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please help!; Assuming additivity of volumes, what is the concentration of KBr(in moles/L) in a solution prepared by mixing 2.50x10^2 of 0.0600 Molarity of KBr with 5.00x10^2 ml of 0.0830 Molarity?

Chemistry
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is the first one in mL too?
yea the 2.50x10^2 is in ml
I think this is how you do it. 1st: change the mL into L (since M is in mol/L) 2nd: calculate the number of moles of KBr in each solution (M * L from above) 3rd: add the 2 numbers for moles together 4th: add the 2 numbers for volume together 5th: divide since you now have total moles and total volume (M = moles/L)

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let me know what you get.
alright
just a way to check, your answer should be in-between the two given molarities since you are mixing them together and it should be a little closer to the larger molarity since you start with more of it.
alright i got a 5.2 Molarity does this sound right?
try again. It should be between .6M and .083M since those are the values you started with and you are mixing the same compound. What did you get for your 2 new volumes?
^sorry, .06M not .6
.25L and .5L
great! now what about for the moles of KBr found in each solution (molarity * liters = moles)?
i got .15 and 0.0415....
the first one should be .015...
alright hold on
ya your right its .015
alright! now when you add up your volumes what do you get? also what do you get when you add up the moles?
alright then i got .075 as final answer
correct! .0565mol/ .75L
thanks so much!!!
your very welcome!

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