Answer:
%[tex]Br=75.7[/tex]%
Explanation:
Hello,
From the AgBr precipitate, one could compute one can compute the bromine grams as follows:
[tex]1.573gAgBr*\frac{80gBr}{188gAgBr}=0.669gBr[/tex]
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:
%[tex]Br=\frac{0.669gBr}{0.8838g}*100[/tex]%
%[tex]Br=75.7[/tex]%
Best regards.
