Equilibrium Review worksheet
... In a rigid 1.00 L laboratory reaction vessel, a technician places 1.00 mol of each of the four substances involved in this equilibrium. The vessel is heated to 650 °C. Determine the equilibrium amount concentrations of each substance, organizing your values in an ICE table. (Hint: use the value of K ...
... In a rigid 1.00 L laboratory reaction vessel, a technician places 1.00 mol of each of the four substances involved in this equilibrium. The vessel is heated to 650 °C. Determine the equilibrium amount concentrations of each substance, organizing your values in an ICE table. (Hint: use the value of K ...
TANNIC ACID
... Consists of gallotannins obtained by solvent extraction from certain natural sources; the substance is not an acid in the chemical sense. The common name "Tannic acid" has been adopted to distinguish the commercial substance from other tannins, such as condensed tannins. These specifications relate ...
... Consists of gallotannins obtained by solvent extraction from certain natural sources; the substance is not an acid in the chemical sense. The common name "Tannic acid" has been adopted to distinguish the commercial substance from other tannins, such as condensed tannins. These specifications relate ...
Kinetics of Oxidation of Benzyl Alcohol with Dilute Nitric Acid
... first studied in detail by Ogata et al.15 They proposed that the reaction proceeded through some intermediate; however, it was not identified. Many other investigators have also studied the oxidation of benzyl alcohol using nitric acid.17-21 However, the published information suffers from two limita ...
... first studied in detail by Ogata et al.15 They proposed that the reaction proceeded through some intermediate; however, it was not identified. Many other investigators have also studied the oxidation of benzyl alcohol using nitric acid.17-21 However, the published information suffers from two limita ...
Reaction Systems Engineering II (part 1)
... rH° = (–230.0) – (–285.8) = 55.8 kJ mol–1. rS° = –10.8 – 69.9 = –80.7 J K–1 mol–1. rG° = rH° – TrS° = 55.8 – 298(–80.7) / 1000 = 79.85 kJ mol–1 Kw = exp(–rG° / RT) = exp[–79.851000 / (8.3145298)] = 1.0110–14 mol2 kg–2 2) rG° = rH° – TrS° = 55.8 – 348(–80.7) / 1000 = 83.88 kJ mol–1 Kw = ...
... rH° = (–230.0) – (–285.8) = 55.8 kJ mol–1. rS° = –10.8 – 69.9 = –80.7 J K–1 mol–1. rG° = rH° – TrS° = 55.8 – 298(–80.7) / 1000 = 79.85 kJ mol–1 Kw = exp(–rG° / RT) = exp[–79.851000 / (8.3145298)] = 1.0110–14 mol2 kg–2 2) rG° = rH° – TrS° = 55.8 – 348(–80.7) / 1000 = 83.88 kJ mol–1 Kw = ...
Joint Symposium of Waseda University and Peking University
... electrode and covered it with an organic (or an aqueous) solution. In this manner, a conventional three-electrode potentiostat can be used to study ionizable drugs transfer process at a liquid /liquid interface. Physicochemical parameters such as the formal transfer potential, the Gibbs energy of tr ...
... electrode and covered it with an organic (or an aqueous) solution. In this manner, a conventional three-electrode potentiostat can be used to study ionizable drugs transfer process at a liquid /liquid interface. Physicochemical parameters such as the formal transfer potential, the Gibbs energy of tr ...
aq - Moodle@FCT
... The volume of NaOH solution is given in the problem. Therefore, we need to find the number of moles of NaOH to solve for molarity. From the preceding equation for the reaction between KHP and NaOH shown in the text we see that 1 mole of KHP neutralizes 1 mole of NaOH. How many moles of KHP are conta ...
... The volume of NaOH solution is given in the problem. Therefore, we need to find the number of moles of NaOH to solve for molarity. From the preceding equation for the reaction between KHP and NaOH shown in the text we see that 1 mole of KHP neutralizes 1 mole of NaOH. How many moles of KHP are conta ...
