The role of the IMF in helping poor and debt-troubled countries Research Paper

The role of the IMF in helping poor and debt-troubled countries - Research Paper Example The IMF is tasked with the role of assisting nations suffice their budgets or invest in areas that positively influence the social welfare or the economy of a country. Most of the support activities that IMF undertakes are backed by global policies. An example is the IMF helps less developed nation to improve health care, education and security which is in line with the United Nations Millennium Development goals. Another reason why the IMF helps poor nation is to stabilize nations in economic and social terms. A debt ridden country has a high probability of social disorder, thus more economic failures (Ghosh, Zalduendo, Thomas, Ramakrishnan, Kim &Joshi 2008). Globalization has ensured that all nations are interlinked economically, thus, failure of an economy to meet its obligations can impact on the global economy, which might lead to a crisis. Despite the fact that the IMF is tasked to issue loans to member countries, there are several conditions that must be met as part of eligibility criteria. First, a country is supposed to design a program that is supposed to address the problem that has made it resort to the IMF for assistance (IMF Factsheet). Before the IMF considers a loan, country must identify the causes of the deficit and the strategies that are in place to mitigate the situation. In deficits situations, the IMF holds that countries should engage in activities that try to resolve the deficit since it affects other economies of the world. Second, the International Monetary Fund should be allowed to assess the programs that are under its support. This is meant to ensure that funds are channeled to the intended functions. Moreover, the process also ensures that the intended benefits are realized. The assessment programs are mostly conducted in less developed nations due to inefficiencies, misappropriation and also embezzlement of funds by few individuals. The institution employs structural benchmarks that are meant to protect IMF interests in the

Stonehenge :: essays research papers

I. On Salisbury Plain in Southern England stands Stonehenge, the most famous of all megalithic sites. Stonehenge is unique among the monuments of the ancient world. Isolated on a windswept plain, built by a people with no written language, Stonehenge challenges our imagination. The impressive stone circle stands near the top of a gently sloping hill on Salisbury Plain about thirty miles from the English Channel. The stones are visible over the hills for a mile or two in every direction. Stonehenge is one of over fifty thousand prehistoric "megalithics" in Europe. As Stonehenge is approached, the forty giant stones seem to touch the sky. Most of the stones stand twenty-four or more feet high. Some stones weigh as much as forty tons. Others are smaller, weighing only five tons. At first glance, the stones may seem to be a natural formation. But a closer look shows that only human imagination and determination could have created Stonehenge. II. The Stonehenge today looks quite different from the Stonehenge of old. Wind and weather have destroyed a little of Stonehenge over the ages. People have destroyed much more. Today, less than half of the original stones still stand as their builders planned. Many of the once upright stones lie on their sides. Religious fanatics, who felt threatened by the mysteries posed by Stonehenge, knocked over many of the standing stones. They toppled some of the huge stones, which then split into pieces; they buried others. Other stones were "quarried" over the centuries as free building material and hauled away. Even into this century, visitors have come with hammers to carry away a chip of stone with them. III. Only in recent years have the stones been protected from the huge amounts of people that see them every year. No longer can anyone roam among the stones. Too much damage, intentional or not, has been done by the hundreds of thousands of visitors. Today, tourists are even prevented from walking between the stones for fear that the millions of footsteps every year might make the stones unstable. IV. The twelfth-century English writer and historian, Geoffrey of Monmouth, first recorded Merlin's building of Stonehenge in his famous book History of the Kings of Britain. Geoffrey claimed that his book was a translation of "a certain very ancient book written in the British language." However, no other scholar or historian knows of the existence of such a book. According to Geoffrey, the great stones were brought from Ireland to England to mark the burial place of a group of slain British princes.

