Asked by Lisa
Determine the molarity of nitrate ions in a solution prepared by mixing 25.0 mL of 0.50 M Fe(NO3)3 and 35.0 mL of 1.00 M Mg(NO3)2 Please explain fully do not know where to start?
Answers
Answered by
DrBob222
You start with the definition of molarity. M = moles/L.
moles Fe(NO3)3 = M x L = ??
moles NO3^- will be 3x that.
moles Mg(NO3)2 = M x L = ??
moles NO3^- will be 2x that.
Then M NO3^- = total moles NO3^-/total volume in L.
moles Fe(NO3)3 = M x L = ??
moles NO3^- will be 3x that.
moles Mg(NO3)2 = M x L = ??
moles NO3^- will be 2x that.
Then M NO3^- = total moles NO3^-/total volume in L.
(0.50)(25.0) = 12.5
(1.00)(35.0) = 35
47.5/60 = 0.79M is my answer but my book says
1.79M is the correct answer
(1.00)(35.0) = 35
47.5/60 = 0.79M is my answer but my book says
1.79M is the correct answer
Answered by
DrBob222
<b>You didn't follow any of my instructions. I have typed my bolded instructions below each of your lines.</b>
0.50)(25.0) = 12.5
<b>moles Fe(NO3)3 = M x L = ??
moles NO3^- will be 3x that.
M x L = 0.50M x 0.025 L(not 25 mL) = 0.0125 moles Fe(NO3)3
Then nitrate is 3x that or 0.0125*3 = 0.03750 moles nitrate ion</b>
(1.00)(35.0) = 35
<b>moles = M x L = 1.00M x 0.0350 L (not mL) = 0.0350 moles Mg(NO3)2.
Then nitrate is twice that or
2*0.0350 = 0.0700 M in nitrate.</b>
47.5/60 = 0.79M is my answer but my book says
<b>M nitrate = total moles/total volume in L. (0.0375+0.0700)/0.06L = 1.79 M in nitrate.</b>
1.79M is the correct answer
<b>So it is</b>
0.50)(25.0) = 12.5
<b>moles Fe(NO3)3 = M x L = ??
moles NO3^- will be 3x that.
M x L = 0.50M x 0.025 L(not 25 mL) = 0.0125 moles Fe(NO3)3
Then nitrate is 3x that or 0.0125*3 = 0.03750 moles nitrate ion</b>
(1.00)(35.0) = 35
<b>moles = M x L = 1.00M x 0.0350 L (not mL) = 0.0350 moles Mg(NO3)2.
Then nitrate is twice that or
2*0.0350 = 0.0700 M in nitrate.</b>
47.5/60 = 0.79M is my answer but my book says
<b>M nitrate = total moles/total volume in L. (0.0375+0.0700)/0.06L = 1.79 M in nitrate.</b>
1.79M is the correct answer
<b>So it is</b>
Answered by
Anonymous
so I times 0.0125 by 3 b/c of the 3 at the end of the bracket?
and I times 0.0350 by 2 b/c of the 2 at the end of the bracket?
but why because when i'm doing my molar mass i do include that last # in the molar mass so why do we times it by that #?
and I times 0.0350 by 2 b/c of the 2 at the end of the bracket?
but why because when i'm doing my molar mass i do include that last # in the molar mass so why do we times it by that #?
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