds. Rinse with D-water. Apply Safarin (Counterstain) for 1 minute. Blot dry with bibulous paper. MICROORGANISMS USED: E. coli, B. cereus, S. aureus E. coli (mixture) RESULTS/DATA USED: E. coli cell shape was bacilli (rod) with a diplobaccillus arrangement. The color was pink because it was Gram negative. B. cereus cell shape was bacilli (rod) with a diplobacillus arrangement. The color was purple because it was Gram positive. S. aureus E. coli (mixture) cell shape was cocci (spherical) with a staphylococcus arrangement. We will write a custom essay sample on Lab Report or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page The color was mostly purple with some noticeable pink but the mixture was Gram positive. CONCLUSIONS E. coli is Gram negative, B. ereus is Gram positive, S. aureus E. coli mixture is Gram positive. REVIEW QUESTIONS: Question 1: What are the advantages of differential staining procedures over the simple staining technique? Answer: Simple stains are used to just give color to microbes on slides. Differential stains tell the chemical composition of organisms. Source: http://www. bmb. psu. edu/courses/micro107/notes/staining. htm Question 2: Cite the purpose of each of the following reagents in a differential staining procedure. Answer: a. Primary stain: Passes the color of the stain to all of the cells. b. Counterstain: Used to stain red the cells that have been decolorized (Gram – cells). c. Decolorizing agent: removes the primary stain so that the counterstain can be absorbed. d. Mordant: Increases the cells’ affinity for a stain by binding to the primary stain. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/ Pages 73 74   Question 3: Why is it essential that the primary stain and the counterstain be of contrasting colors? Answer: Cell types or their structures can be distinguished from one another on the basis of the stain that is retained. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/ Pages 73   Question 4: which is the most crucial step in the performance of the Gram staining procedures? Explain. Answer: Decolorization is the most crucial step of the Gram stain. Over-decolorization will result in lost of the primary stain causing Gram positive organisms to appear Gram negative. Under-decolorization will not completely remove the CV-I (crystal-violet-iodine) complex, causing Gram negative organisms to appear Gram positive. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/ Pages 74   Question 5: Because of a snowstorm, your regular laboratory session was cancelled and the Gram staining procedure was performed on cultures incubated for a longer period of time. Examination of the stained Bacillus cereus slides revealed a great deal of color variability, ranging from an intense blue to shades of pink. Account for this result. Answer: The organisms lost their ability to retain the primary stain and appear to be gram-variable. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/ Pages 74 LAB EXPERIMENT NUMBER 12 PURPOSE: The purpose of the Acid fast stain is to identify the members of the genus Mycobacterium, which represent bacteria that are pathogenic to humans. Mycobacteria has a thick, waxy wall that makes penetration by stains extremely difficult so the acid fast stain is used because once the primary stain sets it cannot be removed with acid alcohol. This stain is a diagnostic value in identifying these organisms. MATERIALS: * Bunsen burner * Hot plate * Inoculating loop * Glass slides * Bibulous paper * Lens paper * Staining tray * Microscope METHODS: 1. Prepared a bacterial smear of M. smegmatic, S. aureus, a mixture of M. smegmatic S. aureus 2. Allowed 3 bacterial slides to air dry then heat fixed over Bunsen burner 8 times. 3. Set up for staining over the beaker on hot plate, flooded smears with primary stain-crystal fuchsin and steamed for 8 minutes. 4. Rinsed slides with water 5. Decolorized slides with acid alcohol until it runs clear with a slight red color. 6. Rinsed with water 7. Counterstained with methylene blue for 2 minutes 8. Rinsed slides with water. 