1. For the demo experiment, the balanced chemical equation is as follows: (NH4)2Cr2O7(s)=Cr2O3(s)+N2(g)+4H2O(g). After the lightning of Ammonium dichromate, Chromium (III) oxide was formed while the Nitrogen and Water escaped into the atmosphere in a gaseous phase. Ammonium dichromate((NH4)2Cr2O7) gave rise to Chromium (III) oxide (Cr2O3), Nitrogen Gas(N2) and water (H2O) In terms of microscopic level, the ratio between reactants and products is as follows. One mole of Ammonium dichromate will give rise to one mole of 1 mole of Chromium (III) oxide and 1 mole of Nitrogen gas and 4 moles of Water is gaseous phase.
We concluded that the rate of hydrolysis of (CH3)3CCl is directly proportional to water content in the solvent mixture. Aims of experiment • Determine the rate constants for hydrolysis of (CH3)3CCl in solvent mixtures of different composition (50/50 V/V isopropanol/water and 40/60 V/V isopropanol/water) • Examine the effect of solvent mixture composition on the rate of hydrolysis of (CH3)3CCl Introduction With t-butyl chloride, (CH3)3CCl, being a tertiary halogenoalkane, it is predicted that (CH3)3CCl reacts with water in a nucleophilic substitution reaction (SN1 mechanism), where Step 1 is the rate-determining step. The reaction proceeds in a manner as shown
(0.01 moles of NaOH) x (1 mole Ca(OH)2/ 2 moles of NaOH) = 0.005 moles of Ca(OH)2 Tube 1: (0.0020 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.002 moles of Ca(OH)2 (0.002 moles of Ca(OH)2) x (74.08 grams/mole) = 0.1 grams = theoretical yield Tube 2: (0.0035 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.004 moles of Ca(OH)2 (0.004 moles of Ca(OH)2) x (74.08 grams/mole) = 0.3 grams= theoretical yield Tube 3 (0.0050 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.005 moles of Ca(OH)2 (0.005 moles of Ca(OH)2) x (74.08 grams/mole) = 0.4 grams =theoretical yield Tube
purified through preparative LC as described above and finally characterized as phloretin and phloridzin (Fig. 1). Compound 1 3-(4-hydroxyphenyl)-1-(2,4,6-trihydroxyphenyl)propan-1-one or phlorizin was obtained as amorphous powder, mp 2620C. The UV/Visible spectrum of the compound showed λmax at 225 and 285 nm. ESI–MS m/z 297 [M+Na]+ in positive ion mode and 273 [M-H] in negative ion mode for molecular formula C15H14O5; 274.
s q. s q. s q. s q. s q. s Table no.1 Formulation Table Table No.2: Physicochemical properties of the prepared buccal patches of Metoprolol Tartrate Formulations Folding Endurance* Tensile Strength* (%) Percentage Elongation* (%) Percentage Moisture absorption* (%) Percentage Moisture loss*(%) Drug Content* (%) F1 315 ± 2.516 38.72±1.10 143±1 3.89±0.67 1.5±0.82 91.64±1 F2 326 ± 2.081 46.71±1.505 122±1 4.19±0.75 1.4±0.97 94.60±1.174 F3 328 ± 2.645 42.43±0.735 121±1 5.11±0.94 2.1±0.65 93.25±0.960 F4 318 ± 1.527 49.34±1.01 142±1 3.75±0.81 1.3±0.23 97.80±0.990 F5 345 ± 2.081 46.93±2.16 125±1 4.69±1.32 2±0.71 90.60±0.995 F6 358 ± 2.081 51.31±1.00 134±1 3.45±0.84 1.5±0.84 89.03±1.000 *Data expressed as mean ± SD
Table 2: Effect of Pronase Treatment of the Phosphoprotein Derived from the Synaptosome-Enriched Fractiona5. Treatment Total dpm in the Total dpm in the supernatant ( S ) phosphoprotein residue [S/(S + R)] 100 after digestion of after digestion the phosphoprotein ( R ) 33P 32P 33P 32P 33P 32P Pronase 180 684 30 173 86 78 Control 8 72 773 3 8 Alkaline Phosphatase 1571 886 176 109 90 89 Control 224 122 1445 1053 13 10
The IUPAC nomenclature for haloperidol is 4-[4-(4-chlorophenyl)-4-hydroxypiperidin-1-yl]-1-(4-fluorophenyl)butan-1-one. Figure 1: 2-D structure of haloperidol (taken from https://pubchem.ncbi.nlm.nih.gov/compound/3559) Tacke et al (2008) analysed haloperidol in order to determine the physicochemical properties of the drug. At a pH of 7.4, it was discovered that haloperidol has a distribution coefficient (Log D) of
From the NMR spectrum provided showed for the 3,5-diphenylisoxazoline, chemical shifts at 3.364 ppm and 3.781 ppm for the diastereotopic methylene hydrogens and chemical shift 5.731 ppm for the methine hydrogen, from the article indicated from the computer simulation that the diastereotopic methylene hydrogens chemical shifts 3.346 ppm and 3.783 ppm  and chemical shift 5.741  for the methine hydrogen. With the simulation of 3,4-diphenylisoxazoline chemical shift 5.1 for the diastereotopic methylene hydrogens and chemical shift 4.5 for the methine hydrogen . All three hydrogens produced doublets of doublets. This showed from the data provide and information from the article that the final product produced was in fact the 3,
To find chemical equilibrium, the following chemical equation is used in the experiment: Fe3+(aq) + SCN-(aq) FeSCN2+(aq). When iron (III) and thiocyanate react, thiocyanoiron (III) is produced. When the concentration of all ions at equilibrium are known, the equilibrium constant can be calculated by dividing the equilibrium concentration of the reactant by the equilibrium concentration of the products. In this experiment, four equilibrium systems containing different concentrations of three different ion types (Fe(NO3)3, KSCN-, and distilled water) are made and used to determine equilibrium concentrations. The equilibrium concentrations are used to calculate the concentration that all of the components of the chemical equation are at equilibrium.
The cylinder with the sebacoyl chloride was 27.14 grams and the cylinder with hexamethylenediamine was 36.14 grams. We then calculated the mass of the reactants, which is found when we found the difference of the weight of the cylinders before and after sebacoyl chloride and hexamethylenediamine were added. The total mass of the reactants was equal to 41.35 grams. After we calculated these results, we started to create the chemical reaction. We put the sebacoyl chloride and the hexamethylenediamine in separate beakers, but then slowly added the hexamethylenediamine to the beaker with the sebacoyl chloride in it.