In the test for a reducing sugar, if it changes to a red orange color is it known as a precipitate as it comes out of the solution and forms solid particles dispersed around the water. The test of non-reducing sugar is striving as a result of if there is any sucrose presents it is broken down into those monosaccharides, which can be proved for using the common reducing sugar test. A positive result indicates that non-reducing sugars are present on the original sample. Sucrose fermentation it involves inoculating of sucrose broth with inoculating loop. Usually done for the differentiation of Enterobacteriaceae species.
Once iodine was dropped onto the circle labeled “ Saliva”, it transformed into a white/ yellow Colour, due to it granting the starch to break down properly, it transformed white as a result of there being no starch, hence it executed its main action. Once iodine was dropped onto the circle, which was labeled “HCL+ saliva”, was divided into two colours, white and navy blue, this arose because the starch did not get broken down. The enzyme got denatured with the extension of hydrochloric acid of hydrochloric acid. The acid found in hydrochloric acid obtains a low level oh pH (2), causing it not to be broken down, rather to be denatured, confirms amylase can’t continue it’s activity without a particular high amount of acid, (that is found in stomach acid). My hypothesis is incorrect due to the fact that my prediction is
Discussion: End point in this acid-base titration experiment refers to the point where the chemical reaction has reached its conclusion and no additional titrant should be added. The end point of this experiment can be obtained when the indicator used changes colour. For example colourless to light pink when phenolphthalein is used and red to orange and subsequently yellow when methyl orange is used. Equivalence point, also known as stoichiometric point in an acid-base titration refers to the point where the reacting acid and base is in equal proportions. In the graph plotted, pH against volume of NaOH, the graph becomes nearly vertical at the equivalence point.
Biuret test is a test which is utilized to indicate unhydrolyzed proteins. When there are peptides in a solution, a copper (II) ion forms violet-coloured coordination complexes in an alkaline solution. The biuret test can be utilised to analyse the concentration of proteins due to peptide bonds that occur with the same frequency per amino acid inside the peptide. In this experiment, the colour changed to purple to indicate the presence of protein. The pH was found to be 7, which is in the range of a healthy person’s pH (which is 7.4).Benedict`s solution is made up of alkaline copper sulphate and sodium citrate (blue in colour) (Danson and et al, 1996).
Therefore, acid fat, lactose fermentation, mannitol fermentation were not needed to be performed, because they are selective to a specific thing. As a result, Unknown bacteria “W” was concluded to be gram stain positive, endospore positive, bacilli shape (rod shape), and arranged in chains (strepto-). Test Purpose Reagents Observation Results Gram Stain To determine gram reaction of bacteria. Crystal violet, Grams iodine, Alcohol, and Safranin Purple streptobacilli bacteria was observed. Gram positive rods.
1. How pure is your sample? When analysing our sample under UV light we could see if our sample was pure. We labelled the sample with 1= which was our sample, 2= the pure aspirin sample, 3= the salicylic acid. As you can see from my sample that our aspirin sample contains a small amount of pure aspirin and a lot of salicylic acid.
In our case, if the substance changed to a light brown color, the test was negative and the substance contained simple carbohydrates (like glucose), and if the substance changed to a dark brown or black color, then the test was positive and the substance contained complex carbohydrates (like starch). Since the substance with Alkaline water remained dark brown, and the control substance changed to a light shade of brown, this meant that the
CHAPTER 3 EXPERIMENTAL PROCEDURE 3.1 Materials & Reagents Adsorption of Cu (II) was studied using Montmorillonite-K 10, procured from Sigma Aldrich. It is a very soft phyllosilicate group of minerals that typically form in microscopic crystals, forming clay. Its Cation Exchange Capacity, CEC was 119 meq/100 g .The elemental composition of this clay was [Al1.47Fe0.29Mg0.23][Al0.076Si3.29]O10(OH)2 as reported by the supplier. This MMT was used as such without any further purification. Tri-Octyl Amine(TOA) was the product of Tokyo Chemical Industry Co., Ltd. Japan, which had a purity of about 98% and was procured from Sigma Aldrich.
This test tube lacked hydrogen peroxide in order to keep the reaction from occuring. In two test tubes labeled “Substrate”, we mixed 0.2 ml of hydrogen peroxide with 0.1 ml guaiacol and 4.7 ml of distilled water. We also labeled two test tubes “enzyme” using a glass marking pen and filled them with 1.0 ml of turnip extract and 4.0 ml of distilled water. After the test tubes were prepared, we put the blank test tube into a cuvette and put it into the spectrophotometer in order to zero it out. While one group member set the spectrophotometer to zero, another mixed one enzyme test tube with one substrate tube and observed a change in it’s color.
Blank solution and oil solution were prepared and stored in the dark. Then, they were titrated with Na2S2O3 and volume of titrant was determined for blank and oil solution. Finally, iodine value was determined by using volumeaof titrant. B) Determination of Peroxide Value In this experiment, peroxide value of sunflower oil was tried to determine. After preparing a solvent mixture, it was titrated with sodium thiosulphate but during titration time color change was not observed.
My Citrate (CIT) result was turquoise so that meant the test was positive, and the Hydrogen Sulfide (H2S) had no black precipitate so it was negative. The Urea (URE) and Tryptophane Deaminase (TDA) results were both an orange color, which meant they were both negative. For Indole (IND), my result was yellow so it was negative. My result was colorless for the Voges Proskauer (VP) test so it was negative. The Gelatin (GEL) test result had no diffusion of pigment so that showed it was negative.
More specifically, this lab was met in terms of gaining an understanding in separating an acid, base and neutral compound from a mixture and identify through melting point. Overall, the experiment was successful as the acid (benzoic), base (5-chloro-2- methoxyaniline) and neutral (biphenyl) compounds were correctly identified. The separation of mixtures compounds to give pure components is of great importance in chemistry and in specific in organic chemistry. Many synthetic reactions give mixtures of products and it is important to isolate the wanted compound with a precise methodology of extraction and purification. Identification of the compound can always be identified by melting point
This lab’s end result was to correctly identify each unknown solution using prior knowledge of chemical properties and the results of the first experiment conducted. Unknown solution D was the only colored solution, being blue while the others were clear. This made it easy to then match D up to Copper Sulfate because of its color. As unknown A and B were added together, lots of gaseous bubbles formed and revealed the fact that that reaction was the reaction between Hydrochloric Acid and Sodium Carbonate because it was the only reaction that produced a gas release. Unknown A and C produced the only yellow, brown precipitate just as the reaction between Sodium Carbonate and Silver Nitrate had previously.
When determining the solubility of malonic acid in different solvents both water and methyl alcohol were found to be polar when mixed with malonic acid. Hexane however was insoluble. Lastly biphenyl was mixed with water and was found to be insoluble, methyl alcohol was determined to be partially soluble.Hexane on the other hand was the only soluble solvent for biphenyl 2. Part B. of this experiment determined the solubility of different alcohols in hexane or water. Alcohols 1-Octanol, and 1-butanol were both found to be soluble in hexane while methyl alcohol was determined to be insoluble.
This test was conducted for the purpose of selective and differential whether or not my organism can tolerate high salt concentrations. It is based on the mannitol fermentation. The phenol-red indicator helps to identify the bacteria. Upon viewing my results it was determined that my colonies did not ferment mannitol. This was easily observed because they remained translucent.