Chemistry design: Fuel energy lab Research question: What is the energy density of cyclohexane‚ and how does it compare to the energy density of normal hexane? Variables: Independent variable – The independent variable in this experiment will be the type of fuel that will be used. This will be either Hexane or Cyclohexane. Each type will be given to us and labelled so it will be quite simple to control this variable. This cannot be measured using any apparatus so the lab assistant must be entrusted
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spinach using column chromatography. Spinach was dehydrated using ethanol‚ and the pigments were extracted with dichloromethane. The spinach extracts were dried using CaCl2. Then‚ the solid pigments were run through a column using a non-polar solvent‚ hexane. The polar absorbent material in the column separated the different pigments by allowing the least polar molecules to travel through the column faster than the more polar molecules. The different pigment layers were collected‚ dried‚ and their weights
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Beverly Abstract: Plant pigments were separated and concentrated from a crude spinach extract through the use of column chromatography and an eluatropic series of hexanes‚ hexane/acetone‚ and methanol. The pigments were analyzed using thin layer chromatography with a 30% ethyl acetate/hexane developing solvent. Introduction: Chromatography is a technique used to separate a mixture of two or more components based on differences in their physical properties. It can be used
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containing salt‚ sand‚ and benzoic acid can be separated using the separate components’ solubility properties (ability to dissolve in water). Because it is known that organic compounds (such as benzoic acid) are very soluble in organic solvents‚ a hexane: ethanol mixture was used to dissolve the acid. Also‚ because it is known that sodium chloride dissolves in water‚ water was used to dissolve the NaCl. Along with separating the components of a heterogeneous mixture‚ the percent recovery of each substance
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of solute between immiscible solvents * IODINE CRYSTALS + WATER + N-HEXANE the N-hexane turned to color violet. The method used to separate that mixture is Extraction. Extraction is the distribution of a solute in two immiscible solvents. The substances that separated are H2O and Iodine crystals with N-hexane. The iodine crystals have a very low solubility in water and it dissolves quickly when it combines with N-hexane and when it mixed. Separation of components with different volatilities
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chromatography. Once the Lycopene is isolated‚ IR spectroscopy will determine its percentage actually obtained by chromatography. Procedure: A massed sample of 1.012g of tomato paste was placed in a 125mL Erlenmeyer flask. To the flask‚ 5mL of 50:50 hexane-acetone was added into the flask. After the 50:50 addition the flask was capped and shaken. After the settling of the sediments‚ the liquid portion was pipetted in a 50mL Erlenmeyer flask. The process of separating the liquid from the solid was
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water b. Naphthalene in water c. Amphetamine in ethyl alcohol d. Aspirin in water e. Succinic acid in hexane f. 1-Decanol (n-decyl alcohol) in water 2. Predict whether the following pairs of liquids would be miscible or immiscible. It may be helpful to look up some of the chemical structures. (3pts) a. Benzene and water b. Water and methyl alcohol c. Hexane and benzene d. Methylene chloride and benzene e. Water and toluene f. Ethyl alcohol and isopropyl alcohol
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Scheme 5.5: KMnO4 is the oxidant in this reaction that oxidizes the alcohol to the ketone. The CuSO 4 is there to support the KMnO4. The oxidant reacts with the secondary alcohol which forms benzophenone and MnO2. Then the melting point of the hexanes will be identified. The reactant is a secondary alcohol therefore there would be no over-oxidization. The purity of the benzophenone by recrystallization will be analyzed using thin layer chromatography (TLC) analysis. Two TLCs is done in this experiment
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The TLC plate remained blank showing proof of no compound other than the hexane itself. This error was expected since in the first part of the experiment‚ oxidation of fluoreneto Fluorenone was supposed to be done to a 50-50 level. As mentioned earlier‚ the oxidation had carried out longer than expected and hence the sample contained
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Determine mole fractions: H2O ⇒ 1.34 mol / (1.34 mol + 0.543 mol) = 0.71 C2H5OH ⇒ 0.543 mol / (1.34 mol + 0.543 mol) = 0.29 Example #3: A solution contains 10.0 g pentane‚ 10.0 g hexane and 10.0 g benzene. What is the mole fraction of hexane? Solution: 1) You need to determine the moles of pentane‚ hexane and benzene: to do this‚ you need the molecular weights. Here are the formulas: pentane: C5H12
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