Question #586b5

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Question #586b5

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Answer 1 · 2017-08-06T00:28:41

Here's what I get.

For convenient reference, let's list the relative masses of these species.

$M_{r} : 14.01 m 17.03 m 46.01$ $M_text(r):14.01color(white)(m)17.03color(white)(m)46.01$
$m m m N m m l {NH}_{3} m l {NO}_{2}$ $color(white)(mmm)"N"color(white)(mml)"NH"_3color(white)(ml)"NO"_2$

Ammonia

(a) mg $N$ $"N"$ to mg ${NH}_{3}$ $"NH"_3$

$9 mg N \times \frac{{17.03 mg NH}_{3}}{14.01 mg N} = {11 mg NH}_{3}$ $"9 mg N" × "17.03 mg NH"_3/(14.01 "mg N") = "11 mg NH"_3$

$9 mg N/L = {11 mg NH}_{3} /L$ $"9 mg N/L" = "11 mg NH"_3"/L"$

(b) mg $N$ $"N"$ to mg ${NO}_{2}$ $"NO"_2$

$9 mg N \times \frac{{46.01 mg NO}_{2}}{14.01 mg N} = {30 mg NO}_{2}$ $"9 mg N" × "46.01 mg NO"_2/(14.01 "mg N") = "30 mg NO"_2$

$9 mg N/L = {30 mg NO}_{2} /L$ $"9 mg N/L" = "30 mg NO"_2"/L"$

Nitrate

(a) mg $N$ $"N"$ to mg ${NH}_{3}$ $"NH"_3$

$0.5 mg N \times \frac{{17.03 mg NH}_{3}}{14.01 mg N} = {0.6 mg NH}_{3}$ $"0.5 mg N" × "17.03 mg NH"_3/(14.01 "mg N") = "0.6 mg NH"_3$

$0.5 mg N/L = {0.6 mg NH}_{3} /L$ $"0.5 mg N/L" = "0.6 mg NH"_3"/L"$

(b) mg $N$ $"N"$ to mg ${NO}_{2}$ $"NO"_2$

$0.5 mg N \times \frac{{46.01 mg NO}_{2}}{14.01 mg N} = {1.6 mg NO}_{2}$ $"0.5 mg N" × "46.01 mg NO"_2/(14.01 "mg N") = "1.6 mg NO"_2$

$0.5 mg N/L = {1.6 mg NO}_{2} /L$ $"0.5 mg N/L" = "1.6 mg NO"_2"/L"$

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Question #586b5

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