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Find the pH of a 1.00-L aqueous solution containing 12.43 g of tris (FM 121.14) plus 4.67 g of tris hydrochloride (FM 157.60).
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- Calculate Qsp for calcium fluoride (Ksp = 3.9 ×10–11) when 180.0 mL of a 7.95×10-3 M solution of Ca(NO3)2 is added to 310.0 mL of a 6.90×10-3 M solution of KF.1) Titrant 5 mg H2O/ mL Resolution 2.5 µg H2O/step Detection limit is 125 µg H2O for a 5 g sample (= 25 ppm) 2) Titrant 2 mg H2O/ mL Resolution 1 µg H2O/step Detection limit is 50 µg H2O for a 5 g sample (= 10 ppm) You have a 10 g sample which is expected to contain 200 µg H2O. True or false? 1. Titrant 5mg/mL is the preferred system over 2. Titrant 2mg/mL for this sample. True FalseIn the spectrofluorometric analysis of quinine, which of the following statements/phrases can be considered a factor level? A. concentration of 2ppm of iodide B. concentration of quenchers C. use of quenchers D. use of iodide
- To which sample solution the spike will be add? the distillate>130C the mass standard the distilled <130CIron(III) hydroxide (MM = 106.87 g/mol) has a Ksp of 6.3 x10-38. If 35.9 g Fe(OH)3 is stirred into 6,034.63 L H2O, how many micrograms of iron(III) hydroxide will dissolve?Report your answer to the nearest whole number.A student prepared their sample for Trial 2. They added 2.30 mL of 0.002M KSCN, 5.16 mL of 0.002M Fe(NO 3) 3, and 2.51 mL of water. What is the initial concentration of KSCN in the trial?
- Find the ksp of PbCl2 5.0 g od pbcl2 100ml of h2oAnswer both questions please it wouldn't hurt 1) For the following write out the Ksp expression: Note: If the ion doesn't have a superscript add in "1" and start with the cation then the anion. Example of format: CuCl2 (s) Ksp = [Cu^2+]^1[Cl^1-]^2 (a) CaF2 (s) Ksp = []^[]^ (b) Al2S3 (s) Ksp = []^[]^[]^ 2) Calculate the silver ion, of a solution prepared by dissolving 1.00 g of silver nitrate and 10.0 g of KCN in sufficient water to make 1.00L. Kf = 1.00 x 1021 Ag+(aq) + 2 CN- (aq) ↔ [Ag(CN)2]- (s)In which of the following mixture(s) would CaF2(s) (Ksp = 4.0 x 10-11) form? 30 mL of 0.00020 M Ca(NO3)2 + 20 mL of 0.00125 M NaF 20 mL of 0.00020 M Ca(NO3)2 + 30 mL of 0.00125 M NaF 10 mL of 0.00020 M Ca(NO3)2 + 40 mL of 0.00125 M NaF Group of answer choices I only II only I and II I and III