A 2.500g sample containing Ce++ was treated with excess iodate to precipitate Ce(IO3)4. The precipitate was collected, dried, and ignited to produce g of CeO2. What was the original weight percent of Ce in the original solid?(15) containing only C. Hand maybe Nwas
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- The thiourea in a 1.563 g sample.of an organic material was extracted into a dilute H2SO4 solution and titrated with 36.43 mL of o.o09284 M Hg²* via the reaction below. 4(NH2)½CS + Hg2 - [(NH2),CS],Hg²• What is the percentage of (NH2)2CS (Molar Mass = 76.12 g/mol) in the sample?What is the solution of this? The given answer is 74.19 mL ; 74.19 mLGive a clear explanation handwritten answer..give detailed answer of given opinions with handwritten
- A 0.8965g ore sample is dissolved in nitric acid and then filtered. The aluminum is present in solution as AP". The solution is made basic with ammonium hydroxide, NH,OH, and the aluminum hydroxide, Al(OH)» (FW 78.004), precipitates. This gel is filtered in a porous glass crucible, rinsed with dilute ammonium hydroxide, ignited, cooled in a desiccator, and weighed. The resulting alumina, Al;O; (FW 101.94), weighed 0.1605 g. Why is the solution filtered after acid dissolution? Why rinse with ammonium hydroxide solution? What chemical transformation takes place during ignition? (Show the balanced chemical equation) • Why use a desiccator during cooling? Calculate the weight percent Al (AW 26.9815) in the sample. Al(OH)) -> Al;O, + H0Give a clear handwritten answer with explanation..give clear answer and give correct answer5.2 A 5.199 g sample containing the mineral tellurite was dissolved and then treated with 50.00 mL of 0.03144 M K2Cr2O7. Upon completion of the reaction, titration of the excess Cr20, required 37.28 mL of 0.1083 M Fe* solution. Calculate the percentage by mass of TeO2 (FM 159.60) in the sample. The reactions are 3 TeO, + Cr2O,- +8 H3 H;TeO4 +2 Cr + H;O Cr:O + 6 Fe + 14 H2 Cr + 6 Fe* + 7 H:O
- please answer 2,3 and 45. 0.1500 g sample of chromium ore was dissolved and the chromium oxidized to chromate ion. The solution was treated with 25.00 mL of 0.3500 M AGNO3. The resulting precipitate, from the reaction of the excess and analyte in 1: 1 mole relationship, Ag2CrO4 was removed and discarded. The excess AGNO3 required 30.50 mL of 0.200 M KSCN for titration in a 1:1 mole relationship. Calculate the % Cr2O3 in the ore if the mole relationship between Cr2O3 and CrO42- is 1:2 respectively.The sulfur content of an iron pyrite ore sample is determined by converting it to H2S gas, absorbing the H2S in 10.0mL of 0.00500 M I2, and then back-titrating the excess 12 with 0.00200 M Na2S203. If 2.6mL Na2S20zis required for the titration, how many milligrams of sulfur are contained in the sample? (Mwt of H2S is 34.08g/mol). Reactions: H2S + I2 -------> S+21 + 2H* I2+ 2S20,2- ---------> 21- + S,0,2- I () II !!
- + H (a) CuCN (b) CaCO3 (c) Ag₂CO3 (d) PbBr₂ (e) Y(OH)3 Periodic Table B W Chapter... Supporting Materials Additional Materiala [Ca²+] = 5.2 x 104 M, [CO32- ] = 6.5x 10-6 M * G Please use the values in the resources listed below instead of the textbook values. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated. [Cu] = 1.9x10-10 M, [CN] = 1.9x10-¹0 M [Ag+] = 1.8x10-5 M, [CO32] = 2.7x10-² M [Pb²+] = 1.2x 10-2 M, [Br] = 2.4x10-2 M [Y³+] = 3.5x 10-16 M, [OH-] = 6.6x 10-³ M Constants and Factors №. webassign.net ● F 893 Supplemental Data L C C a Ⓒ + D rCOMPLEX PREPARED (preparation #3 ) cuclz H20 Starting metal salt (MXx,_H;O) Copper (T)chloride dinydrate Metal (Cu or Co) cu Halide (Cl or Br) CL Amine (en, bu, pz, or dien) bu (1] Yield of the Complex Mass of crystalline product (Exp. Yield) 29.682 (2a) Determination of Copper Calculate the mols of Na,S:O, to reach the end point, and then calculate the moles of Cu from the stoichiometry. Finally, calculate the % Cu in the sample. Show your calculations on the next page. Trial 2 Trial 1 O.103g 20.5mL Mass of product,g 0.19 21.ML Vol. of Na;S:O, to reach the end point, ml. Molarity of NazS;0, (from bottle) 0.0100M 0.0002 05 0.000 205 0.000 210 O.000 210 Mols of NazS:O, Mols of Cu Mass of Cu, g 0.0130 0.0133 % Cu in product Ave % Cu Calculations: O.100 M = mol 0.0 20SL 63.55 = 9 0.0002DS %3D O 10100 M = mol o021 니 63.55 = 0.00021 %3D 853