Calculate the iodine value of talisay seed oil using 0.25 g of the oil as a sample. The oil consumed 7.2 ml of 0.1095 N sodium thiosulfate while the blank titration at the sample condition used 24.8 ml of the sample thiosulfate solution. Calculate the iodine value of the sample.
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Calculate the iodine value of talisay seed oil using 0.25 g of the oil as a sample. The oil consumed 7.2 ml of 0.1095 N sodium thiosulfate while the blank titration at the sample condition used 24.8 ml of the sample thiosulfate solution. Calculate the iodine value of the sample.
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- 1. A 1.2-gram sample of lanolin was treated with Wij’s solution and excess potassium iodide solution. The liberated iodine reacted with 30 ml of 0.1 N sodium thiosulfate solution. If the iodine value was determined as 12.69, what is the volume used in blank titration? 2. A fat sample with combination of acids contain standard hydrochloric acid for blank and sample with 8mL and 5mL respectively. The normality of the standard hydrochloric acid is 0.93N and the weight of the sample is 3 grams. Calculate the saponification value. 3. A 3.50-gram sample of Streptomycin powder was tested for its water content. If the water equivalence factor of the KF reagent was 4.6, what is the percentage water content of the sample if 9.2 ml of the KF reagent was used? 4. A 500mg oil sample is taken from a conical flask and is dissolved in 50mL distilled alcohol. An indicator is added and is then titrated against 0.112N KOH until a slight pink color appears. It took 17.6mL of the titrant to reach the…0.9563 g KHCO3 (M.r.(KHCO3) = 100.12 g/mol (mg/mmol)) have measured, and 50.0 cm3 stock solution have been prepared from it. 10.00 cm3 individual samples from it was titrated with HCl titrant in order to determine its concentration. The average HCl consumption was 19.18 cm3. The same HCl solution with were used to determine the concentration of NaOH titrant. The base consumption for 20.00 cm3 HCl was 21.35 cm3 NaOH. 1. concentration of HCl 2. concentration of NaOH100 mL water sample was subjected to DO analysis using the Winkler method. The initial reading of the titrant (0.025 M Na2SO3) in the buret was 2.51 mL. The endpoint was obtained and the final reading was 7.07 mL. A blank was also titrated and 0.75 mL of the titrant was consumed. Compute for the amount of DO (mg/L) in the sample.
- The total cation content of natural water is often determined by exchanging the cations for hydrogen ions on a strong acid ion-exchange resin. A 25.00 mL sample of a natural water was diluted to 100.00 ML with distilled water, and 2.06 g of a cation - exchange resin was added. After stirring, the mixture was filtered and the solid remaining on the filter paper was washed with three 15.00 mL portions of water. The filtrate and washings required 16.30 mL of 0.0282 M NaOH to give a bromocresol green end point. a) Calculate the number of millimoles of cation present in exactly 1.00 L of sample. b ) Report the results in terms of milligrams of CaCO3 per liter. Only typed solution.Approximately 6.0mL of concentrated perchloric acid (70%) was transferred to a bottle and diluted with about 1.0L of water. A sample containing 251.5 mg of primary standard Na2B4O7.10H2O required 27.41 mL of the HClO4 solution to reach the methyl red end point. Calculate the molar concentration of the HClO4 (Fwt = 382g/mole)A double indicator analysis of a soda ash sample was performed. The student weighed 0.4375 g of the soda ash sample and dissolved it to make 100.0 mL solution. The students then took a 25 mL aliquot of the sample solution, then added 25 mL H2O and performed double indicator titration. The phenolphthalein endpoint required 10.27 mL 0.04027 M HCl titrant. The sample was further titrated using Methyl Orange and took a total of 15.97 mL of the titrant to reach the endpoint. Given that the molar weight of Na2CO3 - 105.99 g/mol, NaHCO3 = 84.007 g/mol, and NaOH - 39.997 g/mol answer the following: a. What are the components of the soda ash sample? (The constituents of the sample came from Na2CO3, NaHCO3, and NaOH). b. What are the moles of each component in the titrated 25 mL aliquot? c. What is the percent composition in (% w/w) of each component in the soda ash sample?
