The chemiluminescence part of the experiment, we had to make four solutions labeled as ‘stock solution A, solution A, stock solution B, and solution B’. For the ‘stock solution A’ we put the luminol product, (0.242 g) in a 25 mL Erlenmeyer flask and dissolve it with 2 mL of 3M NaOH. Then we took 1 mL of the ‘stock solution A’ and diluted in 9 mL of water using a 50 mL beaker. Solution A. For the ‘stock solution B’ we mixed 4 mL of potassium ferricyanide solution and 4 mL of hydrogen peroxide solution using a 25 mL Erlenmeyer flask. Then for Solution B we diluted 4 mL of ‘stock solution B’ with 16 mL of water using a 50 mL beaker. The luminesce solution of the experiment, was obtain by diluting 3 mL of ‘solution A’ with 16 mL of water using a 150 mL beaker. Then in a dark room, we poured ‘solution …show more content…
A: The compound that is being oxidized in the luminol synthesis reaction is Na2S2O4, (Sodium Hydrosulfite. Since sodium hydrosulfite is acting as the reducing agent, in this experiment. 2. What is the purpose of acetic acid in the luminol synthesis reaction? How would the amount of the luminol product obtained at the end of the reaction be affected if the number of moles of NaOH added at the beginning of the reaction and the acetic acid added at the end of the reaction added were reversed? A: We know that with the addition of acetic acid in the luminol synthesis since it will protonate the dianion while reaction in the luminol synthesis so it will allow the product to separate completely from water. If we were to add NaOH at the beginning of the reaction, while adding acetic acid towards the end of the reaction, it wouldn’t affect it since we would still get a luminol produced. 3. In the luminol synthesis reaction, why was cold water used for rinsing the filtration step? How could the efficiency of the filtration be affected if hot water was used
Next, about 10 mL of both solutions, Red 40 and Blue 1, were added to a small beaker. The concentration of the stock solution were recorded, 52.1 ppm for Red 40 and 16.6 ppm for Blue 1. Then, using the volumetric pipette, 5 mL of each solution was transferred into a 10 mL volumetric flask, labelled either R1 or B1. Deionized water was added into the flask using a pipette until the solution level reached a line which indicated 10 mL. A cap for the flask was inserted and the flask was invented a few times to completely mix the solution. Then, the volumetric pipette was rinsed with fresh deionized water and
The unknown compound was first reacted with an acid. To begin, 0.50 grams of KCl was mixed with 5 mL of water. Then, 1 mL of 6 M H2SO4 was added to the solution. Secondly, the unknown compound was reacted with a base. Exactly 0.50 grams of KCl was mixed with 5 mL of water, and 1 mL of 1 M NaOH was added to the solution next.
Because of this, three different products (as previously mentioned) are potentially formed.1 The compound created from the reaction can be analyzed to determine
Benzyne Formation and the Diels-Alder Reaction Preparation of 1,2,3,4 Tetraphenylnaphthalene Aubree Edwards Purpose: 1,2,3,4-tetraphenylnaphthalene is prepared by first producing benzyne via the unstable diazonium salt. Then tetraphenylcyclopentadienone and benzyne undergo a diels-alder reaction to create 1,2,3,4-tetraphenylnaphthalene. Reactions: Procedure: The reaction mixture was created. Tetraphenylcyclopentadienone (0.1197g, 0.3113 mmol) a black solid powder, anthranilic acid ( 0.0482g, 0.3516 mmol) a yellowish sand, and 1,2-dimethoxyethane (1.2 ml) was added to a 5-ml conical vial.
The reaction to synthesize benzocaine was known as a Fisher esterification reaction. The Fisher esterification was reaction between alcohol and carboxylic acid in the presence of acid. The reaction was used to form an ester. In the experiment, sulfuric acid acted as a catalyst and necessary for this reaction to occur. There was a change between the –OH group of carboxylic acid to an –OCH2CH3 group in the reaction.
