Partial Rate of Diffusion of Potassium Dichromate

Partial Rate of Diffusion of Potassium Dichromate ABSTRACT Diffusion is the process where molecules spread into spaces. It was observed that the formation of ammonium chloride was near the hydrochloric acid meaning the ammonia diffused faster. One factor that affects the rate of it is the molecular weight of a substance. If molecular weight affects the rate of diffusion, then, the higher or lower the molecular weight the slower or faster the rate of diffusion. To test this factor, a drop of potassium dichromate, potassium permanganate, and methylene blue were placed on each wells of the petri dish containing agar water gel. After thirty minutes, the partial rate and average rate of diffusion are calculated and the results showed that the substance with the lowest molecular weight, potassium permanganate, diffuses fastest. The molecular weight affects the rate of diffusion, the lighter the molecular weight the slower the rate of diffusion and vice versa. The molecular weight of the substance is inversely proportional to the rate of diffusion. INTRODUCTION Diffusion is the process where the molecules of solid, liquid or gas spread spontaneously to occupy a space. Gas molecules are the fastest to diffuse since their molecules are scattered from one another. The molecules of diffusion moves randomly and they spread into available space (Mendoza E, 2003). Hydrochloric acid is formed by mixture of hydrogen ion and chlorine ion. According to the Lewis’ theory of acids and bases, hydrochloric acid is an acid because an element (chlorine) was paired to a hydrogen ion and when it is dissolved in water, it produces hydrogen ion (Ebbing and Gammon, 2011). A 0.1M of hydrochloric acid has a pH level of 1 meaning it is strong enough to dissolved iron nails (Mendoza E, 2003). Hydrochloric acid has a molecular weight of about 36.46 g/mol and can be found inside our stomach because it helps on the digestion of the food we eat (Reece et al, 2005). Ammonium hydroxide is a base because it produces hydroxide ion when dissolves in water. The ammonium hydroxide decomposes forming ammonia and water. In a decomposition reaction, a compound turns into simpler substances or elements. The ammonia has a molecular weight of about 17.031 g/mol and water has about 18g/mol (Mendoza E, 2003). When ammonium hydroxide breaks down into ammonia and water, since hydrochloric acid is an acid, when it is dissolved in water, it will produce hydrogen ion. On the other hand, the chlorine ion(C ) will react with ammonia (N). This reaction is called synthesis reaction (HCl + N O → NCl + +O) (Mendoza E, 2003). From the reaction, ammonium chloride is produced by the synthesis of ammonia and chlorine ion. The ammonia and hydrochloric acid spread to available space until they meet. After they met, a reaction of white powder is formed. The ring is near to the hydrochloric acid compare to the ammonia (France C. 2014) There are many factors that affect the rate of diffusion. When heat is applied, the molecules moves quicker, making the solute dissolves faster. The diffusion rate also increased by stirring where molecules move faster between the molecules of water. Diffusion is also affected by the density of the solvent where diffusion is slower when it has a higher density (Lozano Sandico, 2003). Molecular weight also affects the rate of diffusion. Lighter particles diffuse faster than heavy particles It has an inverse proportionality where the smaller the size of the particles, the faster the rate of diffusion (Tro, 2008). The difference in concentration can affect the rate of diffusion too. The distance of diffusion also affects the rate of it where it takes time to diffuse a particle for farther place. Permeability is also a factor of diffusion where if the substance does not permit a substance to pass through it, thus, there will be no diffusion (Meyertholen E., 2014) Based on the observation, the ammonium hydroxide decomposed into ammonia (N) and water (O), while the hydrochloric acid decomposed into hydrogen ion() and chlorine ion (C). A single displacement had occurred from the ammonia and hydrochloric acid, forming ammonium chloride (NCl). This substance appeared in the tube in the presence of a white smoke. With this, data were obtained to know how molecular weight affects the diffusion of a certain substance. As seen in table 4.1, comparing the ratio of