...the region of wavenumbers greater than 1500 cm-1 . Look for likely functional groups keeping in mind, the degree of unsaturation and molecular formula. e.g, if there is one double bond and an oxygen, it could be a carbonyl; if there is no double bond and an oxygen, it will be either an ether or an alcohol; if there are 4 double bonds, there might be an aromatic ring. Step 3: See the PMRand 13C NMR spectrum.Check that the functional groups infered by IR and NMR are consistent or not. Step 4: Draw possible structures consistent with the unsaturation, IR, and NMR data. Check...
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...Sloper CVC The Customer Value Company 48 St Mary's Road, Long Ditton, Surrey KT6 5EY, United Kingdom Tel. +44 7768 861920; e-mail: andrew.sloper@customervalue.co.uk February 2005 Ashridge Business School UK - http://www.ashridge.org.uk The Impact of Project Portfolio Management on Information Technology Projects Abstract The ever-increasing penetration of projects as a way to organise work in many organisations necessitates effective management of multiple projects. This has resulted in a greater interest in the processes of project portfolio management (PPM), with more and more software tools being developed to assist and automate the process. Much of the early work on PPM concentrated on the management of IT projects, largely from the perspective of the management of resources and risk. Many of the recent articles have been by vendors of the software, promoting the value of the PPM process. However, the claims made in those articles are typically only supported by anecdotal evidence. In this paper, we assess whether there...
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...Torres 1 Luis A. Torres Group #11 USC Chemistry 322b Formal Lab Report 6th November 2015 I. II. Enzymatic Resolution of 1-Phenylethanol and Diastereomer Analysis Objective/Abstract Enzymatic transesterification reaction was performed to study the resolution of diastereomers using 1H-NMR analysis. The stereo-selectivity of acylase I, an enzyme, for a 50:50 racemic mixture of 1-phenylethanol was determined. In the first of a two-step reaction, 1-phenylethanol was reacted with vinyl acetate with the help of acylase I to form an ester, unreacted 1-phenylethanol, and vinyl alcohol. The unreacted 1phenylethanol was separated from the ester by column chromatography and confirmed by thin-layer chromatography (TLC). In the second reaction, the unreacted 1-phenylethanol was reacted with (R)-(-)-acetoxyphenylacetic acid to form a diastereomer ester. In the latter reaction, four different 1-phenylethanol samples were used in order to compare 1HNMR data of the resulting diastereomer esters and determine which enantiomer of the 50:50 racemic mixture was preferred by acylase I. Those four samples were: (1) racemic 1-phenylethanol, (2) unreacted 1-phenylethanol, (3) (R)-1-phenylethanol, and (4) (S)-1phenylethanol. After 1H-NMR analysis, it was found that the (S)-1-phenylethanol was preferred by acylase I. III. Introduction and Background Information Scheme 1: Reaction of 1-phenylethanol with vinyl acetate in the presence of the enzyme acylase I to produce 1-phenylethyl...
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...increase. I also predicted that there would be an upper temperature limit. The experiment proved part of my hypothesis correct and part of it incorrect. I was correct in predicting that the rate would increase as the temperature increase, but I was incorrect in hypothesizing that there would be an upper limit. My hypothesis for the larvae experiment was that the level of aerobic respiration would increase as the temperature increased and there would be an upper temperature limit. My hypothesis was proved to be correct, for the rate of respiration continued to increase until the upper limit was hit and the rate started to decline. There were no unexpected results from our group, but I noticed that group six had some unexpected data in the class larvae table. The only explanation I can make out of it is that the group made a math error. I do not have any suggestions to make this experiment better. Other factors that can affect the rate of cellular respiration are amount of available nutrients, because this allows more energy to be produced from the cell with an increase in amount of nutrients. Another factor is the state of the cell, such as the difference between working and dormant cells or the difference between plant and animal cells. From our data, we noticed that the temperature had an effect on the germinated Pisum sativum seeds. As the temperature increased, the rate of cellular respiration was doubled. At 45 degrees, the rate still increased, but it just barely...
