Thursday, May 17, 2012
Evaporation and Intermolecular Attrations
First thing we did was get probe 1 and probe 2, wrapped square pieces of filter paper and tightened them with rubber bands. In the methanol container we put probe 1 in, then we placed probe 2 in the ethanol container. After them being in their for 45 seconds we immedietly taped them at the edge of the counter at 5 cm. Once they were nice and secured we observed and waited for the temperatures were at their lowest. We got the highest temp and the lowest temp and subtracted to figure out the change during evaporation. After the whole process we repeated it with the exact steps and recorded the data.
Tuesday, May 15, 2012
Beer's Law
In the begining we got 4 beakers and labeled them from one to four. The first beaker we put 2 mL of NiSo4, the second beaker got 4 mL, The third got 6 mL, and the fourth one got 8 mL. We then got the pipet and put 8 mL of distilled water in the first beaker, 6 mL in the second, 4 mL in the third, and then 2 mL in the last one. All the beakers then got stirred up with a stirring rod.
Then in the macbook we opened up to "Exp 11 colorimeter". We created our graphs on it and calibrated our colorimeter to help us out during the experiment. When using the colormeter we had to be gentle when handling the cuvette and clean it before we used it. Then we put 3/4 of each solution in the cuvette then copied the results.
Trial Number | 0.40 M NiSo4 (mL) | Distilled H2O (mL) | Concentration |
| 1 | 2 | 8 | 0.08 |
| 2 | 4 | 6 | 0.16 |
| 3 | 6 | 4 | 0.24 |
| 4 | 8 | 2 | 0.32 |
| 5 | ~10 | 0 | 0.40 |
Then in the macbook we opened up to "Exp 11 colorimeter". We created our graphs on it and calibrated our colorimeter to help us out during the experiment. When using the colormeter we had to be gentle when handling the cuvette and clean it before we used it. Then we put 3/4 of each solution in the cuvette then copied the results.
| Trial | Concentration (mol/L) | Absorbance |
| 1 | 0.080 | .124 |
| 2 | 0.16 | .238 |
| 3 | 0.24 | .714 |
| 4 | 0.32 | .740 |
| 5 | 0.40 | .880 |
| 6 | Unkown number 3 | .661 |
| Concentration of Unknown | 0.24 mol/L | |
Thursday, May 3, 2012
acid/base lab
When it comes to acids and bases the similarity that they share is they both occurs on the Ph scale that ranges from 0-14. Other than that the similarities end, on the Ph scale numbers below 7 are considered acids and the numbers above 7 are considered bases. Then the number 7 is considered neutral and is neither an acid or base.
Making *Crystals* BROO!
First as a group we got a beaker and put distilled water in it, at no precise amount. Then we let it bowl over a mini heating stove thing, and once it was very bubbly we added potassium. My trick was to put not too much but not that little in and stir til it evaporated and then keep the process going. Soon enough we just stopped and class was over. We walked into class the next day with a beaker full of crystals not more than half way up but close. With all these crystals each one of us in the group tried to break up big crystals to tie up to a string. Then it was a self lab, I got my crystal and did the same process to a beaker but did not but the crystal in yet. Also I cold died my water with yellow and a few drops of red making it a light orange or a dark yellow. After putting as much dissolved potassium as I wanted I let the substance cool down and then later on added my crystal. Then next day my crystal got bigger and that day I did the same process . . The last day came and this is what my crystal looked like:
Thursday, March 8, 2012
The Hydrates Lab
Before I begin to talk about this experiment may I include that this crystal lab blew my mind! (CuSO4 5H2O) We got blue crystals that weighed 5 grams and put them in a test tube over a flame to heat them up. When they were over the burner they began to turn white because the water was evaporating from the crystals. After they were fully white we weighted them in a cup at 1.73, this means that 3.2 grams of water evaporated. We got distilled water and poured it in the white crystlas and BOOM they turned back to the same exaclty blue color they started with!!
Silver/Copper Replacement Lab
We got a copper wire that was about 30 cm long and coiled it around a pencil forming a spring like structure. We placed the sprinted copper into a tube and made sure it reached from the top to the bottom. When it did, we set it aside for later use. . . We weighted some silver nitrate and it cam to 1.018 grams. We transferred the silver nitrate into the test tube and poured distilled water until the water was 2 centimeters from the top. Then we covered the top of the test tube with parafilm, and you place your thumb on top and shake until the silver nitrate was all dissolved. After that we took the stretchy parafilm off and observed the reaction it made, it set their all week because Mr. Ludwig said that it would take time for it to react. The very next Monday we came back to school and the craziest thing happened!
