Constant Pressure Calorimeter for Heat Capacity

Constant Pressure Calorimeter for Heat Capacity Kanwarpal Brar Purpose: To calibrate a constant pressure calorimeter and use it to determine the heats of the reaction and dissolution of different reactants and to use these heats of the reactions to find the enthalpy of a reaction by hess’s law. Analysis/ calculations: Determine the heat capacity of the coffee cup, Ccal in j*degC for all three trials and calculate the average value. Provide all these values in your report: provide full calculations only for trial . From table 1 Mass of 1.0 M NaOH solution used = 51.67g Mass of 1.0 M HCl solution used = 50.85g Total mass of final solution = 102.52g Initial temperature of reagents = 21.3 deg C Final temperature after neutralization = 27.8 deg C Heat absorbed by calorimeter q1 = C (heat capacity)*deltaT Heat abrorbed by soluction Q2 = heat capacity (C) *mass of the solution (m)*deltaT Heat released by neutralization reaction, Q3 = heat of reaction (delta H)*moles(n)/1mole In this reaction, Delta T=T2-T1 =27.8degC-21.3degC =6.5degC heat capacity of the solution, C=4.02J/g degC (given) mass of the solution, m = 102.52g heat of the reaction, ΔH = -57.3 KJ (given) = -57300 J HCl(aq) + NaOH(aq) > H2O(l) + NaCl(aq) Because HCl and NaCl react 1:1, any one can be used as limiting reagent Molarity of HCl = 1.0 M Volume of HCl = 50.0 ml = 0.0500 L Therefore moles of HCl, n = molarity * volume = 1.0 mol/L * 0.0500 L = 0.0500 mol It is assumed that not heat is lost to surrounding ΔE system = 0 J ΔE system = q1 + q2 + q3 = 0J Q1 = -q2 – q3 C1 * ΔT = -(4.02 J/g degC * 102.52g * 6.5 deg C) – (-57300 J * 0.0500 mol/1 mol) C1 * ΔT = -2678.85 J + 2865 J C1 = 186.15 J/ ΔT C1 = 186.15 J/ 6.5 deg C C1 = 28.64 J/ deg C Trial 1 = 28.64 J/ deg C Trial 2 = 31.09 J/deg C Trial 3 = 29.48 J/deg C Average = 29.73 J/deg C Determine the overall heat of reaction per mole od calcium meatl for the addition of calcium metal. to 1.0 M HCl folloed by the addition of water and b) to water folloed by addition of 1.0 M HCl. In each case, treat the overall reaction as a single process, i.e. instead of determining a delta H value for each step, determine . mass of ca = 0.404 g molar mass of ca = 40.08 g/mol moles of ca, n = mass/molar mass = 0.404 g/ 40.08 g/mol = 0.0100 mol Mass of water used, m = 50.0 g (1ml = 1g) ΔT = Tfinal – Tinitial ΔT = 30.5 – 21.4 deg C = 9.1 deg C ( table 2) Heat of the reaction per mole = -(q of reaction – (Ccal * ΔT))/moles of meatal -(Cwater*m*water*detaT(0*ΔT) /n = -(4.184 J/ degC * 50 * 9.1 degC) /0.0100 mol = -1903.72 J/ 0.0100 mol = -190372 J/mole = -190.372 KJ/mole ΔH = -190.372 KJ/mole b) mass of ca = 0.403g molar mass of ca = 40.08 g/mole moles of ca = mass/ molar mass = 0.400 g/ 40.08 g/mole = 0.00998 mol Mass of water used = 50 g (1ml = 1 g) Temperature difference ΔT = Tfinal – Tinitial ΔT = 30.5 – 20.3 degC = 10.2 degC (table3) Heat of reaction per mole = -q of reaction – (Ccal*ΔT)/mole of metal = -(Cwater*mwater*ΔT-(0*ΔT)/n = -(4.184 J/g degC*50g*10.2 deg)/ 0.00998mole = -2133.84 J/ 0.00998mole = -213811.62 J/mole = -213.81 KJ/mole ΔH = -213.81 KJ/mole Determine deltaEdissolution in J (g salt) for the unknown salt for all three trials and calculate the average value. Provide all of these values in your report, provide full calculation only for trial 1. Unknown salt = C Mass of salt = 4.013g Mass of water = 100g Mass of solution after reaction = 100g + 4.013g = 104.013g ΔT = Tfinal – Tinitial = 27-19.9 degC = 7.1 degC ΔEdissolution = -q of reation = -m*C*ΔT = -4.184 J/g degC*104.013g*7.1 = -3089.85 J ΔEdissolution/g salt = -3089.85 J/4.013g = -769.96 J/g salt Trial 1 = -769.96 J/g salt Trial 2 = -769.87 J/g salt Trial 3 = -754.18 J/g salt Average = -764.67 J/g salt Determine deltaEdissolution in J (g salt) for six salts in table 1. Provide all of these values in your report, provide full calculation only for LiCl. ΔEdissolution = ΔElattice + ΔEcation hydration + ΔEanion hydration ΔEdissolution = 846 KJ/mol + (-506 KJ/mol) + (-377 KJ/mol) from table ΔEdissolution = -37 KJ/mol ΔEdissolution = -37000 J/mol Molar mass of LiCl = 42.39 J/mol ΔEdissolution/ g of salt = ΔEdissolution/ molar mass = -37000 J/mol/ 42.39 g/mol = -873 J/g salt ΔEdissolution for LiCl = -873 J/g salt ΔEdissolution of LiBr = -472 J/g salt ΔEdissolution of NaCl = 51.3 J/g salt