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...Advance Project Management PM587, Week 3 quiz Keller Graduate School of Management Week 3 quiz covers material from Weeks 1 2 and 3 Grading Summary These are the automatically computed results of your exam. Grades for essay questions, and comments from your instructor, are in the "Details" section below. Date Taken: 7/28/2012 Time Spent: 58 min , 25 secs Points Received: 72 / 80 (90%) Question Type: # Of Questions: # Correct: Essay 4 N/A Grade Details - All Questions 1. Question : (TCO A) You work for company Bravo and have been asked to create a presentation for the Board of Directors to explain PPM. Your presentation is to include a comparison and contrast of project portfolio management, program management, and project management. The presentation should also include the benefits of PPM and why the company should implement this program. Below, in your own words, write the narrative for the presentation. Be sure to address all elements of the presentation. Student Answer: Bravo Board of Directors Presentation PPM A PPM stands for Project Portfolio management. It is a management process used to develop a series of projects that are grouped into a portfolio based upon similar strategic objectives of the organization. The Project Portfolio management does not actually manage the project; however the management group creates guidelines and processes, which are then enforced to select the proper projects that will allow the organization...
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...which gives peaks at 1073.94 cm-1, 1330 cm-1, 1515.05, 1600.13 cm-1, 1708.25 cm-1, 2930.82 cm-1and broad peak at 3418.10 cm-1. The mass of the product is 0.013 grams which gives a percentage yield of 29.81%. The melting point of the product is not taken due to minimal product. In the second part of the reaction excess reagents are used and the synthesized product is in very small quantities. The product synthesized is diastereomers of 1-(4-nitrophenyl)-3-oxobutyl 3,3,3-trifluoro-2-methoxy-2-phenylpropanoate. The identity of this compound is confirmed by the following peaks seen in 1H NMR: 8.19 and 7.62 ppm, 5.47 ppm, 2.90, 2.13 ppm, 3.30 ppm, 7.36 – 7.38 ppm. The melting point, IR spectrum, theoretical yield or percentage yield is not found since all of it used in 1H NMR analysis. The product 1H NMR shows a mix of both the diastereomers, but it is difficult to ascertain which one is in excess. Introduction The aldol reaction that was experienced in this lab is the nucleophillic addition of an enolate to a carbonyl group to form a B-hydroxycarbonyl. This reaction is a very powerful method to construct of carbon-carbon bonds. Nature has developed the aldolase enzymes that catalyze biological aldol reactions. Aldolase is the enzyme that produces only one enantiomeric product that is hard to achieve. A simple amino acid L-proline can be used as the aldolase catalyst in asymmetric aldol reaction of acetone with a variety of aldehydes. This method produced 68% and 76% of...
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...ring Hydrogen that is directly adjacent to 1 Hydrogen on the aromatic ring, and that single Hydrogen that causes the initial doublet coupling is surrounded by 2 Hydrogens which can allow that first Hydrogen to couple over 4 bonds and not just 3. The next signal is at about 7.48 ppm and has a multiplicity as a triplet of doublets and integrates to 1 Hydrogen. This is indicative of an aromatic ring Hydrogen that is adjacent to 2 other Hydrogens on the aromatic ring. Of those 2 Hydrogens one is adjacent to two Hydrogens and the other is adjacent to a single Hydrogen. This signal also presents two types of coupling. The third and fourth signals overlap, but the first signal at 7.35 ppm which has a multiplicity of a triplet and integrates to 1 hydrogen, is indicative of a Hydrogen on an aromatic ring that is adjacent to two Hydrogens. The next signal is at 7.28 ppm and has a multiplicity of a doublet which is indicative of an aromatic ring Hydrogen that is adjacent to only a single other Hydrogen on the aromatic ring. The next signal is at 2.97 ppm with a multiplicity of a triplet and integrates to 2 hydrogens, which is indicative of a methine group that adjacent to 2 hydrogens, most likely another methine group. The next signal is at 2.65 ppm with a multiplicity of a...
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...According to the Iodoform test which tests for methyl ketones, A345 turned cloudy in the test tube while the B192 did not have any color change, and A345 is concluded to be the ketone. The chemical formula of A345 is C10H9O. In the A345 HNMR, there are 2 methyl groups present, indicated by the two singlets with an integration of about 3 at around 2.42 ppm and 2.58 ppm. The compound is UV active, and because the degrees of unsaturation is 5, there is a benzene ring. The chemical formula indicates that there is a total of 10 hydrogens, and there are two doublet signals on the HNMR that indicate that there is an integration 2 at 7.86 ppm and 7.26 ppm (the singlet signal at 7.26 ppm is chloroform), which means that the methyl group splits the benzene in half due to its splitting and integration. So the inferred structure of the liquid starting material A345 is a 4-methylacetophenone. The B192 chemical formula is C7H5ClO. In the B192 HNMR, there is a singlet with an integration of 1 at 9.99 ppm that correlates to the aldehyde...
