... Determine the mass of excess water remaining in the alum product at the end of day one. If you hadn’t waited a week to determine the percent yield, how large would the error be in your calculation of percent yield? 1. The amount of excess water remaining in the alum product after one day, which was measured a week later, was 0.588g. If you hadn’t waited a week to determine the percent yield, the mass recorded would be greater than the dried alum due to the water content present. Since the greater mass would suggest a greater abundance of alum, the percent yield of the same day would be higher than the percent yield calculated after a week. Comment on the relative error in your data and the class data that could have lead to the percent yields differing from the theoretical yields. 2. Both my data and the class data average produced a percent yield of less than 100%. These results suggest that alum was lost during the experimental procedure. We know this to be true because of the Law of Conservation of Mass. Some errors that may have led to lower percent yields than the theoretical yield include the loss of alum when transferring the crystals to the final beaker, not retrieving all the alum that was stuck to the filter paper after using the vacuum apparatus, or the spattering of solution that may occur due to overheating when dissolving the aluminum foil. Other factors that can contribute to lower yields may be not dissolving the entirety of the aluminum foil in the solution...
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...the balance and measure and record the mass of the 2.0 mL. Repeat the step four more times like previously ending with 10.0 mL. Again, wash the graduated cylinder, dry it, record the mass of the empty cylinder, and pick a sample that has not been done yet to do next: A, B, C, D, or unknown, and add 2.0 mL and record the mass of your chosen sample five times ending with 10.0 mL. Finally, once more, wash the graduated cylinder, dry it, record the mass of the empty cylinder, and pick the last sample left that has not been done to do next: A, B, C, D, or unknown, (I suggest staying in order and doing the unknown last). Add 2.0 mL using the plastic pipit and record the mass of your chosen sample five times ending with 10 mL. Clean up your lab which includes; taking off safety goggles, washing the graduated cylinder, clean the counter (with fancy spray to be super efficient), and putting equipment back in the right...
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...Purpose of the work: The purpose of this experiment was to experimentally determine the concentration of the unknown solution as well as what molecule the unknown solution was. In order to do that, we found Ka1 and Ka2 of unknown acid and Ka for the salt of the acid. Finally, we compared data to reveal the unknown solution identity and concentration. Experimental measurement and data analysis: The concentration of NaOH we used for this experiment was approximately 0.0804M. For titration of the H3PO4, we had two equivalence points and two half-equivalence points because H3PO4 has three ionizable H+ ions, however it only loses two because by the time it reaches HPO4 the last H+ is not recognized by the water. Therefore, since it can lose two ions it has two equivalence points and two half-equivalence points that give us two Ka and two pKa values. The pKa value is equal to pH at equivalence point and to determine Ka, we took 10-pKa. The 1st equivalence point for H3PO4 was 1.920 pH with half-equivalence points of 0.563 pH. The Ka was 0.274 and the pKa was 0.563. While The 2nd equivalence point for H3PO4 8.811 pH with half-equivalence point of 6.347 pH. The pKa was 6.347 and the Ka was 4.5 x 10-7. The data makes sense as the Ka decreased after the release of the first ion to the second ion and the pH rose with the increased amount of NaOH. The equivalence point for H2PO4 was 8.655 pH with half-equivalence point at 6.416 pH. The pKa was 6.416 and the Ka was 3.84 x 10-7. The data...
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...transferring this heat via conduction. When the lights were switched off they were still collisions therefore there was still heat remaining. This is why the bottle was still warm because the particles still had kinetic energy. How does the composition of the gases in the bottles differ from the composition of gases naturally found in the atmosphere? The gases in the bottles differed from the gases found in the atmosphere in 2 ways:-The gases in the bottle were not produced naturally because of the use of Alker seltzer tablets and they gases in the bottles were trapped and filled into the bottles they were not found naturally. If you increased the concentration of CO2 in the bottle, how might this affect the temperature trend in the lab If the concentration of CO2 increased more heat would be trapped and therefore they would take longer to gain...
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...The basic aqueous layer separated previously was then transferred into a 250 mL beaker, and a 50 mL of HCl was collected in a 100 mL beaker. Using a pipette, HCl was slowly added to the basic aqueous substance to acidify it. The lab TA stated that the solution would turn cloudy once acidified, but this did not happen, despite more than 50 drops being added. When the pH was tested, it showed that the pH was definitely 1, despite the solution not being cloudy. The volume of the NaOH and HCl mixture was 90 mL. The acidic solution was then poured back into the separatory funnel and washed with 25 mL portions of DCM as per the previous procedure. The aqueous top layer from these washes was discarded after use. Due to time constraints, this portion of the experiment could not continue to...
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...Hydrogen Carbonate. Unknown compound #1 is Benzoic Acid and the percentage error is -1.03%. Unknown compound #2 is Sodium Carbonate and the percentage error is -6.48%. Unknown compound #3 is Calcium Oxide and the percentage error is -1.55%. Unknown compound #4 Sodium Sulfate and the percentage error is -1.23%. Unknown compound #5 is Ammonium Sulfate and the percentage error is -0.48%. Unknown compound #6 Potassium Sulfate and the percentage error is -0.56%. Unknown compound #7 Sodium Hydrogen Carbonate is and the percentage error is -0.29%. All of the unknown compounds percentage error to the known compounds range from -0.29% to -6.48%. Molar masses were commonly found in both the unknown and known compounds. Molar mass is related to this lab, because we calculated molar mass in both the unknown and known compounds to match each unknown to its known. The molar masses were not exactly the same, but they were close together, except for unknown compound #2, which was sodium carbonate. It had a difference of 6.87 from the actual and the expert’s number. But in conclusion, molar masses were used to match the unknown to its known. ...
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...The initial mass of aluminum was measured to be 0.22 grams. An estimated mass of alum of 3.866 grams was calculated(equation 3). After the reactions had taken place and the alum was created and filtrated, the total mass of the alum was 2.420 grams. This experimental lead to a percentage yield of 62.58%(equation 4). This low percentage yield means that we produced less alum than what was expected. This percentage yield is around what would be expected due to errors in the procedure. The reason that the experimental mass of alum was less than the theoretical mass was because of various errors in the lab. One possible error for this lab was the purity of the aluminum can. The cans had to have the inside and the outside of them scraped with steel...
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