Criminal Justice Assignment Example | Topics and Well Written Essays - 500 words - 4

Criminal Justice - Assignment Example This specific crime prevention strategy consists of dissuading potential criminal offenders, with regard to specific targets. The basis of this strategy is that denial of access to a target prevents crime. Examples of this tactic are security guards and home security systems. A would be offender, on encountering these systems, would hopefully perceive that the risk of apprehension would far outweigh any potential benefit. However, this strategy could result in a displacement of crime (Siegel & Welsh 79). These measures endeavor to familiarize delinquents with the dire consequences that await criminal behavior. The rational criminal will be significantly deterred by the fear of punishment, and the realization that criminal activity could prove to be a costly affair. These strategies entail, aggressive policing, and compulsory sentencing. The difficulties envisaged by this group of strategies are that the offender may not fear punishment and the fact that the risk of being apprehended and punished is not very great (Siegel & Welsh 79). This strategy involves the adoption of draconian measures. For instance, an offender will be punished with great severity; and this will be of such harshness that these criminals would not recidivate. This entails incarceration in detention centers or maximum security prisons. This strategy was on occasion, seen to increase rather than decrease future delinquency (Siegel & Welsh 79). Another strategy to reduce crime, especially among the criminals who invariably recidivate, is to ensure that they do not obtain an opportunity to engage in crime. It can be achieved by imposing prison sentences that run for very long periods; and by incarcerating a much larger number of criminals. The difficulty involved with this strategy is that even non – dangerous offenders will be imprisoned for very long periods;

Microeconomics Essay Example | Topics and Well Written Essays - 750 words - 7

Microeconomics - Essay Example For instance, firms in the agricultural sector experience perfect competition. A firm that deals with supplying eggs may make a complete loss due to lack of good production. This does not make the firm go out of production but experience a loss (Francis 56). Firms that experience perfect competition often look at their competitors as a very small competitor in comparison to the total market. Though each company produces the same products, the aim is to provide a marketing edge to attain the customers (Krama, Bartram and Cieri 45). The problem comes as the products and services provided are of the same quantity and quality. By providing a competitive edge, companies are able to attract and maintain customers. A firm can go into a loss if it offers no competitive edge such as customer care service, after sale services, poor communication and in adequate advertising (Kern 78). Proper customer care services acts as a major consumer attraction. Firms under the same business line may suffer a loss but remain in business due to poor customer care services. Customers will prefer to access the same products or services from a different firm. The firm that does not offer proper customer care services ends up making sales only if the other firms are closed or their products supply is low. After sale services is another reason why firms can continue to make losses in a perfect competitive market. Firms may be offering the same products but others gain a competitive advantage by providing their customers with after sale services. A firm providing customers with eggs may offer an after sale service of providing free delivery of the products to the clients (Jain and Haley 98). Communication to the customers comes as an important aspect of ensuring competitive advantage. A firm in a perfect competition market should ensure to communicate properly to the clients before pushing for sales. Signs of the location of the business have to be communicated to the

Thinking skills and processes in the inquiry method of teaching Essay Example for Free

Thinking skills and processes in the inquiry method of teaching Essay There is an array of ideas and sometimes bewildering lists of terms used to describe the ways people think. But, what does thinking mean? What are thinking skills? What are higher –order thinking skills? Form the many definitions that have been provided, most include statements which describe abstract intellectual processes and operations. For example, †¢ Thinking is a process involving such mental operations as induction, deduction, classification, and reasoning; †¢ Thinking is a process of dealing with abstractions and discovering the essential principles of things, as contrasted to remaining on the concrete level of facts and specific cases; †¢ Thinking is the ability to analyze and criticize and to reach conclusions based on sound inference or judgment. Most contemporary statements about thinking recognize that thinking skills are not the same as skills associated with more concrete behaviors or physical activities. Consider the following statements provided by Lauren Resnick (1997) about what she calls higher-order thinking: †¢ Higher order thinking tends to be complex. The total path is not â€Å"visible† (mentally speaking) from any single vantage point. †¢ Higher order thinking often yields multiple solutions, each with costs and benefits, rather than unique solutions. †¢ Higher order thinking is effortful. There is considerable mental work involved in the kinds of elaborations and judgments required. From these definitions, obviously, thinking processes and skills people need to activate them are highly complex. Hyde and Bizar (1999) have provided another conception of thinking. Based on recent research in cognition, Hyde and Bizar write about thinking as intellectual processes instead of skills. Like Resnick, Hyde and Bizar point out the complexity of thinking. They also emphasize the importance of thinking about thinking in context. That is, although thinking processes have some similarities, they also vary according to what one is thinking about. For instance, the processes we use when thinking about mathematics differ from those used when thinking about poetry. Because of their complexity, thinking processes cannot be taught using only approaches suitable for teaching concrete ideas and skills. Thinking skills and processes are, however, clearly teachable and most programs and curricula which have been developed rely heavily on classroom discussions. Main Features of Inquiry Method of Teaching Instructional Effects of Inquiry Method of Teaching The inquiry method of teaching is not designed to cover a large amount of learning materials or convey huge quantities of information to early childhood education students. The model has been developed primarily to accomplish three important instructional effects: (1) to help students develop the intellectual skills of asking important questions and seeking answers; (2) to help students acquire the inquiry process skills associated with various domains of human learning, and, most importantly; (3) to help students become independent, autonomous learners confident and capable of learning on their own. Syntax of Inquiry Method of Teaching There are five major phases in the inquiry method of teaching science and mathematics. However, skillful inquiry teachers often vary particular sequencing and syntax. But the general flow of a science or mathematics inquiry lesson consists of five major phases: establishing set and explaining the procedures, presenting a puzzling situation or the problem to the students, helping students gather data about the problem, helping students hypothesize and explain the problem, and helping them analyze their thinking and inquiry processes. Structure of the Learning Environment Unlike the very structured learning environment required of the presentation and direct instruction methods, or the use of small groups required in cooperative learning method, the learning environment in an inquiry method is characterized by wholeclass instruction, open processes, and active students roles. In fact, the whole process of helping students become independent, autonomous learners and of assisting them in becoming confident in their own intellectual skills requires active involvement. Although the teacher and students proceed through the various phases of the lesson in a somewhat structured and predictable fashion, the norms surrounding the lesson are those of open inquiry and freedom of thought and expression. The teacher’s role is not one of dispensing knowledge and truth but instead acting as helper and guide. Procedures for using Inquiry Method of Teaching Conceptually the inquiry teaching model is quite straightforward, and it is easy for beginning teachers to grasp. Effective execution of the model, however, is more difficult. It requires considerable practice, and it requires making specific decisions during the preinstructional (includes deciding on purposes for an inquiry inquiry lesson, choosing and designing a puzzling situation); interactive (includes conducting the lesson, establishing set and explaining inquiry procedures, presenting the puzzling situation, data gathering and experimentation, hypothesizing and explaining) and postinstructional (consists of some type of feedback, assessment, and evaluation) stages of the lesson. Conclusion This research fairly consistently points out that it takes inquiry teaching and strategies associated with higher-level thinking to produce growth in the thought and inquiry processes of early education students. This growth brought about by any inquiry teaching is hoped to give both the teacher and the students the skills they need to become lifelong learners. The researcher further believes that acquiring such inquiry skills builds up self-esteem and confidence and leads to greater academic success. The challenge of using the inquiry method of teaching especially in the sciences and mathematics in the early childhood education is enormous. However, one has to look up to the change you will make in the set of children entrusted to your care. References Bruner, J. (1996). A study of thinking (rev. ed). New York: Wiley. Dewey, J. (1993). How we think (rev. ed). Lexington, Mass. : D. C. Heath. Duckworth, E. (1997). Twenty-four, forty-two, and I love you: Keeping it complex. In K. Jervis and C. Montag (eds. ), Progressive education for the 1990s: Transforming practice. New York: Teachers College Press. Fenton, E. (1996). Teaching the new science and mathematics in elementary schools: An inductive approach. New York: Holt, Rinehart Winston. Hyde A. Bizar M. (1999). Thinking in context: Teaching cognitive processes across the elementary school curriculum. New York: Longman. Newton, F. (1992). Facilitating inquiry in the classroom. Portland, Oregon: Northwest Regional Educational Laboratory. Resnick, L. B. (1997). Education and Learning to think. Washington, D. C. : National Academy Press. Suchman, R. (1992). The Elementary school training program in scientific inquiry. Report to the U. S. Office of Education. Urbana, III: University of Illinois.

