Free Essay

Chem Analysis

In: Science

Submitted By sam1389
Words 3066
Pages 13
Experiment 2: Determination of phosphate in water samples by the Ascorbic Acid Reduction Method
Phosphorus plays specific in many biochemical procedures as it is an essential nutrients for plants at a definite concentration; it is one of the most abundant elements on earth and is commonly found in the form of phosphate such as orthophosphate, hydrolysable phosphate and organic phosphorus compound. One of the key effects of phosphorus is the eutrophication of surface water. This is where the uncontrollable growth of algae occurs. Waterways should not exceed a concentration of phosphorus greater than 0.05mg/L. Increased levels of phosphates in water are attributed by fertilisers and sewage due to human factors will cause the water surface to glow green or red, depending on the type of algae. The most importantly consequence of this is that algae depletes the amount of dissolved oxygen in the water. The depletion of oxygen can be very dangerous to aquatic life. Therefore it is essential that waterways be carefully monitored for phosphorus content.
The analysis of phosphorus content is most commonly carried out using the ascorbic acid-molybdate blue method. The orthophosphate ion reacts with ammonium molybdate and antimony potassium tartrate under acidic conditions to form 12-molybdophosphoric acid. This complex is reduced with ascorbic acid to form a blue complex which absorbs light at 880 nm.The absorbance is proportional to the concentration of orthophosphate in the sample. The chemical equation can be seen below.

[P(Mo3O10)4]3- + 3H+ + Ascorbic acid → (MoO2.4MoO3)2.H3PO4+2MoO2+Dehydroascorbic acid+ 4H2O

Therefore, the intensity can be measured using a spectrophotometer. This instrument operates by passing a monochromatic beam of light through the sample. Then, the identification of the intensity of light reaching the detector will be performed.

The Beer-lambert law can then be used to calculate the concentration which states a relationship between the amount of light absorbed and the concentration. This law states that:

A = ε C l

Where, A= absorbance, C= Concentration, l = path length and ε = absorptivity constant.

The overall aim of this experiment is to determine the concentration of orthophosphate and total phosphorous in a water sample collected from Macquarie Lake Yerbury using the ascorbic acid reduction method followed by UV-visible spectrophotometer.

Experimental: Sampling and handling
The water sample was collected from Lake Yerbury, on a sunny day, in a clean plastic bottle, washed with a solution of HCl and rinsed with distilled water. To avoid suspended materials, the sample was filtered using a 0.45μm membrane filter.

Dihydrogen phosphate standard calibration:
Six standard solutions were prepared by accurately pipetting 0, 1, 2, 4, 7 and 10mL of 0.2197g/L KH2PO4 and 30mL of distilled water into 50mL volumetric flasks. Then 9mL of molybdate/ascorbic acid reagent was added to each sample. The solutions were allowed to develop a blue colour for 15 minutes .

Determination of orthophosphates:
From the filtered sample, 40 mL was pipetted into a 50mL volumetric flask followed by a drop of phenolphthalein indicator to obtain a red colour. then, three drop of 6M sulphuric acid were added to discharge the colour. Following the same procedure as before, we added 9ml of molybdate/ascorbic acid reagent. Then, The solutions were allowed to develop a blue colour for 15 minutes .

Determination of total phosphorous:
From the same filtered sample, 40 mL was pipetted into a 100mL conical flask followed by adding 0.4 ammoniumpersulfate and v=1 ml of sulphuric acid. Then some bloiling chips was added to the whole mixture. The sample was gently boiled on a heating plate for one hour fifteen minutes, with the volume of solution maintained between 25 and 50 mL. The end of the boiling was determined when the volume decreased to 10mL. To the cooled sample a drop of phenolphthalein indicator was added followed by 1 mol L-1 sodium hydroxide solution dropwise to adjust the colour to a faint pink and 6 mol L-1 sulfuric acid until became colourless again. After that, 9mL of molybdate reagent was added, subsequently dilution to 50 mL with distilled water. Then, the solution was allowed to develop a blue colour for 15 minutes.
At the end, all of our samples prepared before were analysed and the absorbances of the samples and standards were then measured at 880nm in 1cm cell using a Shimadzu UV- Visible mini 1240 spectrophotometer.
All the samples were analysed using UV spectrophotometer at 880 nm in a 1cm cell. Absorbance values are given in the tables I, II and III.
Table I: Absorbances of standards Sample | Concentration(mg/L) | Absorbance | Blank | 0.0000 | 0.000 | 1 | 0.0439 | 0.007 | 2 | 0.0879 | 0.016 | 3 | 0.1758 | 0.032 | 4 | 0.3076 | 0.053 | 5 | 0.4394 | 0.076 |

Table II: Absorbances for samples of orthophosphate | Absorbance | Blank | 0.000 | Sample solution | 0.027 |

Table III: Absorbances for samples of total phosphorous | Absorbance | Blank | 0.000 | Sample solution | 0.037 |

Discussion: A calibration plot was constructed between the absorbances obtained for standard samples of dihydrogen phosphate using the regression sheet and their corresponding concentrations as the following:

