A 0.4505 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 0.7514 g. What is the mass percentage of bromine in the original compound

70.98% is the mass percentage of bromine in the original compound

Explanation:

First, we need to convert mass of AgBr to moles with its molar mass (AgBr: 187.77g/mol). Moles of AgBr = Moles of Br. With molar mass of Br-: 79.904g/mol we can find mass of bromide and mass percentage as follows:

0.7514g AgBr * (1mol / 187.77g) = 0.00400 moles AgBr = Moles of Br-

0.00400 moles Br- * (79.904g / mol) = 0.320g Br- in the sample

Mass percentage is:

0.320g Br- / 0.4505g sample * 100

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation: