Abeaker holds xliters of 0.2m albr3. give answers to parts a through din terms of x. a) how many moles of aluminum ions are there in this solution? b) how many moles of bromide ions are there in this solution? c) how much water must you add if want to dilute the original solution to a concentration of 0.02 m?

A)

B)

C)x L - 10*x L

Explanation:

A) and B)

Firs, you need to determine the amount moles of aluminum and bromine in one mole of AlBr3. For that you must look at the subindices that appear after each element in the compound. So, you have 1 mol of aluminum and 3 mol of bromine in one mol of AlBr3.

Now, you have 0.2 M of AlBr3, this means that in 1 L of solution you have 0.2 moles of AlBr3.   And to calculate the amount of moles of aluminum and bromine in 0.2 moles you multiply this coefficient by the amounts previously calculated, these are 1 mol of aluminum and 3 mol of bromine. So you have 0.2 moles of aluminum and 0.6 moles of bromine in 0.2 moles of AlBr3.

So, now you know that in 1 L of solution you have 0.2 moles of AlBr3, in consecuense in 1 L of solution you have 0.2 moles of aluminum and 0.6 moles of bromine.

Now you can apply a rule of three: If in a liter of solution you have 0.2 moles of aluminum, how many moles of aluminum are in x liters of solution?

If in a liter of solution you have 0.6 moles of bromine, how many moles of bromine are in x liters of solution?

C)

First you must determine the amount of moles of AlBr3 in x liters of solution. For that you can apply a rule of three: If in a liter of solution you have 0.2 moles of AlBr3, how many moles of AlBr3 are in x liters of solution?

When you add a solvent like water, the moles of AlBr3 you have before and after adding it are the same. So, knowing that the new concentration is 0.02 M (in 1 L of solution are 0.02 moles of AlBr3) and applying the following rule of three you can determinate the final volume of the solution:

If in a liter of solution you have 0.02 moles of AlBr3, how many liters of solution contain xmoles of AlBr3?

Your initial solution had a volume of x L, so the difference between x L and the previous expression will be the volume of water you added to get the diluted solution: x L - 10*x L

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.