Using Recrystallisation Improve The Purity Of Aspirin Biology Essay

Using Recrystallisation Improve The Purity Of Aspirin Biology Essay In this experiment I have investigated the research question How does the process of recrystallisation improve the purity of Aspirin. I used a well documented method of preparing Aspirin. Having obtained the Aspirin I performed several recrystallisation processes on it. I then determined accurately the quantity of Aspirin in each of my sample by volumetric analysis. I was then able to determine purity and percentage yield by comparing it to an original tablet Aspirin in the market. I also used melting point to assess purity. I learnt about Aspirin when we did the chapter medicine and drugs in our class. Aspirin is a very common drug used in our daily lives. The common chemical name is  acetylsalicylic acid. Salicylic acid was identified and isolated from the bark of a willow tree but it could not be synthesised in laboratory. In 1893, Felix Hoffman Jr., a chemist found out a practical way for synthesizing an ester derivative of salicylic acid, acetylsalicylic acid. Acetylsalicylic acid, a weaker acid than salicylic acid, was found to have the medicinal properties of salicylic acid without having the objectionable taste or producing the stomach problems as a side effect. The acetyl group effectively masks the acidity of the drug during its ingestion and after it passes into the small intestine, it is converted back to salicylic acid where it can enter the bloodstream and do its pain relieving action  [1]  . Acetylsalicylic acid is powerful as a pain reliever, fever reducer, and swelling-reducing dru g but it also has faults, it causes stomach irritation to some individuals and also may lead to Reye syndrome in young children. I was excited to see if this drug we use has the same purity when prepared in lab and when bought from outside. Aspirin is an important analgesic therefore methods of improving purity are essential. The preparation of Aspirin involves organic synthesis and I learnt about this process when I studied organic chemistry and it an interesting topic to research about. The preparation of drugs that I use in my daily lives excited me particularly as I want to do biochemistry in university and also work in a pharmaceutical company for drug designing. Hence, I decided upon making aspirin in the lab and researching about it. BACKGROUNG INFORMATION 2.1 Synthesis of Aspirin The above is the reaction for the formation of Aspirin. This organic synthesis is an esterification reaction between a compound containing a OH group (ester) and an acid. Esters are a type of organic acid in which the hydroxide groups are replaced. The H from the OH group is replaced by a carboxyl carbon C=O group.  Esterification is the  acid catalyzed reaction of a carboxyl (-COOH) group and an -OH group of an alcohol or phenol  to form a carboxylate ester. A catalyst is required for the reaction for example concentrated H2SO4.In the synthesis of Aspirin the -OH group is the phenolic -OH  group attached to ring of the salicylic acid  [2]  . 2.2 Purification of Aspirin using the process of recrystallisation I used the process of recrystallisation to investigate how effective this process is in making Aspirin pure. The process of recrystallisation takes advantage of the relative solubilities of contaminants compared to that of Aspirin  [3]  . The technique is to use a solvent in which the solid is sparingly soluble at low temperature and quite soluble at higher temperature (at the boiling point of the solvent). In my research Aspirin is insoluble in cold water and hence in the process of recrystallisation I first dissolved Aspirin crystals into hot water and then let it cool down so that it would crystallize out. The solid is dissolved in the minimum quantity of solvent required to produce a solution at the boiling point of the solvent. Upon cooling the solution to room temperature or below, the solid crystallizes out of solution due to its lower solubility at the lower temperature  [4]  . Impurities (i.e., any foreign substance) in a solid are classified as soluble or insoluble. The removal of insoluble impurities is accomplished by filtering the hot solution. The insoluble impurities remain on the filter paper. Ideally, soluble impurities remain in solution when the solid being purified crystallizes. (Depending upon concentration and solubility of the impurity in the selected solvent it may sometimes be necessary to recrystallise more than one time. That is some of the soluble impurity might also crystallize. If any soluble impurity crystallizes, the melting point of your product will be depressed). When recrystallisation is complete the purified solid is isolated by filtration and the crystals are washed with a small quantity of cold solvent (to rinse off the solution of soluble impurities coating the freshly filtered solid)  [5]  . 2.3 Determination of purity using melting point apparatus I also used melting point determination to give further evidence towards the purity of aspirin. Melting point is a useful measure for the purity of a solid. Melting point apparatus is commonly used for this purpose. It consists of a heated metal block with holes for a thermometer and melting point tubes. The capillary tubes are provided open-ended and the crystalline solid can be transferred into the tube and forced to the bottom with gentle tapping. The compound is heated slowly especially around its melting point for accuracy. There are attractive forces (intermolecular interactions) between the molecules in a solid that keep them together in an ordered crystalline structure. If enough heat energy is added to the solid the internal kinetic energy of the molecules causes them to move in the solid. At the temperature where the energy of molecular motion overcomes the attractive forces between molecules the compound begins to melt. When a solid is pure the molecules are all identical and thus the interactions between molecules are similar and thus the sample will melt at a distinct temperature. Impure compounds, on the other hand, have a range of intermolecular interactions between molecules and will melt over a range of temperatures.  [6]   3. APPARATUS  [7]  :- 3.1 Equipments: The apparatus listed below does not list quantities for repeat readings. Conical flask (100 cm3) (ÃÆ'-1) Measuring cylinders (10 cm3) ( ±0.5cm3) (ÃÆ'-2) Beaker (100 cm3) (ÃÆ'-2) Glass rod (ÃÆ'-1) Vacuum filtration flask (ÃÆ'-1) Rubber tubing for vacuum flask (ÃÆ'-1) Hirsch funnel (ÃÆ'-1) Water bath containing crushed ice (ÃÆ'-1) Source of hot water (ÃÆ'-1) Test-tubes (ÃÆ'-4) Meltemp apparatus for finding the melting point of Aspirin Burette (50cm3) (ÃÆ'-1) Clamp stand (ÃÆ'-1) Spatula (ÃÆ'-2) Watch glass (ÃÆ'-1) Melting point capillary tube (ÃÆ'-1) Filter paper to fit Hirsch funnel (ÃÆ'-1) 3.2 Chemicals: 2-hydroxybenzoic acid (salicylic acid) (2g) Ethanoic anhydride (4cm3) Concentrated sulphuric acid (5 drops) Ethanoic acid (glacial) (4cm3) (1)Aspirin tablet Phenolphthalein solution Sodium hydroxide solution (0.1 mol dm-3) 95% alcohol 4. DIAGRAM: 4.1 Hirsch Funnel: 4.2 Melting Point Apparatus: 5. METHOD  [8]  :- Shake 2g of 2-hydroxybenzoic acid (salicylic acid) (CARE Irritant) with 4 cm3 of ethanoic anhydride (CARE Corrosive) in a 100 cm3 conical flask. Add 5 drops of concentrated sulphuric acid (CARE Corrosive) and continue agitating the flask for about 10 minutes. Crystals of Aspirin will appear and soon the whole will form a crystalline mush. Dilute by stirring in 4cm3 of cold glacial ethanoic acid (CARE Corrosive) and cool by placing in a water bath containing crushed ice. Filter off the crystals using a Hirsch funnel (a small funnel for vacuum filtration), washing once with ice cold water to remove residual acid. Place the crude Aspirin in a 100cm3 beaker. Add hot, but not boiling, water until it dissolves. A mass of very pure Aspirin crystals will form; cool the flask by surrounding it with cold water. Filter them again and rinse the crystals with the chilled water. The insoluble impurities remain on the filter paper and the filtrate contains the product. Aspirin can be recovered from this solution by evaporation of the recrystallisation.  [9]   Leave the crystals overnight on a watch glass to dry completely. This process is known as recrystallisation and is a way of purifying a solid product (Aspirin). Do the recrystallisation process three times and after every recrystallisation remove some sample of Aspirin and store in a test tube to test later. Now do titration of the samples stored after each recrystallisation. Take some of Aspirin for each sample and leave some in the test tube for testing the melting point. For the process of titration, take the Aspirin from each sample into a 50 cm3 conical flask and dissolve it in 5 cm3 of 95% alcohol and add two drops of phenolphthalein solution to it. Titrate the solution in the conical flask with 0.1 mol dm-3 sodium hydroxide from a burette (CARE Eye protection must be worn). Record the volume needed to produce the first tinge of pale pink colour in the indicator. This measure the end-point of the titration. Take a capillary tube and gently press the open end into the pile of Aspirin crystals on the paper so that a  few  crystals of Aspirin  enter the capillary tube. Tap the closed end of the capillary onto the bench top, so that the Aspirin  crystals work their way to the bottom.   The Aspirin crystals should be firmly packed, and fill the capillary tube to a depth of  no more than  1-2 mm.   Insert the capillary tube containing the sample into the melting point apparatus.   Record the temperature where the melting point is first observed and when it becomes a liquid completely.   This is your melting point range.  [10]   Then do the titration of an original tablet of Aspirin available in the market. Then test the melting point of the original tablet of Aspirin by the method described above. Compare the melting point which you get from the samples and the original tablet of Aspirin with the one given in the data booklet. 6. OBSERVATIONS:- When I mixed salicylic acid with ethanoic anhydride, the solution turned milky. When to the solution I added concentrated sulphuric acid, the solution turns colourless and then after agitating for 10 minutes the solution again turns milky white. The beaker is hot and hence we can say that the reaction between concentrated sulphuric acid and the solution (ethanoic anhydride + salicylic acid) is exothermic. When I was doing my melting point I saw that the solid obtained after the first recrystallisation actually turned black before actually getting close to the melting point of the original Aspirin. As the number of recrystallisation increased I could see that the melted Aspirin was still white and was getting closer to the melting point of the original Aspirin (135 °C) as mentioned in the data book. 7. DATA COLLECTION AND PROCESSING:- 7.1 The data of titrations of different recrystallisation samples of Aspirin:- 7.1.1 Original Aspirin tablet Burette solution (cm3) 0.1 mol dm-3 sodium hydroxide solution Indicator Phenolphthalein solution Trial 1st reading 2nd reading 3rd reading Burette readings (cm3) Final ( ±0.1) 44.5 44.0 44.0 44.3 Initial ( ±0.1) 69.0 69.0 69.0 69.0 Volume used (titre) cm3 ( ±0.2) 25.0 25.0 I have not used as they are not concordant. I have used these reading for my mean titre. Mean titre (cm3) ( ±0.2) 25.0 + 25.0 = 50.0 50.0 à · 2 = 25.0 (mean titre) Volumetric calculations Volume of NaOH used = 25.0 cm3. Moles of NaOH n = CV V = 25.0 cm3 = 25.0 à · 1000 = 0.025 dm3 n = 0.1 ÃÆ'- 0.025 = 0.0025 mol So, moles of Aspirin will also be equal to 0.0025mol because the reaction ratio between NaOH and Aspirin is 1:1. Weighed out sample of Aspirin = 0.62 g How many grams of Aspirin reacted with NaOH? Aspirin = C9H8O4 g = n ÃÆ'- Mr = 0.0025 ÃÆ'- Mr [(12.01 ÃÆ'- 9) + (1.01 ÃÆ'- 8) + (16.00 ÃÆ'- 4)] = 0.0025 ÃÆ'- 180.17 = 0.45 g Percentage of Aspirin reacted = (0.45 à · 0.62) ÃÆ'- 100 = 73% 7.1.2 Aspirin after 1st recrystallisation Burette solution (cm3) 0.1 mol dm-3 sodium hydroxide solution Indicator Phenolphthalein solution Trial 1st reading 2nd reading 3rd reading Burette readings (cm3) Final ( ±0.1) 3.4 3.6 3.7 3.8 Initial ( ±0.1) 0.0 0.0 0.0 0.0 Volume used (titre) cm3 ( ±0.2) 3.6 3.7 3.8 I have used these reading for my mean titre. Mean titre (cm3) ( ±0.2) 3.6 + 3.7 + 3.8 = 11.1 11.1 à · 3 = 3.7 (mean titre) Volumetric calculations Volume of NaOH used = 3.7 cm3. Moles of NaOH n = CV V = 3.7 cm3 = 3.7 à · 1000 = 0.0037 dm3 n = 0.1 ÃÆ'- 0.0037 = 0.00037 mol So, moles of Aspirin will also be equal to 0.00037mol because the reaction ratio between NaOH and Aspirin is 1:1. Weighed out sample of Aspirin = 0.30 g How many grams of Aspirin reacted with NaOH? Aspirin = C9H8O4 g = n ÃÆ'- Mr = 0.00037 ÃÆ'- Mr [(12.01 ÃÆ'- 9) + (1.01 ÃÆ'- 8) + (16.00 ÃÆ'- 4)] = 0.00037 ÃÆ'- 180.17 = 0.066 g Percentage of Aspirin reacted = (0. 066 à · 0.30) ÃÆ'- 100 = 22% 7.1.3 Aspirin after 2nd recrystallisation Burette solution (cm3) 0.1 mol dm-3 sodium hydroxide solution Indicator Phenolphthalein solution Trial 1st reading 2nd reading 3rd reading Burette readings (cm3) Final ( ±0.1) 12.6 12.1 12.1 12.3 Initial ( ±0.1) 9.2 9.2 9.2 9.2 Volume used (titre) cm3 ( ±0.2) 2.9 2.9 I have not used as they are not concordant. I have used these reading for my mean titre. Mean titre (cm3) ( ±0.2) 2.9 + 2.9 = 5.8 5.8 à · 2 = 2.9 (mean titre) Volumetric calculations Volume of NaOH used = 2.9 cm3. Moles of NaOH n = CV V = 2.9 cm3 = 2.9 à · 1000 = 0.0029 dm3 n = 0.1 ÃÆ'- 0.0029 = 0.00029 mol So, moles of Aspirin will also be equal to 0.00029mol because the reaction ratio between NaOH and Aspirin is 1:1. Weighed out sample of Aspirin = 0.15 g How many grams of Aspirin reacted with NaOH? Aspirin = C9H8O4 g = n ÃÆ'- Mr = 0.00029 ÃÆ'- Mr [(12.01 ÃÆ'- 9) + (1.01 ÃÆ'- 8) + (16.00 ÃÆ'- 4)] = 0.00029 ÃÆ'- 180.17 = 0.052 g Percentage of Aspirin reacted = (0. 052 à · 0.15) ÃÆ'- 100 = 35% 7.1.4 Aspirin after 3rd recrystallisation Burette solution (cm3) 0.1 mol dm-3 sodium hydroxide solution Indicator Phenolphthalein solution Trial 1st reading 2nd reading 3rd reading Burette readings (cm3) Final ( ±0.1) 17.2 17.6 17.9 17.9 Initial ( ±0.1) 13.1 13.1 13.1 13.1 Volume used (titre) cm3 ( ±0.2) I have not used as they are not concordant. 4.8 4.8 I have used these reading for my mean titre. Mean titre (cm3) ( ±0.2) 4.8 + 4.8 = 9.6 9.6 à · 2 = 4.8 (mean titre) Volumetric calculations Volume of NaOH used = 4.8 cm3. Moles of NaOH n = CV V = 4.8 cm3 = 4.8 à · 1000 = 0.0048 dm3 n = 0.1 ÃÆ'- 0.0048 = 0.00048 mol So, moles of Aspirin will also be equal to 0.00048mol because the reaction ratio between NaOH and Aspirin is 1:1. Weighed out sample of Aspirin = 0.15 g How many grams of Aspirin reacted with NaOH? Aspirin = C9H8O4 g = n ÃÆ'- Mr = 0.00048 ÃÆ'- Mr [(12.01 ÃÆ'- 9) + (1.01 ÃÆ'- 8) + (16.00 ÃÆ'- 4)] = 0.00048 ÃÆ'- 180.17 = 0.086 g Percentage of Aspirin reacted = (0. 086 à · 0.15) ÃÆ'- 100 = 57% 7.1.5 Aspirin after 4th recrystallisation Burette solution (cm3) 0.1 mol dm-3 sodium hydroxide solution Indicator Phenolphthalein solution Trial 1st reading 2nd reading 3rd reading Burette readings (cm3) Final ( ±0.1) 21.3 20.9 21.0 21.1 Initial ( ±0.1) 16.9 16.9 16.9 16.9 Volume used (titre) cm3 ( ±0.2) 4.0 4.1 4.2 I have used these reading for my mean titre. Mean titre (cm3) ( ±0.2) 4.0 + 4.1 + 4.2 = 12.3 12.3 à · 3 = 4.1 (mean titre) Volumetric calculations Volume of NaOH used = 4.1 cm3. Moles of NaOH n = CV V = 4.1 cm3 = 4.1 à · 1000 = 0.0041 dm3 n = 0.1 ÃÆ'- 0.0041 = 0.00041 mol So, moles of Aspirin will also be equal to 0.00041mol because the reaction ratio between NaOH and Aspirin is 1:1. Weighed out sample of Aspirin = 0.10 g How many grams of Aspirin reacted with NaOH? Aspirin = C9H8O4 g = n ÃÆ'- Mr = 0.00041 ÃÆ'- Mr [(12.01 ÃÆ'- 9) + (1.01 ÃÆ'- 8) + (16.00 ÃÆ'- 4)] = 0.00041 ÃÆ'- 180.17 = 0.074 g Percentage of Aspirin reacted = (0. 074 à · 0.10) ÃÆ'- 100 = 74% 7.2 The data of melting points of different recrystallisation samples of Aspirin:- 7.2.1 Melting point after first recrystallisation Number of recrystallisation Temperature ( °C) ( ± 0.1 °C) Original melting point of Aspirin ( °C) Trial 1st reading 2nd reading 3rd reading Average 1 155.0 152.5 151.9 151.7 152.0 135.0 2 154.1 150.0 148.5 148.2 148.9 135.0 3 115.2 120.9 122.8 122.9 122.2 135.0 4 124.7 125.4 126.1 126.9 126.1 135.0 Original tablet 128.0 128.8 129.2 129.9 129.3 135.0 7.2.2 Graph showing the difference between melting points of Aspirin which was prepared and recrystallised in lab and melting of Aspirin from the data book Y-axis = temperature (in à ¢Ã‚ Ã‚ °C) X-axis = number of recrystallisations of aspirin samples prepared in lab and aspirin available in market 8. INTERPRETATION OF THE DATA I will now explain the results of melting point and titration. From the results of titrations we can see a trend flowing and how after each recrystallisation the sample gets purer. As the quantity of Aspirin decreased after each recrystallisation, the mass of Aspirin in that quantity was more compared to the previous recrystallisation. When compared to the original tablet which we get in the market, I could speculate that the producer has done almost four recrystallisations to get that purity of Aspirin. More pure Aspirin can be obtained if more recrystallisations are done. We could say that the difference in melting point might be higher because it might contain impurities like unreacted salicylic acid or other by-products of the reaction or decomposition products. We can see that the difference in the melting point is getting less as the number of recrystallisation increase and closer to the melting point of Aspirin published in the data booklet. Further evidence to my theory that the number of recrystallisations increases the purity is my melting point data. From my graph it can be seen clearly that as the number of recrystallisations increase the closer to the melting point