...trial 1 was more accurate and precise than trial 2. Introduction: The purpose of this experiment was to determine the heat of combustion under constant volume and excess oxygen. We identify ∆H°comb, heat of combustion, as the change in heat when the reaction between oxygen and two moles of DMP at a temperature of 298K and a pressure of 1 atm. The reaction for the combustion is expressed as…. 2 (l) + 39 O2 (g) → 32 CO2 (g) + 22 H2O (l) (1) A Parr bomb calorimeter was used in the experiment. As starting the experiment the volume was constant in the process. We can also say that changes in internal energy and molar enthalpy related with chemical changes are independent of pressure. Having a good understanding of the principles of calorimetry and thermodynamics would be beneficial while performing the experiment. The First Law of Thermodynamics states that energy is neither created nor destroyed, but remains constant in the universe. The change in internal energy of a system, ΔU, at a constant volume is the main thermodynamic property and is expressed by the following equation: ΔU = qv + w (2) Where, qv and w represent the heat and...
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...After recording the mass and volume of four types of liquids and an unknown liquid, one must find the density of the samples by graphing the data found. One must also identify the unknown concentration of a saltwater solution by using their graph. Finally, the percent error of the unknown concentration will be calculated where one will evaluate the precision and accuracy of the experiment. Materials: balance, 10.0 mL graduated cylinder, liquid samples: A, B, C, D, and unknown, plastic pipit, safety goggles, computer to record data, paper towels, and water to rinse. Procedure: Put safety goggles on and gather materials listed above. Find the mass of the graduated cylinder alone Choose a sample to start with: A, B, C, D, or unknown, (recommend going from A to unknown but it depends on what liquid samples are available first). Start by measuring 2.0 mL of your chosen sample. Place the graduated cylinder on the balance and measure and record the mass of the 2.0 mL. Hint: When finding mass, you should read the volume at the bottom part of the meniscus for more accuracy. Add another 2.0 mL of the same liquid to the same graduated cylinder making it a total of 4.0 mL. Place the graduated cylinder on the balance and measure and record the mass of the 4.0 mL. Again, add an additional 2.0 mL of the same liquid to the same graduated cylinder making it a total of 6.0 mL. Place the graduated cylinder on the balance and measure and record the mass of the 6.0 mL… Repeat the step adding...
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... Determine the mass of excess water remaining in the alum product at the end of day one. If you hadn’t waited a week to determine the percent yield, how large would the error be in your calculation of percent yield? 1. The amount of excess water remaining in the alum product after one day, which was measured a week later, was 0.588g. If you hadn’t waited a week to determine the percent yield, the mass recorded would be greater than the dried alum due to the water content present. Since the greater mass would suggest a greater abundance of alum, the percent yield of the same day would be higher than the percent yield calculated after a week. Comment on the relative error in your data and the class data that could have lead to the percent yields differing from the theoretical yields. 2. Both my data and the class data average produced a percent yield of less than 100%. These results suggest that alum was lost during the experimental procedure. We know this to be true because of the Law of Conservation of Mass. Some errors that may have led to lower percent yields than the theoretical yield include the loss of alum when transferring the crystals to the final beaker, not retrieving all the alum that was stuck to the filter paper after using the vacuum apparatus, or the spattering of solution that may occur due to overheating when dissolving the aluminum foil. Other factors that can contribute to lower yields may be not dissolving the entirety of the aluminum foil in the solution...
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...The percentage yield acts as a guideline to tell the successfulness of the synthesis experiment. A low percentage yield enable d means that the conditions were not most favorable and could be better. The percentage yield of the synthesized aspirin obtained in this experiment is 13.8%, which is considered as relatively low. Some errors like random human errors and mechanism errors may have occurred. One of the sources of error might occur during the addition of ice water to the mixture for complete recrystallisation. There is a possibility that the ice water is not at its optimal temperature. The temperature of the ice water will increase because it is placed under room temperature for a period of time when waiting for the Erlenmeyer flask to cool down. Failure for the ice water to have optimal temperature will then cause the amount of crystallized aspirin obtained decrease as it is possible that some of the salicylic acid were not able to crystallize. Hence, the percentage yield of the synthesis could be low as the mass of crystallized aspirin decreases. Furthermore, the mechanism error could occur when spatulas, beakers, Erlenmeyer flask and filter papers were used in the process of transferring the product. It is possible that some residue of the product could be stuck on them when we were transferring the product from filter paper into the beaker for weighing purposes. It is also possible that there are more residues left in the Erlenmeyer flask when the mixture was poured...
