Jump to content

shafaifer

Members
  • Posts

    8
  • Joined

  • Last visited

Everything posted by shafaifer

  1. Thank you very much, Charon. Yes, that is what I am asking.
  2. The LOD is defined as the minimum concentration of an analyte required, wherein the analytical signal is significantly different from the blank signal (background signal). Am I right in this: The lower LOD-value, the higher analyte concentration. In an AAS study of the mass of calcium carbonate in potsherds (first dissolving the sherds in concentrated nitric acid, then measuring an absorbance-value, calculating a concentration for calcium --> mCaCO3(s) is obtained), what use is it to know what the minimum concentration of calcium is, wherein the analytical signal is significantly different from the blank signal? If this cannot be answered, what use is it in general in analytic chemistry?
  3. Is something here correct? Let us say you have this salt: Na3PO4, to this salt you add HCl(aq): Na3PO4(s) + 3HCl(aq) --> 3NaCl(aq) + PO43-(aq) + 3H+(aq) (equation 1) Now phosphate reacts with added molybdate, then added HCl(aq): PO43-(aq) + 3NH4+(aq) + 12Mo2+(aq) --> (NH4)3[PMo12O40](s) (equation 2) According to a lab manual, the product on the right side of the reaction in Eq. 2 is correct. The manual only says that phosphate reacts with molybdate and creates the product in equation 2 - not what the reactions in question look like. I cannot balance the left side of Eq. 2, how is it done? Best regards, Shafaifer
  4. Is Eq. 1 wrong (the one Elite Engineer answered)? I got Eq. 2 from a lab manual written and corrected by Ph.d's.
  5. Basic conditions. The experiment was not conducted under STP. The temperature in the laboratory that day was 295 - 298 K, I am not sure about the pressure.
  6. What is/are the product/(s) of the following reaction: Co(OH)2(s) + OH-(aq) (Eq. 1) I can inform you of the earlier reaction leading up to this one: [Co(H2O)6]2+(aq) + 2OH-(aq) Co(OH)2(s) + 6H2O(l) (Eq 2) The precipitate formed in (Eq. 2) is blue. This precipitate then turns red/grey as conc. NaOH(aq) is added. What is/are the product/(s) formed in equation 1? Best regards, Shafaifer
  7. Hello, I have these reactions (they should be correct): CaCO3(s) + HNO3(aq) --> H2O(l) + Ca2+(aq) + CO2(g) + NO3-(aq) Eq. 1 Calcium cations will now form a complex with water according the following reaction: Ca2+(aq) + 6H2O(l) --> [Ca(H2O)6]2+(aq) + 6H2O(l) Eq. 2 Now comes the combustion: This is a project of AAS (atomic absorption spectroscopy), and I want to measure the absorbance for calcium when irradiated with a hollow cathode lamp. I am only interested in calcium so I ignore water, nitrate ions and CO2(g). As a combustion agent / oxidising agent, C2H2(g)/O2(g) is used: [Ca(H2O)6]2+(aq) + C2H2(g)/O2(g) --> Eq. 3 What is the product of this equation? I am sure solid calcium is formed (thus reduced). But how can it be reduced when the compounds used for the combustion are oxidizing agents? I am considering C2H2(g)/O2(g).
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.