User:Sarah Burkhard/Notebook/CHEM 481 Polymers/2016/09/14: Difference between revisions
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0.275 g * 97 ml/3 g bentonite = 8.89 ml bentonite solution | 0.275 g * 97 ml/3 g bentonite = 8.89 ml bentonite solution | ||
we did not account for the change in volume . we kept the amount of HCl moles constant, which in effect diluted the concentration when using 10 ml of the solution for the beads formation. In following experiments we keep both volume and moles HCl constant; the only variable is water added. | |||
==Notes== | ==Notes== |
Revision as of 10:12, 21 September 2016
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Protocol3 wt% bentonite solution instead of NaMT powder for the following protocol: Create a solution of 2.5 g PVA, 0.275 g NaMT, and 30 mL 10% HCl 3 wt% = 3 g bentonite / 97 ml water 0.275 g * 97 ml/3 g bentonite = 8.89 ml bentonite solution we did not account for the change in volume . we kept the amount of HCl moles constant, which in effect diluted the concentration when using 10 ml of the solution for the beads formation. In following experiments we keep both volume and moles HCl constant; the only variable is water added. NotesFor centrifuge: make sure they have equal weight. For malachite green: plot a graph and check for linear fit. Find r^2 . films: with clay more brittle, but stronger half of film will be dried in oven. mass before oven of clay film: 1.24545 g. clear film mass (glut): 0.53 g . watchglass is labeled "A". weight 1: 49.35200 g of beads 2: 53.903 equal weight: 48.666 48.65 centrifuge machine: RC 6 in Balance Room 3rd floor
take ethyl acetate out 34 ml starting point for beads titrate down until buret reads 44 ml add 1 ml glutaraldehyde ONCE, stir for 7 minutes at 320 rpm add 7 ml sodium bicarbonate replace glutaraldehyde to prevent sticking together ? Friday: 20 percent clay mass of whole solution = 30 ml Hcl = 30 g 20 percent clay in PVOH. 2.5 g PVOH = 80 percent 2.5/.8 = total mass = 3.125 g multiply by 0.2 (20 %) to get mass of clay in g : 0.625 g . divide by 0.03 to get solution: 0.625 g / 0.03 g/ml = 20.83 ml bentonite solution 3 g / 100 ml solution = 0.03 g/ml attention: the HCl will be diluted by the water added with the clay solution. Report the new concentration. Through adding 20.83 ml bentonite solution (97 ml H20/ 100 ml bentonite soluution) we effectively added 20.2 ml H20. This will dilute our 30 ml 10% HCL by 1/3 , so to 6.66 % HCl. weight of clay film to be crosslinked: 1.43208 g Xray room 302, Rigaku
Which samples did you test? Did you use the same sample holder for each? expect broad diffraction: means no pattern in "slit width" or distance between atoms diffraction angle (twice diffraction angle). refracted so you get 2 beta
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