The molecular equation is 3Ba(NO3)2(aq) + 2(NH4)3PO4(aq) ==> Ba3(PO4)2(s) + 6NH4NO3(aq)The spectator ions are NH4^+ and NO3^-
Overall reaction including spectator ions: BaCl2(aq)+K2SO4(aq) --> BaSO4(s)+2KCl(aq) Ionic Reaction: Ba+2(aq) + 2Cl-(aq) + 2K+(aq) + SO4-2(aq) --> BaSO4(s) + 2K+(aq) + 2Cl-(aq) Note that the BaSO4(s) is insoluble. Net Ionic: Ba+2(aq) + SO4-2(aq) --> BaSO4(s) The net ionic equation eliminates the ions that just stay in solution and do not contribute to the precipitate of the barium sulfate.
it is a strong electrolyte because it occurs in strong bases and that's why it is bonded between two different elments
9 moles (there are four oxygen atoms for every mol of BaSO4, so you multiply 2.25 by 4)
Barium nitrate is soluble in water.
This is barium nitrate, an inorganic ionic salt.
Overall reaction including spectator ions: BaCl2(aq)+K2SO4(aq) --> BaSO4(s)+2KCl(aq) Ionic Reaction: Ba+2(aq) + 2Cl-(aq) + 2K+(aq) + SO4-2(aq) --> BaSO4(s) + 2K+(aq) + 2Cl-(aq) Note that the BaSO4(s) is insoluble. Net Ionic: Ba+2(aq) + SO4-2(aq) --> BaSO4(s) The net ionic equation eliminates the ions that just stay in solution and do not contribute to the precipitate of the barium sulfate.
When barium nitrate (Ba(NO₃)₂) and potassium chromate (K₂CrO₄) are mixed in a solution, a double displacement reaction occurs, resulting in the formation of barium chromate (BaCrO₄), which is a yellow precipitate, and potassium nitrate (KNO₃), which remains in solution. The reaction can be represented as: [ \text{Ba(NO}_3\text{)}_2 + \text{K}_2\text{CrO}_4 \rightarrow \text{BaCrO}_4 (s) + 2 \text{KNO}_3 ] The formation of the bright yellow precipitate is often used as a demonstration of a precipitation reaction.
it is a strong electrolyte because it occurs in strong bases and that's why it is bonded between two different elments
Correctly it should be written as 'Ba(NO3)2 ' and it is barium nitrate. Notice the use of brackets and the '2' , to indicate that there are two nitrate anions combined to the one barium cation. NB As you gave it, it does not make sense as a chemical formula.
To find the amount of AgCl formed, we first need to calculate the limiting reagent. This is done by converting the given masses of AgNO3 and BaCl2 to moles, determining the mole ratio between them, and then the limiting reagent based on the smaller value. Once the limiting reagent is determined, use the mole ratio from the balanced chemical equation to find the moles of AgCl formed and then convert that to grams.
To find the mass of 0.625 moles of Ba(NO3)2 (barium nitrate), first calculate its molar mass. The molar mass of Ba(NO3)2 is approximately 137.33 g/mol (for Ba) + 2 × (14.01 g/mol for N) + 6 × (16.00 g/mol for O), totaling about 261.34 g/mol. Thus, the mass of 0.625 moles is 0.625 moles × 261.34 g/mol ≈ 163.35 grams.
9 moles (there are four oxygen atoms for every mol of BaSO4, so you multiply 2.25 by 4)