...Xian China National Center for Atmospheric Research, Boulder, CO, USA H I G H L I G H T S • • • • The cases of haze formation in Beijing, China were analyzed. The effects of RH on PM2.5 concentration and visibility were studied. Gas-phase to particle-phase conversion under different visibility was analyzed. With high RH, the conversion SO2 to SO4= accounted for 20%. a r t i c l e i n f o Article history: Received 25 July 2014 Received in revised form 5 September 2014 Accepted 24 September 2014 Available online 7 October 2014 Editor: P. Kassomenos Keywords: Beijing Hazes Visibility PM2.5 PBL Secondary particle formation a b s t r a c t The causes of haze formation in Beijing, China were analyzed based on a comprehensive measurement, including PBL (planetary boundary layer), aerosol composition and concentrations, and several important meteorological parameters such as visibility, RH (relative humidity), and wind speed/direction. The measurement was conducted in an urban location from Nov. 16, 2012 to Jan. 15, 2013. During the period, the visibility varied from N 20 km to less than a kilometer, with a minimum visibility of 667 m, causing 16 haze occurrences. During the haze occurrences, the wind speeds were less than 1 m/s, and the concentrations of PM2.5...
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...Therefore cells size is wanted to be relatively small. Box A's surface area to volume ratio is the smallest and therefore will be very sufficient for diffusion to occur at a high rate. Box B's surface area to volume ratio is in between Box A and Box C and so therefore will not be very sufficient but diffusion can still occur at a medium rate. Box C has the largest surface area to volume ratio out of the three boxes and so diffusion will be unable to occur at a high rate. We can conclude that Box A would have the most beneficial cell shape as it has the smallest surface are to volume ratio out of the three boxes. Essay Feedback You have written an intelligent answer Emerald. Discussing the SAVR and also the shapes of each in reference to the data can earn full credit here. Points earned: 3 I found this question difficult because I didn’t understand referencing the CELL shapes and the SAVR to make a connection between the two, I was only able to reference the size of the boxes. I think for future reference I should discuss more about the “cells shape” to be able to make a clear reference and connection between the cells and the boxes. I think highlighting key points in the question can be a good strategy for me to use in the future in order for me to understand and answer the question fully. Question 2 (Worth 5 points) A student designed an experiment to test whether different concentration gradients affect the rate of diffusion through dialysis tubing. Four...
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...Project 13: Finding Concentrations and Transmittance Through Exploring Various Sodas By Kathleen Browne (browne8@clemson.edu) Emma Brandberg (ebrandb@clemson.edu) Dajonia Jackson (dajonij@clemson.edu ) Junkai Yang (junkaiy@clemson.edu ) Chemistry 101, Section 314 Group 2 Instructor: Mahsa Foroughian October 3rd Goals Understand how to properly read transmittance and absorbance levels using a Spec 20/Spec 200. Then,use food coloring to make solutions of different colors to distinguish the relationship between color, absorption, and wavelength. Analysis how the potassium permanganate reflect on the absorption of light, and its relationship to its percent transmittance and wavelength. Investigate how each beverage react with phosphoric acid. Next, Use ammonium vanadomolybdate (AVM) solution as a reagent to produce the color that represents the wavelength of phosphoric acid. Take the KH2PO4 to make a calibration curve that represents a range typical of the concentration of phosphoric acid in colas. Concentration and absorbance represents a linear relationship. Using this, a calibration curve can be created to determine the concentration of unknown solutions....
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...6) Discussion Procedure one: This very high surface area to volume ratio allowed the phenolphthalein to diffuse very quickly across the membrane in the cylinder shaped agar. Procedure two: The water diffuses away from the NaCl solution in every scenario, leading to the conclusion that it has the highest water potential of the solutions tested. Because a greater proportion of water diffused away from the sucrose than the glucose, it can be hypothesized that the sucrose has a lower water potential than glucose. Because water diffused to the glucose and away from the 5% ovalbumin solution, we can also hypothesize that the glucose has a lower water potential than the 5% ovalbumin solution. In order from greatest to least water potential, the...
