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- Gypsum (CaSO4) is a common percipitant in water desalination. CaSO4 <=> Ca2+ + SO42- where Ksp=10-4.6. Assuming that: [Ca2+]=2x10-2 M; [SO42-]=2x10-3M: a) Find Qsp or IAP for the given water b) Find the solubility index (SI) and determine whether CaSO4 is under, super, or at saturation in water. Feel free to make any assumptions you wish, as long as they're stated.Objective: To determine the percent NaCl (w/v) of a water sample 5.00 mL sample diluted to 50 mL solution 0.1000 M AgNO3 Volume of AgNO3 trial 1=15.30 mL trial 2=14.90 mL trial 3=15.10 mlA mixture of pure BaCO3 and pure Na2CO3 weighs 1.000 g and has the total neutralizing power of 15.37 meq of CaCO3. Calculate the percentage of combined CO2 in the mixture and the weight of Li2CO3 that has the same neutralizing power as 1.000 g of the above mixture. Note: include up to 4 decimal places Please explain each step
- 19 g of unknown organic sample was dissolve in 640 mL of Dicloromethane (DCM). The boiling point of benzene was increased by 3.78oC. Determine the molecular weight of the unknown sample? Kb of DCM = 2.42oC/m Bb of benzene = 39.6 oC density of benzene = 1.33 g/mL at 25 °C Round your answer to the nearest whole number, no units required.How to prepare the solutions listed below Solution 1) 25 mL 7 M sulfuric acid and the respective dilution of this solution to prepare 50 mL 2 M sulfuric acid in deionised water. Sulfuric acid Mw = 98.079 g mol-1 Sulfuric acid density = 1.84 g mL-1 Sulfuric acid purity = 98% Solution 2) 100 mL 0.02M potassium permanganate potassium permanganate Mw = 158.03 g mol-1 potassium permanganate purity = 99%2. Show by calculation how you would prepare the following solutions.2.1 200ml stock solution of 1M NaOH (mwt: 40g/mol)2.2 Perform a 1:10 dilution of the stock solution prepared in 2.1 above to a final volume of 50ml.2.3 Perform a 1:50 dilution of the stock solution prepared in 2.1 above to a final volume of 100ml.2.4 Calculate the final concentration of NaOH in 2.2 and 2.3 above.
- Detailed calculations on how to prepare the solutions listed below Solution: 50 mL 1 M oxalic acid Oxalic acid Mw = 90.03 g mol-1 Oxalic acid purity = 98% Solution: 10 mL 3 M sulfuric acid Sulfuric acid Mw = 98.079 g mol-1 Sulfuric acid density = 1.84 g mL-1 Sulfuric acid purity = 98% Solution: 10 mL saturated potassium oxalate potassium oxalate (monohydrate used) Mw = 184.23 g mol-1 Solubility of potassium oxalate in water at 25°C = 360 mg mL-1 Solution :20 mL 3% hydrogen peroxide Hydrogen peroxide is commercially available as a 32% solution.Gypsum (CaSO4) is one of common precipitants in brackish water desalination processes. CaSO4 ⇋ Ca2+ + SO42- Ksp = 10-4.6Assume a water containing: [Ca2+] = 10-2.1 M; [SO42-] = 10-3.2 M.(a) Find Qsp or IAP for the given water. (b) Find the solubility index (SI) and determine whether CaSO4 is under, super, or at saturation in the water.A mixture of pure BaCO3 and pure Na2CO3 weighs 1.000 g and has the total neutralizing power of 15.37 meq of CaCO3. Calculate the percentage of combined CO2 in the mixture and the weight of Li2CO3 that has the same neutralizing power as 1.000 g of the above mixture. Instructions: Include up to 4 decimal places See sample photo for the format of solution
- D. Preparation of 3% (vol/vol) Alcohol Solution1. Compute for the volume of denatured ethanol needed to prepare 25 mL of 3% (vol/vol) solution. Note that the concentration of the denatured alcohol stock solution is 98%.2. Explain in detail how the solution will be prepared in the laboratory.1. A 50.00 mL stock CaCl2 solution was diluted with water in a 250.0-mL volumetric flask. A 25.00 mL aliquot of this was further diluted to a final volume of 100.0 mL. Finally, 20.00-mL aliquot of the resulting solution was analyzed and found to contain 0.1500 M CaCh2 Solve for the concentration of the stock solution in M. 2. A 50.00 mL stock CaCl2 solution was diluted with water in a 250.0-mL volumetric flask. A 25.00 mL aliquot of this was further diluted to a final volume of 100.0 mL. Finally, 20.00-mL aliquot of the resulting solution was analyzed and found to contain 0.1200 g CaCl2 Solve for %w/v CaCl2 in the stock solution.Why do you multipy the # of H2S mols by 4, when 4 corresponds to the reactory amount of O2?The logic there is somewhat confusing to me.I understand how/why you found out the mols of the equation but dividing the sample's MM by the equation confuses me.