9. Blot dry using bibulous paper and examine under oil immersion MICROORGANISMS USED: * Mycobacterium   smegmatic * S. aureus * A mixture of S. aureus M. smegmatic RESULTS AND DATA USED: 1. M. smegmatic, a bacilli bacteria that colored pink resulting in acid fast. 2. S. aureus, a cocci bacteria that colored blue resulting in non acid fast. 3. M. smegmatic S. aureus resulted in both acid fast non acid fast. CONCLUSION The conclusion to the acid fast stain is that S. aureus lacks a cellular wax wall causing the primary stain to be easily removed during decolorization, causing it to pick up the counterstain-methylene blue. This results in a non acid fast reaction, meaning it is not in the genus Mycobacterium. M. smegmatic has a cellular wax wall causing the primary stain to set in and not be decolorized; this results in an acid fast reaction meaning it is in the genus Mycobacterium. REVIEW QUESTIONS Question 1: Why must heat or a surface-active agent be used with application of the primary stain during acid-fast staining? Answer: It reduces surface tension between the cell wall of the myobacteria and the stain. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 79 Question 2: Why is acid-alcohol rather than ethyl alcohol used as a decolorizing agent? Answer: Acid-fast cells will be resistant to decolorization since the primary stain is more soluble in the cellular waxes than in the decolorizing agent. Ethyl alcohol would make the acid fast cells non-resistant to the decolorization. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/ page 79 Question 3: What is the specific diagnostic value of this staining procedure? Answer: Acid-fasting staining represents bacteria that is athogenic to humans Question 4: Why is the application of heat or a surface-active agent not required during the application of the counter stain in acid-fast staining? Answer: The counter stain methylene blue is only needed to give the stain its color. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 79 Questi on 5: A child presents symptoms suggestive of tuberculosis, namely a respiratory infection with a productive cough. Microscopic examination of the child’s sputum reveals no acid-fast rods. However, examination of gastric washings reveals the presence of both acid-fast and non-acid fast bacilli. Do you think the child has active tuberculosis? Explain. Answer: Yes, the child may have active tuberculosis. Although, acid-fast microorganisms are not easily removed and non-acid fast are. Tuberulosis represents bacteria that are pathogenic to humans, the stain is of diagnostic value identifying these organisms. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 79 LAB EXPERIMENT NUMBER 13 PURPOSE: The purpose of this experiment is to identify the difference between the bacterial spore and vegetative cell forms. The vegetative cells are highly resistant, metabolically inactive cell types. The endospore is released from the degenerating vegetative cell and becomes an independent cell. MATERIALS: * Bunsen burner * hot plate * staining tray * inoculating loop * glass slides * bibulous paper * lens paper * microscope METHODS: 1. The spore stain (Schaeffer-Fulton Method) is performed on a microscopic slide by making an individual smear of the bacteria on slide and heat fixing until dry. . Flood the smears with malachite green and place on top of a beaker of warm water on a hot plate, allowing it to steam for 5 minutes. 3. Remove the slide and rinse with water. 4. Add counter stain safranin for 1 minute then rinse again with water and blot dry with bibulous paper. MICROORGANISMS USED: * S. aureus * S. aureus B. cereus mix RESULTS/DATA USED 1. B. cereus- green spores, pink vegetative cells, endospore located in center of cell 2. B. cereus S. aureus- green spores, pink vegetative cells, endospore located in center of cell   CONCLUSION: An endospore is a special type of dormant cell that requires heat to uptake the primary stain. To make endospores readily noticeable, a spore stain can be used. In using a microscope, under oil immersion, you will be able to identify the color of the spores, color of the vegetative cells and be able to locate the endospore in certain bacteria like S. aureus and B. cereus. REVIEW QUESTIONS Question 1: Why is heat necessary in spore staining? Answer: The heat dries the dye into the vegetative cell of the spore. Source: Microbiology Lab Manual, 8th edition, Cappuccino Sherman, p. 5 Question 2: Explain the function of water in spore staining. Answer: The water removes the excess primary stain, while the spores remain green the water   rinses the vegetative cells that are now colorless. Source: Microbiology Lab Manual, 8th edition, Cappuccino Sherman, p. 85 Question 3: Assume that during the performance of this exercise you made several errors in your spore-staining procedure. In each of the following cases, indicate how your microscopic observations would differ from those observed when the slides were