Dileepa Galagedera 02/14/14 Chemistry 106LB – 03 Ms. C. Chrissian Lab 2 – A Library for Chemists * Part I - Warm Up Questions; * Graph #1: a plot made from an equation. * Graph #2: a plot made from supplied data and an equation derived from the plot. * Graph #3: A plot made from an equation; If the value of y is 0.12, the value of x would be (0.12 + 0.01)/ 0.008, or 16.25. * Graph #4 – A straight line plot generated from a made up equation. * Graph #5 – An equation generated from a made up straight line plot.
pyridinium hydrobromide perbromide type of stationary phase column length column temperature rate flow of the carrier gas List the 4 general factors that affect the separation obtained on a gas chromatograph What specific technique is used to collect/isolate your purified unknown compound at the end of the recrystallization experiment? suction filtration 14 of 22 4/16/12 9:15 PM StudyBlue Flashcard Printing of Lab Final 2211L UGA
|Cloudy milk liquid|Rapid |Red-center |Fizzled |No change | | | | |Powder remains. Odor noted.| |boiling/slight odor|white with |Chemical | | | | | | | | |faded blue |Reaction | | | | | | | | | |Powder to |
The Cover Sheet to the Confidence Report South Street Seaweed Seaport Tabulated test Results Name of Test Requirements Result Test for Iodine If the mineral oil turns a shade of purple (either dark, or light such as pink), then the Tincture is positive for iodine. The solution became light lavender/pink. This shows that the tincture is positive for iodine. Test for Iodide ion A few things can happen: 1) A white precipitate only, indicating chloride ion is present predominantly (Na Cl in salt water). 2) A yellow precipitate only, indicating iodide ion is present predominantly.
Determine the percent yield of this reaction, showing all steps of your calculation. (3 points) heoretical yield of H2 gas: (1.156 x 10^-3 moles)(1 mole H2/ 1 mole Mg) = 1.156 x 10^-3 moles Theoretical mass: (1.156 x 10^-3 moles)(2.02 g/mole) = 2.335 x 10^-3 g Using ideal gas law: P = (1.1 atm)(760 torr / 1 atm) - 19.8 torr = 816.2 torr V = 0.026 L T = 295 K Solve for n: n = PV/(RT) n = (816.2 torr)(0.026
1A-Water/MSG: When shaken the water and MSG they mostly dissolve creating an orange gold color, the substance foams at the top from the vigorous shaking action yet still some MSG residue remains at the bottom of the test tube, it was unclear if the reason was because we did not mix it enough or rather we added too much MSG to the amount of water given. 1A-Alcahol/MSG: These two substances barely mix together leaving a transparent yellow tint to the alcohol as well as MSG that was unable to dissolve at the bottom of the test tube. We concluded that this residue was unable to be dissolved in this liquid because the amount of MSG remaining in the bottom of the test tube was around the same amount that was originally placed into the alcohol. 1B- Alcohol/Vegetable Oil: Created a foamy top layer, was semi unclear but as to the best of my observations the alcohol nearly completely dissolved the oil, leaving a foggy appearance to the once clear
Powder B, when water was added, it only became watery. When vinegar was added, the liquid sunk to the bottom and none of the outcomes changed. With iodine, it had a color change with some bubbles but shortly after it went back and stayed the color brown. Powder C, it was not affected by water but by vinegar, it changes into bubbles and fizzes up and has a chemical reaction. With iodine it changes to a brown color and becomes clumpy.
These conditions are more suitable for ionic bonds. That is why the sucrose and salicylic acid did not have good results, because they have covalent bonds. There were a few sources of error that could be improved in the experiment. For example, when the substance was poured into the test tubes, some of the compound remained stuck to the weighing paper. If we used a stick resistant material for this step, the whole 1g would be accounted for.
The formula excludes the absorbance of the “yellow” solution because almost all of the bromothymol blue is in the HBB form in the acidic “yellow” solution, resulting in the yellow color and making its absorbance at 616 nm almost 0, or negligible. The formula uses the absorbance of the “blue” solution because its λmax is 614.9nm, which is very close to 616nm.This means that in the “blue” solution, most of the bromothymol blue is in the form BB-, displaying a blue color that absorbs the light at 616 nm, thus resulting in a higher recorded absorbance. Below are the recorded absorbance values at 616
ANS: B PTS: 1 DIF: L2 REF: p. 276 | p. 277 | p. 278 OBJ: 9.2.2 Apply the rules for naming and writing formulas for compounds with polyatomic ions. STA: SC.912.P.8.7 | SC.912.N.1.1 | SC.912.N.1.6 BLM: application 58. ANS: C PTS: 1 DIF: L2 REF: p. 268 | p. 278 OBJ: 9.2.2 Apply the rules for naming and writing formulas for compounds with polyatomic ions. STA: SC.912.P.8.7 | SC.912.N.1.1 | SC.912.N.1.6 BLM: