the Molecular equation would be AgNO3 + Na2CrO4 =yields= Ag2CrO4 + NaNO3 On the product side the silver cromate is insoluble, therefore it will form a precipitate remember everything is aquous (aq) except the silver cromate it is solid (s) Net Ionic equation is Ag(aq) + CrO4 -2 (aq) > Ag2CrO4(s)
balance equation of barrium chloride to sodium chromate
Ag2CrO4 should form a clay red precipitate which doesn't dissolve in acid.
Agno3+Nacl-
Equation: NaI + AgNO3 ----> NaNO3 + AgI
Produces Silver iodide precipitate and Sodium nitrate
the word chemical equation islead nitrate + sodium chromate --> sodium nitrate + lead chromate
the word chemical equation islead nitrate + sodium chromate --> sodium nitrate + lead chromate
its already balanced
AgNO3+NaBr--->NaNO3+AgBr
87
balance equation of barrium chloride to sodium chromate
Ag2CrO4 should form a clay red precipitate which doesn't dissolve in acid.
NaCl+AgNO3=NaNO3+AgCl Because all the elements in this equation have a +1 or -1 charge, no coefficents are needed
The net ionic equation is Ag+(aq) + I- --> AgI(s) The sodium (Na+) and nitrate (NO3-) are spectators as always.
The equation for the reaction between silver nitrate and sodium iodide is AgNO3 + NaI -> AgBr + NaNO3. The gram formula masses are 169.87 for silver nitrate, 149.89 for sodium iodide, and 84.99 for sodium nitrate. Therefore, 1.7 g of silver nitrate constitutes 1.7/169.87 or 0.010 formula mass of silver nitrate and 1.5 g of sodium iodide constitutes 1.5/149.89 or 0.010 mole of sodium iodide, to the justified number of significant digits. The reaction equation shows that the number of formula unit masses of each reactant and product are the same, so that there will be 0.85 g of sodium nitrate produced, to the justified number of significant digits.
Agno3+Nacl-
Silver chloride and sodium nitrate.