A student weighed 0.2440 grams of sample of a recently synthesized cobalt amine bromide compound. After dissolving in 120mL of acidified water, she added 2.55 grams of AgNO3 to precipitate any free Br-. The precipitate was found to weigh 0.3700 grams. Calculate the %Br- in the sample. MW Cl- 35.45 g/mol; MW Br- 79.91 g/mol; MW Ag 107.87 g/mol

The bromine percentage is 49.42 %

Explanation:

Mass of AgBr precipitated =   0.31 g

Molecular weight of AgBr =  187.77 g mol - 1

moles of AgBr =   0.31 g by 187.77 g mol - 1 =   0.0016509 moles

moles of Br ion =  0.0016509 moles

Mass of Br ion =   0.0016509 moles into 79.904 g mol - 1 =  0.1319 g

The Total mass of compound = 0.2669 g

Percentage of bromine = 0.1319 by 0.2669 = 49.42 percent

Hence the bromine percentage is 49.42 %

your answer is c) plasma

~*"auba14"*~

the answer is c.the valence of the atoms combine .