You dissolve 150.0 grams of a mixture of MgCl, and KCI in water, add a solution of excess AGNO3, and precipitate all of the chloride ion as AgCl. After filtration and drying, you find that you have formed 335.0 grams of AgCl. What is the mass in grams of KCI that was in the original 150.0 grams of mixture?
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- Malic acid (C4H6O5) is a dicarboxylic acid that naturally occurs in the production of bignay wine. Maria was tasked to find out the acid content in a 200.0 mL wine sample by titrating it with a standardized solution of KOH. He was able to establish the working concentration of the KOH solution by using 39.12 mL of it to titrate a primary standard of KHP that weighed 3.021 g. The grams and %w/w of Malic acid present in the 200.0 mL wine sample were then determined by titrating a 20.00 mL aliquot using 48.95 mL of the standardized KOH solution. C4H6O5 (aq) + 2KOH (aq) ↔ K2C4H4O5 (aq) + 2H2O (l) Assume the density of the wine sample is 1.000 g/mL MM of KHP = 204.22 g/mol MM of C4H6O5 = 134.0874 g/mol ? What is the molar concentration of the prepared KOH solution?…Malic acid (C4H6O5) is a dicarboxylic acid that naturally occurs in the production of bignay wine. Maria was tasked to find out the acid content in a 200.0 mL wine sample by titrating it with a standardized solution of KOH. He was able to establish the working concentration of the KOH solution by using 39.12 mL of it to titrate a primary standard of KHP that weighed 3.021 g. The grams and %w/w of Malic acid present in the 200.0 mL wine sample were then determined by titrating a 20.00 mL aliquot using 48.95 mL of the standardized KOH solution. C4H6O5 (aq) + 2KOH (aq) ↔ K2C4H4O5 (aq) + 2H2O (l) Assume the density of the wine sample is 1.000 g/mL MM of KHP = 204.22 g/mol MM of C4H6O5 = 134.0874 g/mol What is the molar concentration of the prepared KOH solution?…25.0mL of a 0.515 M K2S solution is mixed with 30.0 mL of 0.833 M HNO3 acid solution to give the following reaction: K2S(aq) + 2HNO3(aq) → 2KNO3(aq) + H2S(g) H2S is an unwanted by-product in a pulp and paper industry. To capture H2S gas, it is bubbled through a NaOH solution to produce Na2S with a yield of 94%. H2S(g) + 2NaOH (aq) → Na2S (aq) + 2H2O(l) The mass of H2S(g) that was processed (in kg) if 10.76 kg of Na2S was collected is
- A stock solution of potassium permanganate (KMn04) was prepared by dissolving 13.0g KMn04 with DI H20 in a 100.00-ml volumetric flask and diluting to the calibration mark. Determine the molarity of the solutionBy pipet, 5.00 mLmL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask and diluted to the calibration mark. Determine the molarity of the resulting solution.In the standardization of HCl using pure anhydrous sodium carbonate as the primarystandard for methyl orange as an indicator, 1.0 mL HCl was found to be equivalent to 0.05gof sodium carbonate (MW =106). The normality of HCl is:
- A chemist attempts to prepare some very pure crystals of borax by dissolving100g of Na2B4O7 in 200 grams of boiling water. He then cools the solution slowly until some borax crystallizes out. Calculate the grams of borax recovered in the crystals per 100 grams of initial solution (Na2B4O7 plus water), if the residual solution at 55oC after the crystals are removed contains 12.4% Na2B4O7.What is the percent yield or percent recovery of Na2B4O7. Show process diagram and solution.A chemist attempts to prepare some very pure crystals of borax by dissolving100g of Na2B4O7 in 200 grams of boiling water. He then cools the solution slowly until some borax crystallizes out. Calculate the grams of borax recovered in the crystals per 100 grams of initial solution (Na2B4O7 plus water), if the residual solution at 55oC after the crystals are removed contains 12.4% Na2B4O7.What is the percent yield or percent recovery of Na2B4O7.Tartaric acid, H2C4H4O6, is a diprotic acid that naturally occurs in the production of wine. Jose was tasked to find out the acid content in a 100.0 mL wine sample by titrating it with a standardized solution of NaOH. He was able to establish the working concentration of the NaOH solution by using 35.21 mL of it to titrate a primary standard of KHP that weighed 3.001 g. The grams and %w/w of Tartaric acid present in the 100.0 mL wine sample were then determined by titrating a 25.00 mL aliquot using 43.56 mL of the standardized NaOH solution. H2C4H4O6 (aq) + 2NaOH (aq) ↔ Na2C4H4O6 (aq) + 2H2O (l) Assume that the density of the wine sample is 1.000 g/mL MM of KHP = 204.22 g/mol MM of H2C4H4O6 = 150.087 g/mol What is the molar concentration of the prepared NaOH solution?…
- The concentration of arsenic in an insecticide can be determined gravimetrically through its precipitation as MgNH4AsO4. After the formation of the precipitate, it must be ignited for total conversion to Mg2As2O7, which is then cooled and weighed. Considering that a sample of 1,627 g of the insecticide produced 106.5 mg of Mg2As2O7, determine the% (m/m) As2O3 in the insecticide. Given the molar masses: Mg2As2O7 = 310.447 g / mol and As2O3 = 197.841 g / molA 19.51 ݃ sample of impure methylamine, which contains 72.58% (by mass) of CH3NH2 , isreacted with 30.81 ݃ of pure oxygen gas:4CHଷNHଶ(g) + 9Oଶ(g) ⟶ 4COଶ(g) + 10HଶO(ℓ) + 2Nଶ(g) In another experiment, this impure methylamine was used as follows: An unknown mass of the impure compound is dissolved in enough water to make 500.0 ݉ܮof solution. 20 ݉ܮ of this solution was transferred by pipette to a clean 250 mL volumetric flask andmade up to the mark. The molarity of the CH3NH2 in the final solution was determined to be 0.103 M.Determine the mass of CH3NH2 present in the original amount of impure compound used tomake this solution.A quantitative analysis for ethanol, C2H6O, can be accomplished by a redox back titration.Ethanol is oxidized to acetic acid, C2H4O2, using excess dichromate, Cr2O72–, which is reduced toCr3+. The excess dichromate is titrated with Fe2+, giving Cr3+ and Fe3+ as products. In a typicalanalysis, a 5.00-mL sample of a brandy is diluted to 500 mL in a volumetric flask. A 10.00-mLsample is taken and the ethanol is removed by distillation and collected in 50.00 mL of anacidified solution of 0.0200 M K2Cr2O7. A back titration of the unreacted Cr2O72–requires 21.48mL of 0.1014 M Fe2+. Calculate the %w/v ethanol in the brandy