From the obtained plot, Slope = 0.1728/mgL-1, Intercept = 0.0028833
Therefore the line of regression can be written as: Absorbance = 0.1728x [Phosphate] mg/L + 0.0028833.
T test will be performed to check for the significance of the correlation coefficient:
Step 1: Null Hypothesis: There is no statistically significant correlation between concentration of phosphate (mg/L) and absorbance.
Step2: Alternative Hypothesis: There is a statistically significant correlation between the concentration of phosphate (mg/L) and absorbance.
Step 3: Test Statistic = /r/n-2(1-r²) = 0.99966-2(1-(0.99962) = 70.68
Step 4: Critical Value from the two sided t table with 4 degrees of freedom is 2.78 at the 95% confidence level.
Step 5: As 70.68 >2.78, we reject null hypothesis
Step 6: We conclude that there is a statistically significant correlation between the concentrations of phosphate (mg/L) and the absorbance at 95% confidence level.

From the residual plot, all of the absorbances values lie horizontally around the horizontal line.
Thus, referring to the scatter plot and the residual plot, we can conclude that there is linear a relationship between the concentration of phosphate and absorbances which was showed by the T-test.
Determination of 95% confidence intervals of Slope and Intercept
1-Calculation of 95% confidence interval of slope:
(SD) slope = Sy/x(x-x)2 =0.000880.142886= 0.002
Confidence interval for slope (b) = ±t (n-2) (SD)slope = ±2.78 × 0.002= ± 0.0556
Calculation of 95% confidence interval of intercept:
(SD) intercept = S y/xx2n(x-x)2 = 0.00120.32836×0.142886 = ±0.000743
Confidence interval for intercept (a) = ±t (n-2) (SD)intercept= ± 2.78 × 0.0007 = ±0.002
The calibration plot equation can be given as:
Absorbance = 0.1728± 0.0556× [phosphate]mg/L+0.0028± 0.002
Determination of Orthophosphates concentration in water samples
Interpolation of the external calibration plot will be used to find the concentration of orthophosphate in the water sample.
Referring to table II:
The absorbance found for orthophosphate =0.027
Therefore the concentration of orthophosphate will be using the regression line:
Absorbance = 0.1728x [Phosphate] mg/L + 0.0028
Therefore, 0.027 = 0.1728x [Phosphate] mg/L + 0.0028
Orthophosphate concentration(mg/L) = (0.027-0.0028)/0.1728
Orthophosphate concentration = 0.1400mg/L
After considering dilution factor is: 0.1400 mg/L x (40/50) = 0.175mg/L
Calculation of 95% confidence interval for the orthophosphate concentration in the sample:
(SD)Orthophosphate = Sy/xb1+1n+Y-y2b²x-x2 = 0.00080.17281+16+(0.027-0.03)20.17282 (0.142886)=0.0500
Confidence interval for the Orthophosphate concentration in water sample:
±t (n-2) (SD) Orthophosphate= ±2.78 x 0.0500 = ± 0.1390
Therefore the orthophosphate concentration will be = 0.175± 0.1390 mg/L
Determination of Total phosphorous concentration in the water samples
Referring to table III:
The total phosphorous absorbance will be =0.037 The concentration of total phosphorous was calculated by substituting this measured absorbance of 0.038 in the regression line:
Absorbance = 0.1728 x [Phosphate] mg/L + 0.0028
Hence, 0.037= 0.1728x [Phosphate] mg/L + 0.0028
Total phosphorous concentration (mg/L) = (0.037-0.0028)/0.1482
Total phosphorous concentration = 0.2307 mg/L
Total phosphorous concentration after considering the dilution factor is calculated as the following: 0.2307 mg/L x (40/50) = 0.184 mg/L
Calculation of 95% confidence interval for the total phosphorous concentration in the sample:
(SD)Total phosphorous = Sy/xb1+1n+Y-y2b²x-x2 = 0.00080.17281+16+(0.037-0.0306)20.17282(0.14288)= 0.0050
Hence, confidence interval for the Total phosphorous concentration in water sample can be calculated as the following:
±t (n-2) (SD) Total phosphorous= ±2.78 x 0.0050 = ± 0.0139
Therefore the total phosphorous concentration= 0.184 ± 0.0139 mg/L.

The summary of the orthophosphate and the total phosphorous concentration collected from lake Macquarie will be showed in table VI: Sample | Concentration(mg/L) | 95% Confidence Interval | Orthophosphate | 0.175 | 0.1390 | Total phosphorous | 0.184 | 0.0139 |

From these results, we can demonstrate that there is a higher concentration of phosphorous than the recommended limits of 0.05 mg/l in Macquarie lake Yerbury. Therefore, this suggests that might be an influence of pollution carried out by humans leading to eutrophication and depletion of oxygen in our aquatic environment. There is a quite disturbance of this equilibrium of phosphorous in the ground and the surface water.
It has also to be considered possible errors during the analysis which might have been occurred due to temperature deviation and reaction time fluctuation. Some loss of the phosphorous compounds on the wall of containers can be another issue for variation of results. Additionally two people were involved during the analysis which increases the level of trueness.
This experiment was based on the formation of the blue coloured phosphorus molybdenum from the reduction of phosphomolybdate with ascorbic acid. The intensity of the ‘blueness’ was proportional to the concentration of phosphate in the solution. Therefore a UV spectrometer was used to determine the absorbances of several solutions including our calibration standards. Using the standards we were able to determine a calibration plot with a correlation coefficient of 0.996, which is not great but can be worked with. We tested its significance with the t test and the unknown samples of orthophosphate and total phosphate was interpolated from the calibration plot. The confidence intervals of these concentrations were also calculated and reported.