prepared correctly. Answer: a. ) You used acid-alcohol as the decolorizing agent. The alcohol would wash out all coloring from the bacteria. Source: Microbiology Lab Manual, 8th edition, Cappuccino Sherman, p. 85 b. ) You used safranin as the primary stain and malachite green as the counterstain. Safranin will absorb to vegetative cells and not endospores since you need heat for   endospores to form and malachite green will not absorb without heat but it will to   vegetative cells. Source: Microbiology Lab Manual, 8th edition, Cappuccino Sherman, p. 85 c. You did not apply heat during the application of the primary stain. Without heat, the endospores will not form and it will not penetrate the spore to color   the vegetative cell. Source: Microbiology Lab Manual, 8th edition, Cappuccino Sherman, p. 85 Question 4: Explain the medical significance of a capsule. Answer: The capsule protects bacteria against the normal phagocytic activities of the host cells. Source: Microbiology Lab Manual, 8th edition, Cappuccino Sherman, p. 87 Question 5: Explain the function of copper sulfate in this procedure. Answer: It is used as a decolorizing agent rather than water, washes the purple primary stain out of the capsular material without removing the stain bound to the cell wall, the capsule   absorbs the copper sulfate and will appear blue. Source: Microbiology Lab Manual, 8th edition, Cappuccino Sherman, p. 88 LAB EXPERIMENT NUMBER 44A PURPOSE: The purpose of this experiment is to identify the best chemotherapeutic agents used for infectious diseases. S. aureus is the infectious disease used for this experiment. MATERIALS: * Sensi-disc dispensers or forceps * Bunsen burner * sterile cotton swabs * glassware marking pencil millimeter ruler METHODS: Using the Kirby-Bauer antibiotic sensativity test method is used. This method uses an Antibiotic Sensi-disc dispenser, which placed six different types of antibiotics on an Mueller-Hinton agar plate, infected with S. aureus. The antibiotics are in the form of small, round disc, approximately 5mm in diameter. The anitbiotics are placed eve nly away from each other on the S. aureus infected Mueller-Hinton agar plate and incubated at 37 degrees Celcius for up to 48 hours. After the completed incubation time, any area surrounding the antibiotic disc which shows a clearing or an area of inhibition is then measured. Measurements are taken from the diameter of each antibiotic area of inhibition. This measurement will determine which of the antibiotics is best to be used against the specific organism. (In this case, S. aureus)   MICROORGANISMS USED: S. aureus ANTIBIOTICS USED: Ticarcilin Erythomycin Clindamycin Gentamicin Vancoymycin Lmipenem RESULTS/DATA USED: A chart showing the measurements of each antibiotic is used to determine its effectiveness. The three different types of ranges are: Resistant (Least useful) Intermediate (Medium useful) Susceptible (Most useful) The following results are: Zone Size Ticarcilin 25mm (Susceptible) Erythomycin 20mm (Intermediate) Clindamycin 20mm (Intermediate) Gentamicin 15mm (Susceptible) Vancoymycin 13 mm (Susceptible) Lmipenem 21 mm (Susceptible) CONCLUSION: 4 of the 6 antibiotics above can be effectively used against inhibiting this organism (S. aureus). This information would be passed on to the provider of the infected patient, so the patient can be given the antibiotic chosen by their provider and recover from this infection. LAB EXPERIMENT NUMBER 46B PURPOSE: The purpose of this experiment is to evaluate the effectiveness of antiseptic agents against selected test organisms. MATERIALS: The materials used are five Tryticase soy agar plates. METHODS: 24-48 hours Trypticase soy broth cultures of E. coli, B. cereus, S. aureus and M. spegmatic. MICROORGANISMS USED: The microorganisms used were E. coli, B. cereus, S. aureus and M. spegmatic. RESULTS/DATA USED: The data collected in this experiment shows chlorine bleach having the broadest range of microbial activity because it has the strongest ingredients. Tincture of iodine and hydrogen peroxide seems to have the narrowest range because the contents aren’t as strong. CONCLUSION: The Agar Plate-Sensitivity Method shows the effectiveness of antiseptic agents against selected test organisms. The antiseptic exhibited microbicidal activity against each microorganism. REVIEW QUESTIONS: Question 1: Evaluate the effectiveness of a disinfectant with a phenol coefficient of 40. Answer: A disinfectant with a phenol coefficient of 40 indicates the chemical agent being more effective than the phenol. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 302 Question 2: Can the disinfection period (exposure time) be arbitrarily increased? Explain. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 302 Answer: Yes, the longer the disinfection period the greater the antimicrobial activity. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page 302 Question 3: A household cleanser is labeled germicidal. Explain what this means to you. Answer: A household cleaner labeled germicidal means something that would and the killing of germs on a surface. Source: Microbiology – A Laboratory Manual 4th Edition/ James G. Cappuccino, Natalie Sherman/ 2008/page`302 Lab report free essay sample Introduction: Fractional crystallization is one of the experimental techniques used to separate or purify mixture. Fractional crystallization makes use of the differences of solubility to separate the components of a mixture. Goals: Applying fractional crystallization to separate a mixture of salicylic acid and copper sulfate pentahydrate into its components. Calculate the percent of salicylic acid and copper sulfate pentahy- drate in the mixture. Procedure: Block/McKelvy/with contributions from Georgia Perimeter Faculty (2012) Laboratory Experiments for Chem 1211L 1212L Cengage Learning (7th ed. ) p. 2324 (Separation of a mixture into its components by fractional crystallization) Data and result: Unknown number9 Mass of unknown + beaker69. 1607 g Mass of beaker66. 6636 g Mass of unknown mixture2. 4971 g Salicylic Acid Crystallization Mass of weighing paper + Salicylic acid1. 7468 g Mass of weighing paper0. 9446 g Mass of Salicylic Acid0. 8022 g Calculation of percent salicylic Acid32. 13 % in original mixture: % salicylic acid = (m Salicylic Acid : m unknown mixture) x 100% = (0. We will write a custom essay sample on Lab report or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page 8022 : 2. 4971) x 100% = 32. 13% CuSO4. 5H2O Crystallization Mass of weighing paper + CuSO4. 5H2O2. 6484 g Mass of weighing paper1. 0127 g Mass of CuSO4. 5H2O1. 6357 g Calculation of percent CuSO4. 5H2O65. 504% in original mixture: % CuSO4. 5H2O = (m CuSO4. 5H2O : m unknown mixture) x 100% = (1. 6357 : 2. 4971) x 100% = 65. 504% Percent of original sample not recovered 100% ( 32. 13% +65. 504%) = 2. 37% Comments concerning appearance and purity: The salicylic acid’s appearance are pure, fine, off-white needles. The copper sulfate pentahydrate ’s appearance first observing are contaminated, white coat outside blue crystals. Prelab question: 1. Explain briefly why CuSO4 is more soluble in water than in an organic (nonpolar) solvent. Why is the reverse true for salicylic acid? As the rule of thumb says â€Å"like dissolve like† : CuSO4 is a polar substance, therefore, it is more soluble in polar solvent (water is a polar solvent) than in nonpolar solvent. Salicylic acid is a nonpolar substance, therefore, it is less soluble in polar solvent (water is a polar solvent) than in nonpolar solvent (such as gasoline). 2. Suggest reasons why the solubility of most solid substances in liquid solvents increases with temperature, whereas the solubility of gases decreases with temperature. As the temperature is increasing: The particles in solid substances, which are normally in fixed positions and close together, become more active and free to move so it increases the chance for particles in liquid solvents to contact to the particles in solid substances. Therefore, the solubility of most solid substances in liquid solvents increases with temperature. The particles in gases, which normally have complete freedom of motion and are far apart, become more disordered and move faster so it decreases the chance for particles in liquid solvents to contact to the particles in gases. Therefore, the solubility of gases decreases with temperature. 3. E. D. Student used 2. 498g of a mixture of copper sulfate pentahydrate and salicylic acid in the performance of this experiment. E. D. recovered 1. 184g of salicylic acid and 1. 367g of copper sulfate pentahydrate. What is wrong with E. D. ’s results? How might this error have been avoided? In E. D. ’s results, the total mass of separate substances he recovered, 2. 551g, is 0. 053g higher than the mass of the original mixture, 2. 498g. His recovered substances may be contaminated. The mass of the copper sulfate pentahydrate he weighed may be including the mass of ethanol. To avoid this error, we should wait for a while for the ethanol to evaporate completely then weigh the copper sulfate pentahydrate again. Discussion: We did experiment to separating the mixture of Salicylic Acid and CuSO4. 