- Please explain and show calculations on how to do it. A student was given a white powdered solid unknown sample mixture made-up of sodium carbonate and sodium hydroxide. He took 0.3721 g of dissolved in 250.0 mL distilled water. In each trial he took 50.00 mL of unknown sample solution and conducted double titration as shown in the videos against 0.1000 M HCl. He did 5 trials and his results are given below: Trial V1 (mL) V2 (mL) 1 10.00 2.00 2 3 4 5 Calculate % NaOH and % Na2CO3 for trial#1.A 1.250 g sample of cheese was subjected to a Kjeldahl analysis to determine the amount of protein. The sample was digested, the nitrogen is oxidized to NH₄⁺, and was then converted to NH₃ with NaOH, and distilled into a collection flask containing 50.00 mL of 0.1050 M HCl. The excess HCl is back titrated with 0.1175 M NaOH, this required 21.65 mL to reach the bromothymol blue end point. A. How many moles of N is present in the cheese sample? B. Report the %N of the cheese sample. C. Report the %protein of the cheese sample, assuming that there are 6.4 grams of protein for every gram of nitrogenA 1.250 g sample of cheese was subjected to a Kjeldahl analysis to determine the amount of protein. The sample was digested, the nitrogen is oxidized to NH₄⁺, and was then converted to NH₃ with NaOH, and distilled into a collection flask containing 50.00 mL of 0.1050 M HCl. The excess HCl is back titrated with 0.1175 M NaOH, this required 21.65 mL to reach the bromothymol blue end point. What is the reaction for the digestion of the limestone sample A. 2 NaOH + CaCO₃ ⇌ Na₂CO₃ + Ca(OH)₂ B. 2 HCl + CaCO₃ ⇌ CaCl₂ + H₂O + CO₂ C. CaCO₃ ⇌ CaO + CO₂ D. NaOH + HCl ⇌ NaCl + H₂O
- A 1.250 g sample of cheese was subjected to a Kjeldahl analysis to determine the amount of protein. The sample was digested, the nitrogen is oxidized to NH₄⁺, and was then converted to NH₃ with NaOH, and distilled into a collection flask containing 50.00 mL of 0.1050 M HCl. The excess HCl is back titrated with 0.1175 M NaOH, this required 21.65 mL to reach the bromothymol blue end point. What is the reaction for the back titration for this analysis? a. NH₄⁺ + NaOH ⇌ NH₃ + H₂O + Na⁺ b. NaOH + HCl ⇌ NaCl + H₂O c. NH₃ + HCl ⇌ NH₄⁺ + Cl⁻ d. N₂ + 3 H₂ ⇌ 2 NH₃ How many moles of N is present in the cheese sample? a. 544 x 10⁻³ b. 706 x 10⁻³ c. 250 x 10⁻³ d. 794 x 10⁻³ Report the %N of the cheese sample. a. 851% b. 884% c. 033% d. 736%A serum sample of 3mL is analyzed for nitrogen by modified Kjeldahl. The sample is digested, distilled, and 14mL of standard of HCl is needed to titrate the ammonium borate. The HCl was standardized using 0.30g (NH4)2SO4. 30mL acid was required in titrating for standardization. What is the concentration of nitrogen in the serum (weight/volume)?A piece of Gold weighing 12,359 Kg is suspected of being contaminated with Iron. To perform an instrumental analysis and To confirm whether or not it contains Fe, a portion of the sample (0.954 g) is taken from the piece and dissolved with 25 mL of aqua regia. Heats up For its complete dissolution, it is cooled and made up to 100 mL. A 10 mL aliquot is taken from this solution and made up to 50 mL. From This last solution is given the appropriate treatment to visualize Fe+2, for which the 1,10-phenanthroline reagent is added. (it forms a complex that is red in color) and is taken to a visible spectrophotometer and with a 12 mm cell a absorbance of 0.45. Previously, a calibration curve of Fe+2 was obtained under the same instrumental conditions obtaining the following data: (view table) Calculate the purity of the gold piece, assuming impurities only due to Fe.