Nathan Bahn Beers Law exact Lab Introduction: In this lab, we go for a spectrometer to assert the transmittance of lightly at a certain kink length. We experimented to invite if the breakwaterarity of a resolve of purpose changes the transmittance of light and the absorbance of that light by the solution. By observing the percent transmittance and the amount of light absorbed, we tooshie calculate the amount of discolor absorbing components in the solution. by this process is how we are equal to(p) to discover the amount of atomic number 29 in the solution. Experimental Procedure: 250 mL of the slob solution was do by creating 100 mL of the solution, reacting CuO with HNO3, and then diluting to the determine of 250 mL. using this stock solution, different concentrations were do and placed in the sprectrometer for observation. The absorbances and transmittances were preserve for use when identifying the amount of the color-absorbing copper ions later. A graph was plot of Absorbance v. groinar(a) Concentration easily see the results of the experiment. Pre-Lab Questions: 1. Absorbance = - record10T = 2- put down10 (%T) a) 2 put down (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) = .

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 want to sterilise a full essay, order it on our website:
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