A Gas-phase Electrochemical Reactor for Carbon Dioxide
... usually titania) the deposition of 10 % Pt nanoparticles on CNT (sample Pt10-PS) increased H2 generation with respect to not noble metal (Fe10-PS) as it is shown in Figure 1. On the contrary, carbon monoxide, which is the main intermediate in liquid fuel generation, was produced in higher concentrat ...
... usually titania) the deposition of 10 % Pt nanoparticles on CNT (sample Pt10-PS) increased H2 generation with respect to not noble metal (Fe10-PS) as it is shown in Figure 1. On the contrary, carbon monoxide, which is the main intermediate in liquid fuel generation, was produced in higher concentrat ...
Data Analysis
... the above sentence because three-dimensional plots, representing, for example, points with coordinates x, y, and z, can be represented in oblique projection on a twodimensional surface by computer techniques. Wherever possible, the data from an experiment should be plotted at an early stage, even if ...
... the above sentence because three-dimensional plots, representing, for example, points with coordinates x, y, and z, can be represented in oblique projection on a twodimensional surface by computer techniques. Wherever possible, the data from an experiment should be plotted at an early stage, even if ...
Wet Chemical Etching
... DU - T·DS < 0), which corresponds to the condition T·DS > DU. Therefore, the nobler the metal (the higher the required energy DU for etching), the higher the temperature and/or gain in entropy (e. g. by an increase of spatial degrees of freedom when changing from the solid into the liquid or gaseous ...
... DU - T·DS < 0), which corresponds to the condition T·DS > DU. Therefore, the nobler the metal (the higher the required energy DU for etching), the higher the temperature and/or gain in entropy (e. g. by an increase of spatial degrees of freedom when changing from the solid into the liquid or gaseous ...
Lecture 11 - U of L Class Index
... Sodium is still produced by electrolysis, although molten NaCl is now used. Common chemical species are not strong enough reducing agents to convert Na+ to the metal, so electrolysis is the only viable method of preparation. Potassium can be made by electrolysis also, but there are problems with thi ...
... Sodium is still produced by electrolysis, although molten NaCl is now used. Common chemical species are not strong enough reducing agents to convert Na+ to the metal, so electrolysis is the only viable method of preparation. Potassium can be made by electrolysis also, but there are problems with thi ...
Liquid–liquid extraction
Liquid–liquid extraction (LLE) consists in transferring one (or more) solute(s) contained in a feed solution to another immiscible liquid (solvent). The solvent that is enriched in solute(s) is called extract. The feed solution that is depleted in solute(s) is called raffinate.Liquid–liquid extraction also known as solvent extraction and partitioning, is a method to separate compounds based on their relative solubilities in two different immiscible liquids, usually water and an organic solvent. It is an extraction of a substance from one liquid into another liquid phase. Liquid–liquid extraction is a basic technique in chemical laboratories, where it is performed using a variety of apparatus, from separatory funnels to countercurrent distribution equipment. This type of process is commonly performed after a chemical reaction as part of the work-up.The term partitioning is commonly used to refer to the underlying chemical and physical processes involved in liquid–liquid extraction, but on another reading may be fully synonymous with it. The term solvent extraction can also refer to the separation of a substance from a mixture by preferentially dissolving that substance in a suitable solvent. In that case, a soluble compound is separated from an insoluble compound or a complex matrix.Solvent extraction is used in nuclear reprocessing, ore processing, the production of fine organic compounds, the processing of perfumes, the production of vegetable oils and biodiesel, and other industries.Liquid–liquid extraction is possible in non-aqueous systems: In a system consisting of a molten metal in contact with molten salts, metals can be extracted from one phase to the other. This is related to a mercury electrode where a metal can be reduced, the metal will often then dissolve in the mercury to form an amalgam that modifies its electrochemistry greatly. For example, it is possible for sodium cations to be reduced at a mercury cathode to form sodium amalgam, while at an inert electrode (such as platinum) the sodium cations are not reduced. Instead, water is reduced to hydrogen. A detergent or fine solid can be used to stabilize an emulsion, or third phase.