Answer:
M=0.380 M.
Explanation:
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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:

Now, we compute the total moles of bromide:

Then, the total volume in liters:

Therefore, the concentration of total bromide is:

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if you were to fold a piece of paper into a bunch of pieces.
Explanation:
its just a different shape but it's still the same piece of paper
The correct answer is uneven
Energy has been added to the system for the reaction to occur.