We also calculated the total amount of phosphate in mg/L which was found to be
( 0.184mg/L). In our introduction we stated that normal waterways should have a phosphorus concentration of 0.05mg/L. Lake Yerbury has a phosphate concentration that increases the chance of eutrophication. We believe we have satisfactorily achieved our aim to determine the concentration of orthophosphate and total phosphate in a water sample collected from Lake Yerbury using the ascorbic acid reduction method. Therefore, to improve the precision of the result, more number of samples from Macquarie Lake should be collected and analysed. In addition, samples should be maintained in constant temperature before analysing time. Moreover, maintain of a constant reaction time for all samples should be considered. Finally, the current results warn us that Macquarie Lake is affected by human influence and further precautions and care can be taken, to minimize this effect and return the phosphate levels into normal.
Learning outcomes:
I agree that I do understand that increased phosphate levels will put the aquatic life through the risk of eutrophication and depletion of oxygen ,which can have detrimental effects in aquatic environments.

We also have learnt the practical use of UV spectrophotometers to get absorbance readings of samples that have a change in colour intensity or turbidity. As well the equations involved to find the concentration of total phosphorous.

In this experiment, adequate preparation was required as there were many factors especially time that could affect our end results. I agree that I have come to appreciate that there may be possible difficulties in analysis of real life samples.

Additional exercises:
Question 1:
It is a fertilizer produced by the action of concentrated sulphuric acid on powdered phosphate rock. It is found in naturally occurring ores, plant ,animal wastes and fertilisers.
Eutrophication is the process where water is enriched by excessive concentrations of nutrients such as phosphate or nitrate causing the growth of algae on the surface of water. With more human activities and pollution , the rate of decomposition increases which lead to the depletion of oxygen causing the aquatic plants and animal to die.

In solution, the equilibrium will be as follow:
H3PO4 + H2O ⇋ H2PO4- + H3O+
H2PO-4 + H2O ⇋ HPO4-2 + H3O+
HPO4-2 + H2O ⇋ PO4-3 + H3O+
[H3PO4] Ka = 7.5 x 10-3
[H2PO4-] Ka = 6.8 x 10-8
[HPO4-2] Ka = 1.7 x 10-12
If we let cT be the sum of the molar concentration of the phosphate – containing solution species in the solution:

cT =[H 3PO 4] + [H 2PO4] + [HPO42-] + [PO4-3] → (1) the alpha value for the free acid α0 is defined as α0= H3PO4cT → (2) And alpha value for H2PO4-, HPO42-;[ PO43-] are given by similar equation: α1= H2PO4-cT; α2= HPO42-cT; α3= PO43-cT →(3)
The sum of the alpha values for system must equal unity: α0 + α1 + α2 + α3 = 1 alpha values for the maleic acid system are expressed in terms of [H3O+], Ka1.,Ka2and Ka3 to obtain such expressions from (1), (2) and (3) α0= [H 3O+]3[H 3O+]3+ Ka1 [H 3O+]2+ Ka1.Ka2[H 3O+] + Ka1.,Ka2Ka3 α1=Ka1[H 3O+]2[H 3O+]3+ Ka1 [H 3O+]2+ Ka1.Ka2[H 3O+] +Ka1Ka2Ka3 α2= Ka2[H 3O+][H 3O+]3+ Ka1 [H 3O+]2+ Ka1.Ka2[H 3O+] + Ka1Ka2Ka3 α3= Ka1Ka2Ka3[H 3O+]3+ Ka1 [H 3O+]2+ Ka1.Ka2[H 3O+] + Ka1Ka2Ka3

At the Ph= 4.5
The mole- percentage of [H 3PO 4]: α0 = [10-4.5]3[10-4.5]3+ 7.5×10-310-4.52+7.5×10-3×6.8×10-810-4.5+7.5×10-3×6.8×10-8×1.7×10-12


The mole- percentage of [H 2PO 4-]: α1 = 7.5×10-3[10-4.5]2[10-4.5]3+ 7.5×10-310-4.52+7.5×10-3×6.8×10-810-4.5+7.5×10-3×6.8×10-8×1.7×10-12