5H2O successfully. At the first filtration, we recovered the pure salicylic acid crystal and weighed it easily. At the second filtration, we recovered the contaminated copper sulfate pentahydrate ( observe the white coat outside the blue crystals) so it takes longer time to wait for the ethanol to evaporate then we can weigh the crystals. We calculated the percent of recovered salicylic acid and copper sulfate pentahydrate . The percent of original sample not recovered is small. Lab Report free essay sample ?Beginning with science : 06 Matter and Energy Kevin Edwards. Introduction :You will be exploring three scenarios and conducting observations on the physical and chemical changes in matter: The Iced Tea Debate, Salty Soup, and Fire Bug. Objective : To demonstrate the differences between physical and chemical changes while observing the laws of conservation of matter and energy. Procedure : Watch each part of the experimental demonstration and make predictions about what will happen in each scenario. 1. View the three scenarios listed below. You must observe and analyze these as part of the lab. The Iced Tea Debate The Salty Soup The Fire Bug 2 Use the DATA chart provided for recording observation based on three demonstrations you will watch 3 Watch each part of the experimental demonstration and make preditions about wht will happen in each scenario . Record your preditions and observations in the Data an Observations section of your labatory report format. Problem: How can matter and energy be described in avarity of systems? Analysis : Use your data table to identify the physical and chemical changes observed in the demonstrations as follows. We will write a custom essay sample on Lab Report or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page Analysis Use your data table above to identify the physical and chemical changes observed in the demonstrations as follows: 1. Complete Data Chart above filling in Scientific Methods for each demonstration. 2. Use the drop-down menu options to record any physical and/or chemical changes observed. 3. Once you completed the data chart, complete pre-made paragraphs below by writing an analysis that includes details on how the Law of Conservation of Matter and the Law of Conservation of Energy are supported by the experimental demonstrations. Use the following reflection questions to guide you in writing the analysis:  · Was matter or energy lost or destroyed in any of the scenarios?  · Did your predictions match the second portion of the demonstrations?  · What phase changes did you observe?  · What kinds of energy were input and output in each of the scenarios? Paragraph 1 How the Law of Conservation of Matter is supported by the experimental demonstrations: In the law of conservation particles and materials are neither created nor destroyed . It was similar towards the experimental demonstrations because nothing changed or destroyed in the processes . Kinetic and potential energy was used in the experiments. Paragraph 2 How the Law of Conservation of Energy is supported by the experimental demonstrations: Experiments do not consome the energy but conserve by changing them from one form to the other . This allows all energy to remain constant so for each experiment diffrent form occured but never changed the element of the experiment Fourth Scenario Describe a fourth scenario in which either the â€Å"Law of Conservation of Matter† or the â€Å"Law of Conservation of Energy† could be observed. Using as many sentences as needed, describe how an experiment could be set up to further explore your recorded observation. The goal is to show understanding of the concepts in the lesson. I choose law of Conservation of Energy Suppose a Hard mini ball is placed in an automatic air power sealed containerYou will notice that no matter how much you will leave the ball bunching the ball will never stop bouncing. That happens because the ball develops both potential and kinetic energy,the sum of those two give us the mechanical energy which remains constant. Scientific Method The Iced Tea Debate The Salty Soup The Fire Bug Predictions—What do you think will happen? Well the ice tea should freeze and it will have kept its same chemicals I thinkThe soup would have got salty from the Heat extracting the salt from the spices and making it expand from liquid