Can someone explain to me how to write an ionic formula? the simplest way. for example, use barium fluoride.?
821637
1 break up the compuond into ions or ions and radicals in this case there are no radicals.\n2 place the charges of each ion to the upper right of the symbol and viola e.g. \nBa2+  Fl-
708448
This deals with Bronsted-Lowry (B-L) acids and bases, and conjugate-base pairs \n\nB-L acids: donate H+ ions: STRONG ACIDS dissociate into ions\nH20 + HCl --> Cl- + H+  (H+ is the same as hydronium [H3O+])\nso it can be written as H20 + HCl --> Cl- + H3O+ \nThis looks like it donates a H+ ion, but it dissociates because it is a STRONG acid.  \n\nwk acid: HA= wk acid; A- = conjugate base\nHA + H2O ---> H3O+ + A-\n\nB-L bases: accept H+ ions: STRONG bases dissociate\nNaOH + H2O --> H2O + Na+ + OH-\n\nwk base: B = wk base; HB = conjugate acid\nB- + H2O --> HB + OH-\n\nThat's all you need to know to write all eqns.
729557
_NaOH(aq) + _Fe(ClO4)3(aq) => _Fe(OH)3(s) + _NaClO4(aq)\n\n3  NaOH(aq) + 1  Fe(ClO4)3(aq) => 1  Fe(OH)3(s) + 3  NaClO4(aq)\nThe coefficients of one could be excluded but I put one which is also equivalent.\n\n_((NH4)2)SO4(aq) + _Pb(NO3)2(aq) => _PbSO4(s) + _NH4NO3(aq)\n\n1  ((NH4)2)SO4(aq) + 1  Pb(NO3)2(aq) => 1  PbSO4(s) + 2  NH4NO3(aq)\n\nThese are only the balanced equations.\n\nThe following is the ionic equations:\n(Na)+ (aq) + (OH)- (aq) + (Fe)3+ (aq) + (ClO4)- (aq) => Fe(OH)3(s) + (Na)+ (aq) + (ClO4)- (aq)\nand\n(NH4)+ (aq) + (SO4)2- (aq) + (Pb)2+ (aq) + (NO3)- (aq) => PbSO4(s) + (NH4)+ (aq) + (NO3)- (aq)\n\nThe ionic equations are already balanced.  It is simply the breaking down of the equations into ions.  The solid remains intact while the compounds in aqueous solution become ions.\n\nThe + and - signs are the charge of the ions.  I can't find any other way to write them so I separated the compound with parentheses.  Sorry if it's confusing.\n\nThe net ionic equations are "simplified" ionic equations.  Basically, the things that are repeated on both sides get crossed off.  What is remaining is simply the net ionic equation.\n\n(OH)- (aq) + (Fe)3+ (aq) => Fe(OH)3(s)\nand\n(SO4)2- (aq) + (Pb)2+ (aq) => PbSO4(s)
298301
Hmm. This is exactly what I've been doing for the past few weeks. I'm assuming these are aqueous solutions of the three, right?\n\nWell, both magnesium and strontium form precipitates with hydroxides, so you could add some sodium hydroxide and centrifuge it down. To the liquid remaining, add some kind of sulfate, like potassium sulfate, because barium precipitates with sulfates, forming BaSO4.\n\nTo the solid containing strontium and magnesium, add ammonia to make the solution basic, then add sodium carbonate. The magnesium should complex with the carbonate and fall out, forming Mg4(CO3)3(OH)2. Then, whatever's left is strontium.
164971
BaCl2 + Na2SO4 --> BaSO4 + 2NaCl is the chemical reaction. \nNet ionic equation: Ba(2+) + SO4(2-) --> BaSO4 \nTotal ionic equation: Ba2+ 2Cl- + Na+ + 2SO4(2-) --> BaSO4(s) + 2Na+ + Cl-\nBarium sulfate is a classic precipitate.  \n\nAll of the masses of these compounds can be calculated by adding the individual ions' masses together, which is the same as atomic masses on a periodic table.   \nBarium chloride: 208.25 g/mol\nSodium sulfate: 142.04 g/mol (anhydrous)\nBarium sulfate: 233.43 g/mol \nSodium chloride: 58.4 g/mol
976855
Ions are basically charged atoms. The formula of an ion can be found out if you have the valency of the atom. You can figure out the valency of any atom by finding the group at which the atom is located in the periodic table. For example, Sodium (Na) is located in group 1A of the periodic table. Hence the valency is +1. The formula of the sodium ion would be Na+1.\n\nHope that helps!
910345
The formula would be Al2(CO3)3; the numbers are subscripts. \n\nLearn your charges for the polyatomic ions and for the other elements, it will really help.
