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.
Calculate the value of Ke for this system. 2 H2S (g) === 2 H2 (g) + S2 (g) [1.1(10-4] 7. At a given temperature, the following system has an equilibrium constant, Ke, of 0.27. C(g) + B(g) === 2 E(g) The system was established by placing 8.00 moles of C and 5.0 moles of B in a 4.0 L vessel. Calculate the concentration of all substances at equilibrium.
A mean less than 0.480: | 25 percent | 9c. Random numbers. A mean between 0.450 and 0.550: | 92 percent | 9d. Random numbers. A mean between 0.480 and 0.530: | 60 percent | 10.
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.
| 56 | | 13 | 48 | | 57 | | Questions: 2. The freezing point of paradichlorobenzine is approximately 53.2 °C but my graph illustrates it as 51°C. (Note this is due to the different atmospheric pressures and concentration of paradichlorobenzine which causes some minor variation.) 3. The melting point of paradichlorobenzine is approximately 53.5 °C but my graph illustrates it at 51°C.
0.05mol/6M=8.3*10-3 L=8.3mL stock solution c. 100mL-8.3mL=91.7mLwater Add 91.7 water to 6M stock solution to prepare 0.5M acetic acid. Exercise 8: a. 42.35 - 0.55 = 41.8 mL b. The moles of EDTA4- : 0.0189M*(41.8*10-3)L=7.9*10-4mol c. Zn2+(aq)+EDTA4-(aq)—Zn(EDTA)2-(aq) The ratio Zn2+ and EDTA4- is 1:1 The moles of Zn2+= the moles of EDTA4-=7.9*10-4mol d. 7.9*10-4mol*65.39g/mol=0.0517g Zn e.
Abstract The focus of this experiment was to analyze the kinetics of a nucleophilic substitution. A mixture of 0.3622-M 1-bromopropane and 0.3622-M potassium hydroxide in an 90:10 ethanol/water solvent provided the reactants for a SN2 reaction to occur in a temperature controlled bath at 50.0˚C. The disappearing reactant was found by titrating timed aliquots during the reaction and then measuring the concentration of hydroxide. The k-value was found to be 0.0202 M-1Min-1. Using the linear form of the Arrhenius equation the activation energy was calculated to be 19.9 kcal/mol.
Find the difference between -27 and -32. a. -59 b. -5 c. 59 d. 5 21. What is 5 less than -21. a. 26 b.
(2 points) Mg(s) + 2 HCl(aq) → H2(g) + MgCl2(aq) 2. Determine the partial pressure of the hydrogen gas collected in the gas collection tube. (3 points) partial pressure H2 = total pressure - vapor pressure of water = 746mmHg - 19.8mmHg = 726mmHg 3. Calculate the moles of hydrogen gas collected. (4 points) n = 125 4.
The following data were obtained when a sample of barium chloride hydrate was analyzed as described in the Procedure section. Calculate (a) the mass of the hydrate, (b) the mass of water lost during heating, and (c) the percent water in the hydrate. Mass of empty test tube 18.42 g Mass of test tube and hydrate (before heating) 20.75 g Mass of test tube and anhydrous salt (after heating) 20.41 g. Mass of the Hydrate is 2.33g. Loss (H2O) is 0.34g. Percent H2O in Hydrate is equal 0.34/2.33=14.6% 3.