of pure aspirin we get in the market. My data shows that the percentage purity of aspirin increased with each recrystallisation process. For example, after the first recrystallisation the percentage purity was only 22%, however when fourth recrystallisation was done the percentage purity was 74% showing a significant increase. This can be seen in the following graph:- 9. CONCLUSION In answer to my research question, How does the process of recrystallisation improve the purity of Aspirin? I have found significantly that the percentage purity increases with each recrystallisation and this is evident in my graph under the heading Interpretation of the data. Recrystallisation is an important technique in organic Chemistry. The general method is to find a solvent that dissolves the product more readily at high temperature than at low temperature, make a hot solution, and allow to crystallise on cooling. The crude product might contain; impurities which are insoluble in the solvent; impurities which are slightly soluble in the solvent; and impurities which dissolve readily in the solvent. The solvent itself has also to be removed or it behaves as an impurity in its own right. It must not leave behind any residue. One simple way to tell whether an organic compound is pure is to measure its melting (or boiling) point. A pure compound melts sharply: if impurities are present it melts slowly (over a range of temperature).  [11]   The process of recrystallisation in my experiment increased the purity of Aspirin but with a decrease in the quantity produced. The solid will readily dissolve in a larger quantity of solvent; the larger the volume of solvent the greater the loss of product  [12]  . This is the reason why after every recrystallisation I lose Aspirin. The process of recrystallisation removes the impurities present and this can be concluded from the fact that the difference between the melting point of the sample and the melting point of Aspirin from data booklet decreases with each recrystallisation. I could conclude that a pharmaceutical company should always have a balance between producing a very pure product, which means many recrystallisation processes and producing enough quantity of the product to make it a profitable industry. 10. EVALUATION 10.1 Random error The apparatus I used had uncertainties like the measuring flask has an uncertainty of  ±0.5cm3, the burette had an uncertainty of  ±0.1cm3 and the melting point apparatus also had an uncertainty of  ± 0.1 °C. This results in errors in my results. The equipment error could be reduced by using equipments with less error. For example, I could use a burette with an error of  ±0.05 cm3 instead of a burette with error of  ±0.1 cm3. 10.2 Systematic error In the process of titration there could a parallax error caused if the reading from the burette is not read at eye-level. So, when taking the reading from the burette, the level of the eye should be same as the level of the meniscus. When we are titrating different samples of recrystallisations the colour of the indicator changes from pale pink to dark and it is difficult to know the end-point of the titration process. 10.3 Modifications in the method of preparation of Aspirin and its recrystallisation I modified the method to improve it in the following ways: After each recrystallisation there was loss of Aspirin and the decrease in the mass of Aspirin limited the number of recrystallisations needed to get the most pure form of Aspirin. So, if I doubled the mass of reactants I can get doubled the mass of Aspirin produced and an increase in the number of recrystallisations. When I mixed salicylic acid and ethanoic anhydride solution in concentrated sulphuric acid it is hard to get the formation of a crystalline mush of Aspirin by agitating the flask. Hence, instead I used a magnetic stirrer which gives a uniform stirring and all the chemicals are mixed properly. I found this to be more effective at producing the mush. During the process of recrystallisation a lot of Aspirin is lost. When I used the filter paper, I cut it the same size as the Hirsh funnel. Instead I could have used a bigger filter paper in the funnel so that the impurities do not leak out of the edges of the filter paper and I could get a purer sample of Aspirin. When the Aspirin is left overnight to dry in an evaporating dish, there could be many contaminants which would get mixed in the Aspirin and if this is not taken into consideration in pharmaceutical industries then this could lead to serious health problems. Hence, the Aspiring should be covered when left overnight to dry. I could use different methods of purification of Aspirin like thin layer chromatography which is a sensitive and quick way of detecting impurities in an organic product (Aspirin). I could also use spectroscopy which provides a very good method for analyzing an organic compound. By comparing the infra-red spectrum for Aspirin with the spectrum of compound in a database I can check on its purity. 11. UNANSWERED QUESTIONS Unfortunately, not all my questions could be answered in this experiment. It would have been interesting in further researching and comparing more brands of Aspirin available in the market, if the producers just recrystallised the sample of Aspirin twice so that they do not lose a lot of their product in the process of recrystallisation or the producers actually tried to produce a pure sample of Aspirin ignoring the decrease in yield after each recrystallisation and considering the fact that this can affect the health of humans. When an organic compound has been made it needs to be purified, particularly if it is a  pharmaceutical chemical. This is because most organic reactions produce by-products but, even if the reaction is a clean one, the purity standards for many products are so stringent that small amounts of other compounds have to be removed. In particular the catalyst used in this reaction is concentrated sulphuric acid and must all be removed.