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...Purpose of the work: The purpose of this experiment was to experimentally determine the concentration of the unknown solution as well as what molecule the unknown solution was. In order to do that, we found Ka1 and Ka2 of unknown acid and Ka for the salt of the acid. Finally, we compared data to reveal the unknown solution identity and concentration. Experimental measurement and data analysis: The concentration of NaOH we used for this experiment was approximately 0.0804M. For titration of the H3PO4, we had two equivalence points and two half-equivalence points because H3PO4 has three ionizable H+ ions, however it only loses two because by the time it reaches HPO4 the last H+ is not recognized by the water. Therefore, since it can lose two ions it has two equivalence points and two half-equivalence points that give us two Ka and two pKa values. The pKa value is equal to pH at equivalence point and to determine Ka, we took 10-pKa. The 1st equivalence point for H3PO4 was 1.920 pH with half-equivalence points of 0.563 pH. The Ka was 0.274 and the pKa was 0.563. While The 2nd equivalence point for H3PO4 8.811 pH with half-equivalence point of 6.347 pH. The pKa was 6.347 and the Ka was 4.5 x 10-7. The data makes sense as the Ka decreased after the release of the first ion to the second ion and the pH rose with the increased amount of NaOH. The equivalence point for H2PO4 was 8.655 pH with half-equivalence point at 6.416 pH. The pKa was 6.416 and the Ka was 3.84 x 10-7. The data...
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...Use the data given and developed graph to answer the questions on the attached worksheet. Investigation Questions: Describe the general trends that you see in the temperature over time: As time increased the temperature of the gases also increased. As the gases gain more energy the temperature also increased. Did one gas warm more quickly than the other? YES H2O increased the fastest. Was the increase in temperature gradual or were there changes in the slope of the line? The rise in temperature was gradual. Which gas had the greatest change in temperature while heating? H2o had the greatest increase in temperature(9.3 difference). How did the cooling of the gases compare to the warming? Because the gas that trapped the most heat held the most heat and took longer to release that heat .Which gas appeared to hold the heat the longer of the two tested? H20 held the heat the longest. Recall that temperature is a measure of kinetic energy of molecules. Explain, in terms of kinetic energy, why the bottles remained warm after the light source was turned off or the bottles were shaded. Because the particles of the individual gases gained heat energy this was then converted into kinetic energy and started to collide with each other transferring this heat via conduction. When the lights were switched off they were still collisions therefore there was still heat remaining. This is why the bottle was still warm because the particles still had kinetic energy. How does...
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...The basic aqueous layer separated previously was then transferred into a 250 mL beaker, and a 50 mL of HCl was collected in a 100 mL beaker. Using a pipette, HCl was slowly added to the basic aqueous substance to acidify it. The lab TA stated that the solution would turn cloudy once acidified, but this did not happen, despite more than 50 drops being added. When the pH was tested, it showed that the pH was definitely 1, despite the solution not being cloudy. The volume of the NaOH and HCl mixture was 90 mL. The acidic solution was then poured back into the separatory funnel and washed with 25 mL portions of DCM as per the previous procedure. The aqueous top layer from these washes was discarded after use. Due to time constraints, this portion of the experiment could not continue to completion (i.e to creating the thin film at the bottom), so organic solution was sealed, labeled and set aside for the next lab. The waste was disposed of as per instructions by the TA, and all equipment was washed and stored appropriately. 9/8: Continuing from the previous lab, the solution was uncovered and poured into a separatory funnel held by a ring stand. A beaker was placed under the separatory funnel, and 25 mL water was added to the organic solution in the separatory funnel. A greased glass stopper was placed on top; the same depressurizing procedure as before was used. Once there was no more pressure within the separatory funnel, the separatory funnel was replaced in the ring...