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...The concentration versus absorbance graph yielded a straight line with R2 = 0.995. The limitation of pipetting skill can be seen clearly in the absorbance value of 1:250 solution, which slightly goes off the trendline. A possible error was that the required amount of liquid was not fully released to the tube. However, based on the linearity of the graph, the pipetting and dilution skills were acceptable. The DH5α in LB plate resulted in no bacterial growth, which indicates that some sources of error happened during the streak. Since there was no ampicillin presented, the bacteria should completely grow on the plate. One reason of error is that the loop used to streak the plate was too hot, which potentially killed the bacteria. Another possible...
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...for equilibrium were as follows: Kc (run#1) = 620; Kc (run#2) = 599; Kc (run#3) = 545. In effort to distinguish between the species of iodine the distribution coefficient ratio between two immiscible solvents (dichloromethane and water) was determined at multiple concentrations. I. Introduction The purpose of this experiment is to determine the equilibrium constant of iodine species at various concentrations. The equilibrium equation of the iodine species is as follows: I2+I_=I3- (1) The equilibrium constant is calculated as: KC=(I3-)I2I- (2) Where the equilibrium constant KC is calculated from the concentrations of the species and does not take into consideration the activities of the species. Knowing Kc will indicate if the reaction favors products (if Kc>1) or if it favors products (Kc<1). Sodium thiosulfate is used to determine the concentration of Iodine in solution via titration in the following reaction: 2S2O32-+ I3- → S4O62-+ 3I- Before titration excess KI is added to each solution to ensure that I2 is converted completely to I3-. This is done to ensure that molecular iodine is not loss to evaporation while in the aqueous phase (I3- is non-volatile). However, this method only exposes the total amount of iodine dissolved in the aqueous solution and not the concentrations of individual species. That determination can be aided by examination of the heterogeneous equilibrium existing between the aqueous layer and an organic layer comprised of...
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...Experiment 5 Titration of a Strong Acid with a Strong Base Part I: The purpose of this experiment was to prepare an aqueous solution of NaOH, and to determine the concentration of the secondary standard NaOH solution by titrating it with a solute of primary standard, KHP. Another objective of this experiment was to learn how to plot a 2nd derivative graph using LoggerPro, and determining the concentration of the secondary standard from the 2nd derivative data. The primary standard was KHP and the secondary standard was NaOH. The difference between the 2 types of standards is that the primary standard is a has powerful reactants and isn't sensitive to the the environment and the secondary standard is something that is determined will react with a highly pure primary standard that can be standardized. It was necessary to standardize the NaOH solution because NaOH absorbs moisture from the air, making the compound not 100% NaOH, so to obtain precise concentrations, the NaOH needed to be standardized with KHP. The molarity of the NaOH solution was determined by dividing the average number of moles of NaOH (determined by taking the moles of KHP and using the 1:1 ratio) and diving it by the average volume (number of L) at the indicator point. The difference between the equivalence point and the indicator point is that the indicator point is where the indicator changes color, and the equivalent point is the place in a titration where the amount of indicator added neutralizes...
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...Set out materials on a sizeable, open surface. Choose and create three to five concentrations of the Original, Diet, and Caffeine Free Coke. These will be used to make the solutions for which will have Daphnia placed into them to allow heart rate measurement. (How to create concentrations: For example (in mL), a twenty five percent solution would be represented in the ratio 1:4. The twenty five percent would consist of the liquid that is desired to be used; in this case, it is soda. The remaining seventy five percent would be water. In a basic 10 mL solution, 2.5 mL would entail the soda, and 7.5 mL would be comprised of water.) For each of the concentrations that have been chosen, measure the heart rate of four to seven individual Daphnia....
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...PROPERTIES OF SOLUTIONS [PROBLEMS ONLY] Units of solution concentration ➢ Molarity (M) = # of moles of solute per liter of solution [pic] ▪ Mass percent (weight percent) = percent by mass of the solute in the solution [pic] ➢ Mole fraction (() = ratio of the number of moles of a given component to the total number of moles of solution. Mole fractiona = (a [pic] ➢ Molality (m) = # of moles of solute per kilogram of solvent NOT temperature dependent. Represents a ratio of solute:solvent molecules at all times. [pic] Exercise 1 Various Methods for Describing Solution Composition A solution is prepared by mixing 1.00 g ethanol (C2H5OH) with 100.0 g water to give a final volume of 101 mL. Calculate the molarity, mass percent, mole fraction, and molality of ethanol in this solution. Exercise 2 Calculating Various Methods of Solution Composition from the Molarity The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent, molality, and normality of the sulfuric acid. Exercise 3 Differentiating Solvent Properties Decide whether liquid hexane (C6H14) or liquid methanol (CH3OH) is the more appropriate solvent for the substances grease (C20H42)...
