The apparent reaction rate was calculated using the equation, kapp= (1/∆t) x ([S2O32-]/[S2O82-]) which resulted in apparent rate constants of 5.66 x 10-5 s-1, 6.1958 x 10-5 s-1, 6.0356 x 10-5 s-1. The actual concentration was calculated using the basic chemical equation, C1V1 = C2V2. In order to find the order of reaction a a graph of log rate vs. log [S2O82-] was drawn, and was found that the results gave a zero order reaction But in reality the reaction order in [I-] and [S2O82-] is in first order each, although [I-] is kept at a constant volume throughout the reaction therefore the overall reaction is pseudo- first order. -d [S2O82-] = kapp [S2O82-] Dt In the second part the rate constant was found using the equation k = (1/∆t) x ([S2O32- ]/[Iodine][S2O82-]). Where it resulted to values of 3.990602 x 10-3 s-1, 4.653278 x 10-3 s-1, 5.944044 x 10-3 s-1, 7.499958 x 10-3 s-1, 7.499958 x 10-3 s-1, 9.84554 x 10-3 s-1, for flasks 4, 5, 6, 7, 8.
a. 1200.0 b. 8.00 c. 22.76 x 10-3 d. 731.2204 Question 3 Which value has more significant figures? 2.63 x 10-6 or 0.0000026 Question 4 Express 4,610,000 in scientific notation. a. with 1 significant figure b. with 2 significant figures c. with 3 significant figures d. with 5 significant figures Question 5 Express 0.0003711 in scientific notation.
Date 5/8/2013 Introduction to Organic and Biochemistry CHM 132 Unit 15 Lab (Body Fluids) Buffers Report Sheet Record the properties of each of the flasks on the table below: |Property |Flask 1 |Flask 2 |Flask 3 |Flask 4 | |Contents | | | | | |Initial pH |7.0 |7.0 |7.88 |7.88 | |pH after adding strong acid |2.32 | |6.16 | | |pH after adding strong base | |11.68 | |11.79 | 1. Compare what happen to the pH of flask 1 to what happened to the pH of flask 3 when HCl was added. The pH of flask 1 dropped much more than that of Flask 3. 2. Which substance, water or the buffer does a better job of maintaining pH when small amounts of strong acid are added?
| | | Mass A | Mass B | Mass B - A | | | Liquid | Volume(mL) | GraduatedCylinder (g) | GraduatedCylinderwith liquid (g) | Liquid (g) | Densityg/mL | %Error | Water | 5 | 15.8 | 20.8 | 5 | 1 | 0 | Isopropyl alcohol | 5 | 15.8 | 19.5 | 3.9 | 0.78 | 0.76 | .76 Data Table 5: Magnet – Measurement Method. | Object: | Mass(g) | Length(cm) | Width(cm) | Height(cm) | Volume(cm3) | Density(g/cm3) | Magnet | 4.0 | 2.5 | 0.6 | 0.6 | 0.9 | 0.225 | Data Table 6: Displacement method. | Object | Mass(g) | Initial volume of graduated cylinder (mL) | Final volume of graduated cylinder (mL) | Object Volume (mL) | Density(g/mL) | Magnet | 4.0 | 8 | 10 | 2 | 2
Isolation of Caffeine from Tea CHEM243A 2-23-11 Abstract: The purpose of this experiment was to isolate caffeine from a tea bag. Once isolated, the caffeine was purified using sublimation. The percent recovery was calculated to be 83.3%. The melting point of the original substance was 51ºC-139ºC. Since it was inaccurate, the melting point of another group’s sample was recorded.