It looked so pretty i didn't want to mess it up but we had to in order to get to the next step. Nika shook the tube and poured the substance in a filter we made with paper that weighted .906 grams. We took the coiled copper out and placed it on the side on a paper towel it weight 3.17 grams. Then we let that sit for a day so it could filter out all the way out. The next day the silver was dry in the paper filter and we weighted it and it was 2.17 grams.
It looked so pretty i didn't want to mess it up but we had to in order to get to the next step. Nika shook the tube and poured the substance in a filter we made with paper that weighted .906 grams. We took the coiled copper out and placed it on the side on a paper towel it weight 3.17 grams. Then we let that sit for a day so it could filter out all the way out. The next day the silver was dry in the paper filter and we weighted it and it was 2.17 grams.
Sunday, February 12, 2012
measurement and calculation
sample 1:
CuO--> Cu= 1 Cu/ total> 100= 64/80= 80%Cu/20%O
O= 1*16
sample 2:
Cu2O--> 2*64 128 128/144> 100= 89%Cu/ 11%O
1*16 16
144 g/mol
Step by step process. . .
Write a balance chemical equation and identify the knowns and the unknowns.
25.0 50.0g ---> ?g
P'4(s) + 50'2(g) ---> P'4O'1O(s)
Determine the number of moles of the reactants by multiplying each mass by the conversation factor that relates moles and mass, the inverse of molar mass.
25.0 gP'4 * 1 mol P'4/123.9 g P'4 = .202 mol P'4
50 g O'2 * 1 mol O'2/32 g O'2= 1.56 mol O'2
Calculate the actual ratio of available moles of 0'2 and available moles of P'4.
1.56 mol O'2/.0202 mol P'4 = 7.72 mol O'2/ 1 mol P'4
Determine the mole ration of the two reactants fomr the balanced chemical equation.
5 mol O'2/ 1 mol P'4
Because 7.72 mol O'2 is available but only 5 mol is needed to react with 1 mol P'4, O'2 is in excess and P'4 is the limiting reactant. Use the moles of P'4 to determine the moles of P'4O'10 that will be produced.
Multiply the number of moles of P'4 by the mole ratio of P'4O'10 (the unknown) to P'4 (the unknown)
.0202 mol P'4 * 1 mol P'4O'10/1 mol P'4 = .202 mol P'4 O'10
and then i will add the yellow paper from the scanner in class. . .
CuO--> Cu= 1 Cu/ total> 100= 64/80= 80%Cu/20%O
O= 1*16
sample 2:
Cu2O--> 2*64 128 128/144> 100= 89%Cu/ 11%O
1*16 16
144 g/mol
Step by step process. . .
Write a balance chemical equation and identify the knowns and the unknowns.
25.0 50.0g ---> ?g
P'4(s) + 50'2(g) ---> P'4O'1O(s)
Determine the number of moles of the reactants by multiplying each mass by the conversation factor that relates moles and mass, the inverse of molar mass.
25.0 gP'4 * 1 mol P'4/123.9 g P'4 = .202 mol P'4
50 g O'2 * 1 mol O'2/32 g O'2= 1.56 mol O'2
Calculate the actual ratio of available moles of 0'2 and available moles of P'4.
1.56 mol O'2/.0202 mol P'4 = 7.72 mol O'2/ 1 mol P'4
Determine the mole ration of the two reactants fomr the balanced chemical equation.
5 mol O'2/ 1 mol P'4
Because 7.72 mol O'2 is available but only 5 mol is needed to react with 1 mol P'4, O'2 is in excess and P'4 is the limiting reactant. Use the moles of P'4 to determine the moles of P'4O'10 that will be produced.
Multiply the number of moles of P'4 by the mole ratio of P'4O'10 (the unknown) to P'4 (the unknown)
.0202 mol P'4 * 1 mol P'4O'10/1 mol P'4 = .202 mol P'4 O'10
and then i will add the yellow paper from the scanner in class. . .
Monday, February 6, 2012
How much water is in popcorn. . . .
In this lab we were suppost to find the mass of water loss from the certain amount of popcorn and calculate what percent of the original mass was the water given off in the popping. First we got a 1000-ml beaker and covered at the bottom with vegetable oil. Put foil over the top and it weighted 221 1/2 grams. Then we added exactly 28 corns and then it weighted at 225 grams. Pocked holes into the top of the foil then we lit that baby up. After all the popcorn was popped we let it set to cool and then weighted it at 2221/2 grams. We got that 1.9% of water was given of the popping. mm then we got some popcorn for everyone:) it was a very good day lol
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