most silver compounds have low solubility.
AgF is soluble in water, because Fluorine is very electronegative.(Electronegativy is the tendancy to attract the bonding electrons in a covalent bond.) Water is a polar molecule, which means that for a compound to be soluble, it must also be polar. As fluorine is the most electronegative out of all the elements, it makes the AgF molecule slightly polar, which makes it soluble in water. Chlorine, Bromine and Iodine are not as electronegative, therefore they don't create a polar molecule. They instead form precipitates.Yes, AgBr is insoluble in water.
You can use the Ksp (the solubility product) to determine the solubility.
Ksp = 5.0 x 10^-13 and the equilibrium equation is AgBr(s) Ag+ + Br-
This video explains it in great detail:
www . khanacademy.org/science/chemistry/acid-base-equilibrium/copy-of-solubility-equilibria-mcat/v/solubility-from-the-solubility-product-constant
I had the experiment and i knew it before AgCl dissolves in very small amounts in amounts, i mean almost none by saying small.
Nope! It's insoluble
No.
In acids.
AgBr is the chemical formula of silver bromide.
Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] First check that the given equation is balanced ... it isn't ... so the first thing to do is balance the equation: balancing Na: 2Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] and everything is now balanced so we've got the balanced equation molar mass AgBr = 107.87 + 79.90 = 187.77 g/mol mol AgBr available = 42.7 g AgBr x [1 mol / 187.77 g] = 0.2274 mol AgBr from the balanced equation the mole ratio AgBr : Na2S2O3 = 1 : 2 so mol Na2S2O3 required = 0.2274 mol AgBr x [ 2 mol Na2S2O3 / mol AgBr] = 0.455 mol Na2S2O3 (to 3 sig figs)
Silver Bromide
Silver will have a +1 and bromate is -1 so they combine in a 1:1 ratio. The formula would be AgBrO3.
AgBr
AgBr is the chemical formula of silver bromide.
AgBr is the chemical formula (not symbol) of silver bromide.
Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] First check that the given equation is balanced ... it isn't ... so the first thing to do is balance the equation: balancing Na: 2Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] and everything is now balanced so we've got the balanced equation molar mass AgBr = 107.87 + 79.90 = 187.77 g/mol mol AgBr available = 42.7 g AgBr x [1 mol / 187.77 g] = 0.2274 mol AgBr from the balanced equation the mole ratio AgBr : Na2S2O3 = 1 : 2 so mol Na2S2O3 required = 0.2274 mol AgBr x [ 2 mol Na2S2O3 / mol AgBr] = 0.455 mol Na2S2O3 (to 3 sig figs)
Silver Bromide
Element
Compound
Silver will have a +1 and bromate is -1 so they combine in a 1:1 ratio. The formula would be AgBrO3.
Silver Bromide
AgBr
Its yellow in colour
Образуется растворимое комплексное соединение: AgBr + 2 NH4OH -----> [Ag(NH3)2]+ + Cl- + 2 H2O.
AgCl.