During recrystallization, an orange solution of a compound in hot water was treated with activated carbon and then filtered through fluted paper. On cooling, the filtrate gave gray crystals, although the compound was reported to be colorless. Explain why the crystals were gray and describe steps that you would take to obtain a colorless product. answer
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During recrystallization, an orange solution of a compound in hot water was treated with activated carbon and then filtered through fluted paper. On cooling, the filtrate gave gray crystals, although the compound was reported to be colorless. Explain why the crystals were gray and describe steps that you would take to obtain a colorless product. answer
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- A researcher reconstituted a vial of 750 mg Cefuroxime Sodium Powder for Injection with 6mL of sterile water for injection. The reconstituted solution was dark amber-colored solution. The package insert states that solution colors range from clear to yellow depending on concentration, diluent, and storage conditions. The researcher was then hesitant to give the patient the solution due to its unusual dark color. Five portions were taken from a batch of cefuroxime sodium. Prior to testing in the instrument, each part was subjected to one of the following conditions: Conditions Specifications Temperature Portion 1: 8°C ± 2°C Portion 2: 30°C ± 2°C Portion 3: 40°C ± 2°C Light Portion 4: Kept in the dark Portion 5: Exposed to direct sunlight 1. Of the several solutions prepared, which absorbance value results should be compared with each other to answer the questions of the pharmacist? Explain your answer.Guide Questions[1] What type of precipitate is BaSO4?[2] Why is filter paper preferred over a sintered glass filter?[3] What made an ashless filter paper ashless?[4] Why is a porcelain crucible used in this procedure?[5] Why do we need to precipitate from a hot solution?[6] Why is the precipitating agent added all at once?[7] What is the purpose of adding 6M HCl in the sample solution?[8] Why do we need to test the washings with AgNO3?[9] What is ignition?[10] What is the product after the precipitate was ignited?[11] What are the advantages of performing ignition?[12] Cite the sources of errors for this experiment and how to circumvent these problems.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 water sample was analyzed for Fe content using the iron-phenanthroline method. The following data were obtained from the analysis:Analyze the table of some common inorganic precipitating agents and answer the following questions: Zn and Sn ions coprecipitate when they exist in solution to form mixed oxides. Suggest a precipitating agent that precipitate one of the oxides a. only A silver chloride precipitate is precipitated in gravimetric analysis. Which solution would you recommend to wash the precipitate: AGNO3 or HNO3. Justify your answer. b. Precipitating Element precipitated agent NH3(aq) Be (BeO), Al (A12O3), Sc(Sc203), Cr (Cr203), Fe (Fe203) Ga (Ga2O3), Zr (Zr02), In (In2O3), Sn (SnO2), U (U3O8) H2S Cu (CuO), Zn (ZnO or Zn2O4), Ge (GeO), As (As203 or As205), Mo (Mo03), Sn (SnO2), Sb (Sb2O3 or Sb2O5) (NH4)2S Hg (HgS), Co (Co304) (NH4);HPO4 Mg (Mg2P2O7), Al (AIPO4), Mn (Mn2P2O7), Zn (Zn2P2O7), Zr (Z12P207), Cd (Cd2P207), Bi (BİP04) H2SO4 Li, Mn, Sr, Cd, Pb, Ba (all as sulfates) H;C¿O4 Ca (СаО), Sr (SrO), Th (ThО2) HCI Ag (AgCl), Hg (Hg2C12), Na (NaCl), Si (SiO2) AGNO3 CI (A£CI), Br (AgBr), I (Agl) (NH4);CO3…Draw a schematic diagram using this information. Preparation of the solution Take about 25-30 ml of distilled water in a beaker and add powdered crude copper sulphate while stirring the solution so that the powder dissolves. Keep adding the powdered sample till a little of it remains undissolved, even if it is stirred thoroughly. Now add 2-3 ml of dil. H2SO4 to make the solution clear. This prevents the hydrolysis of copper sulphate. Filtration of the solution Take a funnel and fix a filter paper in it and clamp it to a stand. Place a china dish under the funnel. The stem of the funnel should touch the wall of the china dish to avoid the solution splashing out. Pour the solution into the funnel over a glass rod and collect the filtrate in a china dish. The insoluble impurities are left in the filter paper as residue. Concentration of the filtrate to crystallisation point Heat the china dish on a sand bath till the solution is reduced to about one-third of its original volume. To…
- X ray films consist of a flexible plastic base coated in a gel that contains light sensitive silver halide crystals (i.e. silver bromide and silver chloride). these films are exposed to X ray radiation to produce images of the inside of a subject. Once the films are no longer needed, it is common practice to recycle the film to recover the silver. This process is done by shutting the film and placing it in a large cyanide solution bath. The silver is precipitated out as silver cyanide. a. What can you conclude about the water solubility of silver cyanide? b. Using the knowledge of limiting reactants, how can the process of be optimized to ensure that the maximum amount of silver is recovered?A mixture of solid chlorides was known to contain some or all of the following cations: Ag ,Al^3+Cr^3+,Ca^2+,and/or Zn^2+ . A student performed the following steps: 1.Water was added to the solid and the mixture was stirred well. After several minutes, some of the solid did not dissolve. The mixture was centrifuged and the supernatant was decanted from the solid (solution A). 2. NH4CI(aq) was added to solution A followed by NH3(aq) until the solution was basic to litmus. A gray-green precipitate was formed. The mixture was centrifuged and the supernatant (solution B) was decanted. The solid dissolved in NaOH(aq) and H2O2(aq) to form a light yellow solution. The solution was heated to dryness to remove the H2O2 and the solid was redissolved in HCI(aq). When ammonia was added to make the solution basic, a light yellow solution formed and no solid precipitated. 3. (NH4)2C204(aq) was added to solution B, and a precipitate formed. The mixture was centrifuged and the supernatant (solution C)…The calcium and magnesium in a urine sample were precipitated as oxalates. A mixed precipitate of CaC2O4 and MgC2O4 resulted and was analyzed by a thermogravimetric procedure. The precipitate mixture was heated to form CaCO3 and MgO. This second mixture weighed 0.0433 g. After ignition to form CaO and MgO, the resulting solid weighed 0.0285 g. What was the mass of Ca in the original sample? 2.42 x 10^-3 g Ca 2.76 x 10^-3 g Ca 6.29 x 10^-3 g Ca 1.57 x 10^-3 g Ca
- 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 =16.1. The substance to be analyzed – is white, highly soluble in water salt. Addition of barium chloride or silver nitrate (in presence of nitric acid) doesn`t give any precipitation. Reaction with zinc uranyl acetate leads to formation of yellow crystals, which have tetrahedral geometry. Addition of antipyrin in acid medium changes color to red. What kind of substance is this? Explain you answer, provide chemical equationsA 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. NaOH