the distances from the two substances up to the smoke over the total distance, the ratio of the ammonia are bigger than the ratio of the hydrochloric acid. But on the first trial, the ratio of hydrochloric acid over total distance is bigger by 0.2cm compared to the ammonia. Using the formula, the ratio of the ammonia over the ratio of hydrochloric acid is used to get the average ratio. With ammonia having a molecular weight of about 17.031 g/mol and the hydrochloric acid having a molecular we ight of about 36.46 g/mol, the lighter the molecular weight of the substance, the faster the diffusion is. However, the molecules of the substances on the observation cannot be seen. Table 4.1. The ratios of the distances from the hydrochloric acid over total distance, from the ammonia over total distance, and ammonia over hydrochloric acid. Trial Distance(cm) (d) Total Distance (D) Ratio d HCl d N 1 19.2 18.7 37.9 0.51 0.49 0.96 2 15.0 20.5 35.5 0.42 0.58 1.38 3 16.8 20.0 36.8 0.46 0.54 1.17 4 17.5 18.5 36.0 0.49 0.51 1.04 This study aimed to determine how molecular weight affects the rate of diffusion. The specific objectives are: To explain the relationship between molecular weight and the rate of diffusion of a certain substance. To prove if molecular weight affects the rate of diffusion, then, the higher or lower the molecular weight the slower or faster the rate of diffusion. MATERIALS AND METHODS To find out if the molecular weight of a substance affects the rate of diffusion, a petri dish of agar water gel with three wells, potassium dichromate (C), potassium permanganate (KMn), and methylene blue were used. The diameter of each well is measured in millimetre using a ruler. Then, a drop of potassium dichromate, potassium permanganate, and methylene blue were dropped on each wells of the petri dish containing agar water gel. It was immediately covered to prevent it from drying. At a three-minute interval for thirty minutes, by lifting the petri dish, the diameter of the coloured area is measured. Using the collected data in the setup, a line graph was used to compare the rate of each substance’s diffusion. RESULTS AND DISCUSSION On the experiment, potassium dichromate, potassium permanganate, and methylene blue in agar-water gel inside the petri dish were used to determine the partial rate and the average rate of diffusion. After 30 minutes of observing, the diameter of the potassium permanganate has the biggest among the rest which is 14, compared to the potassium dichromate which is 12 and to methylene blue which is 11. The average rate of diffusion is calculated by the formula: Where: df= final diameter do= initial diameter t= total time While the partial diffusion rate is calculated by the formula Where: dx=diameter at a given time dx-1= diameter immediately before dx tx= time when dx measured tx-1= time immediately before tx Table 4.2. Rate of diffusion of potassium dichromate, potassium permanganate, and methylene blue in agar-water gel inside the petri dish for 30 minutes. Time (Minute) Diameter (mm) Potassium Permanganate (MW 158g/mol) Potassium Dichromate (MW 294g/mol) Methylene blue (MW 374g/mol) 0 3 3 3 3 6 5 5 6 8 7 6 9 10 8 7 12 11 9 7 15 12 9 8 18 12 10 9 21 13 10 9 24 13 11 10 27 13 11 10 30 14 12 11 Table 4.3. Partial rate of diffusion of potassium dichromate, potassium permanganate, and methylene blue in agar-water gel inside the petri dish for 30 minutes. Time elapsed (minute) Partial rates of diffusion(mm/min) Potassium Permanganate (MW 158g/mol) Potassium Dichromate (MW 294g/mol) Methylene blue (MW 374g/mol) 3 1.00 0.67 0.67 6 0.67 0.67 0.33 9 0.67 0.33 0.33 12 0.33 0.33 0.00 15 0.33 0.00 0.33 18 0.00 0.33 0.33 21 0.33 0.00 0.00 24 0.00 0.33 0.33 27 0.00 0.00 0.00 30 0.33 0.33 0.33 Average rate of diffusion (mm/min.) 0.367 0.300 0.267 Figure 4.3. A bar graph comparing the average rate of diffusion of potassium dichromate, potassium permanganate, and methylene blue in agar-water gel inside the petri dish. Figure 4.4. A bar graph comparing the partial rate of diffusion of potassium dichromate, potassium permanganate, and methylene blue in agar-water gel inside the petri dish. SUMMARY AND CONCLUSION To test the hypothesis, an experiment was performed to test if the molecular weight affects the rate of diffusion, then, the higher or lower the molecular weight the slower or faster the rate of diffusion. An observation using a glass rod to measure the