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...intensity of the transmitted radiation leaving the medium. T is usually expressed as percent transmittance, %T: %T = I/Io x 100 The relationship between percent transmittance (%T) and absorbance (A) is given by the following equation: A = 2-log (%T) On most spectrophotometers two scales are present, %T and Abs. Absorbance has no units (Why?) and varies from 0 to 2 (linear region for most substances is from 0.05 to 0.7). The Beer-Lambert Law states that Abs is proportional to the concentration (c) of the absorbing molecules, the length of light-path through the medium and the molar extinction coefficient: A = εcl where: ε = molar extinction coefficient for the absorbing material at wavelength in units of 1/(mol x cm) c = concentration of the absorbing solution (molar) l = light path in the absorbing material (l=1 cm for our purposes) The Beer-Lambert Law may not be applicable to all solutions since solutions can ionize/polymerize at higher concentrations, or precipitate to give a turbid suspension that may increase or decrease the apparent absorbance. Further, the Beer-Lambert Law is most accurate between Abs...
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...See discussions, stats, and author profiles for this publication at: http://www.researchgate.net/publication/223357366 Cloud-point extraction of lead in saliva via use of non-ionic PONPE 7.5 without added chelating agent ARTICLE in TALANTA · FEBRUARY 2000 Impact Factor: 3.51 · DOI: 10.1016/S0039-9140(99)00252-0 · Source: PubMed CITATIONS DOWNLOADS VIEWS 77 369 333 4 AUTHORS, INCLUDING: Maria Fernanda Silva Liliana P Fernández National University of Cuyo Universidad Nacional de San Luis 17 PUBLICATIONS 363 CITATIONS 43 PUBLICATIONS 540 CITATIONS SEE PROFILE SEE PROFILE Available from: Maria Fernanda Silva Retrieved on: 08 September 2015 Talanta 51 (2000) 123 – 129 www.elsevier.com/locate/talanta Cloud point extraction of lead in saliva via use of nonionic PONPE 7.5 without added chelating agents Marta O. Luconi, M. Fernanda Silva, Roberto A. Olsina, Liliana P. Fernandez * ´ Area of Analytical Chemistry, National Uni6ersity of San Luis, CONICET, San Luis 5700, Argentina Received 4 March 1999; received in revised form 13 August 1999; accepted 16 August 1999 Abstract A new micelle-mediated phase separation of metal ions to preconcentrate trace levels of lead as a prior step to its determination by flame atomic spectroscopy has been developed. The methodology is based on the cloud point extraction of lead with PONPE 7.5 in the absence of chelating agent. The chemical variables affecting the sensitivity ...
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...CALCULATION EXERCICES To the LABORATORY EXPERIMENTS IN MEDICAL CHEMISTRY Edited by: Zoltán Matus Compiled by: Péter Jakus László Márk Anikó Takátsy Pécs, 2007 Table of content: Introduction 3 1. Stoichiometry I. Balancing equations 4 2. Stoichiometry II. Calculation exercises 9 3. The gaseous state 13 4. Concentrations of solutions 17 5. Calculations connected to titrimetry 28 6. Electrolytic dissociation 33 7. Dilute solutions 38 8. Hydrogen ion concentration, pH, buffers 44 9. Heterogeneous equilibria. Crystallisation, solubility product, partition coefficient 55 10. Thermochemistry 64 11. Electrochemistry 67 2 INTRODUCTION The chapter is devoted to helping the students practice the most important topics of General Chemistry. The order of the sections follows the schedule of the lectures and seminars, and their volume indicates the importance of the topic. Each section begins with a few solved problems. They represent the minimum requirement at the exam. The worked-out solutions are not the only ones. For an easier self-checking, the numerical results of the unsolved calculation exercises are given in parentheses after each question. Sources: 1.) Laboratory experiments in medical chemistry, ed. György Oszbach, Pécs, 1998. 2.) Villányi Attila: Ötösöm lesz kémiából, (6. ed.) Mőszaki Könyvkiadó, Budapest., 1999 3.) Charles E. Mortimer...