783799
wait a minute.....i know this one.........calcium doesn't have multiple ionic charges so there's only Calcium (I) or just plain calcium.\nCaCl \nUnlike iron and copper and others calcium only has one ionic charge\n\nYou don't use prefixes with ionic compounds, however you do\n with molecular compunds\nGO COWBOYS!!!!!!!!!!!!!!!!! in your face indy 9-1
141058
sodium sulfte is soluble, but barium sulfate would form as a precipitate. Thus\n\nBaNO3 + Na2SO4 -> BaSO4 + 2 NaNO3, or simply\n\nBa(2+) + SO4(2-) -> BaSO4
861843
Start with the Ca then balance the number of PO4 ions, then balance the Oxygen and the hydrogens should balance themselves out.  Its just trial and error. Monkey with it until it works.
454203
The Solubility Product, Ksp, is defined as (for the hypothetical compound MmAn, which dissolves in water according to the equation:\n\nMmAn (s) ---->  mM(+)(aq) + nA(-)(aq)\n\nis:  Ksp = [M(+)]^m * [A(-)]^n\n\nSo, for the dissolution of Barium fluoride:\n\nBaF2 ---> Ba(+2)  + 2 F(-)\n\nthus Ksp = [Ba]*[F]^2\n\nIf [F] = 1.5E-02, then [Ba] = 1/2*[F], or 7.5E-03\n\nKsp  = [7.5E-03]*[1.5E-02]^2 = 1.7E-06\n\nThe textbook value of Ksp for barium fluoride is 1.84E-07, so if your value for the concentration of fluoride in solution is from an experiment, it is a bit high.
814242
Barium Hydroxide is a weak alkali which is not very soluble. If you try to neutralise it with dilute sulphuric acid you will get a precipitate of Barium sulphate. This means you won't be able to neutralise solid barium hydroxide just the solution. To neutralise either the solution or the solid use dilute hydrochloric acid.\nBa(OH)2  +  2HCl  =  BaCl2 +  2H2O\n\nIf you add carbonate ions CO3 2- to a solution of barium hydroxide Barium carbonate will be precipitated. Again carbonate ions (as in sodium carbonate solution) will not completely react with solid barium hydroxide.
867056
barium: 1s22s22p63s23p64s23d104p65s24d105p66s2\n\nradium: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s2\n\nsry if it looks a bit confusing without using exponetnts but im sure u could figure it out
964369
Huge order. Let's see how far it can be served.\n\n1. Monoatomic negative ions: take the element name (sometimes shortened) and add "-ide" (as in hydride, nitride, oxide, sulphide, phosphide, chloride...)\nPositive ions: element-name, charge (if several charges are possible), ion (sodium ion, iron II ion...)\n\n2. Polyatomic ions:\nComplexes with various ligands list the ligands and the central ion/atom.\nOxygen as ligands has its own nomenclature, depending on oxidation state. E.g. Sulphate, sulphite, perchlorate, chlorate, chlorite, hypochlorite...\n\n3. Binary molecular compounds: Halogen, oxygen and sulphur compounds usually have the correspondent "-ide" as if you had ionic compounds. Usually the more electropositive element is used as center.\n\n4. Acids: Oxygen acids have a similar nomenclature to oxygen ions.
825330
Ionic compound is usually non-metal reacting with metals.\nThat is positive ion forming a compound with negative ion.\nWe call the compound is ionic if the elements we've used have positive or negative charge.\n\nConvalent compunds have different story because they are sharing the electrons and they don't really have any charge.\n\nI don't know what you mean by exactly what kinds of elements make ionic compounds because almost all of them do! Except all the noble gases far right side of table will not react at all (Where there is argon).\n\nIt is that simple
674948
HF + H2O => F + H3O+\nNH4+ + H2O => NH3 + H3O+\nH2CO3 + H2O => HCO3- + H3O+\nHNO2 + H2O => NO2 + H3O+\n\nbasically, the oxonium ion is formed in each one, when a proton is donated to water.
973278
I'm assuming you're talking about ionic compounds like CaCl2 or KNO3....In order to name things from their formulas, you MUST know the common ions.  I'm sure you have a list of a table of ions you are expected to know.  Once you know all of those ions and their names, you can go from a formula to a name or the other direction quite easily.  But you've got to learn those ions first. \n\nIn naming from a formula, you name the cation (usually a metal) first, and then name the anion.  So, Na2SO4 is sodium sulfate.\n\nCa3(PO4)2 would be calcium phosphate.\n\nYou need to learn the parts before you can name the compounds.\n\nGood luck at this point...