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Ksivusya [100]
2 years ago
15

In the summer of 2016, the city of Columbus Dublin Road Water Treatment plan exceeded the regulatory level of nitrate, which is

10 ppm. What is the regulatory level concentration in:
a. mg NO3-/L
b. moles NO3-/L
c. mg-N/L

Alkaline water from a well in Owyhee County, Idaho contains carbonate (CO32-) at 24 mg/L and bicarbonate (HCO3-) at 111 mg /L. What is the total carbon concentration in mg C/L?
Chemistry
1 answer:
Verizon [17]2 years ago
8 0

Answer:

a) 10 mg NO_3^-/L

b) 1.61*10^{-4}mol NO_3^-/L

c) 2.26 mg N/L

Carbon: C=26.64 \frac{mg C}{L}

Explanation:

<u>Nitrate</u>

First of all, is important to know that:

1 ppm=1 mg/L

a) 10 ppm of nitrate (NO_3^-) is equal to 10 mg NO_3^-/L

b) The molecular weight of nitrate is 62 g NO_3^-/mol

10 mg NO_3^-/L=0.01 g NO_3^-/L

\frac{0.01 g NO_3^-/L}{62 g NO_3^-/mol}=1.61*10^{-4} mol NO_3^-/L

c) Nitrate has 14 mg of N per 62 mg of NO3

10 mg NO_3^-/L*\frac{14 mg N}{62 mg NO_3^-}=2.26 mg N/L

<u>Carbon</u>

Carbonate has 12 mg of C per 60 mg of CO_3^{-2}

Bicarbonate has 12 mg of C per 61 mg of HCO_3^{-}

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^{-}}

C=26.64 \frac{mg C}{L}

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tia_tia [17]

Answer:

d = 0.9 g/L

Explanation:

Given data:

Number of moles = 1 mol

Volume = 24.2 L

Temperature = 298 K

Pressure = 101.3 Kpa (101.3/101 = 1 atm)

Density of sample = ?

Solution:

PV = nRT     (1)

n = number of moles

number of moles = mass/molar mass

n = m/M

Now we will put the n= m/M in equation 1.

PV = m/M RT   (2)

d = m/v

PM = m/v RT ( by rearranging the equation 2)

PM = dRT

d = PM/RT

The molar mass of neon is = 20.1798 g/mol

d = 1 atm × 20.1798 g/mol / 0.0821 atm. L/mol.K × 273K

d = 20.1798 g/22.413 L

d = 0.9 g/L

4 0
3 years ago
Please help i will give brainliest!!! thank you =)
Anastasy [175]

Answer:

Answer choice B. 2

Explanation:

7 0
2 years ago
What is the name of the compound Ca(NO3)2 ?
natulia [17]

Answer: B) calcium nitrate

5 0
2 years ago
Chemical energy is released when molecular bonds are formed.<br> True<br> False
sveticcg [70]
The answer is false
6 0
2 years ago
If you have 12.5g of fluoride and 16.2g of sodium, which is the limiting reactant and how sodium fluoride in grams is your theor
Korvikt [17]

Answer:

F2 is the limiting reactant

27.6 grams of NaF is produced.

Explanation:

Balance the equation first.

2Na+ F2 ---> 2NaF

To find the limiting reactant, solve for how much NaF can be produced with Na and F2

12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)

=0.658 moles NaF

16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)

=0.705 moles NaF

Since F2 produced the least NaF, F2 is the limiting reactant.

Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.

0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF

27.6 moles of NaF would be theoretically produced.

8 0
3 years ago
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