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...is given as: Qd = 500- P MR = 500 - 2Q a) Determine the monopolist's profit-maximizing price and output. MC = ATC = 50 MR = MC MR = 500 – 2Q 2Q = 500 - 50 2Q = 450 Q = 450÷2 =Q=225 maximizing quantity output 225 = 500 – P P=$275 profit max price. b) Calculate the monopolist's profit. TR-TC (P X Q) - (ATC X Q) = Q(P-ATC) 225×(275) =$61,875 c) What is the Lerner Index for the industry? L = (P-MC)÷P (275-50)÷275 225÷275= L=.818 DQ 3. There are two industries, A and B; and each industry consists of four firms. However, each of the four firms in industry A has a 25% market shares while those firms in industry B have 80% 10%, 5% and 5% market respectively.. a) Calculate the three- and four-firm concentration ratios for each industry. Industry A = 25 + 25 + 25 +25= 100 Industry B = 80+ 10 + 5 + 5 = 100 b) Calculate the Hirfindahl Hirschman Index for each industry. Industry A = 25^2 + 25^2 +25^2 +25^2 625 + 625 + 625 + 625 =0.25 Industry B = 80^2 + 10^2 + 5^2 + 5^2 6400 + 100 + 25 + 25 = .655 c) Are these industries equally concentrated? Explain your answer. No, Industry A 25% concentrated and industry B is higher concentrated at...
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...Method Number: EPA 200.7 Revision Number: 3.0 Date: September 3, 2010 Clinical Laboratory of San Bernardino, Inc. Standard Operating Procedure for the Determination of Metals by Inductively Coupled Plasma Spectroscopy 1. SCOPE AND APPLICATION: 1.1 This method provides procedures for the determination of dissolved elements in ground waters, surface waters, and drinking water supplies. It may also be used for the determination of total recoverable element concentrations in these waters and wastewaters. Dissolved elements can be determined after suitable filtration and acid preservation. Acid digestion procedures are required prior to the determination of total recoverable elements. To reduce potential interference, dissolved solids should be < 0.2% (w/v). Prepared samples may require dilution prior to analysis to avoid physical interferences. The method is applicable to the following analytes: Chemical Abstract Services Registry Numbers (CAS-No.) 7429-90-5 7440-39-3 7440-42-8 7440-70-2 7440-48-4 7440-50-8 7439-89-6 7439-95-4 7439-96-5 7439-98-7 7440-41-7 7440-43-9 7440-47-3 7440-02-0 7440-22-4 7440-09-7 7631-86-9 7440-23-5 7440-66-6 Dectection Limit of Reporting (DLR) 50 ppb 100 ppb 100 ppb 1 ppm 10 ppb 50 ppb 100 ppb / 40 ppb 1 ppm 20 ppb / 4 ppb 10 ppb 1.0 ppb 1.0 ppb 10 ppb 10 ppb 10 ppb 1 ppm 0.5 ppm 1 ppm 50 ppb 1.2 1.3 1.4 Analyte Aluminum (Al) Barium (Ba) Boron (B) Calcium (Ca) Cobalt (Co) Copper (Cu) Iron (Fe) Magnesium (Mg) Manganese (Mn) Molybdenum (Mo) Beryllium...
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...25% of 12. Etc. But do you know enough about percentages? Is a percentage the same thing as a fraction or a proportion? Should we take the difference between two percentages or their ratio? If their ratio, which percentage goes in the numerator and which goes in the denominator? Does it matter? What do we mean by something being statistically significant at the 5% level? What is a 95% confidence interval? Those questions, and much more, are what this book is all about. In his fine article regarding nominal and ordinal bivariate statistics, Buchanan (1974) provided several criteria for a good statistic, and concluded: “The percentage is the most useful statistic ever invented…” (p. 629). I agree, and thus my choice for the title of this book. In the ten chapters that follow, I hope to convince you of the defensibility of that claim. The first chapter is on basic concepts (what a percentage is, how it differs from a fraction and a proportion, what sorts of percentage calculations are useful in statistics, etc.) If you’re pretty sure you already understand such things, you might want to skip that chapter (but be prepared to return to it if you get stuck later on!). In the second chapter I talk about the interpretation of percentages, differences between percentages, and ratios of percentages, including some common mis-interpretations and pitfalls in the use of percentages. Chapter 3...
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