Weighed quantity of dry potassium bromide 1,5560 gr (M=167,000 g/mol) dissolved in water with receiving of 250 ml of the solution. Calculate titre of this solution corresponding to arsenous acid anhydride (M=197,841 g/mol). Express numerical result with an accuracy of: X,XXX 10* 2.
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- If a 3g sample of zinc oxide, 95% ZnO, were treated with 50ml of 1.1230N sulfuric acid in the usual way, what volume of 0.9765N sodium Hydroxide would be required in the residual titration?What is the pH of the solution after mixing 0.188 g of Mg(OH)2 (MW=58.321 g/mol) with 18.1 mL of 0.0173 M HCl? The resulting solution was diluted to 100 mL. Round your calculated value for pH to two figures to the right of the decimal point.During a titration, the initial reading of a burette was recorded at 11.5 ml and the final reading is recorded at 34.2 ml. How much titrant was used in the analysis?
- The protein content of wheat flour can be determined reasonably accurately by multiplying the percentage of nitrogen present by 5.7. A 2.06-g sample of flour was taken through a Kjeldahl procedure and the ammonia produced was distilled into a boric acid solution. If this solution required 34.70 mL of 0.174 N HCl for titration to the methyl red end point, what is the a) % Nitrogen and b) % protein in the flour? (Use 1:1 stoich ratio between N and HCl)Use the following atomic masses (in g/mol): Mg = 24.31; H = 1; S = 32.06; O = 16; Na = 23; Cl = 35. 45; Ca = 40.08; C = 12.01; N = 14.01; Mg = 24.31 5.A sample of Chlorpheniramine Maleate (99.82%) weighing 0.502g was assayed by non-aqueous titrimetry and was found to be equivalent to 22.2mL of perchloric acid. Calculate for the normality of perchloric acid. Each mL of 0.1N perchloric acid is equivalent to 19.54mg of Chlorpheniramine Maleate.Express the concentration of acetic acid in both samples as % by mass of acetic per 100mL of solution given : %= mass/ volumex100% Given: NaOH vs CH3COOH Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 20.1 mL 20.1 mL 20.3 mL Titration 1 0.0 mL 20.9 mL 20.9 mL Titration 2 0.0 mL 20.0 mL 20.0 mL To find the average volume Average volume = 20.1 mL + 20.9 mL + 20.0 mL320.1 mL + 20.9 mL + 20.0 mL3 = 20.3 mL Concentration of Vinegar = Volume of NaOH * Concentration of NaOHVolume of vinegarVolume of NaOH * Concentration of NaOHVolume of vinegar = 20.3 mL * 0.0647 M5.0 mL20.3 mL * 0.0647 M5.0 mL = 0.2627 M Concentration of NaOH = Volume of H2SO4 * Concentration of H2SO4Volume of NaOHVolume of H2SO4 * Concentration of…
- mass of KIO3 = 0.9290g titrated volume Na2S2O3 = 28.88mL standardisation sodium thiosulfateMr. Clean recently bought a laboratory-grade sodium carbonate from a chemical company known as Brand X. He was supposed to use it in the production of detergents. Unfortunately, he was scammed by the company. He suspected that he purchased a crude sodium carbonate so he tasked the Quality Assurance Department to determine the components of the purchased chemical. The chemist assigned to analyze the sample used double indicator method. For the standardization of HCl titrant, 0.1025 g Na2CO3 of 99.5% purity (FW: 106.00) required 8.20 mL of the titrant to reach the phenolphthalein endpoint. FW: NaOH (40.00), NaHCO3 (84.01), Na2CO3 (106.00) a. What is the molarity of the titrant? The chemist obtained a 3.150 g sample and dissolved it in distilled water to produce a 50.0 mL solution. An aliquot of 10.00 mL was obtained and diluted in a 100.0 mL volumetric flask. A 50.00-mL aliquot of the diluted sample was taken and it required 25.70 mL of titrant for the methyl orange endpoint, while…A 0.4126-g sample of primary-standard Na2CO3 was treated with 40.00 mL of dilute perchloric acid. The solution was boiled to remove CO2, following which the excess HClO4 was back - titrated with 9.20 mL of dilute NaOH. In a separate experiment, it was established that 26.93 mL of the HClO4 neutralized the NaOH in a 25.00-mL portion. Calculate the molarities of the HClO4 and NaOH.
- Twenty sodium salicylate tablets labeled 325 mg were dispersed in sufficient water to make 200.0 mL. A 15.0–mL aliquot of the filtrate was titrated to a bromophenol blue endpoint in the usual way by 32.11 mL of 0.1000 N hydrochloric acid. Calculate the amount of sodium salicylate in each tablet, and from that, the percentage of the labeled amount. (NaC7H5O3 MW = 160.11 g/mol) Amount of sodium salicylate in each tablet = % labelled amount =Bay Water Titration The concentration of Cl- in ocean water is about 500-600 mM. The baywater is diluted by a factor of 12.5 using a 20.00 mL volumetric pipet and a 250 mL volumetric flask. Using a 15.00 mL volumetric pipet, 15.00 mL is transferred into three clean Erlenmeyer flasks. 10 mL of 1% dextrin solution, 20 mL of de-ionized water, and 3-4 drops of indicator are added and titrated each with the AgNO3 solution. From the procedure, find: Dilution factor Volume of diluted bay water Then calculate: The [Cl-] of diluted bay water The [Cl-] of bay water Consider this information: [AgNO3] = 0.04043177 trial # Veq 1 13.91 2 13.73 3 13.9 4 13.86 5 13.87 6 13.84 average 13.85167 [Cl-]Bay = ([AgNO3](Veq(ave))/Vdil bay water pipeted) (Vflask/Vbay water)Silver nitrate can be standardized using primary standard KCl. A dried sample of analytical grade KCl of mass 0.918 g was dissolved and diluted to 250.0 mL. Repeat 10.00-mL aliquots of the potassium chloride solution were titrated with the silver nitrated solution. The mean corrected titration volume was 8.98 mL Calculate the molarity (M) of the silver nitrate solution.