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ladessa [460]
3 years ago
8

A 0.8838-g sample of an ionic compound containing bromide ions and an unknown metal cation is dissolved in water and treated wit

h an excess of AgNO3. If 1.573 g of a AgBr precipitate forms, what is the percent by mass of Br in the original compound?
Chemistry
2 answers:
Anestetic [448]3 years ago
7 0

Answer:

%Br=75.7%

Explanation:

Hello,

From the AgBr precipitate, one could compute one can compute the bromine grams as follows:

1.573gAgBr*\frac{80gBr}{188gAgBr}=0.669gBr

As long as there was an excess of silver nitrate, one knows that into the 1.573 g of AgBr, 0.669 g correspond to the bromine that was initially contained into the 0.8838-g sample, thus, the percent is computed as follows:

%Br=\frac{0.669gBr}{0.8838g}*100%

%Br=75.7%

Best regards.

Alex17521 [72]3 years ago
4 0
Wow this is really hard dude
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