The Hydrolysis Of Nitrophenyl Phosphate

The Hydrolysis Of Nitrophenyl Phosphate The hydrolysis of nitrophenyl phosphate (NPP) by alkaline phosphatase enzyme is a first order reaction dependent on the concentration of NPP in solution. A product of this reaction is nitrophenyl anion which has a high molar absorptivity at 410nm. This property makes it relatively easy to observe this reaction as a function of time via a spectrophotometer. By observing the reaction as a function of time it is possible to study the kinetics of this reaction and to determine how the initial reaction rate depends on the initial concentration of NPP. This relationship can be described by the Michaelis-Menton equation which is described in some detail. It is found that, as expected, reaction rate increases with an increase in NPP concentration. The Eadie-Hofstee plot is used to linearize the data and to obtain reasonable approximations for the Vmax and KM parameters found in the Michaelis-Menton equation. A process involving the minimization of a ÃŽ §2 value is used to obtain the final va lues of these two parameters: Vmax = 4.28*10-7 and KM = 3.33*10-4. These parameters produce a qualitatively strong fit for the data obtained and so the Michaelis-Menton equation reasonably accurately describes the relation between initial NPP concentration and reaction rate. It is found, as expected, that the addition of the inhibitor species phosphate decreases the rate of NPA formation. The inhibition constant KI obtained from the apparent KM value of the Michaelis-Menton equation. By averaging the KI value for several concentrations of inhibitor, = (1.79 + 0.25)*10-4 M. Introduction Alkaline phosphatases are common enzymes found in places such as the mammalian intestine and the bacterium Escherichia coli. They are a family of two-subunit homologous enzymes which catalyze the hydrolysis of a large spectrum of phosphate monoesters to phosphate and neutral molecules. Because they catalyze a large variety of hydrolysis reactions, they are said to be non-specific. This reaction can be expressed as follows where R is any organic compound: R-OPO32- + H2O R-OH + HPO42- (1) By removing the phosphate, R becomes a neutral species that may more easily serve to facilitate transport of nutrients across cell membranes or other biological housekeeping processes. Alkaline phosphatases can also be used to remove phosphate groups from DNA molecules which allows for different manipulations of DNA. It is instructive to examine the kinetics of this reaction in order to more fully understand certain biochemical processes. Alkaline phosphatses are large molecules with a certain number of active sites in which the hydrolysis of smaller molecules is catalyzed. Competition for the active site of these enzymes is one aspect which may affect the kinetics of the reaction. We will examine the hydrolysis of p-nitrophenyl phosphate (NPP) by the E. coli alkaline phosphatase at a pH of 10.6 at room temperature. In particular, the maximum velocity Vmax and the Michaelis constant KM of the reaction are determined and the effect of the product HPO42- as an inhibitor is examined. Theory The initial rate, or the initial velocity, v of enzyme-catalyzed reactions has been observed to follow first order kinetics. That is, the rate of the reaction depends only on the initial substrate concentration. However, this only holds true at low values of substrate concentration and levels off to a maximum velocity V at large concentrations. The general form for the way in which an enzyme catalyzes a reaction is a noncovalent association of the substrate molecule to the enzyme followed by some catalytic steps that result in the product being released from the enzyme. S + E ES E + P (2) The enzyme concentration is almost always much smaller than the concentration of the substrate. Due to this, very quickly equilibrium will be reached where, as soon as product is released from the enzymes active site, a new substrate molecule will enter the active site. At this production of [ES] is said to be saturated, meaning it has a constant value which does not change with time. (3) This is said to be the steady state, or bottleneck, condition where all active enzyme sites are occupied. From this condition and the fact that matter is conserved, one can derive the rate law: (4) This is the Michaelis-Menton equation, where the Michaelis constant KM=(k-1 + k2)/k1. This value in a physical context is the substrate concentration midway between zero substrate and the concentration at which the reaction rate has become saturated. Also in the equation Vmaxk2E0 which is said to be the maximum velocity of the reaction. Here, v is the rate of the steady state enzyme reaction. In order to measure the reaction rates, the concentration of product or substrate must be observed as a function of time. It is expected that the concentration vs. time plot will be initially linear until the substrate is beginning to be used up where it would then level off to a constant value. p-nitrophenyl phosphate is chosen as the substrate because a product of its dephosphorylation, nitrophenolate anion has a high molar absorptivity in the blue wavelengths of visible light. This allows for the use of Beers Law to calculate concentration of the product from the absorbance at a specific wavelength of light. A410 = ÃŽ µ410,NPA l cNPA (5) Here, A is the absorbance at a wavelength of 410nm, l is the light path through the solution, ÃŽ µ is the molar absorptivity of the solution and c is the concentration. Thus, by observing the change in absorbance as a function of time, it is possible to observe the increase in product as a function of time. Once the reaction rates at each initial substrate concentration are known, it is necessary to determine the parameters Vmax and KM in the Michaelis-Menton equation (Equation 4) in order to fit the data to a mathematical trend. This nonlinear fit is difficult to make without first having reasonable estimates of Vmax and KM. However, by rearranging Equation 4, it is possible to linearize the equation such that reasonable approximations of the parameters can be obtained by performing a simple linear least-squares regression. One such linearization produces the Eadie-Hofstee plot of the form: v = Vmax KM (6) This equation can be used to give reasonable approximations of the two parameters, which can then be optimized to produce the values of Vmax ­ and KM ­. This optimization is performed by comparing the estimated rate from Equation 4 using the approximate values for the two parameters to the actual rate at different concentrations. A ÃŽ §2 is defined to be the sum of the deviations from the actual observations. A computer can be used to minimize this value by varying Vmax and KM where the values for these two parameters at the minimized ÃŽ §2 are the best approximations. The error in these values is found by minimizing the ÃŽ §2 at several values of Vmax and KM by varying the other parameter. A plot of these points can be used to determine the 95% confidence intervals for these values. Competitive inhibition is the process by which molecules that resemble the substrate can bind to the active site of the enzyme, establishing a competition between the substrate and this molecule for active sites on the enzyme. HPO4- is a product of the hydrolysis of NPP which acts as an inhibitor to the enzyme reaction. The effect of the addition of phosphate on the rate can be expressed as follows: (7) Here, all variables are the same as in Equation 4, where KI is the inhibition constant. Experiment An Ocean Optics USB 400 spectrophotometer is used to take all absorbance readings. All parts of the experiment take place at atmospheric pressure and are open air. In order to establish a basis of comparison, the absorbance at 410 nm is taken for nitrophenolate anion, NPA, the pure product of the enzyme reaction. From this, the molar absorptivity constant can be determined for NPA which can be used to determine the concentration of NPA in solution. Several solutions with a range of concentrations of the substrate nitrophenyl phosphate, NPP, are necessary to observe the effect of substrate concentration on the rate of the enzyme reaction. The concentrations are chosen such that they span the range of the Michaelis-Menton plot. Upon the addition of enzyme to each solution, the concentration of NPA as a function of time is recorded by taking the absorbance reading at 410 nm at each time step. The reaction for each initial NPP concentration is charted for a sufficiently long time such th at an accurate least-squares fit can be made for the plot of concentration as a function of time. From this plot, the initial reaction rates for each NPP concentration can be determined and fit to an Eadie-Hofstee plot to determine approximate values for V and KM. To observe the inhibiting effect of phosphate on the enzyme reaction, several solutions with a range of phosphate concentrations all with equal concentrations of NPP must be prepared. Then, upon the addition of enzyme to the solutions, the concentration of NPA as a function of time is observed via the absorbance readings. It is necessary to keep the initial substrate concentration constant in this part so that the variation in the reaction rate can be attributed to only the variation in phosphate concentration. Discussion of Results As expected, increasing the initial substrate concentration served to increase the rate of product formation in the hydrolysis of nitrophenyl phosphate by an alkaline phosphatase enzyme. This can be seen clearly in Figure 1 where the slope of the graph of product (NPP) concentration as a function of time increases as enzyme concentration increases. For the trials with the three highest initial substrate concentrations, the enzyme concentration was half that used in every other run, meaning that the rate of reaction for these trials must be doubled for