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Jusaquikie

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

  • one year ago
  • one year ago

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  1. Jusaquikie
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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.

    • one year ago
  2. aaronq
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    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.

    • one year ago
  3. Jusaquikie
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    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.

    • one year ago
  4. aaronq
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    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

    • one year ago
  5. Jusaquikie
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    so 3/.5

    • one year ago
  6. Jusaquikie
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    the answer is .370 M just not getting it.

    • one year ago
  7. aaronq
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    yeah thats right

    • one year ago
  8. aaronq
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    so 3/.5 this is wrong, but the answer is 0.37 M

    • one year ago
  9. Jusaquikie
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    i understand now thank you

    • one year ago
  10. aaronq
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    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

    • one year ago
  11. aaronq
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    okay good stuff

    • one year ago
  12. Jusaquikie
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    .08 + .105 / .5

    • one year ago
  13. Jusaquikie
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    thank you for your time it is appreciated.

    • one year ago
  14. aaronq
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    no problem

    • one year ago
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