0.1M 0.05M 0.025M 0.0125M 0.00625M Use equation C1V1 = C2V2 to calculate the amount of water needed to dilute the solution. 10ml of 0.1M KI + 10ml of deionized water = 20ml of 0.05M KI Measuring the time for the colour change to occur Using a pipette, add 1ml of H2O2, HCl, starch solution and 3ml of S2O3 into a cuvette. Set up the spectrophotometer and put the cuvette into the machine Data collection starts now, allow 5 seconds.
Glacial acetic acid and acetic anhydride were added to the mixture while refluxing, which converted the lime colored solution into a clear mixture. The flask was cooled in an ice bath and the solution
The mixture was then distilled. When the temperature was reached to about 59℃, half vial of distillate (1V) and 1 mL of the liquid residue (1L) were collected. For 61.0℃, the distillation was then continued. Samples (2V, 2L) were taken at about 61.0℃.
Throughout the experiment, copper was altered a total of 5 times, but after the final chemical reaction, solid, elemental copper returned. Each time the solution changed color, a precipitate formed, or when gas appeared, indicated that a chemical reaction was occurring. For the first reaction, copper was added to nitric acid, forming the aqueous copper (II) nitrate (where the copper went), along with liquid water, and
11) After you have prepared the dilutions, clean the outsides of the cuvettes with a paper towel. 12) Place the blank tube (tube 0) in the spectrophotometer. Since distilled water has no color it will not absorb any light so the absorbance number would be zero and this done to test the absorbance scale on the Spectrophotometer for the purpose of having it calibrated correctly. 13) Set the spectrometer to a wavelength of 530 nanometers. 14) Place the cuvettes (numbers 1-6) with the appropriate substance and record it’s reading in the data table.
Purpose The purpose of this experiment is to determine K, the rate constant k1 of the forward reaction divided by the rate constant k-1, as well as Ymax, which is the maximum number of moles of acetic acid that can be adsorbed on the surface of the charcoal, per gram of charcoal. Methods
Experiment #1 Isolation of Caffeine from Tea Date: 11.09.14 Prepared by: Alibek Abilev Purpose The aim of the experiment was to isolate crude caffeine from tea leaves by using liquid – liquid extraction with methylene chloride, purify the crude substance by performing sublimation and determine the melting point of both crude and pure caffeine. Safety Lab coat, goggles, gloves. Methylene chloride is a carcinogenic substance, therefore should be kept in a well-ventilated place.
The final product, luminol, produced light therefore this signifies that the reaction worked. It produced a neon blue, green color that showed that the reaction was fluorescence instead of being phosphorescence which produces a red color. The reaction lasted milliseconds which suggests that the reaction was again fluorescent because phosphorescent light lasts seconds before the light distinguishes because it stores absorbed energy for a longer time and is not spin paired. With this, more energy is required to flip spins and go back which makes the light last longer when jumping from the excited singlet to excited triplet state. However, with fluorescence it goes from an excited singlet to ground singlet state, because it must go downhill to
Use these results to determine the product concentration, using Beer-Lambert’s Law: A= ɛCl (where A is the absorbance, ɛ is the molar absorptivity, C is the product concentration and l is the length of solution that the light passes through). Calculate the product concentrations at every minute for 10 minutes for all 7 of the test tubes using Beer-Lambert’s Law. Plot a graph of product concentration vs. time and then use the gradients of the 7 test tubes to determine the velocities of the reaction. After calculating the velocities, plot a Michaelis-Menten graph of velocity vs. substrate concentration.
The chemical equation for this experiment is hydrochloric acid + sodium thiosulphate + deionised water (ranging from 25ml to 0ml in 5ml intervals) sodium chloride + deionised water (ranging from 25ml to 0ml in 5ml intervals) + sulphur dioxide + sulphur. As a scientific equation, this would be written out as, NA2S2O3 + 2HCL + H2O (ranging from 25ml to 0ml in