distance on how far the particles of the hydrochloric acid and ammonia travelled until both of them had a chemical reaction producing ammonium chloride. Ammonia that has 17.031 g/mol travelled faster than the hydrochloric acid that has 36.46 g/mol. The experiment was done to support the observation because it is more clearly and visible to the naked eye because the diffusion can be observe in this setup. The potassium permanganate having a diameter of 14 is the biggest among them. The average rate of diffusion of each substances was calculated and the result is potassium permanganate has the fastest rate of 0.367mm/min. potassium permanganate has the lightest molecular weight of 158g/mol. Based from the setup that was performed, the substance that has the lightest molecular weight has the fastest rate of diffusion. This study confirmed the hypothesis where the molecular weight affects the rate of diffusion, the lighter the molecular weight the slower the rate of diffusion and vice versa. The molecular weight of the substance is inversely proportional to the rate of diffusion. However, some errors were made during the experiment like the unequal amounts of the substances placed in the petri dish, methylene blue was spilled from the well, and the three substances are not measured at exactly every three minutes. This study needs further research because there are also other factors other than the molecular weight that could affect the rate of diffusion. It is recommended to research and study other factors that can affect the rate of diffusion. LITERATURE CITED Cain, M.L., Jackson, R.B., Minorsky, P.V., Reece, J.B., Urry L.A., and Wasserman, S.A.2011. Campbell Biology 9th Edition. USA: Pearson Education Inc. p. 53. Ebbing,D. and Gammon, S. D. 2009. General Chemistry Enhanced Edition. USA: Cengage Learning Inc. p. 143-144. France C. 2014. Elements, Compounds and Mixtures: Information retrieval. http://gcsescience.com/e17- ammonium-chloride-reversible.htm> Accessed October 12, 2014. Lozano L.F. and Sandico P.M. C.2003. Science and Technology for the Future II. Makati City: Diwa Learning System Inc. p. 110. Mendoza, E.E.2003. Phoenix Science Series Chemistry. Quezon City: Phoenix Publishing House, Inc. p. 163, 244, 299, 305. Meyertholen E. 2014. Diffusion II: Information retrieval. http://austincc.edu/emeyerth/ diffuse2.htm> Accessed October 13, 2014. Tro, N. J.2008. A Molecular Approach.Oregon, USA: GEX Publishing Services. p. 229.   Ã‚  

Importance of Biodiesel

IMPORTANCE OF BIODIESEL The world’s accessible oil reservoirs are gradually depleting due to a burgeoning demand of fossil fuels. Owing to the enormous dependency of transport vehicles running on gasoline engines, the development of bio-gasoline may well reduce the dependence of the fuel market on fossil fuels. Biofuel development has gained the attention of researchers in recent years owing to the rate of depletion of fossil fuels. Several processes are currently employed in the conventional production ofdifferent biofuels: the production of biodiesel is catalytically performed either through the transesterification of triglycerides using alcohol or the deoxygenative ecofining of triglycerides in a nonalcoholic environment; bio-oil is produced by the pyrolysis of biomass; bio-ethanol is produced by thefermentation of sugars obtained from starch or cellulosic based biomass, while bio-gasoline is producedfrom the catalytic cracking of triglycerides. The present work investigates the suitability of mesuaferrea (nahar) oil seed as a source of alternative fuel in CI engines. Non-edible straight vegetable oil obtained from mesuaferreaseeds was extracted using a solvent extraction technique. The catalytic cracking mesuaferrea seed oil to biofuel was studied in a microwave reactor with selected catalysts at . 500-600? C The thermo-physical and chemical properties of straight mesuaferrea oil were determined. The flash point, kinematic viscosity, and density of the straight mesuaferrea oil were found to be much higher than diesel, though the calorific value was found to be lower. The variation in thermo-physical and chemical properties of various blends of straight mesuaferrea oil and diesel were studied after fraction of biofuel products may be obtained at different temp. The present study also report on the use of nahar oil as a potential source of biodiesel that can be used as a partial substitute for fossil fuels.

