Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

they're asking you for the concentration (I'm assuming in molarity) of the NO3^- ion.

so M= (total number of moles in solution)/0.5L

Answer 5

so 3/.5

Answer 6

the answer is .370 M just not getting it.

Answer 7

yeah thats right

Answer 8

so 3/.5 this is wrong, but the answer is 0.37 M

Answer 9

i understand now thank you

Answer 10

so from the first you have 0.2M (0.2L) = 0.04 moles x 2 = 0.08 moles
from the second you have 0.35M(0.3 L) = 0.105 moles

M=(0.08+0.105)/0.5L

Answer 11

okay good stuff

Answer 12

.08 + .105 / .5

Answer 13

thank you for your time it is appreciated.

Answer 14

no problem