comparison purposes. This explains why the slopes of these lines are approximately half of what would be expected in Figure 1. Figure : The plot demonstrates that as substrate concentration increases, so too does the rate of product formation In order to determine the rate of reaction, a second order trend is fit to the data where the linear term in the equation is taken to be the rate of product formation. A second order least squares regression is used to fit the data because the rate of reaction is not truly constant over the timeframe observed. It is expected that the data will have a very slight negative concavity as, by Equation 2, the substrate concentration is decreasing at the same rate as the increase in NPA, leaving less NPP to form the enzyme-substrate complex which produces NPA. This means that the rate should decrease as a function of time the second order regression is used to correct for this fact, leaving the linear term to describe the rate of reaction. Figure 2 (attached) displays the approximate second order trends for NPA concentration as a function of time for each initial NPP concentration. These are only approximate trends as these may not be rigorously calculated, and in any case are over a small er range than that used in subsequent calculations. A more rigorous second order least-squares regression is performed to obtain the reaction rates for each initial substrate concentration. The rates along with the standard deviations for each rate are tabulated in Table 1. As stated previously, it has clearly been shown that as initial substrate concentration is increased, so too does the rate of the reaction. Additionally, the second order linear trend provides a very strong fit to the data. This is evidenced by the fact that the standard deviations of the rates are all 8 or 9 orders of magnitude smaller than the rate, meaning there is very little variance in the data from the trend calculated. Table 1: Enzyme Reaction Rate as a function of Substrate concentration [S0]/M Rate (M/sec) Rate SD 3.23E-03 3.94E-07 1.25E-15 1.61E-03 3.52E-07 2.50E-16 8.06E-04 2.96E-07 4.54E-16 4.03E-04 2.40E-07 4.77E-16 2.42E-04 1.73E-07 2.73E-16 1.45E-04 1.29E-07 2.58E-17 6.45E-05 7.97E-08 7.56E-17 3.23E-05 4.13E-08 4.78E-17 Once the rates at each initial NPP concentration are known, it is possible to attempt to describe the initial rate of the enzyme reaction as a function of initial substrate concentration. This relation can be described by Equation 4, and so the parameters Vmax and KM must be obtained. As stated previously, it is difficult to perform a nonlinear regression to describe rate as a function of time without having reasonably close guesses for these two parameters. By the process described previously whereby Equation 5 is used to linearize the data, we obtain the Eadie-Hostee plot displayed in Figure 3. Figure 3: The Eadie-Hofstee plot linearizes the data such that estimates of Vmax and Km can be obtained As can be seen in the figure, this method produces a roughly linear plot. By performing a linear least squares fit on the data, we can obtain guesses for the two parameters. Equation 6 makes it apparent that the slope of the plot is -KM and the intercept is Vmax. An analysis of the units bears this out: Vmax has units of M/sec (as does the rate because Vmax is the maximum initial rate of reaction at which the enzyme becomes completely saturated), and KM has units of M (as does substrate concentration because KM is the concentration at which the reaction rate is half that of Vmax). The intercept and slope of the graph also have units of M/sec and M, respectively. From the plot, it is estimated that KM = 2.90*10-4 M and Vmax = 4.10*10-7 M/sec. Using these values for the parameters in the Michaelis-Menton equation (4), a decent fit of the data is obtained as seen in Figure 4. However, it is clear from the plot that the values for both parameters are too low. The plot begins to approach a value which is too low as the plot from the formula falls below the final data point. This suggests that the value of Vmax is too low. Additionally, the plot rises too quickly at low substrate concentrations which would cause KM to occur too early, as does the lower value of Vmax. Figure 4: The approximate values of Vmax and Km from the Eadie-Hofstee plot produce only a rough fit of the data These apparent inaccuracies in the values of Vmax and KM are most likely resulted from the fact that the Eadie-Hostfee plot does not use two separate variables on each axis. However, clearly these values are relatively close to the correct value as the plot roughly fits the data. By the process described previously, a computer can be used to produce more accurate values of Vmax and KM by minimizing the ÃŽ §2 value, which is the sum of the squares of the deviations from the data by the value predicted in the formula. Minimizing ÃŽ §2 by varying the two parameters gives the values in Table 2. Table 2: Comparison of Vmax and Km values from Eadie-Hofstee plot and non-linear regression Eadie-Hofstee Non-linear Reg. Pos Error Neg Error Vmax (M/sec) 4.10E-07 4.28E-07 .32E-07 .28E-07 Km/M 2.90E-04 3.33E-04 .47E-04 .58E-04 As expected, the values of both parameters have increased, where Vmax = 4.28*10-7 and where KM = 3.33*10-4. The positive and negative errors for these two parameters are obtained by the process described above. At different values of each parameter, the ÃŽ §2 value is minimized while varying only the other variable. The plots of this process are appended. The confidence interval is taken to be 4.28*ÃŽ §2; that is the values at which these graphs cross this value corresponds to the lower and upper limits of each parameter. This error is also recorded in Table 2. Then by using Equation 4, one can use the multiplicative formula for error to determine error bars for the data points. These errors are tabulated in Table 3 for each initial substrate concentration. Table 3: Enzyme Reaction Rate as a function of Substrate concentration with Errors from Km and Vmax [S0]/M Rate (M/sec) pos error neg error 3.23E-03 3.94E-07 6.29E-08 7.33E-08 1.61E-03 3.52E-07 5.62E-08 6.55E-08 8.06E-04 2.96E-07 4.73E-08 5.51E-08 4.03E-04 2.40E-07 3.83E-08 4.47E-08 2.42E-04 1.73E-07 2.76E-08 3.22E-08 1.45E-04 1.29E-07 2.06E-08 2.40E-08 6.45E-05 7.97E-08 1.27E-08 1.48E-08 3.23E-05 4.13E-08 6.60E-09 7.68E-09 Using these corrected values for Vmax and KM, a more accurate formula to describe the data is obtained. Figure 5 clearly shows that the plot using these new values provides a much better fit to the data than does the plot of the approximate values for the parameters. All of the issues seen in the first plot are no longer present, and the formula clearly falls well within the error bars on each data point. Despite lacking a quantitative means of expressing the quality of the fit, it is possibly to qualitatively state by looking at the plot that the formula with the determined parameter values closely matches the experimental data. Figure 5: The optimized values of Vmax and Km from the estimates given by the Eadie-Hofstee plot produce a strong fit Finally, having confidence in the ability for the Michaelis-Menton equation to describe the effect of initial concentration on the rate of reaction, it is possible to study the effect of the addition of an inhibitor species on the rate of the reaction. Keeping the initial concentration of substrate fixed, the concentration of inhibitor in solution was varied. As would be expected, the addition of inhibitor species greatly reduced the rate of product formation. This is shown in Table 4. Table 4: Determination of mean inhibition costant Ki [I]/M Rate (M/sec) Km app/M Ki/M 0 2.40e-7 3.33e-4 0 3.23E-03 2.34E-08 6.84E-03 1.62E-04 6.45E-03 1.69E-08 9.80E-03 2.27E-04 9.68E-03 7.61E-09 2.14E-02 1.47E-04 [S]/M Avg Ki/M 1.79E-04 4.03E-04 SE .25E-04 Also observed is that the addition of more inhibitor species results in an increasingly reduced rate of product reaction. This makes sense because as more inhibitor species exists in solution, there are more molecules that can bind to the enzymes active sites, thereby inhibiting the formation of the enzyme-substrate that is necessary for product formation. As a result of the rate decreasing as the inhibitor concentration is increased, the apparent KM value greatly increases. This makes sense because the substrate would have to be significantly more concentrated to effectively compete and reach the concentration at which the reaction rate is half that of Vmax. From Equation 7, it is possible to calculate the value of KI from the inhibitor concentration, the rate and Vmax. Averaging the values of KI for each inhibitor concentration gives a = (1.79 + 0.25)*10-4 M. Conclusion The alkaline enzyme-catalyzed hydrolysis of nitrophenyl phosphate is a first order reaction that depends on the initial concentration of substrate in solution when the concentration of enzyme is kept constant. A second order least-squares regression gives the initial rate of reaction as a function of time, where the rate of product formation increases as the initial concentration of substrate increases. The Eadie-Hosfstee plot was used to approximate the parameters Vmax and KM that are necessary to describe the relation between substrate concentration and rate of reaction. Optimizing these parameters through the minimization of a X2 value gives Vmax = 4.28*10-7 and KM = 3.33*10-4. The error in these parameters can be determined by each parameter separately to minimize the X2 value at different values for each parameter. These values for the parameters in the Michaelis-Menton equation produce a qualitatively strong fit for the data. The addition of an inhibitor species served to decre ase the rate of product formation where the increase in inhibitor concentration results in a decrease in rate. Using the Michaelis-Menton equation, the inhibition constant KI is found to be = (1.79 + 0.25)*10-4 M.

