Calculate the Normality of the vinegar using the previously given equation. Na = (Nb)(Volumeb) (Volumea) C. Calculate the mass of the acetic acid in grams using the previously given equation. Massa = (Na)(GMWa) D. Calculate the percentage of acetic acid using the previously given equation. % Acid = Massa(g/L) x 100 1000g/L Discussion and Conclusion: Questions: LabPaq question guidelines: Answer questions A and G in the lab manual. Skip questions B, C, D, E and F in the lab manual, and answer these instead: A.
25 ml of diluted unknown acid solution to 100ml beaker by using 25 volumetric pipet. 10ml of deionized water and 3 drops of phenlpthalin indicator the beaker labeled as 3. Potentiometric titration acid solutions 125 ml of NaOH was obtaining in a beaker and 50 ml of NaOH transfer to buret the tip and the meniscus is at below 0 ml. one magnetic stirring bar placed in a beaker contain one of the known solution on a stir. The pH recorded by using pH electrode before adding NaOH.
Calculate the mass of the liquid for each trial. (Subtract the mass of the empty graduated cylinder from the mass of the graduated cylinder with liquid.) * Trial 1 35.46-25.3=10.16 * Trial 2 36.01-25.39=10.62 * Trial 3: 36.41-26.03=10.38 2. Calculate the density of the unknown liquid for each trial. (Divide the mass of the liquid calculated above by the volume of the liquid.)
First, we had to calculate how many grams of copper (II) sulfate we needed to form 100 mL of a 0.200 M solution of copper (II) sulfate. We determined that we needed to use 4.994g of copper (II) sulfate to make the solution. We added distilled water to the 4.994g of copper (II) sulfate in a beaker until it reached 100 mL. Then we put the beaker on a hot plate and added a magnetic stirrer. We determined that the mass of zinc necessary to completely react with the copper (II) ions in the solution was 1.308g.
The theoretical yield of aspirin is 5.77 grams. The percent yield of aspirin is 66.3%, which was calculated by dividing the mass of aspirin (actual yield) by the theoretical yield of aspirin times 100%. A side reaction happened when acetic anhydride and water produced acetic acid. This means that when using water to rinse out the Erlenmeyer flask,
Approximately 3mL of Clear Liquid 2 was carefully poured into the graduated cylinder. The graduated cylinder was placed on the scale again (with the added liquid) and the mass was measured in grams to the nearest .01 gram. The graduated cylinder was then placed on the tabletop and the volume was read to the nearest tenth of a
CHEM 10050 Problem Set #3 Name_________________ 1. Calculate the molarity of normal saline (0.90% w/w NaCl) given the density of this solution to be 1.05 g/ml. Assume 100 g solution. Then: [pic] [pic] [pic] [pic] 2. Calculate the molarity of an aqueous solution resulting when 275 ml of 3.15 M glucose is diluted with 150 ml of water.
Comparing the rate of appearance of C and the rate of disappearance of A, we get[pic]. A) [pic] B) [pic] C) [pic] D) [pic] E) [pic] Answer: A Diff: 1 Page Ref: Sec. 14.2 A flask is charged with 0.124 mol of A and allowed to react to form B according to the reaction A(g) →B(g). The following data are obtained for [A] as the reaction proceeds: [pic] 5) The average rate of disappearance of A between 10 s and 20 s is __________ mol/s. A) [pic] B) [pic] C) [pic] D) 454 E) [pic] Answer: A Diff: 1 Page Ref: Sec.
Three titrations were performed to find the caplets mass. In these titrations we mixed a crushed up ASA caplet with 25mL of isopropyl alcohol. After adding the solution into a 250mLerlenmeyer flask, we added our second solution of NaOH until our solution turned to a light pink color, repeating the process three times. To find the mass of the caplet, we used stoichiometric calculations, based on our volume of solution used to dilute the caplet.
Data See Graphs Calculations The calculations consisted of the deviations to create the pH curve graph. The Ka was calculated by taking the –log of the amount of acetic acid added and squaring it then dividing it by the amount of solution added. The moles for the NaOH was calculated by taking .1 moles and dividing it by 1,000 mL and then multiplying it by the amount of NaOH added. For the Molarity, the mole-mole ratio is used to convert it. Conclusion The pH curve came out zig zagged.