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Gravimetric determination of Calcium as CaC2O4·H2O: Determine the Percent Ca in Unknown for all four trials of the lab experiment. Mass of CaCO3 unknown in stock solution = 1.239 g
Below is the equilibrium equation from the said experiment (use if needed):
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- Determine the percentage Fe in a sample of limonite from the following data:Sample : 0.5000g ; KMnO4 used = 50.00 ml ; 1.000ml of KMnO4 is equivalent 0.005317 g Fe,FeSO4 used = 6.00 ml; 1.000ml FeSO4 is equivalent 0.009200 g FeO ( ans 44.59 %)A student was tasked to perform gravimetric analysis of a soluble sulfate. His unknown sample weighed 0.7543 g. The sample underwentprecipitation using BaCl2 and was digested for overnight. The precipitate was then filtered off to obtain white crystalline precipitate that was collected inan ash less filter paper. In performing constant weighing, he obtained a crucible mass that is 29.9442 g. After burning his samples inside the crucible,the obtained mass was 30.3375 g. Compute for the theoretical % SO3 obtained by the student and the theoretical mass (g) of SO3 that should be obtained by the student using his weighed sampleA student was tasked to perform gravimetric analysis of a soluble sulfate. His unknown sample weighed 0.7543 g. The sample underwentprecipitation using BaCl2 and was digested for overnight. The precipitate was then filtered off to obtain white crystalline precipitate that was collected inan ash less filter paper. In performing constant weighing, he obtained a crucible mass that is 29.9442 g. After burning his samples inside the crucible,the obtained mass was 30.3375 g.29. Compute for the mass (g) of BaSO4 from the experiment.A) 0.3933B) 0.3393C) 0.3133D) 0.3951E) 0.359130. Compute for the experimental mass (g) of SO3 in grams obtained by the student.A) 0.1439B) 0.1349C) 0.1943D) 0.1394E) 0.359131. Compute for the experimental % SO3 obtained by the student.A) 73.21B) 56.33C) 17.89D) 56.89E) 72.8032. Compute for the theoretical % SO3 obtained by the studentA) 0.3933B) 56.37C) 17.33D) 17.89E) 0.425233. Compute for the theoretical mass (g) of SO3 that should be obtained by the student…
- Stock iron(II) solution (200Ug mL-1 Fe) ferrous ammonium sulfate hexahydrate mass= 0.1437g, transfer it to a 100 ml beaker. add 15 ml approx of water and 15m1 'approx of dilute sulphuric acid (2M H2SO.). then transfer FeII to 100 ml flask makeup to the mark with water. calculate the moles of ferrous ammonium sulfate hexahydrate solution in unit ug/mL.50.00 cm3 of a 1.5784 mol.dm-3 solution of potassium hydroxide is transferred to an empty 700.00 cm3 volumetric flask. This flask is made up to the mark with distilled water and then shaken well. The concentration of the potassium hydroxide in this second flask is:Using the percent purity calculations, determine the percent yield of synthesis of aspirin. Part I Synthesis of Aspirin Mass of salicylic acid used (g) 2.029g Volume of acetic anhydride used (mL) 5ml Mass of acetic anhydride used (vol. × 1.08 g/mL) 5.4g Mass of aspirin synthesized (g) 3.256g Part II Melting Temperature Data Melting temperature (°C) 133°C Part III Salicylic Acid Standard Stock Solution Initial mass of salicylic acid (g) 0.210g Moles of salicylic acid (mol) 0.0147 mol Initial molarity of salicylic acid (M) 0.724 M Part III Beer’s Law Data for Salicylic Acid Standard Solutions Trial Concentration (M) Absorbance Water (mL) 1 10 0.301 0 2 7.5 0.219 2.5 3 5.0 0.163 5.0 4 2.5 0.074 7.5 Best-fit line equation for the salicylic acid standards Test of the Purity of the Synthesized Aspirin Initial mass of aliquot of product (g)…
- 5.00 mL of stock solution is diluted to 25.00 mL, producing solution ALPHA. 10.00 mL of solution ALPHA is diluted to 25.00 mL, resulting in solution BETA. 10.00 mL of solution BETA is then diluted to 25.00 mL, producing solution GAMMA. dilution factor for ALPHA from stock solution = 0.167 dilution factor for BETA from ALPHA solution = 0.0476 part c and d?using exactly 5.00 mL of 0.0400 M stock CuSO4 solution. Add 100 mL of water. Data for Part IMass of empty dish: 32.470 g empty dishVolume of 0.0400 M solution: 5.00 mL CuSO4 solution.Mass of dish and 0.0400 M solution: 37.497 g dish and solutionMass of dish and CuSO4 solid: 32.503 g dish and CuSO4 solidCalculations for Part I1. Calculate the mass of solution2. Calculate the mass of solid CuSO4 dissolved in the solution.3. Calculate the number of moles of solid CuSO4 dissolved in the solution.4. Calculate the mass of water evaporated from the solution.5. Calculate the density of solution, (g solution/mL solution).6. Calculate the % by mass, CuSO4 in solution (100 x g CuSO4/g solution).7. Calculate the molality of solution (moles CuSO4/kg solvent).8. Calculate the molarity of solution (moles CuSO4/L solution).9. Given that the true molarity is 0.0400 M, calculate the percent error of your result.