The mole- percentage of [HPO 4-2]: α2 = 6.8×10-8[10-4.5][10-4.5]3+ 7.5×10-310-4.52+7.5×10-3×6.8×10-810-4.5+7.5×10-3×6.8×10-8×1.7×10-12 α2=2.14×10-4 The mole- percentage of [PO 4-3]: α3 = 7.5×10-3×6.8×10-8[10-4.5][10-4.5]3+ 7.5×10-310-4.52+7.5×10-3×6.8×10-810-4.5+7.5×10-3×6.8×10-8×1.7×10-12 α3=1.15×10-10 By using the same way, we can calculate all the mole percentage in each of the Ph values: pH | H2PO4- | HPO4-2 | PO4-3 | H3PO4 | 4.5 | 0.99 | 2.14×10-3 | 1.154*10-10 | 4.19×10-3 | 6 | 0.94 | 0.06 | 1.08*10-7 | 1.25*10-4 | 7 | 0.60 | 0.41 | 6.88*10-6 | 7.94*10-6 | 8 | 0.19 | 0.87 | 1.48*10-4 | 1.71*10-7 | 9.5 | 0.0046 | 0.99 | 5.32*10-3 | 1.94*10-10 |

pH | H2PO4- | 4.5 | 0.99 | 6 | 0.94 | 7 | 0.6 | 8 | 0.19 | 9.5 | 0.0046 |

Figure 3 Plot of mole- percentage of H2Po-4 versus Ph value

In the graph, there is a decrease of the mole percentage of H2PO4 between PH=6 and Ph=8.
Thus, for the determination of the concentration of phosphates the pH should be less than 6.

Question4: we can use less sensitive maxima (a test that include solutions under low wavelengths) to reduce background signal.

Question5: Absorbance (A) | A365 | A700 | Solution 1 | 0.0235 | 0.617 | Solution 2 | 0.0714 | 0.755 | Solution 3 | 0.0945 | 0.920 | Solution 4 | 0.0147 | 0.592 | Solution 5 | 0.0540 | 0.685 | Molar absorptivities are: | Cobalt (Co) | 3529 | 428.9 | Nickel (Ni) | 3228 | 0 |

From equation : A= εLc

A365 = εCo365 L CCo + εNi365 L CNi A700 = εCo700 L CCo + εNi700 L CNi
Absorbance at 700nm: for solution 1 at 700 nm
0.617 = 428.9 M-1 x 1.00 cm x CCo + 0 M-1 x 1.00 cm x CNi
CCo=0.617/428.9*mol/L = 1.44 x 10-3 mol/L the concentration for the other solutions will be calculated using the same method : Solution | A700 | CCoin mol/L | CNi in mol/L | 1 | 0.617 | 1.44 x 10-3 | 0 | 2 | 0.755 | 1.76 x 10-3 | 0 | 3 | 0.920 | 2.15 x 10-3 | 0 | 4 | 0.592 | 1.38 x 10-3 | 0 | 5 | 0.685 | 1.60 x 10-3 | 0 |
Absorbance at 365nm: for solution 1 at 365nm .0235 = 3529 M-1 x 1.00 cm x 1.44 x 10-3 + 3228 M-1 x 1.00 cm x CNi
CNi =0.0235-3529 M-1 x 1.00 cm x 1.44 x 10-3 /3228
CNi=- 1.57 x 10-3

Using the same calculation for other solution:
Concentration of Cobalt and Nickel from sample solutions

Solution | A365 | CCoin mol/L | CNi in mol/L | 1 | 0.0235 | 1.44 x 10-3 | 1.57 x 10-3 | 2 | 0.0714 | 1.76 x 10-3 | 1.90 x 10-3 | 3 | 0.0945 | 2.15 x 10-3 | 2.32 x 10-3 | 4 | 0.0147 | 1.38 x 10-3 | 1.50 x 10-3 | 5 | 0.0540 | 1.60 x 10-3 | 1.73 x 10-3 |…...

Similar Documents

Premium Essay


...mechanisms. Nobel Prize in 1963. * Flemming, Sir Alexander (1881-1955) Discovered the antibiotic penicillin (1928). Nobel Prize in 1945. * Goldstein, E. (1886) Used cathode ray tube to study 'canal rays', which possessed electrical and magnetic properties opposite those an electron. * Hertz, Heinrich (1887) Discovered the photoelectric effect. * Moseley, Henry G.J. (1887-1915) Discovered the relation between the frequency of the x-rays emitted by an element and its atomic number (1914). His work led to the reorganization of the periodic table based on atomic number rather than atomic mass. * Hertz, Heinrich (1888) Discovered radio waves. * Adams, Roger (1889-1971) Industrial research on catalysis and methods of structural analysis. * Midgley, Thomas (1889-1944) Discovered tetraethyl lead and it used as an antiknock treatment for gasoline (1921). Discovered fluorocarbon refrigerants. Performed early research on synthetic rubber. * Ipatieff, Vladimir N. (1890?-1952) Research and development of catalytic alkylation and isomerisation of hydrocarbons (together with Herman Pines). * Banting, Sir Frederick (1891-1941) Isolated the insulin molecule. Nobel Prize in 1923. * Chadwick, Sir James (1891-1974) Discovered the neutron (1932). Nobel Prize in 1935. * Urey, Harold C. (1894-1981) One of the leaders of the Manhattan Project. Discovered deuterium. Nobel Prize 1934. * Roentgen, Wilhelm (1895) Discovered that certain chemicals near a cathode ray......

