AlBr3 * 6H20
we will be happy to critique your thinking or work.
Calculate the percentage of water in each of the following hydrates.
Aluminum bromide hexahydrate
I just need help setting this up. I can work it from there
Would it be:
AlBr * 6H2O
Did I write it right?
3 answers
Al = 26.981
3Br = 239.727
12H = 12.096
6O = 95.994
Total = 374.798
Total H2O = 108.090
108.090/374.798 x 100% = 28.840%
Correct? are my sig figs messed up in anyway?
3Br = 239.727
12H = 12.096
6O = 95.994
Total = 374.798
Total H2O = 108.090
108.090/374.798 x 100% = 28.840%
Correct? are my sig figs messed up in anyway?
It is fine. The sig figures are ok, IF those figs in the individual elements are correct.