CuSO4(aq) + BaCl2(aq) -> CuCl2(aq) + BaSO4(s)
(NH4)2SO4(aq) + BaCl2(aq) → 2NH4Cl(aq) + BaSO4(s). Barium sulfate is the precipitate.
Yes. 2HCl + Ba(OH)2 --> BaCl2 + 2H2O
If an aqueous solution of copper(II) sulfate is contacted with metallic iron, at the least the surface of the iron passes into solution and is replaced by a layer of metallic copper. This is an example of displacement by a metal higher in the electromotive series than the metal it displaces.
Ag+Ba yields no reaction.
The equation (not "formula") is 2 AgNO3 + FeCl2 -> 2 AgCl + Fe(NO3)2 for iron (II) chloride and 3 AgNO3 + FeCl3 -> 3 AgCl + Fe(NO3)3 for iron (III) chloride.
(NH4)2SO4(aq) + BaCl2(aq) → 2NH4Cl(aq) + BaSO4(s). Barium sulfate is the precipitate.
The balanced equation for sodium sulfate (Na2SO4) plus barium chloride (BaCl2) yielding barium sulfate (BaSO4) and sodium chloride (NaCl) is: Na2SO4 + BaCl2 -> BaSO4 + 2NaCl
barium chloride plus sodium sulphate yields barium sulphate plus sodium chloride
Sulfuric acid plus copper (II) nitrate yields nitric acid plus copper (II) sulfate. Sulfuric acid plus copper (I) nitrate yields nitrous acid plus copper (I) sulfate.
Yes. 2HCl + Ba(OH)2 --> BaCl2 + 2H2O
Pb(NO3)2 + CuSO4 -> PbSO4 + Cu(NO3)2
Fe + CuSO4 --> FeSO4 + Cu Iron + Copper (II) Sulfate yields Iron (II) Sulfate and Copper
the answer to your little riddle is magnesium sulphate and hydrogen chloride gas not hydrogen chlorate you nugget
Silver plus barium yields no reaction.
If an aqueous solution of copper(II) sulfate is contacted with metallic iron, at the least the surface of the iron passes into solution and is replaced by a layer of metallic copper. This is an example of displacement by a metal higher in the electromotive series than the metal it displaces.
Balanced equation: CuO + 2HCl --> CuCl2 + H2O Word equation: One mole of copper (II) oxide plus two moles of hydrochloric acid produces (or yields) one mole of copper (II) chloride plus one mole of water.
no