Aliya chem

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University of South Florida, Tampa *

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2045

Subject

Chemistry

Date

Feb 20, 2024

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docx

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3

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Module #6 Post Lab Questions 1. How are you going to analyze your data statistically? My partners and I will use a table and a line graph to statistically examine our data. The Alcohol Initial, Final Weight and Consumed Alcohol Weight, Water Initial, Final Temperature Weight will be displayed in the table. The line graph, on the other hand, will show the relationship between enthalpy combustion and the number of carbons. Trial Alcohol Initial Weight (g) Alcohol Final Weight (g) Consumed Alcohol Weight Water Initial Temperature Weight (C) Water Final Temperature Weight 8C) 1 21.63g 20.33g 1.30g 21.3C 31.8C 2 20.05g 18.79g 1.26g 21.6C 31.6C Averag e 20.84g 19.56g *1.28g 21.45C 31.7C Alcohol Calculations: 1. 21.63g - 20.33g = 1.3g 2. 20.05g - 18.79g = 1.26g 3. 20.84 - 19.56 = 1.28g Energy: 1. Δ Н = (150)(4.186)(10) = 6279 2. 62791(1.28/88.15) = 432417.07 Methanol - 1 Carbon Trial Alcohol Initial Weight (g) Alcohol Final Weight (g) Consumed Alcohol Weight (g) Water Initial Temperature Weight (C) Water Final Temperature Weight(C) 1 21.53g 18.79g 2.74g 21.2C 31.2C 2 18.79g 15.29g 3.50g 21.4C 31.4C Average 20.16g 17.04g *3.12g 21.3C 31.3C Calculations: 1. 21.53g - 18.79g = 2.74g 2. 18.79g - 15.29g = 3.5g 3. 20.16-17.04 = 3.12g 2. Are there any factors in your experimental design that you now realize you did not control? Waiting for the calorimeter to cool down before inserting a new beaker of water reduces the time between trials. 4. Which alcohol produces the most energy per mole? Because pentanol has the most carbons in its structure, it is the best choice. Breaking bonds requires more energy in structures with more carbons since there are more bonds. We
discovered a positive association between the number of carbons and the heat of combustion in our investigations, which were conducted utilizing experimental procedures and multiple trials to provide more exact data. Δ Н =mc Δ Т; Δ Н comb = Δ Н/moles of burned alcohol Energy for Pentanol: 1. Δ Н = (150)(4.186)(10) = 6279 2. 6279441.28/88.15) = 432417.07 Energy for Methanol: 1. Δ H = (150)(4.186)(10) = 6279 2. 62794(3.12/32.04) = 64,480.5
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