One published value for Keq for FeSCN2+ formation is approximately 113 at 20oC. Based on this value, does equilibrium favor the formation of FeSCN2+, or does it favor the reverse reaction? Briefly justify your answer.
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One published value for Keq for FeSCN2+ formation is approximately 113 at 20oC. Based on this value, does equilibrium favor the formation of FeSCN2+, or does it favor the reverse reaction? Briefly justify your answer.
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- Use the data below to determine the % by mass of Fe to TWO decimal places, if a 0.02 M KMnO4 solution was used for the titration. gH * + MnO 4 + 5Fe 2+ -> SFe 3+ + Mn 2+ + 4H 20 (MM Fe = 56 a/moll mass of iron compound (g) 0.604 initial buret reading (mL) 0.94 Tinal buret reading (mL 16.21Calculate the solubility (in g/L) of silver chromate in water at 25°C if the Ksp for Ag3PO4 is 1.75 × 10-18. Ag3PO4(s) <==> 3Ag+ (aq) + PO43- (aq); Ksp = 27s4, s = _________ mol/L x MM of Ag3PO4 = _________ g/L.A 1.250 g sample of cheese was subjected to KKjeldahl analysis to determine the amount of protein. the sample was digested and the nitrogen is oxidized to NH4+ and was then converted to NH3 with NaOH., and distilled into a collection flask containing 50.00 mL of .1050 M HCl. The excess HCl is back titrated with .1175M NaOH, this required 21.65 mL to reach the bromothymol blue endpoint. What is the reaction for the back titration for this analysis and the %N of cheese sample.
- A piece of Gold weighing 12,359 Kg is suspected of being contaminated with Iron. To perform an instrumental analysis and To confirm whether or not it contains Fe, a portion of the sample (0.954 g) is taken from the piece and dissolved with 25 mL of aqua regia. Heats up For its complete dissolution, it is cooled and made up to 100 mL. A 10 mL aliquot is taken from this solution and made up to 50 mL. From This last solution is given the appropriate treatment to visualize Fe+2, for which the 1,10-phenanthroline reagent is added. (it forms a complex that is red in color) and is taken to a visible spectrophotometer and with a 12 mm cell a absorbance of 0.45. Previously, a calibration curve of Fe+2 was obtained under the same instrumental conditions obtaining the following data: (view table) Calculate the purity of the gold piece, assuming impurities only due to Fe.Using basic conditions, MnO4- can be used as titrant for the analysis of Mn2+, with both the analyte and the titrant ending up as MnO2. In the analysis of a mineral sample for manganese, a 0.5165-g sample is dissolved, and the manganese is reduced to Mn2+. The solution is made basic and titrated with 0.03358 M KMnO4, requiring 34.88 mL to reach the endpoint. Calculate the %w/w Mn in the mineral sample. Answer: % Mn =Calcium fluoride is considered as a relatively insoluble compound and therefore lime or slakedlime has been considered as a possible material to remove excess fluoride in water of boreholesin certain parts of the country. The solubility product of CaF2 is Ksp = 3 x 10 – 11 and that ofCa(OH)2 isKsp =8x10-61. How much lime can be added to the water to remove 10 mg of F- ion per litre ofborehole water?(The atomic masses are Ca: 40.08; F: 19.00; O: 16; H: 1)
- Calculate the solubility of lead(II) sulfate (Ksp = 2.53x10-8) in a 0.0034 M solution of sodium sulfate. Give your answer to three sig. figs. and in exponential form (e. g. 1.23E-3).A 100.0 mL100.0 mL solution of 0.0200 M Fe3+0.0200 M Fe3+ in 1 M HClO41 M HClO4 is titrated with 0.100 M Cu+0.100 M Cu+, resulting in the formation of Fe2+Fe2+ and Cu2+Cu2+. A PtPt indicator electrode and a saturated Ag∣∣AgClAg|AgCl electrode are used to monitor the titration. Write the balanced titration reaction. titration reaction: Fe3++Cu+⟶Fe2++Cu2+Fe3++Cu+⟶Fe2++Cu2+ Complete the two half‑reactions that occur at the PtPt indicator electrode. Write the half‑reactions as reductions. half‑reaction: ?∘=0.161 V half‑reaction: ?∘=0.767 V Select the two equations that can be used to determine the cell voltage at different points in the titration. ?E of the Ag∣∣AgClAg|AgClelectrode is 0.197 V.0.197 V. ?=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V E=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V ?=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V E=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V ?=0.767 V−0.05916×log([Cu+][Cu2+])−0.197 V E=0.767…25.00 mL of a solution containing KIO4 is treated with excess potassium iodide. IO4- + 8H+ + 7I- → 4I2 + 4H2O The resulting solution is titrated with 0.15 M sodium thiosulfate pentahydrate. The average titre was 21.76 mL. I2 + 2S2O32- → 2I- + S4O62- What is the concentration of the sodium thiosulfate solution as a percentage (w/v)
- In an experiment to determine the concentration of H2SO4 in a brand of toilet cleanser, 10.0 cm3 of the cleanser was first diluted to 250.0 cm3 with distilled water. 25.0 cm3 of the diluted cleanser were titrated with 0.320 mol dm–3 NaOH solution, using methyl orange as indicator.If 29.2 cm3 of NaOH solution is used, what is the concentration of H2SO4 in the undiluted toilet cleanser?a. 2.30 mol dm–3b. 4.60 mol dm–3 c. 9.20 mol dm–3 d. 11.50 mol dm–3A 200.00 mL solution of 0.00105 M AB4 is added to a 270.00 mL solution of 0.00245 M CD5. What is pQsp for AD4? pQsp = -log(Qsp)A mineral in a fine state of division (0.6324 g) was dissolved in 25.0 mL of 4.0 mol / L boiling HCl and diluted with 175.0 mL of H2O containing two drops of methyl red indicator. The solution was heated to 100 ° C and a heated solution containing 2.00 g of (NH4) 2C2O4 was added slowly to precipitate CaC2O4. Next, NH3 6.0 mol / L was added until the indicator changed from red to yellow, indicating that the liquid was neutral or slightly basic. After slow cooling for 1 hour, the liquid was decanted, the solid transferred to a crucible and washed five times with 0.10 wt% (NH4) 2C2O4 solution, until no Cl- was detected in the filtrate with the addition of AgNO3 solution. The crucible was dried at 105 ° C for 1 hour and then taken to an oven at 500 ° C ± 25 ° C for two hours. The mass of the empty crucible was 18.2311 g. The crucible mass with CaCO3 (s) was weighed 5 times to an average of 18.5467 g. Determine the percentage, by mass, of Ca in the mineral. Ca (40.078 g / mol); C (12.01078…