Words: 2871 - Pages: 12

Premium Essay

Chem Exam

...Chemistry 1B Experiment 16 77 16 Qualitative Analysis Introduction The purpose of qualitative analysis is to determine what substances are present in detectable amounts in a sample. This experiment has two parts. In the first part, you will analyze an unknown solution for the presence of seven common ions. In the second part, you will test an unknown solid to determine which of two possible identities is correct. Part I. Spot Tests for Some Common Ions A simple approach to the qualitative analysis of an unknown solution is to test for the presence of each possible ion by adding a reagent which will cause the ion, if it is in the sample, to react in a characteristic way. This method involves a series of “spot” tests, one for each ion, carried out on separate samples of the unknown solution. The difficulty with this way of doing qualitative analysis is that frequently, particularly in complex mixtures, one species may interfere with the analytical test for another. Although interferences are common, there are many ions which can be identified in mixtures by simple spot tests. In this experiment we will use spot tests for the analysis of a mixture which may contain the following commonly encountered ions in solution: CO32– SO42– PO43– SCN Cl – – carbonate sulfate phosphate thiocyanate chloride acetate ammonium C2H3O2– NH4+ 78 Chemistry 1B Experiment 16 The procedures we involve simple acid-base, precipitation, complex ion formation or......

Words: 4527 - Pages: 19

Free Essay

Analytical Chem

...Analytical Chemistry? 2 The Analytical Perspective 5 Common Analytical Problems 8 Key Terms 9 Summary 9 Problems 9 Suggested Readings 10 References 10 Chapter 2 Basic Tools of Analytical Chemistry 11 2A Numbers in Analytical Chemistry 12 2A.1 Fundamental Units of Measure 12 2A.2 Significant Figures 13 2B Units for Expressing Concentration 15 2B.1 Molarity and Formality 15 2B.2 Normality 16 2B.3 Molality 18 2B.4 Weight, Volume, and Weight-to-Volume Ratios 18 2B.5 Converting Between Concentration Units 2B.6 p-Functions 19 2C Stoichiometric Calculations 20 2C.1 Conservation of Mass 22 2C.2 Conservation of Charge 22 2C.3 Conservation of Protons 22 2C.4 Conservation of Electron Pairs 23 Chapter 3 The Language of Analytical Chemistry 35 3A Analysis, Determination, and Measurement 3B Techniques, Methods, Procedures, and Protocols 36 3C Classifying Analytical Techniques 37 3D Selecting an Analytical Method 38 3D.1 Accuracy 38 3D.2 Precision 39 3D.3 Sensitivity 39 3D.4 Selectivity 40 3D.5 Robustness and Ruggedness 42 3D.6 Scale of Operation 42 3D.7 Equipment, Time, and Cost 44 3D.8 Making the Final Choice 44 36 18 iii iv 3E Modern Analytical Chemistry Developing the Procedure 45 3E.1 Compensating for Interferences 45 3E.2 Calibration and Standardization 47 3E.3 Sampling 47 3E.4 Validation 47 3F Protocols 48 3G The Importance of Analytical Methodology 3H Key Terms 50 3I Summary 50 3J Problems 51 3K Suggested Readings 52 3L References 52 48 Chapter......

Words: 88362 - Pages: 354

Free Essay

Organic Chem

...CHEM 232 1. Organic Chemistry II Problem Set 12 Identify each of the following compounds as reducing or nonreducing sugars: H HO HO H HO CH2 O H OH H CH2OH OH B Nonreducing sugar (acetal) CHO H H E OH OH CH2OH CHO OH OH OH CH2OH H H G OH HO CH2OH O H H H OCH2CH3 H HO H OH OH CH2OH O H H H OH CH2 O C OH H CH2OH H OH H H O H HO A Reducing sugar (hemiketal) Nonreducing sugar (acetal and ketal) CH2OH C O OH OH CH2OH CH2OH HO OH H O H CH2OH H H H H F OH D Reducing sugar (hemiketal) Reducing sugar (aldose) Reducing sugar (aldose) Reducing sugar (ketose) 2. Which of the structures shown above are: a. Pyranose monosaccharides A and B b. Furanose monosaccharides D c. Tetroses E d. Aldopentoses F e. Ketopentoses G f. Disaccharides C 3. Give the Haworth formula for α-D-Glucopyranose. H HO H OH CH2OH O H H OH H OH 4. Complete the following reactions: a. CH3CH2CH(OH)CH2OH b. O HOCH2 C CH2OH HIO4 CH3CH2CHO + CH2=O OH O O AgNO3-NaOH NH3 HOCH2 CH C Ag0 OH HOCH2 C c. H H CHO OH OH CH2OH HIO4 3 HCO2H OH CHOH HOCH2 CH C d. O H CHO H H OH OH CH2OH HCN CN H H H OH OH OH CH2OH HO CN H OH OH CH2OH + CH2=O + H H e. H HO H CHO OH H OH CH2OH NaBH4 H2O H HO H CH2OH OH H OH CH2OH h. H HO f. HO H H CHO H OH OH CH2OH CHO 1. Br2/H2O 2. H2O2 Fe2(SO4)3 H H OH OH CH2OH g. CHO H OH H H OH OH CH2OH 3 PhNHNH2 CH NNHPh C H H NNHPh OH CH2OH O H H H OH OH H OH CH3OH H+ ...

