1 / 7

A.P. Chemistry. Gravimetric Analysis using PPT reaction information…a review and extension

A.P. Chemistry. Gravimetric Analysis using PPT reaction information…a review and extension. 1965 to date   Clad Coinage Composition:  75% copper, 25% nickel  Weight:  2.27g  Diameter:  17.9mm     Edge:  Reeded.

agamemnon-elena
Download Presentation

A.P. Chemistry. Gravimetric Analysis using PPT reaction information…a review and extension

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. A.P. Chemistry. Gravimetric Analysis using PPT reaction information…a review and extension 1965 to date   Clad CoinageComposition:  75% copper, 25% nickel Weight:  2.27g Diameter:  17.9mm    Edge:  Reeded Numismatics is the scientific study of money and its history in all its varied forms. "All science is either Physics or stamp collecting." - Ernest Rutherford physicist and Noble Laureate.

  2. This Roosevelt dime is different though. Pre-1965 Composition:  silver and copper… but how much of each typically???

  3. A type of quantitative analysis in which the amount of one species in a material is determined by isolating and massing it A.P. Chem Gravimetric Analysis: PURPOSE: Procedure: 1. 2. 3. 4. 5. 6. 7. Calculations: To determine the silver composition of a pre-1965 dime Data Mass dime 2.8357 g Dissolve the dime in HNO3 Ag +NO3- + H+ --> Ag + + NO + H2O Add HCl Ag+ + Cl- --> AgCl Mass of Filter and Sample = 4.1860 g (Filter paper = 0.7942 g) Filter and dry Mass AgCl AgCl = 3.3918 g % Ag = 75.265 % Find % comp. of AgCl Find mass of Ag Ag = 2.5528 g 6.) 107.87 / 143.32 = 0.75265 7.) 3.3918 x 0.75265 = Answer: ( 2.5528 g / 2.8357 g ) x 100% = 90.024%

  4. What volume of 6.0 M HCl would have been needed to completely precipitate out all of the silver ions from the solution if the concentration of the silver ion was 0.473 M and there was 50.0 mL of the solution?

  5. What volume of 6.0 M HCl would have been needed to completely precipitate out all of the silver ions from the solution if the concentration of the silver ion was 0.473 M and there was 50.0 mL of the solution? 50.0 ml Ag+ 0.473 mol = 0.0237 mol Ag+ 1000.0 mL Ag+ + Cl- --> AgCl Ag+ + HCl(aq) --> AgCl + H+ Ag+ + H++ Cl---> AgCl + H+ 0.0237 mol HCl 0.0237 mol Ag+ 1.0 mol HCl = 1.0 mol Ag+ 0.0237 mol HCl 1000.0 mL HCl = 4.0 mL HCl sol’n 6.0 mol HCl 50.0 ml Ag+ 0.473 mol Ag+ 1 mol HCl 1000.0 mL HCl = 1 1000.0 mL Ag+ 1 mol Ag+ 6.0 mol HCl

  6. What mass of Fe(OH)3 is produced when 50.0 mL of 6.0 M LiOH sol’n reacts with an excess of 2.0 M Fe(NO3)3 sol’n? Fe(NO3)3(aq) + LiOH(aq)→ Fe(OH)3(s) + LiNO3(aq) 3 3 50.0 ml LiOH 6.0 mol LiOH 1 mol Fe(OH)3 106.8 g Fe(OH)3 = 1 1000.0 mL LiOH 3 mol LiOH 1.0 mol Fe(OH)3 11 g Fe(OH)3

  7. What volume of 6.0 M HCl would have been needed to completely precipitate out all of the silver ions from the solution if the concentration of the silver ion was 0.473 M and there was 50.0 mL of the solution? What volume of 6.0 M HCl would have been needed to completely precipitate out all of the silver ions from the solution if the concentration of the silver ion was 0.473 M and there was 50.0 mL of the solution? What mass of Fe(OH)3 is produced when 50.0 mL of 6.0 M LiOH sol’n reacts with an excess of 2.0 M Fe(NO3)3 sol’n? What mass of Fe(OH)3 is produced when 50.0 mL of 6.0 M LiOH sol’n reacts with an excess of 2.0 M Fe(NO3)3 sol’n?

More Related