Repeat the titration until there are two titres within 0.1cm3 of each other. Record results in a suitable table. Results: Rough Titre: 7.653 First Run: 6.553 Second Run: 6.453 Third Run: 6.553 Calculations: During the titration, iron(II) ions are oxidised to iron(III) ions and manganate(VII) ions are reduced to manganese(II) ions. The equation is as follows: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) ? 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) The above equation shows that one mole of manganate(VII) ions reacts with 5 moles of iron(II) ions in acid solution.
The absorption spectrum is measured using a spectrophotometer and the data is graphed in Excel. The peak of the line is used to find Vmax of Fe2+. Vmax is used to find the moles of Fe2+ and ligand. The unknown n is a ratio of moles ligand divided by moles Fe2+. Results and Discussion For the first part of the experiment (Part A), five different 100 mL volumetric flasks were each filled with 1,2,3,4 and 5 mL of iron (II) solution.
pyridinium hydrobromide perbromide type of stationary phase column length column temperature rate flow of the carrier gas List the 4 general factors that affect the separation obtained on a gas chromatograph What specific technique is used to collect/isolate your purified unknown compound at the end of the recrystallization experiment? suction filtration 14 of 22 4/16/12 9:15 PM StudyBlue Flashcard Printing of Lab Final 2211L UGA
With the use of this technique we placed chlorine, bromine, and iodine into solutions containing chloride, bromide, and iodide. In the reaction the free halogen (X2) oxidizes the other halide ion (Y-) and gets reduced by gaining electron(s). In table 3, chlorine was the strongest oxidizing agent and iodine was the weakest oxidizing agent. Since chlorine was the strongest oxidizing agent it will react more and the weak agent will react less. This explanation can be demonstrated in table 3 also because the results of the reactions demonstrates that chloride reacted more by the color of the product compared to the color of chloride in the mineral oil.
A) is neutralized by water B) is surrounded by water molecules C) reacts and forms a covalent bond to water D) aggregates with other molecules or ions to form a micelle in water Answer: B Page Ref: Section 3 11 9) Which would you expect to be most soluble in water? A) I B) II C) III D) IV Answer: A Page Ref: Section 3 10) Solutes diffuse more slowly in cytoplasm than in water because of A) the higher viscosity of water. B) the higher heat of vaporization of water. C) the presence of many crowded molecules in the cytoplasm. D) the absence of charged molecules inside cells.
Reactions Lab David Vaghari INSTRUCTOR: Dr. Chernovitz Monday, July 23, 2012 Oxygen Production Introduction In this lab, potassium chlorate will be decomposed producing oxygen gas and potassium chloride. The hypothesis is that the reaction will yield 3.916 grams of oxygen gas. Materials Test tube 10 grams potassium chlorate Bunsen burner Procedure Step 1. Obtain a test tube, place a 10 gm of potassium chlorate. Step 2.
Which substance, water or the buffer does a better job of maintaining pH when small amounts of strong acid are added? The buffer does a much better job at maintaining pH. 3. Compare what happen to the pH of flask 2 to what happened to the pH of flask 4 when NaOH was added. Both flask’s pH went up 4.
Synthesis of Methyl Stearate Post-Lab Submitted by Matthew Sharma TA: Evan Determining the Limiting Reagent Methyl Oleate (MW= 296.49 g/mol) Amount used = 1.141g (1.141g)(1mol / 296.49g) = 0.003848 mol ∴ The limiting reagent for the reaction is methyl oleate because hydrogen is used in excess Theoretical Yield Methyl Stearate (0.003848 mol)(298.504g/mol product) = 1.1486 g methyl stearate Percent Yield Given the 1:1 stoichiometry of the reaction: 100% x (0.2551g product / 1.1486g theoretical yield) = 22.2% Conclusion A 22.2% yield of synthesized methyl stearate was obtained via the catalytic hydrogenation of methyl oleate in the presence of the catalyst 10% palladium on carbon. The product
CHEM 1412 SAMPLE FINAL EXAM PART I - Multiple Choice (2 points each) _____ 1. In which colligative property(ies) does the value decrease as more solute is added? A. boiling point B. freezing point and osmotic pressure C. vapor pressure D. freezing point and vapor pressure _____ 2. What is the molarity of a solution prepared by dissolving 25.2 g of CaCO3 in 600 mL of solution? A.
Lab 4: Determination of Percent by Mass of the Composition in a Mixture by Gravimetric Analysis Introduction Thermal gravimetric analysis is used to determine the percent by mass is used to determine the percent by mass of a component in a mixture. When a mixture is heated to an appropriately high temperature, one component in the mixture decomposes to form a gaseous compound. The mass of this particular component is related to the mass of the gaseous compound. In this experiment, the percent by mass of sodium hydrogen carbonate (NaHCO3) and potassium chloride (KCl) in a mixture will be determined. Experimental First, we weighed 2 samples, each has 1 gram of NaHCO3-KCl mixture Second, we put the samples in 2 crucibles (A and B) and weighed them.