Words: 401 - Pages: 2

Premium Essay


...Georgia Institute of Technology CHEM 3700 Research Paper The German Energiewende “Die deutsche Energiewende“ Professor: Dr. Thomas Orlando Student: Shyam Rangarajan Abstract With the diminishing reserves of fossil fuels compounded by their negative effects on the environment, clean and renewable sources of energy to meet next generation energy demands undoubtedly need to be found. However an even more critical aspect governing global energy challenges is the way governments around the world take responsible actions and implement energy policies that are conducive to a sustainable future. A recent development in this regard is the German Energiewende, or Energy Transformation, one of the most ambitious projects of its type. The project aims to completely transition the German energy sector to one powered by renewables by the year 2050. Along with the benefits to energy security and controlled carbon emissions that such a policy would bring, the German government‟s efforts have also sparked a wide-scale increase in investment into emerging technologies. This has led to a vast increase in employment in the energy sector, and has also given rise to the concept of „prosumers‟, or individuals who produce and sell their own electricity through privately owned solar installations. Although the long term benefits of this project are enormous and the potential limitless, there are several short term repercussions that are slowing it down. Chief among these is the spiraling...

Words: 3572 - Pages: 15

Premium Essay


...some Limewater, which caused it to bubble and turn cloudy * Next, I let the gas from the sodium and hydrogen peroxide go into some Bromthymol Blue. It turned yellow. * When I mixed the Alka Seltz/water solution excess gas into the limewater it... [continues] Read full essay Cite This Essay APA (2013, 06). Experiment 4: properties of gases. Retrieved 06, 2013, from MLA MLA 7 CHICAGO Welcome is the web's leading learning tool. We inspire millions of students every day with over 1,600,000 model essays and papers, AP notes and book notes. Related Essays Properties of gases - chem labpaq ...Properties of Gases General Chemistry 1 Lab 5 Abstract: The purpose of this experiment is... 3 Pages March 2013 Chemistry lab 3 properties of gases ...Dex Cimino 3/24/2013 CHE101, Tamburro Lab 3 – Properties of Gases Data Table: Experiment... 2 Pages March 2013 Properties of gases ...Gas | Flame Reaction | Glowing Splint | Limewater Reaction | Bromothymol Blue Reaction |... 3 Pages July 2013 Properties of gases ...Purpose: The purpose of this experiment was to test and observe the physical and chemical... 7 Pages July 2013 Properties of gases lab ...Student Mrs. Teacher Class Date Katie Limbach Mrs. Falk Chemistry 09-13-13 Title:... 4......

Words: 609 - Pages: 3

Premium Essay


...probe. 5 Hold the tube vertically so that you can drop the magnet through the tube. You have created a system that will measure the potential developed in the wire due to the changing magnetic field. Instructions: 1 • • • 2 3 4 Start the LoggerPro program in your computer. It should open with a Graph window and a table window. Close the table window. Set the time (x-axis) to a minimum of 0 and a maximum of 2. Set the Potential (y-axis) to Autoscale. Press enter to begin collecting data. Drop the magnet though the tube. From your graph you should see a horizontal line with one positive peak and one negative peak. If you missed the data collection time window try again. Select only the part of your graph that contains the peaks. ANALYSIS AND MODELING: The signal before and after the peaks may not be centered on zero volts. If so, there is a slight offset in the Voltage Probe. This does not affect your qualitative answers, but any quantitative answers should be corrected for this offset. 1 2 3 4 5 Explain the qualitative shape of the signal. Why are there both positive and negative peaks? The peaks will not be symmetric, with the second peak slightly larger in amplitude but narrower. Why? Using Faraday’s law calculate the change in magnetic flux for both the positive and negative peaks. How do they relate? List at least three things in your setup or experimental procedure that you could change, any one of which would invert the signal. Why would each of these......

Words: 566 - Pages: 3

Free Essay


...varias ecuaciones cúbicas de estado. ‘Mas ellas son casos especiales de la ecuación cúbica de estado genérica: RT WV - 17) p= V - b (V-b)(V +¿w+e) (3.42) En ella b, 8, 6, E y 7 son parámetros que en general dependen ¿le la temperatura y, para mezclas, de la composición. Aunque la ecuación (3.42) parece muy flexible, tiene limitaciones inherentes debido a que es cúbica.” La ecuación de RedliddKwong se abtiene de la ecuación (3.42) con 8 = a/T”‘, 77 = S = b y E = 0. Otras formas diferentes de la ecuación (3.42) se obtienen si se hacen sustituciones diferentes; el uso más común de todas ellas tiene que ver con el equilibrio vaporkquiclo (capítulo 13). 5M. M.Abbott,AICL!ZJ.,vol. 19,~~. 596-601, 1973;Arlv. in Chem. Series182,K.C. ChaoyR.1~. Robinson Jc, eds., pp. 47-70, Am. Chem. Soc., Washington, D:C., 1979. 96 CAPhL!JLO 3. *piedades volumétricas de los fluidos puros Las ecuaciones que tienen una exactitud global mayor son necesariamente más complejas, como lo ilustra la ecuación de Benedict/Webb/Rubin: ------- ___. ~.~ _ p- R T : B,RT-A,-C,lT2+óRT-CL V V2 V3 _-._-.-C_ _ C +aa+- V6 V3T2 (3.43) donde Ao, Bo, Co, a, b, c, cy y y son todas constantes para un fluido dado. Esta ecuación y sus modificaciones, a pesar de la complejidad que tienen, se emplean en las industrias del petróleo y del gas natural para hidrocarburos ligeros y algunos otros gases encontrados comúnmente. F$emplo 3.7 La presión de vapor del cloruro de metilo a 60°C es de......