The Ka and Molar Mass of a Monoprotic Weak Acid

The Ka and Molar Mass of a Monoprotic Weak Acid Chemistry Lab 152 Professor: James Giles November 7, 2012 Abstract: The purpose of this experiment was to determine the pKa, Ka, and molar mass of an unknown acid (#14). The pKa was found to be 3. 88, the Ka was found to be 1. 318 x 10 -4, and the molar mass was found to be 171. 9 g/mol. Introduction Acids differ considerable as to their strength. The difference between weak and strong acids can be as much as 10 orders of magnitude.Strong acids dissociate more completely than weak acids, meaning they produce higher concentrations of the conjugate base anion (A-) and the hydronium cation (H30+) in solution. HA(aq) + H20 (( A- + H3O+ With the following formula the degree to which an acid dissociates (Ka) can be calculated and given a numerical value. Ka = [A-][H3O+] / [HA] Ka is the conventional way of measuring an acid’s strength. The purpose of this experiment was to determine the Ka of an unknown acid, along with its pKa and mol ar mass. Experimental The unknown acid for this experiment was #14.The experiment began with the preparation and standardization of NaOH solution. It was calculated that 2. 00 grams of NaOH pellets were needed to prepare 0. 5 L of 0. 1 M NaOH solution. The solution was then standardized by conducting three titration trials. It was calculated that 0. 7148 grams of KHP were necessary to neutralize 35 mL of the 0. 1 M NaOH. Three samples of KHP were weighed approximating this number (Table 1). Each sample was mixed with 40 mL of deionized water and 2 drops of phenolphthalein in 3 Erlenmeyer flasks. Each flask was then titrated with the NaOH to a light pink endpoint.The volumes of NaOH were recorded, averaged, and the standardized. The molarity of the NaOH was found to be 0. 0981. Assuming a molar mass of 100 g/mol, it was calculated that 0. 3930 g of acid was needed to neutralize 40 mL of the standardized NaOH solution. This amount was weighed out on an electronic balance to full preci sion and added to a clean 250 mL beaker. The acid was first diluted with 10 mL of isopropanol and then 90 mL of water. A pH meter was immersed in the acid solution and an initial pH reading of 2. 61 was recorded.A buret filled with the NaOH solution was incrementally added to the acid solution and the changing pH values were recorded (Table 2). As the pH meter approached the equivalence point the amount of NaOH added each time was reduced. As the Table 2 shows, the pH rose significantly with the addition of little NaOH over this interval. This information was plotted using Graphical Analysis producing a titration curve graph of pH vs. NaOH (Graph 1). Additional calculations and graphs were produced to help identify the equivalence point: ? pH/? V vs. NaOH (Graph 2) and Vtotal x 10-ph vs. NaOH (Graph 3) Tables and CalculationsPreparation of 500 mL of 0. 1 M NaOH M = moles / volume 0. 1 M NaOH = moles NaOH / 0. 5 L H20 = 0. 05 moles NaOH 0. 05 moles NaOH x 39. 986 g/mol NaOH = 1. 99 g NaOH Preparation of KHP 0. 1 M NaOH = moles NaOH / 0. 035 mL NaOH = . 0035 moles NaOH 0. 0035 moles KHP x 204. 233 g/mole KHP = 0. 7148 g KHP Table 1: NaOH Titration Trials |Trial |KHP |NaOH (to titrate to endpoint) | | |(grams) |(mL) | |1 |0. 7159 |35. 75 | |2 |0. 7147 |35. 65 | |3 |0. 7149 |35. | | | | Avg. 35. 66 | Standardization of NaOH 0. 0035 moles NaOH / . 03566 mL NaOH = 0. 0981 M NaOH Table 2: pH vs. NaOH Values |NaOH |pH |NaOH |pH |NaOH |pH |NaOH |pH | |(mL) | |(mL) | |(mL) | |(mL) | | |0 |2. 61 |19. 2 |4. 54 |22. 15 |6. 56 |25. 4 |9. 74 | |2 |2. 94 |19. 4 |4. 58 |22. 2 |6. 2 |25. 9 |9. 82 | |4 |3. 18 |19. 6 |4. 61 |22. 25 |6. 87 |26. 4 |9. 96 | |5 |3. 3 |19. 8 |4. 65 |22. 3 |6. 98 |26. 9 |10. 02 | |6 |3. 4 |20 |4. 68 |22. 35 |7. 06 |27. 4 |10. 11 | |7 |3. 49 |20. 2 |4. 72 |22. 4 |7. 14 |28. 4 |10. 21 | |8 |3. 58 |20. 4 |4. 77 |22. 5 |7. 24 |29. 4 |10. 1 | |9 |3. 66 |20. 6 |4. 84 |22. 6 |7. 44 |31. 4 |10. 46 | |10 |3. 73 |20. 8 |4. 88 |22. 7 |7. 58 |33. 