ΔEdissolution of NaBr = 0 J/g salt ΔEdissolution of KCl = 228 J/g salt Discussion : In the experiment, a simple constant-pressure, coffee cup calorimeter was calibrated using an acid-base neutralization reaction. the calculated specific heat of calorimeter was then used to determine the heats of reactions and dissolutions of other chemical compounds. A simple constant pressure calorimeter was produced out of two styroform cups. The cups were covered with a plastic lid with a hole in centre. While erformiing the acid-base neutralization reaction, the temperature of both acid and base were measure using PH metre temperature probe. The temperature were about each other. When HCl was added to NaOH no visible change was observed while adding the acid. But the temperature of the soluction was rise after the acid was added. This showed that the reaction between HCl and NaOH was exothermic reaction. after that mass of the final solution was measured. The second objective was to find the heat of the reaction per mole of calcium metal, while following the hess’s law provided in the lab manual. This was done in two different trials. First the calcium metal was added to 50.0 ml of 1.0M HCl and then 50ml of water. When calcium was added to HCl it reacted vigorously creating bubbles. The highest temperature recorded was almost double the initial temperature. When water was added to this solution, no visible change was observed, but temperature was dropped by 10 degC. The overall process was still an exothermic reaction the heat of the reaction was calculated to be -190.372 KJ/mole In the second trial, the calcium was first added to water. This reaction was similar to the first one. Calcium reacted with the water vigorously. The temperature of the solution was increased showing that is was exothermic reaction. when HCl was added to this solution the temperature was dropped by 3.6 degC. Which was less than the first case. The heat of the raction waw calculated to be -213.81 KJ/mole. The closeness o fthe both results can be explained by the fact that heat of the reaction is a state function, and does not depend on the path of the reaction. this also increases the confidence in the result. The final objective of the reaction was to determine the heat of dissociation of the unknown salt, and thus find the unknown salt by comparing the heat of dissotiation to the heat dissolution of possible salts. This unknown salt code C was white powder form. When unknown salt was added to water, temperature raise by 7.1 degC. This reaction showed that this was a exothermic reaction. the average enthalpy of dissolution of the unknown salt C was calculated to be -764.67 J/g. This value of enthalpy of dissolution corresponded to the calculated value of Lithium chloride, LiCl. A number of experiment errors could have affected the data collected, which includes the accuracy and precission of the instruments used environment conditions. The graduated cylinder was used to measure liquids was accurate to only one decimal place, or could only round the value to .0 or to .5. the measuring balance used to weigh had had high accuracy up to three decimal place, dispite that there was difference in the total weigh of the soluction in all three trials. This shows that may be weighing machine was not accurate. it is also possible that when solution was shaked to mix the reactant some of the solution lost or may be was left over on the cover lid. Or into the walls of cups and glass container. While doing the experiment some liquid was spilled that could be that reason for the difference in the weight. The volume of the solution could be measured by burettes or pipetts for higher accuracy. Overall the results of the experiment calculations were really promising and confident based on the fact that they folled the theory of the experiment. Conclusion: A calorimeter was prepared. The heat capacity of calorimeter was calculated to be 29.73 J/deg C. The heat reaction calcium was found to be -190.372 KJ/mole and -213.81 KJ/mole, in the two trials. the unknown salt had -764.67 J/g salt. The unknown salt C was found to be Lithium chloride. Results of this experiment is promissign and confident. References: Olmsted, john 3; Williams, greg; burk Robert c. Chemistry, 1st Canadian ed; john Wiley and sons ltd: Mississauga, Canada, 2012, pp 511-550

