Silver plus barium yields no reaction.
When silver nitrate is added to barium chloride, a white precipitate of silver chloride forms because silver chloride is insoluble in water. This occurs due to a double displacement reaction where the silver ions from silver nitrate react with the chloride ions from barium chloride to form silver chloride. The remaining solution would contain barium nitrate as the other product of the reaction.
The spectator ions in the reaction between silver sulfate and barium nitrate are nitrate (NO3-) ions and sulfate (SO4^2-) ions. These ions do not participate in the formation of the precipitate (barium sulfate) and remain unchanged throughout the reaction.
The electron configuration for neutral Barium is [Xe] 6s2. Barium plus 2 means it has lost 2 electrons, so the electron configuration for Barium plus 2 would be [Xe].
The equation of the reaction is BaBr2 + 2 AgNO3 -> 2 AgBr + Ba(NO3)2. Therefore, exactly as many bromide ions from barium bromide must be supplied to precipitate any particular number of silver ion from silver nitrate. From the definition of molarity, 100 ml of 52 M solution contains 5.2 moles (preferably called "gram formula units") of silver nitrate. The gram formula unit mass of silver nitrate is 169.87, and each gram formula mass contains equal numbers of silver and of nitrate ions. Therefore, 5.2 gram elemental masses of bromide ions will be required for the precipitation. This amount of bromide ions can be supplied by 5.2/2* or 2.6 gram formula masses of barium bromide, and the gram formula unit mass of barium bromide is 297.14. Multiplying this number by 2.6 shows that 7.7 X 102 grams of barium bromide, to the justified number of significant digits, will be needed.
2 KBr + BaI2 ----> 2 KI + BaBr2
Silver chromate is not soluble in water.
When pure barium is added to a solution of silver acetate, a double displacement reaction occurs. Barium acetate and silver metal are formed as products. Additionally, barium sulfate may also form if sulfate ions are present in the solution.
The spectator ions are Ag+ and (NO3)-.
To balance the reaction between silver and barium, you first need to determine the chemical formulas of their compounds. Silver forms Ag+ ions while barium forms Ba2+ ions. When they react, they form silver nitrate (AgNO3) and barium chloride (BaCl2). The balanced equation is 2AgNO3 + BaCl2 -> 2AgCl + Ba(NO3)2. This is a double displacement reaction.
No.
9.11 g
When barium chloride reacts with silver acetate, a white precipitate of silver chloride is formed, along with barium acetate remaining in solution. This is because silver chloride is insoluble in water, while barium acetate is soluble.
6,36 g of silver chloride are obtained.
When silver nitrate is added to barium chloride, a white precipitate of silver chloride forms because silver chloride is insoluble in water. This occurs due to a double displacement reaction where the silver ions from silver nitrate react with the chloride ions from barium chloride to form silver chloride. The remaining solution would contain barium nitrate as the other product of the reaction.
The spectator ions in the reaction between silver sulfate and barium nitrate are nitrate (NO3-) ions and sulfate (SO4^2-) ions. These ions do not participate in the formation of the precipitate (barium sulfate) and remain unchanged throughout the reaction.
The electron configuration for neutral Barium is [Xe] 6s2. Barium plus 2 means it has lost 2 electrons, so the electron configuration for Barium plus 2 would be [Xe].
Barium chloride; BaCl2