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. Plot a graph of absorbance versus mole fraction of ligand. 3. From the graph, look for the maxima and from the corresponding mole fraction, determine the formula of the complex. Maximum absorbance occur when mole fraction of ligand is 0.65.
(3 marks) (ii) What is the chromatographic mechanism involved? Describe how the mechanism effects separation to the three compounds. (8 marks) (iii) With explanation, describe what would happen to the peaks in the chromatogram when the packing silica with a larger pore size and smaller surface area is used in the above separation. (6 marks) 2. A mixture of Br-, Cl- and PO43- ions was subjected to an anionic column packed with polym-N(CH3)3+OH-, eluted with 9mM sodium carbonate.
Running Head: UNIT 8 Final Project Unit 8 Final Project Tammy Barker Kaplan University SC 155-01 Professor Leonard August 26, 2012 From what I see, technically, this is not a back-titration, because base (antacid) is being added to acid, more base is added to the acid until the equivalence point is reached. Wouldn’t a back- titration involve going past the equivalence point with base and then back titrating with acid until equivalence was reached? Then, with each case I am reacting the antacid and base with .040L (the 40mL) of 0.1M HCl, and molarity times volume giving moles as follows: (.040L HCl)*(0.1 mol HCl/L HCl) = .004 mol HCl. So in each case, in order to get to the equivalence point in this reaction of a strong acid with a strong base, I need to have .004 mol of base to fully react in a 1:1 ratio with the acid that's present. The combination of antacid and base that is added will add up to .004 mol.
Thus, the molarity of the HCl solution can be calculated by dividing the number of moles of HCl by the volume of HCl (in liters) used to neutralize the Na2CO3 . Now that it is a neutralized solution, we are able to use it for the titration of NaHCO3. NaHCO3(aq.) + HCl(aq.) ==> NaCl(aq.)
4. Chlorine, Cl 2, is a strong oxidizing agent found in bleach. 10.0 cm3 of bleach solution was added to 25.0 cm3 of 0.4M Sn2+ solution acidified with sulphuric acid and made up to 100 cm3. 10.0 cm3 of the resulting solution containing the unreacted Sn2+ was titrated against 0.02M K2Cr2O7. Given that 11.40 cm3 of K2Cr2O7 was required for complete oxidation, a) Calculate the numbers of moles of Cr2O7- used.
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.
NaCl 1 Na+ and 1 Cl- are formed). Calculate the osmolarity of that solution. What is the molar concentration of ALL particles? e.g. 0.15 M NaCl solution = 0.15 moles of Na+ atoms + 0.15 moles of Cl- atoms = 0.30 Osmoles In other words, the solution is said to have an osmolarity of 0.30 Osm (or 300 mOsm) Assume the osmolarity of the ICF of body cells to be 0.300 Osm (300 mOsm) 2nd, determine if the solute is a PENETRATING particle or is NON-PENETRATING.
Calculate the exact normality of Na2S2O3 knowing that in this chemical reaction 1 gram-equivalent of K2Cr2O7 react with 1 gram-equivalent of Na2S2O3 (1 mole K2Cr2O7 react with 6 moles Na2S2O3). Determination of peroxide value. Weigh 3.00 g oil (with precision of 0.001 g) into a 250 ml Erlenmeyer flask. Add 10 ml chloroform and swirl to dissolve oil. Add 15 ml acetic acid,
Acid-Base Indicator Properties of Methyl Orange [pic] Procedure Pretreatment before Diazotization 1. Dissolve 1.2 g anhydrous Na2CO3 with 50 ml DI H2O
the total titratable iodide that was present is greater than the concentration of the I2 in the aqueous phase as some is being bound as I3-. This is how the stability constant is calculated. Method: As per lab manual page 89. Results: Organic Layer A: Moles of I2: 0.25/1000*0.95 = 2.375*10^-4/2 = 1.1875*10^-4 KD = [I2]H2O/[I2]Volasil 244 = 0.0118 I2 1.1875*10^-4*0.0118 = 2.2125*10^-6 Titrated S2O3 = 2(I2 + I3-)H2O 2.375*10^-4 = 2(2.2125*10^-6 + I3) I3 = 1.1654*10^-4 Iint = (0.2M * volume)/1000 Iint = 0.2/1000*15 = 3.0*10^-3M Iint = (I- + I3-)H2O 3*10^-3 = I- + 1.1654*10^-4 I- = 0.192M I2 in M/L : 1.1875*10^-4/15 * 1000 = 0.0125M/L I3= 1.1654*10^-4/15 *1000 = 0.007769M/L K= [I3]/[I2][i-] K= 0.007769/(0.0125)(0.192) K= 3.237 Organic Layer B: 0.25/1000*1.12= 2.8*10^-4/2 = 1.4*10^-4 moles of