Objective: The objective of the first lab is to use the graph of Pressure versus Temperature to estimate the value of Absolute Zero. Based on the Gay-Lussac’s Law, pressure is proportional to temperature; we are able to find the Absolute Zero when the pressure reaches zero. Conclusion: Our experimental result is -285.40 C. It means when pressure reaches zero, the temperature should reach -273.15 C, which is the value of Absolute Zero. Our experimental Absolute Zero value is fairly close to the accepted value, and the percentage error is 4.48%. One of the major sources of error in this experiment is that the volume of metal ball will rise while the surrounding water’s temperature rise.
Name: Date: Period: Course: CHEM 510 Ideal Law Practice Set 2.0 Directions: Solve the following problems on a separate sheet of paper using the 5-step method to show your work. All answers must be in proper SIG FIGs. Use the Ideal Gas Law (PV=nRT) to solve each of the following. R = 8.31 (L•kPa)/(K•mol) or 0.082 (L•atm)/(K•mol) 1. Calculate the volume 3.00 moles of a gas will occupy at 24.0 °C and 101.3 kPa.
b. the VO2 of horizontal running is always 1 MET. c. the VO2 increases linearly with running speed. d. none of the above. Ans: c 8. The most common technique used to measure oxygen consumption in exercise physiology laboratories is a. closed-circuit spirometry.
The acceptable change in the temperature was ± 0.05 ºC. The amount of CO2 produced was kept the same through having the same distance of 5 cm in the manometer tube, as only the same amount of CO2 would move through that distance. The time has been measured since agreed level in the manometer tube was reached, in order to avoid the paralax effect. Mass of sugar and solution was controlled by using weighting machine and by keeping the same mass of each solution for different yeast concentration. Apparatus: • Yeast (20g) • Sugar (8 g) • Water (52 g= 52 cm3) • Termometer (0-110 ºC ±0.05 ºC) • Weighting machine (±0.05 g) • Razor blade • Boiling tubes (20.0±0.5 cm3) • Stand, bosses and clamps • Manometer tube • Clip
Theory The height of the air column is equal to 14 the length of the wavelength because the particles on the surface of the water have no displacement, whereas the air particles at the top of the air column have the most displacement. The wave equation below can then be rearranged in order to plot the two variables of this experiment and linearise the data, where the slope equals the speed of sound. v=f×λ λ=v1f Processed Data Frequency (Hz) | Height (m) ±0.005 | 1f (s) | λ (m) ±0.02 | 256.0 | 0.320 | 0.0039063 | 1.28 | 288.0 | 0.290 | 0.0034722 | 1.16 | 320.0 | 0.250 | 0.0031250 | 1.00 |
From hot to cold or cold to hot? -Technically, if it were to be mixed like the example shown in class, there would be at equilibrium. But since the heat from one is 20℃ and the other is 80℃, we ended up with 60℃ as the final temperature. The heat would transfer from cold to hot because if we mixed 50g of the 20℃ and 30g of the 80℃, we would end up having more of the 100g of water at 20℃ and less of the 200g of water at 80℃. (c) If a sample of hot water is mixed with a sample of cold water that has twice its mass, predict the temperature change of each sample?
In the Rankine cycle, heat is added reversibly at a constant pressure but at infinite temperatures. If Tm1 is the mean temperature of heat addition as shown in Fig. 6, then the area under 4 and 1 is equal to the area under 5 and 6. Then, we found the efficiency of this cycle is Rankine = 1 – (T2/Tm1), Where T2 is the temperature of heat rejection. The lower is the T2 for a given Tm1, i.e.
Calculation: The Equation for the reaction is: 5Fe2+ + MnO4- +8H+ → 5Fe3+ + Mn2+ + 4H2O 1. No of moles of MnO2 n = (C*V) V=0.03245 dm3 C= 0.01335Mol dm3 m 0.03245 dm3 * 0.01335 Mol dm3 = 4.33 * 10-4 2. No of Moles of Fe2+ in 25cm3 of the solution. 5Fe2+ + MnO4- +8H+ → 5Fe3+ + Mn2+ + 4H2O . From the above equation, 5 moles of Fe is formed for every 1 mole of Mno4 ∴ The number of moles of Fe= No of moles of MnO4- * 5 = 4.33 * 10-4 * 5= 2.166 * 10-4 3.
Acid Base Titration Purpose: The purpose is to calculate the molarity of a NaOH solution by titrating the base with 5mL of standard HCl solution in each trial. By adding the base with unknown molarity to the acid with 0.10M the molarity of NaOH can be calculated. The base, NaOH, helps bring the pH of the acid, HCl, closer to seven, which neutralizes it. When using the buret the amount of NaOH used is able to be determined. Then by writing a balanced chemical equation and using the titration formula, Nb+Ma+Va=Na+MbVb , the molarity is able to be determined.
Do not distill to dryness. Fractional distillation: The same procedure, just add a fractional tube between the flask and the still head. Data and Observations: Chemicals: acetone | Water | | | For simple distillation Volume | Temperature | 5 ml (acetone) | 60.0 oC | 10 ml | 62.5 oC | 15 ml | 64.0 oC | 20 ml | 65.0 oC | 25 ml | 66.0 oC | 30 ml | 68.0 oC | 35 ml | 70.0 oC | 40 ml | 73.5 oC | 45 ml | 79.0 oC | 50 ml (water) | 90.0 oC | 55 ml | 96.0 oC | 60 ml | 97.0 oC | 65 ml | 97.5 oC | 70 ml | 99.0 oC | For fractional distillation: Volume | Temperature | 5 ml | 58.0 oC | 10 ml | 59.0 oC | 15 ml | 59.5 oC | 20 ml | 59.8 oC | 25 ml | 60.0 oC | 30 ml | 61.8 oC | 35 ml | 62.5 oC | 40 ml | 68.0 oC | 45 ml | 85.0 oC | 50 ml | 96.0 oC | 55 ml | 96.5 oC | 60 ml | 96.8 oC | Summary of results: Figure.1: curve of simple distillation Figure.2: curve of fractional distillation Discussion: As the mixture is heated the temperature rise until it reaches the temperature of the lowest boiling point in the mixture, while the other component, of the