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When the conductivity is at a minimum, what must be true about the amount of Ba(OH)2?
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- A sample is analyzed for chloride by the Volhard method. From the following data, calculate the percentage of chloride present:Weight of sample = 6.0000 g dissolved and diluted to 200 mLAliquot used = 25.00 mL AgNO3 added = 40.00ml of 0.1234MKSCN for back titration = 13.20ml of 0.0930Mpls complete table given the data Determination of Ksp and Molar Solubility1. Add Ca(OH)2 to 250.0 mL distilled water with stirring until equilibrium is achieved. 2. Filter the undissolved precipitate. Measure out 50.0 mL of the supernate into a 250-mL Erlenmeyer flask using a pipette.3. Add a few drops of phenolphthalein indicator and titrate with standardized HCl solution until endpoint is achieved.4. Record the volume the HCl solution used. Perform two more trials.A local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 10.00 mL sample and found that you needed 43.20 mL NaOH to reach the end point. Summary of results: Standardization Sample analysis KHP Weight (g) 0.8053 g Volume of sample 50.00 mL Purity 99.80% NaOH (mL) used 33.20 mL NaOH (mL) used 40.60 mL Determine the following: Molarity of NaOH % (w/v) acetic acid
- Sources of Error Determine the relationship between the observed/apparent value (EX) VERSUS that of the true value (ET) for the quantity being sought by writing either <, >, or = on the space provided TOPIC: Measured mass of the precipitate 1. Filter paper was dried prior to filtration. EX _____ ET TOPIC: Standardization of Titrant 2. Distilled water was not equilibrated to room temperature before the preparation of NaOH titrant. EX ______ ET TOPIC: Determination of Molar Concentration of each component (Double Indicator Titration) 3. No blank correction EX ______ ETAspirin powder = 0.8110g MW of Aspirin = 180g.mol-1 Volume of 0.5N HCl consumed in back titration = 23.50mL Volume of 0.5N HCl consumed in blank titration = 44.50mL Percent purity (USP/NF) = Aspirin tablets contain NLT 90.0% and NMT 110.0% of the labeled amount of aspirin (C9H8O4) What is the calculated weight (in grams) of pure aspirin?..A 50.00 (±0.02) mL portion of an HCl solution required 29.71(±0.02) mL of 0.01963(±0.0032) M Ba(OH)2 to reach an end point with bromocresol green indicator. ? of HCL = 29.71?? ? 0.01963 ???? ??(??)2 ?? ? 2 ???? ??? ???? ??(??)2/ 50.00?? = 0.02333 ? Calculate the uncertainty of the result (absolute error).Calculate the coefficient of variation for the result.
- Knowing that Ksp of PbI2 = 10-⁷ find out if precipitate will form when mixing 100ml of 0.01 N Pb solution (NO3) 2 and 200ml of 0.01 M NaI solutionMake an schematic diagram for procedure below: B. %SO3 determination Transfer all precipitate to the filter paper avoiding any loses during the filtration procedure. One may use a rubber policeman to scrape remaining precipitate in the beaker. Wash the precipitate in the beaker 3x with hot water before transferring to the filter paper. Test the washings with 0.1 M AgNO3 Cloudiness indicates presence of chloride ions; therefore there is a need to wash more the precipitate. Carefully lift the paper out of the funnel, fold it as demonstrated by your instructor and transfer it onto the constant weighed crucible. Dry the crucible cautiously with a small flam, as instructed by your instructor. The flame should be directed at the top of the container, and the lid should be off. Avoid spattering. After drying, char the filter paper by increasing the flame temperature. The crucible should have free access to air. The lid should be kept handy to smother the filter paper in case it catches fire.…I have gotten this wrong 8 times! Please help. See screenshot attached. Thank you! A solution containing a mixture of metal cations was treated as outlined. Dilute HClHCl was added and a precipitate formed. The precipitate was filtered off. H2SH2S was bubbled through the acidic solution. No precipitate formed. The pH was raised to about 9 and H2SH2S was again bubbled through the solution. A precipitate formed and was filtered off. Finally, sodium carbonate was added to the filtered solution and no precipitate formed. What can be said about the presence of each of these groups of cations in the original solution?
- A sodium thiosulfate solution is standardized using pure copper as the primary standard. A sample of copper weighing 0.2624 g is dissolved in acid, excess KI is added, and the liberated iodine was titrated with 42.18 mL of sodium thiosulfate solution. Calculate the molarity of sodium thiosulfate solution. 2Cu2+ + 4I- → 2CuI + I2 I2 + 2S2O32- → 2I- + S4O62-15. A 300.00 mL solution of 0.00165 M A2B5 is added to a 230.00 mL solution of 0.00380 M C2D3. What is pQsp for A2D5?Pre Lab Questions: (Each answer is to be written as a complete sentence) What is the reason for washing the precipitate with water in Step 9? Define precipitate. Define filtrate. In Step 2, what is the purpose for rinsing the stirring rod? read the Procedure to answer the questions Using a balance, mass between 1.50 – 2.00 grams of sodium carbonate in a pre-massed 150mL beaker. Add 20 mL of distilled water and stir thoroughly to make sure all the crystals are dissolved. Rinse the stirring rod with a little distilled water after stirring. Using a balance, mass between 1.50 – 2.00 grams of calcium chloride dihydrate in a pre-massed 50 mL beaker. Repeat Step 2 for the solution in the 50 mL beaker. Pour the calcium chloride solution into the 150mL beaker containing the sodium carbonate solution and stir. Mass a piece of filter paper. Fold the filter paper and place it into the funnel. Wet it with a little distilled water to ensure that it is stuck to the sides of the funnel. Slowly…