# Free Carboxylic acid Essays and Papers

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Corrected Text Slide 1: Aromatic acids Learning Objectives At the end of this class session, you will be able to know about • Preparation and properties of aromatic acids Slide 2: Introduction Aromatic carboxylic acids belong to the group of carboxylic acids. Aromatic acids are white crystalline solids that are soluble in hot water, alcohol and ether. It is partially soluble in cold water. Slide 3: Explanation Preparation of aromatic acids Teacher: Hi, students. Today we are going to learn about

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of these scents are esters. Most of the aromas we know represent a mixture of esters and other molecules like alcohol. The process of making ester is known as esterification. Esters are formed carboxylic acid and a carboxylic acid reacts with alcohol, water is also produced from this reaction. Carboxylic acid contains the –COOH group (Jim Clark, 2003) The general formula for esterification is O O || || R-C-OH + HO-R R-C-O-R + HOH Figure 1: Formula

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bind to substrate in a molecular or formate (dissociated) form. Generally it is assumed that the carboxylic acid group dissociates to give carboxylate and hydrogen, although some theoretical calculations have yielded similar binding energies for both formate and molecular form [23-25]. Furthermore, it has been discussed that dissociated hydrogen may play a role in stabilization of the carboxylic acid layer [26, 27]. Besides, in closed layer of up-right oriented molecules adsorbate-adsorbate interactions

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final weight     percent yield      2,4-DNP     Tollen's test     pathway .42g     67%     positive     negative     oxidation of secondary OH Good Things My experiment went well. I began my experiment with .64g of 2-ethyl-1,3-hexanediol. The molecular weight of this compound is 146.2g/mol. It is converted into 2-ethyl-1-hydroxyhexan-3-one. This compounds molecular weight is 144.2g/mol. This gives a theoretical yield of .63 grams. My actual yield was .42 grams. Therefore, my percent yield was 67%

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range. It was slightly soluble; therefore, there might be a functional group that is polar such as an alcohol. Once again, the IR spectroscopy confirms this since the broad peak is located at around 3400 cm-1. Unknown C was negative for the Chromic acid test so there might be a tertiary alcohol or no alcohol. Since the IR has an alcohol I assume that there might be a tertiary alcohol. After completing the test, I can minimize the options that correspond to the IR and the tests to draw a conclusion

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Extraction To obtain oils from plant materials steam distillation is typically used. Solvent extraction is also used to create a resinoid (resembling characteristics of resin). Solvent extraction is commonly used with volatile substances that may be lost with distillation. Solvent used are pentane, hexane, and mixes of solvents such as ethyl acetate and hexane which is commonly used in place of benzene (see Figure 2). Gas-Liquid Chromatography Gas-Liquid Chromatography is the process of separating

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## The Origins of Life

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such forms as CO2/CO, N2, and H20. Stanley Miller, through experimentation, shows that given an energy source like heat or electric charge it is possible to form reactions that create complex molecules, and through subsequent experiments nucleic acids like adenine were even formed. This is the premise for the “hot” theories of the origin of life. Given there are many derivative possibilities like process evolution, chemoautotrophic, and photoautotrophic origins, the basis is that given an energy

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Which lip balm formulation will provide better moisturization, feel, taste and smell? Hypothesis If each lip balm has different ingredients and formulation, then the results for each lip balm will turn out differently; the moisturization, taste, feel and smell will all be different. Background Research Lip balm was invented by Dr. Charles Brown Fleet. He invented it in the 1880’s as a way to moisturize your lips. Lip balms seal your lips to protect it from losing its own moisture. Lips cannot make

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\caption{Separations r for quadruple star systems} \label{tab:threecols} \end{table} \subsection{Flux ratios} Finally, I determined the flux ratios, the ratios of the peaks in flux, for the components of the sources. To calculate these flux ratios, I have to subtract the background from the peak values before dividing both. I calculated the flux ratios of $2$ sources A and B fr via fr = \frac{\text{peak in flux of source A} - \text{background}}{\text{peak in flux of source B} - \text{background}}

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Fermentation of Apple Cider “Science knows no country, because knowledge belongs to humanity, and is the torch which illuminates the world. Science is the highest personification of the nation because that nation will remain the first which carries the furthest the works of thought and intelligence.” – Louis Pasteur, Great French Scientist (1822-1892). Fermentation is chemical changes in organic substances produced by the action of enzymes. This general definition includes virtually all

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