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Nadya [2.5K]
3 years ago
5

please i really need help, i dont understand any of this, if you could do the work for all and show me how to do it that would b

e amazing please.

Chemistry
1 answer:
AVprozaik [17]3 years ago
8 0
I can hardly see your paper sorry
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Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
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Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

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