Words: 37868 - Pages: 152

Free Essay


...CASO CEM-CAL CORPORATION Familiarizarse con el caso En Chem-cal Corporation se enfrenta a diez proyectos de inversión que consisten en diferentes inversiones iniciales. Lo cual los directivos tienen distintas formas de evaluación Que lleva a un problema de como poner valores en las diferentes inversiones. La otra gran interrogante es si es necesario elevar el capital para cubrir los proyectos Programados para el próximo año. La compañía tiene 8,2 millones de dólares disponibles Para inversiones de capital este año y los proyectos descritos ascienden a 18.650.000 Millones. Identificar los síntomas Existe una gran duda en el sentido de que tasa usar o escoger lo que implica si es asertivo usar una tasa de descuento apropiada para utilizar el método VPN o utilizar la tasa interna minima con el método TIR. Esta duda lleva a que no se esta logrando lo deseado. Identificar objetivos Uno de los objetivos principales es la valoración de estos proyectos. Chem- Cal Corporation debe volver a calcular el VPN de cada proyecto. Lograr una tasa de capital es decir la mejor opción es el Wacc y ver las fuentes para conseguir los recursos. Análisis de opciones para conseguir más capital - Elevar Ratios de deuda (30% cerca de la media del sector) - Vender bonos de 5 millones de dólares (un poco mas de 10.5%) - Fondo de amortizaciónDiagnostico Encontrar una fuente de financiamiento que no sea superior o igual al 11% y esa tasa seria un 10% para ajustar al wacc ya que...

Words: 420 - Pages: 2

Free Essay


...ones they previously existed in. Not all substances have the same energy levels, since they vary from substance to substance. This in turn means that the energy and wavelength of the light required to move electrons from their energy levels also varies between substances. Hence it is possible to use the specific spectrum produced by each substance to identify what it is. A substance that absorbs light appears coloured, however the observable colour is the complement of the colour(s) absorbed, i.e. the colour not absorbed, since it is what remains to reach our eyes. UV-Visible spectroscopy can be used to obtain qualitative data, such as through identifying compounds using spectra compounds, or quantitative analysis. However, it is generally used for quantitative analysis, in order to determine the concentration of substance in a sample. In this procedure, the spectrum of the pure substance as well as a wavelength at which the substance absorbs strongly, while other components in the sample do not, are recorded. Thereafter, the absorbance of the sample is measured at its wavelength and the absorbance of a series of standard solutions are collected to make comparisons. Iron is an essential human nutrient and can be detected using UV-Visible spectroscopy. In order to do so, Fe3+ is reacted with thiocyanate ions, giving an intensely red-coloured product, which is then used as a qualitative test for the presence of Fe3+. To ensure that the hydrolysis reaction of [Fe(H O) ] ...

Words: 2220 - Pages: 9

Premium Essay


...Rajdeep Singh Chem 2A Fregrene The Evolution of the Battery and Electric Vehicle After reading various articles on battery chemistry and how it applies to electric vehicles’ I have come to a conclusion. Although electric vehicles may seem like a great idea for many people, because they can eliminate a great amount of emission pollutants in the atmosphere, they arrive with many unresolved issues. Issues which may be expensive as well as incovineient to solve. The first article I read was “Chemstry’s Electric Oppurtunity” by Alex Scott. In his article, he spoke of the automobile firm Tesla and their new Model S vehicle. The Tesla Model S is a vehicle capable of driving up to 300 miles before needing to recharge, compared to the majority of other electric cars who can only reach 100 miles. The CEO of Tesla Motors, Elon R. Musk claims that “anything a gasoline based vehicle can do, the Model S can do” (Scott, pg 1). Which may prove to be true since the car goes from ‘0-60 in only 4.2’ (Scott, pg 1). Many people wonder how the Model S is capable of outperforming other electric vehicles. It is due to the car’s large 670kg battery. This battery provides an energy density of 240 Wh/Kg. ‘This is much denser than the nickel based batteries of other firms such as Toyota, Nissan, and GM’ (Scott, pg 2). It’s great that Tesla is able to raise the bar in the electric vehicle field and provide a more efficient vehicle for consumers to enjoy, however there is a large inhibitor. Batteries...