4 |10. 58 | |11 | 3. 88 |21 |4. 94 |22. 8 |7. 73 |35. 4 |10. 67 | |12 |3. 91 |21. 2 |5. 02 |22. 9 |7. 89 |36. 4 |10. 75 | |13 |3. 97 |21. 4 |5. 11 |23 |8. 03 |39. 4 |10. 87 | |14 |4. 04 |21. |5. 25 |23. 1 |8. 17 |42. 4 |10. 96 | |15 |4. 11 |21. 7 |5. 32 |23. 2 |8. 38 |44. 4 |11. 02 | |16 |4. 19 |21. 8 |5. 45 |23. 3 |8. 51 | | | |16. 5 |4. 24 |21. 85 |5. 52 |23. 4 |8. 65 | | | |17 |4. 29 |21. 9 |5. 62 |23. 6 |8. 92 | | | |17. 5 |4. 34 |21. 95 |5. 71 |23. 8 |9. 9 | | | |18 |4. 4 |22 |5. 86 |24. 1 |9. 27 | | | |18. 5 |4. 45 |22. 05 |6. 1 |24. 4 |9. 39 | | | |19 |4. 52 |22. 1 |6. 4 |24. 9 |9. 62 | | | Graph 1: pH vs. NaOH Titration Curve [pic] Estimated volume of NaOH at equivalence point based on titration curve: 22. 30 mL NaOH. Table 3: ? pH/? V vs. NaOH Values |NaOH |? pH/? V |NaOH |? pH/? V |NaOH |? pH/?V |NaOH |? pH/? V | |(mL) | |(mL) | |(mL) | |(mL) | | |2 |0. 12 |19. 2 |0. 2 |22. 1 |3. 2 |24. 4 |0. 46 | |4 |0. 12 |19. 4 |0. 15 |22. 15 |3. 2 |24. 9 |0. 24 | |5 |0. 1 |19. 6 |0. 2 |22. 2 |3 |25. 4 | 0. 16 | |6 |0. 09 |19. 8 |0. 15 |22. 25 |2. 2 |25. 9 |0. 28 | |7 |0. 9 |20 |0. 2 |22. 3 |1. 6 |26. 4 |0. 12 | |8 |0. 08 |20. 2 |0. 2 |22. 35 |1. 6 |26. 9 |0. 18 | |9 |0. 07 |20. 4 |0. 35 |22. 4 |1 |27. 4 |0. 1 | |10 |0. 15 |20. 6 |0. 2 |22. 5 |2 |28. 4 |0. 1 | |11 |0. 03 |20. 8 |0. 3 |22. 6 |1. 4 |29. 4 |0. 075 | |12 |0. 06 |21 |0. |22. 7 |1. 5 |31. 4 |0. 06 | |13 |0. 07 |21. 2 |0. 45 |22. 8 |1. 6 |33. 4 |0. 045 | |14 |0. 07 |21. 4 |0. 7 |22. 9 |0. 1 |35. 4 |0. 08 | |15 |0. 08 |21. 6 |0. 7 |23 |1. 4 |36. 4 |0. 04 | |16 |0. 1 |21. 7 |1. 3 |23. 1 |2. 1 |39. 4 |0. 03 | |16. 5 |0. 1 |21. 8 |1. 4 |23. 2 |1. |42. 4 |0. 03 | |17 |0. 1 |21. 85 |2 |23. 3 |1. 4 | | | |17. 5 |0. 12 |21. 9 |1. 8 |23. 4 |1. 35 | | | |18 |0. 1 |21. 95 |3 |23. 6 |0. 85 | | | |18. 5 |0. 14 |22 |4. 8 |23. 8 |0. 3 | | | |19 |0. 1 |22. 05 |6 |24. 1 |0. 4 | | |Graph 2: ? pH/? V vs. NaOH [pic] Estimated volume of NaOH at equivalence point based on ? pH/? V vs. NaOH graph: 22. 30 mL NaOH. Table 4: Vtotal x 10-ph vs. NaOH Values |NaOH |Vtotal x 10-ph |NaOH |Vtotal x 10-ph | |(mL) | |(mL) | | |19. 8 |0. 000443 |21. 6 |0. 000121 | |20 |0. 000417 |21. 7 |0. 000104 | |20. 2 |0. 000385 |21. 8 |7. 70E-05 | |20. 4 |0. 000346 |21. 85 |6. 60E-05 | |20. 6 |0. 000298 |21. 9 |5. 0E-05 | |20. 8 |0. 000274 |21. 95 |4. 30E-05 | |21 |0. 000241 |22 |3. 00E-05 | |21. 2 |0. 000202 |22. 05 |1. 80E-05 | |21. 4 |0. 000166 | | | Graph 3: Vtotal x 10-ph vs. NaOH [pic] Estimated volume NaOH at equivalence point based on Vtotal x 10-ph vs. NaOH graph: 22. 20 mL NaOH Calculating Ka of Unknown Acid pH at ? equivalence point volume: 3. 88 Ka = 10 -3. 88 = 1. 318 x 10 -4 Ka = 1. 318 x 10-4 Calculating the Molar Mass of the Unknown Acid 0. 0981 M NaOH = moles acid / . 02330 L NaOH = 0. 023 moles acid 0. 3930 g acid / 0. 0023 moles acid = 171. 9 g/mol Analysis of Error There is a high degree of agreement among the 3 graphs and therefore a low degree of error in this experiment. According to the Graphical Analysis program, Graphs 1 and 2 indicated that the total volume of NaOH at the equivalence point was 22. 30 mL. Graph 3 indicated a volume of 22. 20 mL, a difference of 0. 1 mL. Discussion Based upon the range of possible values for Ka, anywhere from 3. 2 x 109 for Hydroiodic acid (one of the strongest) to 5. 8 x 10-10 for Boric acid (one of the weakest), this experiment’s unknown acid solution (Ka = 1. 18 x 10-4) falls roughly in the lower quarter of strength. This estimate fits its titration curve. In general, strong acids quickly go from a very low pH to a very high pH, e. g. , 2 to 12, while weak acids quickly go from a lower pH to a higher pH, e. g. , 6 to 10. The unknown solution for this experiment jump from 5 to 10 pH, which is consistent with a Ka of 1. 318 x 10-4 and a weaker acid. References Darrell D. Ebbing and Steven D Gammon, General Chemistry, 9th ed. Cengage Learning: Ohio, 2009. Department of Physical Science—Chemistry, Mesa Community College. The Ka and Molar Mass of a Monop rotic Weak Acid (handout).