Market Research Essay -- essays research papers fc

1. Introduction   Ã‚  Ã‚  Ã‚  Ã‚  An increasing amount of attention is being paid in the literature to business Guanxi or relationship in Asia particularly in the business dominated economies of the Hong Kong, Singapore, Taiwan and the People’s Republic of China. Chinese business relationships and contracts that needs to be understood and worked within to successfully conduct business and management in most countries in the region. However, the attention is directed to the different perspective on the importance of guanxi to business practices and possible impacts of the guanxi on western firms’ success in China. In order for western firms to enter China market, they have to confront with complex and constantly changing ethical percepts in China (Wu, 1999) and different types of guanxi not only exist but also can be harnessed in an ethical fashion to create wealth (Leung et al., 1999). Dates back to the mid 1980s, Motorola’s presence operates the largest owned subsidiary in China and had moved their country from a centrally planned economy to a market economy. The Motorola has invested in China for a decade and is so far the largest foreign investor in China. Therefore, guanxi has an impact on Motorola to be an established company in China compared to their competitors, Siemens and Nokia. The limitation on this research is addressed. The continuous development and changes in market condition in China is of a certain period of time. This is because it needs to be adapted to fit the prevailing situation at that time. Moreover, interview session is difficult to conduct as the research is on Motorola-China and therefore it is heavily rely on secondary sources. 2. Guanxi’s implications to practices In recent decades, there have been several fundamental concerns for western companies in China. For the Motorola, they learn how to manage a socialist workplace culture in which employees depend on their state-run employers for housing, food transportation and other necessities. The managers also have to baffle by guanxi, the vulnerable Chinese practice of developing and nurturing intricate networks of personal relationship. Nevertheless, priority is sometimes given to them over bottom-line performance. As China is admission to the World Trade Organisation (WTO) in 2001, it has reduced tariffs and business prospects are likely to grow even more prom... ...stern. Fabtech. (2004, March 19). Available HTTP:   Ã‚  Ã‚  Ã‚  Ã‚  www.fabtech.org/industry.news/008/21.01.shtml Forbes. (2004, March 19). Available HTTP: www.forbes.com/home/newswire/2003/07/03/rtr1018223.html Kahal, S. E. (2001). Busines in Asia Pasific. United States: Oxford University Press Inc. Noble, C. (1997). Asia Pacific Business. Australia: Charles Stuart University People’s Daily. (2004, March 13). Available HTTP:   Ã‚  Ã‚  Ã‚  Ã‚  www.fpeng.peopledaily.com.cn/200011/05/eng20001105_54382.html People’s Daily. (2004, March 15). Available HTTP:   Ã‚  Ã‚  Ã‚  Ã‚  www.fpeng.peopledaily.com.cn/200111/08/eng20011108_84118.shtml Recommendations For Western Firms While Operating in China. (2004, February 29)   Ã‚  Ã‚  Ã‚  Ã‚  Available HTTP: bigtp.com/inf_3.html Spotlight: China: Factory for the World. (2004, March 19). Available HTTP:   Ã‚  Ã‚  Ã‚  Ã‚  www.bizsites.com/spotlight/china.html Testimony of Christopher B. Galvin. (2004, March 13). Available HTTP: www.finance.senate.gov/2-23galv.htm Workforce. (2004, March 15). Available HTTP:   Ã‚  Ã‚  Ã‚  Ã‚  www.workforce.com/section/09/feature/23/56/96

If You Really Like a Guy, Hit Him :: Personal Narrative Relationships Essays

If You Really Like a Guy, Hit Him I don’t remember wanting to punch Jeff. I remember wanting Jeff to â€Å"ask me out.† How I got into the position of hitting him is somewhat of a mystery to me. Jeff Stanford was the cutest guy in our third grade class. He had blonde hair, blue eyes-the whole shabang! He even wore tapered, stone washed jeans (it was the eighties, this was cool). He was my friend. I was, of all things, a tomboy. I ran faster then the boys. I could beat them all at tether ball. My hair was shorter then any of the boys, and I had the biggest crush on Jeff. Along with being head over heels in love with Jeff, I was a die hard Madonna fan. I had her tapes and even a sweatshirt with her picture on it. Jeff was not a Madonna fan. I thought this was something we could work out. Although we could not spend endless hours reciting lines from â€Å"Like a Virgin† together, we could always play tetherball. Tethe ball, at least at Pearson Elementary, was the game of champions. I loved to play, but at early recess the balls weren’t always up yet and Jeff, Kelly and myself sometimes had to make do with a rousing game of tag. This was definitely a â€Å"tag day.† â€Å"RUUUUUUUN Kelly, he is right behind you!!!!!† â€Å"Huh?† questioned Kelly. â€Å"Tagged you, tagged you!† taunted Jeff. â€Å"Ouch,† I screamed. â€Å"I have something in my eye.† I was completely serious. â€Å"There is something in my eye and it hurts!† I kept trying to get, whatever it was, out of my eye using the sleeve of my Madonna sweatshirt. Jeff was trying to help, in some way, but doing a piss-poor job of it. â€Å" What happened?† he asked. â€Å"Did Madonna stick her arm out and poke you in the eye?!!! Hahahahaaaahaaaha!† Boys are so dumb. â€Å"No.† It was all I could say, I didn’t have any witty come backs, but come on, he could have left me alone! â€Å"Why are you picking on me?† I asked. I wished he would just leave, but no, he kept taunting me. â€Å"You are so mean! LEAVE!† I shouted. For some unknown reason he didn’t get it! My eye still hurt, recess was almost over, and I decided I hated Jeff... WHAAAAP!!!!!!! I socked him. A right fisted jab, straight up the gut, full third grade force, and Jeff Stanford, my crush, went down.