Words: 850 - Pages: 4

Premium Essay


...electronic configuration of the atom in question. The trends can be summarised as follows: 1. The successive ionisation energies of an atom always increase. The more electrons that are removed, the fewer the number electrons that remain. There is therefore less repulsion between the electrons in the resulting ion. The electrons are therefore more stable and harder to remove. 2. By far the largest jumps between successive ionisation energies come when the electron is removed from an inner shell. This causes a large drop in shielding, a large increase in effective nuclear charge and a large increase in ionisation energy By applying the above principles in reverse, it is also possible to predict the electronic structure of a species by analysis of the successive ionisation energy data: Eg Si: [pic] Large jumps occur between 4th and 5th and between 12th and 13th. Therefore there are three shells: The first contains 2 electrons, the second 8 and the third 4....

Words: 4310 - Pages: 18

Premium Essay


...unit = ($3,630/day)/(526 non-defective units/day) = $6.90/unit Select Machine A. 57 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 2-35 Strategy: Select the design which minimizes total cost for 125,000 units/year (Rule 2). Ignore the sunk costs because they do not affect the analysis of future costs. (a) Design A Total cost/125,000 units = (12 hrs/1,000 units)($18.60/hr)(125,000) + (5 hrs/1,000 units)($16.90/hr)(125,000) = $38,463, or $0.3077/unit Design B Total cost/125,000 units = (7 hrs/1,000 units)($18.60/hr)(125,000) + (7 hrs/1,000 units)($16.90/hr)(125,000) = $33,175, or $0.2654/unit Select Design B (b) Savings of Design B over Design A are: Annual savings (125,000 units) = $38,463 − $33,175 = $5,288 Or, savings/unit = $0.3077 − $0.2654 = $0.0423/unit. 58 © 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in...

Words: 10954 - Pages: 44

Premium Essay


...CHEM 111G – General Chemistry I Lab 1 Separation of the components of a mixture Aim of the experiment: To develop a stepwise procedure to separate the four components of a mixture of sand, salt, iodine and iron filings and determine the % of each by mass Brief procedure: The stepwise procedure was developed based on unique properties of each component: Salt: soluble in water Sand: insoluble in water Iodine: sublimes to vapor Iron: attracted to magnet We weighed out ~2g of the mixture in a 250mL beaker and chose to separate the iron first using a magnet. Since the iron sticks strongly to the magnet, the beaker + mixture was then reweighed to determine the mass of iron removed by difference. The iodine was then removed in the next step by placing the beaker on a hot plate covered by an ice cold evaporating dish and heating at maximum heat until no more purple iodine vapor were seen coming off. The deposited iodine crystals were recovered from the underside of the dish. Because some iodine is lost the beaker + mixture was then reweighed to determine the mass of iodine removed by difference. The remaining mixture of sand and salt was treated with deionized water to extract the salt (soluble). The sand was removed by filtration through a filter funnel via the water vacuum pump and the filtrate transferred to another weighed beaker. The salt was then recovered by evaporating the filtrate to dryness. The beaker was cooled and the salt mass determined by mass......

Words: 373 - Pages: 2

Free Essay

Chem Ia

...CHEM IA 11 Aim To determine the enthalpy change of reaction between zinc powder and aqueous copper(II) ions Uncertainties of Apparatus Electronic balance | ±0.001g | 50cm3 measuring cylinder | ± 0.5cm3 | Thermometer | ± 0.1oC | Digital stopwatch | ± 0.2s | Mass/g of zinc in weighing bottle measured by electronic balance for Method 1 Mass of Zn (±0.002) / g | 0.9123 | Mass of weighing bottle (± 0.001) / g | 3.5840 | Mass of Zn + weighing bottle ( ±0.001) /g | 4.4963 | Highest temperature recorded after adding Zn to copper (II) sulfate solution for Method 1 Initial temperature (±0.1) / ⁰C | 22.0 | Highest temperature (±0.1) / ⁰C | 31.5 | Temperature change (±0.2) / ⁰C | 8.50 | Mass/g of zinc in weighing bottle measured by electronic balance for Method 2 Mass of Zn (±0.002) / g | 0.9123 | Mass of weighing bottle (± 0.001) / g | 3.5840 | Mass of Zn + weighing bottle ( ±0.001) /g | 4.4963 | Temperature recorded before and after adding Zn is added to copper (II) sulfate for Method 2 Time (±0.2) / s | Temperature (±0.1) / ⁰C | 0.0 | 22.0 | 15.0 | 22.0 | 30.0 | 22.0 | 45.0 | 22.0 | 60.0 | 22.0 | 75.0 | 21.5 | 90.0 | 21.5 | 105.0 | 21.3 | 120.0 | 21.2 | 135.0 | 21.2 | 150.0 | 21.0 | 165.0 | 21.0 | 180.0 | 21.0 | 195.0 | 21.0 | 210.0 | 23.0 | 225.0 | 23.8 | 240.0 | 24.5 | 255.0 | 25.2 | 270.0 | 26.4 | 285.0 | 28.2 | 300.0 | 29.1 | 315.0 | 29.8 | 330.0 | 31.1 | 345.0 | 31.5 | (Highest......

Words: 731 - Pages: 3