Answer:
a) [tex]10 mg NO_3^-/L[/tex]
b) [tex]1.61*10^{-4}mol NO_3^-/L[/tex]
c) [tex]2.26 mg N/L[/tex]
Carbon: [tex]C=26.64 \frac{mg C}{L}[/tex]
Explanation:
Nitrate
First of all, is important to know that:
[tex]1 ppm=1 mg/L[/tex]
a) 10 ppm of nitrate ([tex]NO_3^-[/tex]) is equal to [tex]10 mg NO_3^-/L[/tex]
b) The molecular weight of nitrate is [tex]62 g NO_3^-/mol[/tex]
[tex]10 mg NO_3^-/L=0.01 g NO_3^-/L[/tex]
[tex]\frac{0.01 g NO_3^-/L}{62 g NO_3^-/mol}=1.61*10^{-4} mol NO_3^-/L[/tex]
c) Nitrate has 14 mg of N per 62 mg of NO3
[tex]10 mg NO_3^-/L*\frac{14 mg N}{62 mg NO_3^-}=2.26 mg N/L[/tex]
Carbon
Carbonate has 12 mg of C per 60 mg of [tex]CO_3^{-2}[/tex]
Bicarbonate has 12 mg of C per 61 mg of [tex]HCO_3^{-}[/tex]
[tex]C=24 \frac{mg CO_3^{-2}}{L}*\frac{12 mg C}{60 mg CO_3^{-2}}+111 \frac{mg HCO_3^{-}}{L}*\frac{12 mg C}{61 mg HCO_3^{-}}[/tex]
[tex]C=26.64 \frac{mg C}{L}[/tex]