Chemistry; Please help

posted by .

Write formulas for the following.
1. potassium hexacyanoferrate(III)
2. sodium hexafluoroaluminate
3. Pentaaquabromomanganese(III) sulfate
4. hexaamminechromium(III) nitrate
5. sodium tetrahydroxochromate(III)
6.trans-dichlorobis(ethylenediamine)cobalt (III) chloride
7.hexaammineruthenium(III) tetrachloronickelate(II)
8.tetraamminecopper(II) pentacyanohydroxoferrate(III)
9. potassium diaquatetrabromovanadate(III)
10.diamminezinc iodide

  • Chemistry; Please help -

    well the first one is K3[Fe(CN)6] T_T i'm looking for it too T.T

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Organic Chem

    Which of the following alkenes do not show cis-trans isomerism?
  2. chemistry

    Explain in word how you would prepae 250mL of a 3.000 M solution of Ba(OH)2 starting form barium hydroxide and water..... 2 equation 2CoI3 + 3 K2S= Co2S3 + 6KI 2 moles of Cobalt (III) iodide + 3 moles of Potassium sulfide = cobalt …
  3. chemistry...

    equation 2CoI3 + 3 K2S= Co2S3 + 6KI 2 moles of Cobalt (III) iodide + 3 moles of Potassium sulfide = cobalt (III) sulfide + 6moles of Potassium Iodide assuming all that can react will react, how many grams of cobalt (iii) sulfide will …
  4. Chemistry...verification

    2 equation 2CoI3 + 3 K2S= Co2S3 + 6KI 2 moles of Cobalt (III) iodide + 3 moles of Potassium sulfide = cobalt (III) sulfide + 6moles of Potassium Iodide assuming all that can react will react, how many grams of cobalt (iii) sulfide …
  5. chemistry

    consider the following balanced chemical reaction: 2 Col3+3 K2S-> Co2S3+6 Kl assuming all that can react will react, how many grams of cobalt (III) sulfate will form when 1318.89 grams of cobalt (III) iodide are combined with 551.35 …
  6. general chemistry

    I need the oxidation half reactions, reduction half reactions, and net ionic equations for the following reactions: 1)magnesium +0.1M zinc sulfate 2)copper+0.1M zinc sulfate 3)zinc+0.1M copper(II)sulfate 4)zinc+3M HCl 5)copper+3M HCl …
  7. chemistry

    One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III) hydroxide from a solution containing rhodium(III) sulfate According to the following balanced chemical equation: Rh2(SO4)3(aq)+ 6NaOH(aq)→ …
  8. chemistry

    One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium (III) hydroxide from a solution containing rhodium(III) sulfate according to the following balanced chemical equation: Rh2(SO4)3(aq)+6NaOH(aq)-->2Rh(OH)3(s)+3Na2SO4(aq) …
  9. chemistry

    Arrange the acids I) hydrogen selenate ion (HSeO− 4 ), pKa = 1.92; II) phosphorous acid (H3PO3), pKa1 = 2.00; III) phosphoric acid (H3PO4), pKa = 2.12; IV) selenous acid (H2SeO3), pKa = 2.46; in increasing order of strengths. …
  10. chemistry

    Arrange the acids I) hydrogen selenate ion (HSeO− 4 ), pKa = 1.92; II) phosphoric acid (H3PO4), pKa = 2.12; III) selenous acid (H2SeO3), pKa = 2.46; IV) phosphorous acid (H3PO3), pKa1 = 2.00; in increasing order of strengths. …

More Similar Questions