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... The data I will collect, is the moles and grams for each unknown compound, calculate the g/mol for the unknown compounds, and calculate the molar mass for the known compounds. To identify the unknown compounds, we first weighed all seven of the unknown bags. To find the weight of the unknown compound in the bag, we first subtracted the weight of the bag by first weighing an empty bag and pressing the zero button to zero out the weight of each for the seven unknown compounds. The mol of each unknown compound was written on each of the bags. To calculate the g/mol for the unknown compounds, we must divide the grams over mol to find the g/mol. To calculate the g/mol for the known compounds, we used the periodic table of elements. We added the atomic numbers for each of the elements of the compound together to find g/mol for the known compounds. Then we wrote all the data in our data table and compared the unknown compounds to the known compounds to find the correct known compounds. Unknown compound #1 is Benzoic Acid. Unknown compound #2 is Sodium Carbonate. Unknown compound #3 is Calcium Oxide. Unknown compound #4 Sodium Sulfate. Unknown compound #5 is Ammonium Sulfate. Unknown compound #6 Potassium Sulfate. Unknown compound #7 Sodium Hydrogen Carbonate. Unknown compound #1 is Benzoic Acid and the percentage error is -1.03%. Unknown compound #2 is Sodium Carbonate and the percentage error is -6.48%. Unknown compound #3 is Calcium Oxide and the percentage error is -1.55%...
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...The initial mass of aluminum was measured to be 0.22 grams. An estimated mass of alum of 3.866 grams was calculated(equation 3). After the reactions had taken place and the alum was created and filtrated, the total mass of the alum was 2.420 grams. This experimental lead to a percentage yield of 62.58%(equation 4). This low percentage yield means that we produced less alum than what was expected. This percentage yield is around what would be expected due to errors in the procedure. The reason that the experimental mass of alum was less than the theoretical mass was because of various errors in the lab. One possible error for this lab was the purity of the aluminum can. The cans had to have the inside and the outside of them scraped with steel wool in order to remove the paint and interior coating of the can. Not all of the paint of the can was able to be removed which would have lead to some of the aluminum no being pure. Also, aluminum cans are made through the recycling of other aluminum cans. It is easy for metals and other materials to contaminate the cans. The next possible error in the lab was through the loss of products. This lab required material to be transferred from one beaker to another which allows for many opportunities to lose material. This value could affect the amount of mass that was left over for the next reaction. This error is also related to the quality of the filtration. The final filtration of the alum crystals from the solution did not work correctly...
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...which was compared with standard functional group frequencies of Aspirin as shown in Table 7.4 Table 7.4: Reported and observed IR frequencies of Aspirin Functional groups (Observed Frequency) (cm-1) (Reported Frequency) (cm-1) O-H-(Stretch) 3350.00 3100-3500 O-H stretch from CO-OH dimer 2991.59 3200-2800 Aromatic C-H stretch 3015.99 3150-3050 Aliphatic C-H Stretch 2898.28 2950-2850 -C=O stretch 1695.82 1700 Aromatic C=C stretch 1548.73 1600-1500 Aromatic C-H bending 713.46 600-850 The Functional group frequencies of Aspirin were in the reported range which indicates that the obtained sample was of Aspirin and was pure. Figure 7.1: FTIR spectrum of pure Aspirin 7.1.5: Differential scanning calorimetry (DSC) The DSC spectrum of pure drug was obtained and is given in Figure 7.2....