The Threat Of Cyber Security - 4382 Words

Internet use has skyrocketed since 2000 to over 2.4 billion users worldwide, with 70% of those users logging on every day (The Culturist, 2013).These days it takes merely minutes for news to travel around the globe. Our airwaves are filled with tiny nuggets of information whooshing past us undetected; that is, until it is detected. In the 21st century, cyber security is an international issue. The threat of cyber warfare is very real and would be devastating beyond any conventional weapon imaginable. With the entire globe becoming more interconnected, an attack on online infrastructure could ground airplanes, control information access, hijack nuclear facilities, and perhaps most terrifyingly, bring the global economy to a†¦show more content†¦From 1997 to 2012, households with internet grew from 18% to 74.8%, where 94.8% of households with computers use it to connect to the interne (U.S. Census. 2014). Computers are not the only source of hardware connection to the inte rnet. Other hardware such as gaming consoles, smart phones, laptops, and tablets are also connecting to the internet as well. Smart phone use for internet connectivity is popular among all age groups 25 and older with 45.3% usage. Among the heaviest users were age groups 25-34 at 88.1% (U.S. Census. 2014). With the growing use of the internet, the nation depends on a safe and stable cyberspace infrastructure for economic prosperity and more importantly, national security. Cyber security: What is it and why is it important? Cyber security generally refers to computer security or IT Security and is applicable to computers, smartphones, the internet and public/private computer networks. ?The field covers all the processes and mechanisms by which computer-based equipment, information and services are protected from unintended or unauthorized access, change or destruction, [? and] natural disasters (Computer Security). 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