Managerial Economics Chapter 5 and 6 Homework Essay

Part A: A firm maximizes profit when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 – L/400 Therefore (40)*(1-L/400) = 20. The solution is L = 200. In turn, Q = 200 – (2002/800). The solution is Q = 150. The firms profit is= PQ – (MC)L= ($40) (150) – ($20) (200) = $2,000 Part B Price increase to $50: Q = Dresses per week L= Number of labor hours per week Q = L –L2/800 MCL=$20 P= $50 A firm maximizes profit when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 – L/400 Therefore (50)*(1-L/400) = 20. The solution is L = 240. In turn, Q = 240 – (2402/800). The solution is Q = 168. The firms profit is ($40) (168) – ($20) (240) = $1,920 Optimal output of the firm would increase from 150 to 168, and labor would increase from 200 to 240, resulting in a decrease in profit to $1,920. Part B inflation in labor and output price: Assuming a 10% increase IN LABOR COST AND OUTPUT PRICE†¦ Q = Dresses per week L= Number of labor hours per week Q = L –L2/800 MCL=$20.20 (20*.10) P= $40.40 ($40*.10) A firm maximizes profit when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 – L/400 Therefore (40.40)*(1-L/400) = 20.20. The solution is L = 200. In turn, Q = 200 – (2002/800). The solution is Q = 150. The firms profit is ($40.40) (150) – ($20.20) (200) = $2,020 Optimal output of the firm would remain the same at 150, and labor would remain the same at 200, however, there would be an increase in profit to $2,020 to correspond to the percentage increase in output price and labor cost (in this example 10%). Part C 25% increase in MPL: The marginal cost of labor would increase by the same percentage amount as price (25%), therefore the Marginal Cost of labor would increase from 20 to 25. Therefore 50 – L/8 =25 and L=200 Output and hours of labor remain unchanged due to the fact that price and cost of labor increase by same percentage amounts ALSO SEE PART B ABOVE INFLATION EXAMPLE I MADE DENOTING 10 PERCENT INCREASE IN LABOR AND OUTPUT. Chapter 5 Question 12 Page 220 Part A: Q = 100(1.01).5(1).4 = 100.50. Compare this to the original of Q=100 and we can determine that Output increases by .5%. The power coefficient measures the elasticity of the output with respect to the input. A 1% increase in labor produces a (.5)(1) = .5% increase in output. Part B: Dr. Ghosh- per my e-mail I was a bit confused with this question based on your lecture notes (as your notes state that BOTH inputs must change for a returns to scale to be determined) , so I have two different opinions. Opinion 1- The nature of returns to scale in production depends on the sum of the exponents, ÃŽ ±+ÃŽ ². Decreasing returns exist if ÃŽ ±+ÃŽ ²Ã‹â€š 1. The sum of the power coefficients is .5 + .4 < 1, the production function exhibits decreasing returns to scale where output increases in a smaller proportion than input. This is reflected in Part A of this problem where a 1% increase in labor (input) results in a .5% increase in output. Opinion 2- BOTH inputs must be changed in the same proportion (according to your lecture notes). Therefore, in this question I am confused. Only one of the inputs are being changed. Does this concept not apply, and is my original answer incorrect? I don’t see any scale where only one of the inputs are changed†¦As such, if both inputs MUST be changed then returns to scale can not be determined for this question as only L was originally changed. Chapter 6 Question 6 Part B Page 265 (part A not required) Demand is P = 48 – Q/200 Costs are C = 60,000 + .0025Q2. Therefore the TR= 48Q-Q2/200, and the derivative MR function would be MR = 48 – Q/100. The firm maximizes profit by setting MR = MC. Therefore, MR = 48 – Q/100 and MC = .005Q. Setting MR = MC (48 – Q/100) = .005Q results in: Q* = 3,200. In turn, P* = $32 (where 48-3200/200). Chapter 6 Question 8 Page 265 CE= 250,000 +1,000Q + 5Q2 $2,000= Cost of Frames and assembly P= 10,000-30Q Part A: Marginal Cost of producing an additional engine†¦ CE = 250,000 +1,000Q +5Q2 MCE = d/dQ (250,000 +1,000Q + 5Q2) =10Q + 1,000 MCCycle=MCEngine +MCframes and assembly; therefore MCCylce = 1,000+ 2,000 +10Q The inverse demand function provided in the text was P= 10,000-30Q TR = (P)*(Q) = (10,000-30Q)*Q =10,000Q – 30Q2 Obtain the derivative of this function to