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...Invitro evaluation and optimization of controlled release chitosan beads of Metformin hydrochloride Manjisa Choudhury, Tanaya Sarma, Samridh Srivastava, Vino S. School of Bio Sciences and Technology, VIT University, Vellore - 632014, Tamil Nadu, India manji.choudhury@gmail.com; +91-9626413610 Abstract Beads of chitosan, in the free and metformin HCl encapsulated forms were prepared by Chitosan/ TPP polyelectrolyte complex. The particles were further stabilized by crosslinking with glutaraldehyde. The size of the particles was found to be in the range of 1.0 to 2.0 mm. Both free and metformin HCL encapsulated Chitosan beads were analyzed by FTIR, Differential Scanning Calorimetry and X-ray diffraction studies. The rate of drug release was investigated by in vitro dissolution studies using UV spectroscopy. The results indicate that these Crosslinked chitosan beads are capable of controlled and sustained release of meformin HCL upto a maximum time of 12 h in PBS at pH 7.4. Keywords: Chitosan, Metformin HCl, Controlled drug delivery, In-Vitro release. Introduction Large number of active compounds is discovered that could serve as therapeutics but very few candidates show clinical success. Their poor activity in vivo is often due to the reason of their low bioavailability. Bioavailability is basically the rate and the extent at which a drug enters and affects the target tissue (Kidane, 2005). When a drug is administered systemically, its bioavailability is...
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...Highlight on My Research Background Overview : Polymeric Composite has emerged as a new technology for innovation of new materials with low cost, high durability, more effectiveness, moreover reduction of environmental pollution. Leather industry is one of the highly environmental pollution creating industry. Solid wastes are being generated from the tanneries 88 metric tons/day on an average basis in my country, Bangladesh. In my research work I have attempted to reuse, recycle the leather waste. Moreover the aim is to reduce the environmental pollution. My research work was conducted to make bioapplicable materials from polyester resin reinforced with scrap leather fiber. The resulting polymer composite has been named “Leather Plastic”. One of the most important objectives of my research work is to use natural leather fiber which is biodegradable. So a definite amount of polyester resin could be replaced by natural fiber, which is very important to our environment. Manufacturing process: The composite was prepared by Wet Layup method. Leather fiber was treated with Unsaturated Polyester Resin and then fabricated and characterized. The matrix was prepared by mixing Unsaturated Polyester Resin with Polyvinyl Alcohol (PVA) solution. The grinded leather fiber was used as the reinforcement. After mixing the matrix with the reinforcement, peroxide was used as a radical initiator to induce polymerizations. After curing period the mechanical properties of the composite was characterized...
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...usually placed in a lower priority compared with hemodynamic or ventilatory support in managing critically ill patients. Under- or over-feeding is not uncommon, but is under-recognized in the intensive care unit. It has been suggested that underfeeding is associated with poor wound healing and immunologic comprise. On the other hand, over-nutrition can cause fatty liver and pulmonary compromise. These complications can possibly translate into morbidities and mortalities. Appropriate nutrition therapy depends on accurate determination of energy expenditure. Critically ill patients may have altered metabolism due to their diseases and the use of multiple medications, causing great difficulties in estimating their energy expenditures. Indirect calorimetry is...
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...I started to think about how I wanted to calorimetry because before we deduced the heat capacity of a calorimeter, but I wanted to think independently and complete a different experiment. I read about a type of calorimetry called soda can calorimetry which seemed enjoyable and fit under the topic of calorimetry in the syllabus. When in the lab I did discover the issue with matches and the setup of my experiment. Primarily, the foods in this paper caught fire quickly, but some of the foods I tried didn't work. I tried fruits like dates and apple, but fruit contains too much water to burn. I tried fatty snack like nuts, neither the peanut nor the hazelnuts caught fire before the match burned out. Even certain carbohydrate-heavy snacks like crackers- I tried filled crackers, Ritz crackers, and Saltines- did not ignite quick...
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...Modernized homes across the planet generally utilize paraffin wax as an alternate source of light in case the power goes out. Understanding the heat of combustion of a lit candle will allow people the knowledge of how effective a candle really is. This lab on the heat of combustion of paraffin wax tests students on their ability to use the previously mentioned ideas and concepts as well as their understanding of calorimetry. The students will calculate the efficiency of a paraffin wax candle and a material of choice by determining the heat of combustion. Fully understanding calorimetry is a very important skill to have whilst analyzing fuels. Calorimetry can be defined as the process of measuring how much heat is released or absorbed in a chemical reaction. The students will be dealing with an endothermic reaction, a reaction taking in heat rather then letting off heat. Calorimetry will be useful when analyzing fuels because the students will be able to collect data on however much heat is being released into the environment after the fuel is burnt....
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