Saturday, November 23, 2013

A Beer's Law Study

Nathan Bahn Beers Law Study Lab Introduction: In this lab, we utilize a spectrometer to observe the transmittance of erupt at a certain wave length. We experimented to see if the breakwaterarity of a firmness of purpose changes the transmittance of elation and the absorbance of that light by the solution. By observing the percent transmittance and the follow of light absorbed, we fag calculate the core of color absorbing components in the solution. by this process is how we are able to discover the amount of atomic number 29 in the solution. Experimental mathematical process: 250 mL of the copper solution was made by creating ascorbic acid mL of the solution, reacting CuO with HNO3, and then diluting to the determine of 250 mL. Using this stock solution, different concentrations were made and placed in the sprectrometer for observation. The absorbances and transmittances were recorded for use when identifying the amount of the color-absorbing copper ions later. A graph was plotted of Absorbance v. seawallar(a) Concentration substantially see the results of the experiment. Pre-Lab Questions: 1. Absorbance = - Log10T = 2-Log10 (%T) a) 2 Log (89.95) = .0460f) 2 Log (28.18) = .5501 b) 2 Log (80.91) = .0920g) 2 Log (20.51) = .6880 c) 2 Log (59.02) = .2290h) 2 Log (17.38) = .7600 d) 2 Log (47.75) = .3210i) 2 Log (14.45) = .8401 e) 2 Log (34.83) = .
bestessaycheap.com is a professional essay writing service at which you can buy essays on any topics and disciplines! All custom essays are written by professional writers!
4580 2. Molar Concentration 1.213 g Cu (1 mol Cu/ 63.546 g Cu) (1 mol [Cu (H2O6)2+]/ 1 mol Cu) (1/ .100 L) = .191 M [Cu (H2O6)2+] a) M2 = (M1V1)/V2M2 = (.191 M [Cu (H2O6)2+])( 1.0 mL)/ (50.0 mL) = .0038 M b) (.191 M [Cu (H2O6)2+]) (2.0 mL)/ (50.0 mL) =! .0076 M c) (.191 M [Cu (H2O6)2+]) (5.0 mL)/ (50.0 mL) = .0191 M d) (.191 M [Cu (H2O6)2+]) (7.0 mL)/ (50.0 mL) = .0267 M e) (.191 M [Cu (H2O6)2+]) (10.0 mL)/ (50.0 mL) = .0382 M f) (.191 M [Cu (H2O6)2+]) (12.0 mL)/ (50.0 mL) = .0458 M g) (.191 M [Cu (H2O6)2+]) (15.0 mL)/ (50.0 mL) = .0573 M h) (.191 M [Cu (H2O6)2+]) (17.0 mL)/ (50.0 mL) = .0649 M i) (.191 M [Cu...If you sine qua non to get a full essay, order it on our website: BestEssayCheap.com

If you want to get a full essay, visit our page: cheap essay

No comments:

Post a Comment

Note: Only a member of this blog may post a comment.