a 200.0 mL solution of .200 M Mg(no3)2 is mixed with a 300.0 mL solution of .350 M NaNO3 what is the concentration of the nitrate ion in the resulting solution

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i did the folowing \[M=\frac{ n }{ V }\]\[.2L*.2 Molarity= .04 grams\]\[.350L*.120 Molarity=.105=grams\] not sure where to go from here.

those would be moles, not grams. M=n/V, you're saying MxL=n, so moles. you in the first Mg(no3)2 , you have 2 times NO3 right? so you have twice as many moles. in the secondm NaNO3, only once so now add them up all together and divide by the TOTAL volume, to find the concentration in terms of molarity.

yes i messed that step up earlier i forgot that i had to m*MM to get grams. but i'm still not sure how to do this problem i get that the total volume would be .5L but what am dividing by this? i know what there are 2 in the first and 1 in the second.

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