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How many grams ofBr-were in a sample that produced 1.000 g of

AgBr precipitate (FM 5 187.77)?

Short Answer

Expert verified

The number of bromide ions generated in a sample that can cause silver bromide precipitation is 0.4255g

Step by step solution

01

Define molarity

The ratio of moles of solute to the volume of the solution is known as molarity (in litres).

Molartiv(M) = Moles of solute(inmol)

02

Find the mass of  AgBr

The amount of bromide ions generated in a sample that can result in silver bromide precipitation.

The number of moles of silver bromide is computed using 1.00 g of AgBr

The number of moles of silver bromide to the number of moles of bromide ion is 1 : 1

Molarmass=MassMole=79.904g/mol×0.00533=0.4255g

Therefore, the amount of bromide ions formed in a sample that can produce silver bromide precipitate is 0.4255 g.

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Most popular questions from this chapter

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