Part I: Density of Unknown Liquid | | Trial 1 | Trial 2 | Trial 3 | Mass of Empty 10 mL graduated cylinder (grams) | 26.10 | 26.15 | 26.05 | Volume of liquid (milliliters) | 8.69 | 8.50 | 8.31 | Mass of graduated cylinder and liquid (grams) | 37.00 | 36.70 | 36.10 | Part II: Density of Irregular-Shaped Solid | Mass of solid (grams) | 38.954 | 39.068 | 42.885 | Volume of water (milliliters) | 49.9 | 49.9 | 50.0 | Volume of water and solid (milliliters) | 54.1 | 54.1 | 55.0 | Part III: Density of Regular-Shaped Solid | Mass of solid (grams) | 27.50 | 26.70 | 27.40 | Length of solid (centimeters) | 5.25 | 5.00 | 4.50 | Width of solid (centimeters) | 3.00 | 4.00 | 3.50 | Height of solid (centimeters) | 2.50 | 3.00 | 2.00 | Calculations Show all of your work for each of the following calculations and be careful to follow significant figure rules in each calculation. Part I: Density of Unknown Liquid 1. Calculate the mass of the liquid for each trial. (Subtract the mass of the empty graduated cylinder from the mass of the graduated cylinder with liquid.) * Trial 1 37.00(g) – 26.10(g) = 10.90(g) * Trial 2 36.70(g) – 26.15(g) = 10.55(g) * Trial 3 36.10(g) – 26.05(g) = 10.05(g) 2.
Experiment 3 Title: Fluid Properties: Density and Viscosity Date: 20th October 2011 Abstract: Introduction: Experimental Procedure: Five liquids were tested for Density measurements. These liquids included Vegetable oil, corn syrup, shampoo and glycerol. For viscosity measurement only three liquids were tested such as shampoo, glycerol and vegetable oil. Part I – Density Measurement The density of the test fluid was found by weighing a known volume of the liquid using the graduated cylinder. The cylinder was weighed empty, and then graduated cylinder was filled to a certain volume according to the graduations on it and weighed again.
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.
Experiment 7 Formula of a Complex Ion by the Continuous Variation Method Objective: To determine the formula of a complex ion by the continuous variation method. Procedure: Refer to laboratory manual. Results and Calculations: 1. Calculate the molarity of both CuCl2 solution and the ethylenediamine solution. Number of moles of CuCl2 used = [pic] =[pic] = 0.127 moles Therefore, Molarity of CuCl2 = [pic] = [pic] = 0.063 mol/dm3 Number of moles of ethylenediamine used = [pic] =[pic] = [pic] = 0.1 mole Therefore, Molarity of ethylenediamine = [pic] = [pic] = 0.05 mol/dm3 2.
The results from the second titration show how much of the excess reagent was used in the first titration and the original analyte's concentration is then determined. C9H8O4+2NaOH → C2H3O2Na+ H2O+Excess NaOH H2SO4+2NaOH → Na2SO4+2H2O Results | 1 | 2 | 3 | 4 | Start Point | 0.0cm³ | 0.0cm³ | 0.0cm³ | 0.0cm³ | End Point | 10.6cm³ | 10.0cm³ | 10.0cm³ | 10.0cm³ | Titre | 10.6cm³ | 10.0cm³ | 10.0cm³ | 10.0cm³ | Table 1: Shows results from the standardisation Titration. | 1 | 2 | 3 | 4 | Start Point | 0.0cm³ | 15.0cm³ | 30.0cm³ | 0.0cm³ | End Point | 14.8cm³ | 30.0cm³ | 45.0cm³ | 15.0cm³ | Titre | 14.8cm³ | 15.0cm³ | 15.0cm³ | 15.0cm³ | Table 2: Shows the results from the back Titration. Weight of weighing boat: 2.091g Weight of weighing boat + aspirin: 2.577g Weight of aspirin: 0.485g Calculations molesNaOH =molesNaOH reacted with aspirin+ molesNaOH reacted with H2SO4 Molar Ratio: NaOH: Aspirin – 2:1 NaOH: H2SO4 – 2:1 Standardisation | Burette | Conical Flask | | H2SO4 | NaOH | Concentration | 0.05 M | X M | Volume | 10 cm³ | 10 cm³ | n | 1 | 2 | m x vn= m1 x v1n1 0.05M × 10cm³1=10cm3xM2 0.5M/cm³=10cm3xM 2