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...1. What ppm values (and standard deviations) did you determine for each type of water? Buxton hall’s water had 42 ppm of CaCO3 present in all three trials conducted. The standard deviation from this test was 0. Kroger water had a 46.2 ppm of CaCO3 in the first trial, 50 ppm in the second trial, and 43.2 ppm in the third trial. The standard deviation for the second water sample was 3.41. Crystal Geyser water had a 16.42 ppm of CaCO3 in the first trial, 21 ppm in the second trial, and 22 ppm in the third trial. The standard deviation for the third sample was 2.97. 2. According to the table provided in the Introduction section, what is the water hardness of each of your samples? According to the table provided, 0-50 CaCO3 ppm is considered...
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...LastName_Lab1 (e.g., Smith_Lab1). * You should submit your document in a Word (.doc or .docx) or Rich Text Format (.rtf) for best compatibility. Exercise 1: Data Interpretation Table 1: Water Quality vs. Fish Population Dissolved Oxygen (ppm) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | Number of Fish Observed | 0 | 1 | 3 | 10 | 12 | 13 | 15 | 10 | 12 | 13 | (hint: ppm stands for “parts per million”) 1. What patterns do you observe based on the information in Table 1? Ans. The quality of fish increase as (ppm) increase to a point while the quality of water continue to increase 2. Develop a hypothesis relating to the amount of dissolved oxygen measured in the water sample and the number of fish observed in the body of water. Ans: If oxygen or ppm in the water increase more fish will be produce 3. What would your experimental approach be to test this hypothesis? Ans. My experimental approach will be gather data, by doing background research, and test experiment by having a independent, and dependent variable. I will analyze my result and make conclusion 4. What are the independent and independent variables? Explain why each was chosen. Ans. My dependable variable will be water with high level of (ppm) which will increase...
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...| Question Type: | # Questions: | # Correct: | Multiple choice | 3 | 3 | Short | 3 | N/A | | Grade Details | | | | 1. | Question: | (TCO A) Which of the following statements about Project Portfolio Management is NOT true? | | Your Answer: | | | The PPM process will force organizations to abandon the preparation of traditional business plans and analyses. | | | | | To keep from falling into the error of accepting faulty assumptions and data, everyone involved in PPM needs to ask hard questions before deciding. | | | | | PPM is still evolving. | | | | | PPM begins with a rational prioritization and selection procedure. | | | | 2. | Question: | (TCO A) Which of the following groups of people are not involved in the implementation of the PPM process? | | Your Answer: | | | Senior management | | | | | Functional managers of the IT department | | | | | Members of the PPM governance council | | | | | The project management office | | | | 3. | Question: | (TCO E) Which of the following is a step in the controlling process? | | Your Answer: | | | Motivating | | | | | Delegating | | | | | Evaluating | | | | | Staffing | | | | 4. | Question: | (TCO E) State the five effective group fundamentals that any team should follow. Taking TWO of the five fundamentals...
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...Available online at www.sciencedirect.com International Journal of Project Management 30 (2012) 608 – 620 www.elsevier.com/locate/ijproman The three roles of a project portfolio management office: Their impact on portfolio management execution and success Barbara Natalie Unger a,⁎, Hans Georg Gemünden a , Monique Aubry a b b Technische Universität Berlin, Chair for Technology and Innovation Management, Straße des 17. Juni 135, Sekr. H71, 10623 Berlin, Germany Université du Québec à Montréal, School of Business and Management, P.O. Box 8888 Downtown Station, Montreal, Quebec, Canada H3C 3P8 Received 27 July 2011; received in revised form 29 November 2011; accepted 26 January 2012 Abstract Project portfolio management offices (PPMOs) are a subset of project management offices (PMOs) that handle collections of multiple single projects and programmes, i.e. portfolios. PPMOs are centralised organisational units that cater to the demands of various stakeholders by performing specialised tasks. They are initiated by their organisation's leadership in response to increasing management challenges originating from project portfolios. Although there has been considerable research on PMOs in general, not only a clear understanding of multi-project PMOs' activity patterns set in specific contexts like project portfolio management, but also both existence and mode of multi-project PMOs' contribution to successful performance are still lacking. By quantitatively analysing PPMOs in 278 portfolios...