find MR: MR=d/dQ =(10,000Q – 30Q2) MR=10,000 – 60Q MR = MC 10,000 – 60Q = 1,000 + 2,000 +10Q 7,000 = 70Q Q=100 (profit maximizing output) P= 10,000 – 30Q =10,000 -30(100) Profit Maximizing Price=7,000 Therefore the Marginal Cost of producing an engine =1,000 + 10Q (q=100 from solving above) =2,000 MCEngine Marginal Cost of Producing a Cycle From equation developed above†¦ MCCycle = 1,000 +2,000 +10Q =1,000 +2,000 + 10(100) =$4,000 MCCycle Part B: Since the firm can produce engines at a Marginal Cost of $2,000, the opportunity to buy from another firm at a greatly reduced Marginal Cost of $1,400 would be sensible. MCEngine=$1,400 MR = MC 10,000 – 60Q = 2,000 +1,400 10,000- 60Q = 3400 Q=110 (profit maximizing output) P = 10,000 – 30(110) =6,700 profit maximizing price Therefore the firm should buy the engine since the engine produced by the firm is more than the engine provided by the other firm. Chapter 6 Question 10 Page 266 Part A: Revenue is P*Q. Obtain Marginal Cost function through 160 + 16Q + 0.1Q2 FOC (derivative of above equation) 16 + 0.2Q= MC From the P= 96 – .4Q we can determine that total revenue = 96Q – .4Q2 and the derivative or FOC is thus 96 – .8Q= MR Set MC = MR 16 + 0.2Q = 96 – 0.8Q Q=80 We solve for P by plugging this into our original equation P= 96-.4(80) P=64 Profit = 5,120 (80*64) – 2,080 (160 + 16*80 + .1(80)2) = $3,040 Part B: C =160 + 16Q + .1Q2 AC= (160+16Q+.1Q^2)/Q MC=d/dQ(160 + 16Q + .1Q2) MC=16 + .2Q AC=MC 160/Q + 16 + .1Q = 16 + .2Q 160/Q = .1Q .1Q2 =160 Q= 40 Average cost of production is minimized at 40 units, she is correct as AC = MC (see below). AC = 960/40 =24 MC = 16 + (.2) ($40) = $24 However, optimal output is Q=80 where MR = MC, therefore her second claim of 40 units as the firm’s profit maximizing level of output is incorrect. P = 96 – .4 (40) P=$80 TR = 80*40 =3,200 C = 160 + 16Q + .1Q2 =960 Profit = Revenue – Cost = 3,200 – 960 = 2,240 therefore output at 80 is greater than the profit at 40. Part C: We learned from part a the single plant cost is $2,080 or (160 + 16*80 + .1(80)2). If two plants were open each producing the minimum level of output detailed in part B (Q=40) then total cost would be (Q)*(AC) = 24*80 = $1,920. You can compare this to the cost in part A of $2,080 and determine it is cheaper to produce using the two plants.

Economic Growth and Poverty Alleviation - 4151 Words

Does Economic Growth lead to Poverty Alleviation? Please compare and contrast very briefly the experiences of China, India and Brazil. What lessons can an African country of your choice learn from these experiences? INTRODUCTION The last few decades witnessed a rapid economic growth in developing countries. However, over 88% of the 1.2 Billion world poor (Olinto et al, 2013) live in these countries. (Appendix: Table 1.1) This phenomenon poses the question if the recent growth has been pro-poor . This essay argues that growth output alone is not sufficient for poverty alleviation; rather complementary measures and policies need to exist to create sustainable pro-poor growth. The essay has been organized as follows: First,†¦show more content†¦In the Lewis Theory (Todaro and Smith, 2011), surplus labour from the rural subsistence sector is gradually transferred to urban industrial sector for higher productivity. The rural-urban migration, in the long run, may result in urban unemployment, wage decrease, loss of agricultural productivity, debt accumulation, monopolization, lack of access to credit and insurance, and social exclusion. Amartya Sen (1999) described economic growth as a crucial means for expanding the substantive freedoms that people value. But in reality, in India, more than 270,000 farmers caught in a debt trap have committed suicide since 1995. (Shiva, 2013; Sainath 2013) According to Gini index , global inequality is 0.7 points today (Milanovic, 2005) while it is between 4.5 – 4.7 for developing countries that saw a proliferation of economic activities in the recent past (Appendix: Figure 2.4 and 2.5) The Kuznet’s curve , on the other hand, tells a different story about inequality. (Kuznets, 1955; Appendix: Figure 2.6). The inequality seen in developing countries is part of the development and will phase out as more growth is achieved. 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