A mixture of NaBr, Nal and NaNO3 weighs 0.6500 g. With AgNO3, a precipitate of the two halides is obtained and is found to weigh 0.9390 g. When heated in a current of C12, the precipitate is converted entirely to AgCl weighing 0.6566 g. What is the %NaNO3 in the original sample?
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- Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory as a primary standard. It has the formula KHC8H4O4. This is often written in the short-hand notation as KHP. If 25.0mL of a potassium hydroxide solution are needed to neutralize 2.26g of KHP, what is the molarity of the potassium hydroxide solution? Potassium hydrogen phthalate sometimes called potassium biphthalate, as shown on this bottle is an acid that is convenient to store and use because it is a solid.A mixture of NaBr, Nal and NaNO3 weighs 0.6500 g. With AgNO3, a precipitate of the two halides is obtained and is found to weigh 0.9390 g. When heated in a current of C2, the precipitate is converted entirely to AgCi weighing 0.6566 g. What is the %NaNO3 in the original sample?If 1.000 ml. of a solution of KMn04 is equiva- lent to 0.1000 millimole of NaCHO2 (sodium formate) in the fol- lowing titration: 3CHO2- + 2MnO- + H2O -> 3CO2 + 2MnO2 + 5OH-, what is the value of the KMnO4 in terms of grams of CaO in the volumetric method for calcium in which that element is precipitated as CaC2 4 .H2O and the precipitate is filtered, dis- solved in dilute H2S04 , and the oxalate titrated with permanganate?
- How many grams of sample which contains 18.00% Na2O (61.98) should be taken for analysis in order to obtain a precipitate of Na3AsO3 (191.89) which weighs 0.3000 g?A 4.912-g sample of a petroleum product was burnedin a tube furnace, and the SO2produced was collectedin 3% H2O2.Reaction:SO2(g)+H2O2→H2SO4A 25.00-mL portion of 0.00873 M NaOH was introducedinto the solution of H2SO4, following whichthe excess base was back-titrated with 15.17 mL of0.01102 M HCl. Calculate the sulfur concentrationin the sample in parts per million.E. Analysis of a mixture consisting of NaOH + NażCO3 + inert matter gives the following data: Sample portions are titrated. With one portion, an end point with phenolphthalein is obtained in cold solution, with 44.52 mL of 0.5000 N HCI. The other portion requires 46.53 mL of the acid for an end point with methyl orange. Calculate the percentage composition of the original sample. 10.00 g. Its aqueous solution is diluted to 250.0 mL and two separate 25.00 mL
- 1. 0.646 g of sample containing BaCl2.2H2O (244.26 g/mol) was dissolved and enoughpotassium chromate was added. After filtering, the precipitate was dissolved in acid andenough KI was added and titrated with thiosulfate. Since 48.7 mL of 0.137 M thiosulfateis used for this, what is the percentage of BaCl2.2H2O in the sample?K2Cr2O7 + 7H2SO4 + 6KI 4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2Adic acid (H2C6H8O4) is an important carboxylic acid compound for the industry in which nylon is manufactured. Adipic acid to be titrated with NaOH according to the following reaction: H2C6H8O4 + 2Na0 -----> 2H2O + Na2C6H8O4 a) Suppose you titrate 0.283g of pure adipic acid, what volume (ml) of 0.1000M NaOH would be needed to reach the equivalence point? b) suppose you do the same titration with 0.304g industrial adipiqje acid (74% w/w). What would be the volume of NaOH required? Please show steps so I will have a better understanding thankssimple steps Find the % of sodium carbonate in a sample if 5.002 g of the sample required 29.10 cm³ of 0.7344 mol dm-³ HCl for complete neutralisation. Make sure to solve the problem using the balanced equation. Solution (a) Determine moles of HC1 (titrant, known) (b) Use a balanced equation relate moles of analyte to moles of titrant (reaction ratio) (c) Determine mass of analyte (determine unknown) and % in sample
- 0.890 g KIO3 was dissolved to give a 80.00 mL Iodate solution. 35.00 mL aliquot reactedwith the iodide to form triiodide. The liberated triiodide was titrated and consumed 11.65 mL of theS2O32- solution to reach endpoint. Calculate the concentration in molarity of the Na2S2O3solution.2.413 g sample contains Na2CO3, NaHCO3 and inert material. This sample was solved in water and diluted to 250.0 ml. 80 ml solution was taken and titrated with 0.09644 N HCl with the indicator of phenolftalein. 6.13 ml HCl was consumed. From a second solution, again 80 ml was taken and titrated with the same HCl solution with the indicator of methylorange. 15.4 ml HCl was consumed. Calculate the percentage of NaHCO3 and Na CO in sample (Na CO =106.0 g/mol, NaHCO =84.02 g/mol)The concentration of CO in air is determined by passing a known volume of air through a tube that contains I2O5, forming CO2and I2. The I2 is removed from the tube by distilling it into a solution that contains an excess of KI, producing I3-. The I3-is titrated with a standard solution of Na2S2O3. In a typical analysis a 4.79-L sample of air is sampled as described here, requiring 7.17 mL of 0.00329 M Na2S2O3 to reach the end point. If the air has a density of 1.23 × 10–3 g/mL, determine the parts per million CO in the air