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...Belgrade, Serbia Received 31 July 2007; received in revised form 26 October 2007; accepted 4 November 2007 Available online 17 November 2007 Abstract Carbofuran toxicity on rats was studied during subchronic exposure. Female and male rats were administered carbofuran in drinking water in concentrations of 25, 100 and 400 ppm for a period of 90 days. Clinical symptoms, water consumption, body weight gain, organ weight, pathological and histopathological changes in the liver and kidneys were observed and biochemical and haematological examinations were carried out. The results obtained show that carbofuran administered to rats caused a significant decrease in water consumption as well as in brain, serum and erythrocyte cholinesterase activities. Statistically significant increases in relation to the control were found in the serum enzyme activities. The haematological data showed that carbofuran had no significant effect on Hb concentration and total RBC, but total WBC showed a significant statistical decrease. The histopathological changes in liver and kidneys were observed. However, cell regeneration in the liver and kidneys was found in all test groups. © 2007 Elsevier B.V. All rights reserved. Keywords: Carbofuran; Water; Rats; Toxicity 1. Introduction Carbofuran is a commonly used carbamate insecticide, nematicide and acaricide effective by contact, stomach and systemic action (Kuhr and Dorough, 1979; Machemer and...
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...Introduction to the Chemical Analysis of Water Samples Each group will collect a fresh water sample from campus according to directions provided form your instructor or TA. You will learn how to use several pieces of equipment and several chemical kits in order to analyze the water sample. Each group may split up the tasks however it sees fit- however, each person in the group is expected to be able to complete any and all of the tests performed, with directions of course! Make sure that you answer the question on the last page of this handout as you work through the lab. You will need to share your answers with the others in your group. Test to Perform | Equipment or Kit Used | pH | Smart Colorimeter | Turbidity | Smart Colorimeter | Conductivity | Vernier LabQuest 2 and Conductivity probe | Ammonia Nitrogen | Smart Colorimeter | Nitrate Nitrogen | Smart Colorimeter | Phosphates | Smart Colorimeter | Dissolved Oxygen | Smart Colorimeter | Fecal Coliform | Coliscan EasyGel | To Begin: You have several small beakers and transfer pipettes on your table. You will need to transfer some of your water samples into these beakers, or other test tubes (found in the kits) in order to complete these tests. Do not put the pH meter or the conductivity probe directly into the large beaker of sample water- this could contaminate the entire sample, thus affecting the results of your other tests. The instructions for the Smart Colorimeter tests are in a...
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...antiretroviral therapy = three-drug cocktail treatment for AIDS OTC means over the counter SSRI = Selective serotonin reuptake inhibitors CFC = Chlorofluorocarbon 1% = 104 ppm = 107 ppb conversion N2 (g) + 3 H2 (g) → 2 NH3 (g) molar mass of ethanol (C2H5OH) = 46.06 g/mol d= st m/s km/hr 1 cm3 =1 ml Micro =1 µ g = 1 x 10 -6 g NANO =1 n g = 1 x 10 -9 g The largest vol of these is 1.0 L 1000 mL 1.0 × 106 µL >>1.0 × 104 cm3 Chemicals causing malformation in newborns are: Teratogenic Chemicals causing changes in DNA are known as: Mutagens A concentration of 21.0 ppm for a gas in air is equal: 2.10 × 10-3 % A concentration of 51.0 ppb for a gas in air is equal: 5.10 × 10-2 ppm Based upon TLV-TWA which poses the most risk: ( answer is the least) Sulfur dioxide (2 ppm) Carbon monoxide (2.5 × 104 ppb) >>Ozone (100 ppb) Nitrogen dioxide (3.0 × 10-4 %) A worker is repeatedly exposed to a concentration of 2.50 × 10 6ppb of a substance. If the TLV-TWA for the substance is 250 ppm, is it a safe work environment for the worker? Yes No (why? 2.50 × 10 6ppb = 2500 ppm > 250 ppm) A worker is repeatedly exposed to a concentration of 2.5 × 10-4 % of a substance. If the TLV-TWA for the substance is 5.0 ppm, is it a safe work environment for the worker? >>Yes No The unit used...
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