water The more heat took in to the body caused more moisture and bugs to come. The less heat from the further away from the fire. Observations—What did you see happening in each demonstration? Nothing was affected from 0 degrees and 20 degrees They measured the water displaced from soup into the orignal amount It Explained the Heat and inner and outer reasons for bugs and body temperure hot and cold from front or back Conclusion—What was demonstrated here? They froze the ice to see if a change would occur in the orginal compound chemical bonds From the measurements of the orignals they found out the amount of water boiled out from heat the salt stayed in and made the intire soup salty The conservation of energy exerted from 1 person to another in diffrent regions of the area Lab Report free essay sample I hypothesize that a chemical reaction has occurred when there is a change in temperature, color. But also when there are bubbles and the evolution of gas. Materials: The materials that were used were Aluminum wire(12 cm), beaker (100 mL), hot plate, copper(II) nitrate (1. 0 M), glass stirring rod, gloves, HCl (1. 0 M), lab apron , lab marker, NaOH (1. 0 M), ruler, safety goggles, test tube (13 mm x 100mm) and test tube rack. Procedure: Safety goggles, gloves, and lab apron were put on. (Only the ones who were doing the experiment) 50mL of water was placed in the 100mL beaker and heated until boiled. And the boiled water will be used as the water bath. The lab marker was used to make 3 marks (1 cm apart). 1. 0 M copper (II) nitrate was added to the first mark on the test tube. 1. 0 M sodium hydroxide (NaOH) was added up to the second mark on the test tube. The solutions were mixed with the stirring rod. We will write a custom essay sample on Lab Report or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page The test tube was put into the water bath. The burner was turned off and the test tube was cooled. 1. 0 M hydrochloric acid (HCl) was added to the third mark on the test tube and was mixed. The 12 cm piece of aluminum was placed in the test tube. The wire was removed from the test tube. Lab station and equipment were cleaned. Observations: I observed that color change occurred and turned blue when Copper (II) nitrate and sodium hydroxide were stirred. Also precipitate was formed. When the test tube was placed in the water bath color change occurred again and turned black. Also there was a temperature change occurred, then the 2 substances on the test tube separated and the leftovers dissolved. When the aluminum wire was placed in the test tube bubbles were produced. Within 5 min. copper atoms formed and were the same color as copper but different shape. Analysis: Some causes of chemical changes are combining chemicals and adding energy. Two ways that energy is involved in chemical change is when you touch the bottom of the beaker to check the temperature. That involves chemical change. Also when we put the aluminum wire; energy was involved to produce the bubbles. Elements that were used or produced were Al, Cu, H2 and the compounds were Cu (NO3)2, NaOH, Cu (OH) 2, CuO, HCl, CuCl2, and AlCl3. Aluminum chloride is in the solution and you can recover it by evaporating the water. The color of solutions of copper compounds is copper color. Soluble: Copper (II) nitrate, sodium hydroxide, hydrogen chloride, copper (II) chloride, aluminum chloride. Insoluble: copper (II) hydroxide, copper (II) oxide, copper metal, aluminum metal. A) sodium nitrate+ copper (II) hydroxide B) copper (II) + water C) copper (II) chloride + water D) copper + aluminum chloride E) hydrogen + aluminum chloride Conclusion: The 4 type of observations that indicate when a chemical change has occurred is formation of a precipitate, color change, formation of a gas, evolution of energy. The type of reactions that I observed in this experiment could be useful in the recycling of copper because copper metal was used in the preparation of the original copper (II) nitrate solution. After several conversions copper metal was again recovered. Advantages: pure metal is obtained from a compound, and the reaction is done in one vessel. Disadvantages: the reaction takes time and may be expensive and the waste products of recovery may cause pollution. Further Questions: 1) If you recover aluminum chloride, would there be a color change? 2) Why did the substances produce copper atoms? 3) How do you know if evolution of energy has occurred?

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