Wordsmith and The Gold Mountain Coat

The poem â€Å"Wordsmith† is talking about a boys father who is constantly working on their house. To perfect it. It mentions how he â€Å" he fills and smooths and sands as filling in all of the empty crevices. † While the boy is watching though I can sense some sadness, maybe for the fact that his father spends more time on the house rather than with him. He might feel as if the imperfections of the house are more important then him. The short story â€Å"The Gold Mountain Coat† is about a father named Sam sing. Sam has an rugged appearance it states that he rarely smiled or even talked for the matter. Sam has two sons Ken and john. He was proud of the fact that when he was unable to look after the chines restaurant anymore he would have to sons to take over. Sam seems very unfriendly. He never visited the other family with his son. His life seemed to be devoted to the restaurant rather that to his family. The similarities between the fathers in the â€Å"Wordsmith† and â€Å"The Gold Mountain Coat† would have to be that they both work very hard. I think that they do it for their families. To show that they care deeply and want to provide a nice place to live and help sustain a good life. In â€Å"Wordsmith† is says that the fathers â€Å"love keeps him moving from room to room†, meaning that he does it out of love although it may come off as if he doesn’t care. In â€Å"The Gold Mountain Coat† it shows that he actually has a lot of love for his family and works hard to provide for them even though it may not seem as it and that the sons are afraid of him. When john asked for a second coat for him and his brother and that his son could grow into it Sam said yes. I think that shows a lot of love. Both fathers are very caring and both do everything out of love.

Organ Lateral Sclerosis A Disease That Affects The...

Amyotrophic Lateral Sclerosis is a disease that affects the Nervous system as well as the Muscular System as well, this disease is defined as a progressive neurodegenerative disease that affects the nerve cells in the brain and the spinal cord. A-myo-trophic is a Greek term when translated it means â€Å"No muscle Nourishment†. This lack of nourishment causes a reaction within the muscles leading to their eventual break down, otherwise known as â€Å"atrophies† which is a simple term meaning that the muscle wastes away. When looking at the term â€Å"Lateral†, this is in reference to the section of spinal cord in which the nerves are being affected. While this area degenerates it will lead to eventual scarring and hardening within this region (this is†¦show more content†¦(Lights,2016),(ALS,2016) The Brain and the Spinal Cord are both components of the nervous system. The nervous system is what carries messages from the brain to the spinal cord and then sends these messages to various locations within the body. The nervous system is comprised of two systems, the first being the Central nervous system. The Central Nervous system is comprised of two main components which are the brain and the spinal cord. The second being the Peripheral Nervous System which is comprised of the autonomic and somatic nervous systems. The Brain is divided into four main parts, the brain stem, the cerebellum, the cerebrum and the diencephalon. Within the brain there are two types of matter the Brain Grey matter, which is responsible for receiving and storing impulses (cell neurons are located in the grey matter), and White Brain Matter, which carries the impulses sent by the Grey Brain Matter. The brain stem is what connects the brain to the spinal cord and is otherwise known as the medulla oblon gata. There are other components that comprise the nervous system such as meninges (Three layers of membrane that cover the brain and spinal cord, they act as protection to the brain and spinal cord against bacteria and microorganisms) and neurons which are comprised of

The Use Of Cognitive Behavioral Therapy ( Cbt ) - 1543 Words

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