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Why a precipitate of manganese dioxide be removed from the potassium permaganate solution before standardization?
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- Why a filter paper cannot be utilized for the removal of potassium permaganate precipitate?Q1)a) In the gravimetric analysis the precipitate should be free from impurities. Give the detailed justification for the above statement by identifying impurities. b) It is required to analyse the metal from its ore sample. Manganese (Mn)present in the ore sample was analysed by forming a precipitate of MnSO4. If 2.98 g of ore sample gives 0.19 g of MnSO4. Calculate the percentage of Mn present in the sample.A 50-mL solution of 0.5005 N standard NaOH was added to a 0.9250-g sample of Aspirin and was boiled for 10 minutes. After cooling a full pipet of phenolphthalein was added to the solution before it was titrated with 24.7 mL of 0.5015 N standard HCl solution until the disappearance of the pink color. The same procedure was carried using a blank which consumed 4.6 mL of the same standard acid. Calculate for the %Aspirin in the sample. Atom weights: C =12, H =1, O =16.
- list the apparatus involved in volumetric or titrimetric analysisA chemist received different mixtures for analysis with the statement that they contained NAOH, NaHCO3, Na,CO3 or compatible mixtures of these substances together with inert material. From the data given, identify the respective materials. Sample 4. The sample was titrated with acid until the pink of phenolphthalein disappeared; this process required 39.96 mL. On adding an excess of the acid, boiling and titrating back with NAOH, it was found that the base was equivalent to the excess acid. O A NazCO3, NaHCO3 OB. NaHCO3 O C. Na2CO3 O D. None of these O E. NaOHWhy are many ionic precipitates washed with electrolyte solution instead of pure water?
- (c) What is a standard solution? What piece of glassware is used to prepare a standard solution and describe how a standard solution of sodium carbonate may be prepared in the laboratory?Part 1: Preparation of the Primary Citric Acid Standard 1. Mass of Citric Acid: 4.05 g 2. Volume of Citric Acid Solution: 0.75 ml (at the equivalence point) 3. Moles of Citric Acid: (Molar Mass = 192.0 g/mol) 4. Molarity of Citric Acid Solution: Part 2: Titration of the Sodium Hydroxide Solution (Show your work on page 4) 1. Volume of Citric Acid at the Equivalence Point 2. Moles of Citric Acid at the Equivalence Point: 3. Moles of NaOH at the Equivalence Point 3 NaOH + H.CaHsO, → Na:CH.O, + 3 H,O 4. Volume of NaOH 10.0 mL 5. Calculated Molarity of NaOH: Do not use the dilution equation to calculate the molarity of the sodium hydroxide. The dilution equation cannot be used when a reaction is occurringAn impure sample containing 2.00 g of sodium carbonate and bicarbonate was taken. It was dissolved in water and then crushed with hydrochloric acid (0.1 N). If the burette reading was at the end point of the phenolphthalein 7.5 ml and at the end of the orange methyl point was 20.0 ml. Calculate the ratio The percentage of sodium bicarbonate in the sample. Note that the molecular weight of sodium bicarbonate is 84 *
- For a Kjeldahl analysis: you weighed 0.0312 g unknown dried blood sample and digested it with concentrated H2SO4. After the distillation into boric acid solution, Methyl Purple indicator changed its color to green. During titration with 0.0110 M H2SO4 it took 15.35 mL of this H2SO4 solution to bring the color of the mixture to grey-blue. a) What number of moles of H2SO4 reacted? b) What number of moles of boric acid reacted with NH3 during distillation? c) What number of moles of NH3 was produced during distillation? d) What was mass of N in the produced NH3? e) What is %N w/w in the original sample?Write measurement of Carbon dioxide evolution during respiration of germinating seeds by the Titration methodWhich among the statements to promote large precipitates is INCORRECT? Having a dilute solution will keep supersaturation ratio small. Temperature affects particle size during crystal growth. The order of addition of the precipitating reagents may affect induction period. It is better to have a faster rate of nucleation than the rate of crystal growth.