We added anhydrous Sodium Sulfate as a drying agent. To complete, we distilled the cyclohexene and collected the product. Knowing this data, we determined the yield % which is 58.5%. This experiment features the dehydration of cyclohexanol and produce cyclohexene. The acid catalyzed dehydration of cyclohexanol with distillative removal of the resulting cyclohexene from the reaction mixture
A solvent is the substance in which the solute is being dissolving. Water Water is chemically H2O. Water molecules are formed when two hydrogen molecules and one oxygen molecule combine. Water is a good solvent due to its polarity. When an ionic or polar compound enters water, it is surrounded by water molecules Salt or Sodium Chloride Salt is a mineral that is composed primarily of sodium chloride (NaCl), a chemical compound belonging to the larger class of ionic salts.
h) A way to make hard water softer is to put an sodium nitrate and create a precipitate to mellow out the reaction. Another way of making it softer is by removing the calcium ions one way of doing that is by boiling the solution to take out some of the ions. Conclusion: Overall, we determined that sodium carbonate, Na2CO3, is the anion that can be used to precipitate the most metal cations. Also, we learned that the anion sodium chloride, NaCl, could be used to remove silver ions from solutions. The stuff that I found interesting was that how many colours you can get when you mix the cations and anions
For example, the reaction |BaCl2•2H2O ( BaCl2 + 2H2O |(2) | is reversible, and if water is added to the anhydrous salt BaCl2, formation of BaCl2•2H2O takes place: |BaCl2 + 2H2O ( BaCl2•2H2O |(3) | The reaction of dehydration of hydrated ferrous sulfate |FeSO4•7H2O ( FeSO4 + 7H2O |(4)
Introduction: You are aware that sugar dissolves in water but oil does not. What factors determine whether one substance will dissolve in another? A solvent is a substance that is capable of dissolving other substances and forming a homogeneous mixture called a solution. The substance dissolved is called the solute and is the component present in the smallest amount. The dissolving process involves a consideration of the relative strength of three intermolecular attractive forces.
Care must be taken when squeezing the pipet bulb on the filter pipet. Too much pressure might cause the filter to leak or fall off. Add about 2 mL of fresh tert-butyl methyl ether to the solid in the RB flask, warm briefly, let the solids settle for a minute, and pipet the liquid to the centrifuge tube as before. Again allow the solids to settle briefly in the centrifuge tube, then filter the liquid through the pressure filtration apparatus, into the same 25 mL Erlenmeyer flask. Doing a rinse such as this helps to ensure that any trimyristin that was left behind in the RB flask and centrifuge tube is not lost, thereby helping to ensure that
The aspirator was turned to medium high, and then the copper was poured onto wetted filter paper. Using distilled water to remove all copper from the beaker. Once completely on filter paper 6mL of acetone was added to the copper to help dry it out. The filter paper was then removed and set down to dry completely. Once dry the filter paper was weighed with the copper on it and subtracted from the original weight to see the amount of copper left after
6) The ether was removed using the rotary evaporator. We discarded the used MGSO4 into solid waste containers. The solid that remained after ether evaporated was the neutral organic compound. 7) We scraped the solid from the flask with a spatula and allowed the solid to dry thoroughly on a piece of filter
For reaction (II) Pb (NO3)2 + 2KI -> PbI2+2KNO3 Lead nitrate is soluble, so it gets written as ions. The same goes for potassium iodide and potassium nitrate. Complete Ionic equation: 2 K+1 + I-1 + Pb2+ + NO32- -> PbI2 + 2K+1 + NO3-1 Net Ionic Equation: Pb+2 + 2 I-1 -> PbI2 Warm-Up Exercise 2 In this lab you will mix 25 mL of 0.05M lead nitrate with 1.4 mL of 0.025M sodium carbonate. After the reaction occurs, you will filter the solution to remove the precipitate. You will then test the remaining solution for excess lead ion and for excess carbonate ion… Imagine that you mix the two volumes and then freeze frame the reaction so the it does not proceed: 1.
Observe the color change while it is being heated. After observing the color change, find the mass and moles of the hydrate. Then find the mass and moles of the water eliminated. And lastly find the mole ratio of water to hydrate. For part 2, do the same thing as part 1 except use an unknown hydrate and calculate the percent mass of water in an unknown hydrate.