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Ksivusya [100]
3 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]3 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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41Ca decays by electron capture. The product of this reaction undergoes alpha decay. What is the product of this second decay re
dimaraw [331]

Answer:

Cl

Explanation:

⁴¹Ca has an atomic number equal to 20, it means that it has 20 protons and 20 electrons ad its neutral state. In the decay by electron capture, it will lose one electron and will become a cation, but the mass (41) and the atomic number will remain the same.

When Ca⁺ undergoes alpha decay, it will lose an alpha particle, which has mass 4 and 2 protons.

⁴¹₂₀Ca⁺ → ³⁷₁₈X⁺ + ⁴₂α

To be stable, X will lose a proton and will become ³⁷₁₇X. The element which has atomic number 17 is chlorine, Cl.

4 0
3 years ago
What is the mole fraction of KCI in a
Margaret [11]

Answer:

Mole fraction of KCl = 0.4056

Explanation:

We'll begin by calculating the number of mole of each compound. This can be obtained as follow:

For NaCl:

Mass NaCl = 0.564 g

Molar mass of NaCl = 58.44 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 0.564 / 58.44

Mole of NaCl = 0.0097 mole

For KCl:

Mass KCl = 1.52 g

Molar mass of KCl = 74.55 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1.52 / 74.55

Mole of KCl = 0.0204 mole

For LiCl:

Mass LiCl = 0.857 g

Molar mass of LiCl = 42.39 g/mol

Mole of LiCl =?

Mole = mass /Molar mass

Mole of LiCl = 0.857 / 42.39

Mole of LiCl = 0.0202 mole

Next, we shall determine the total mole in the mixture. This can be obtained as follow:

Mole of NaCl = 0.0097 mole

Mole of KCl = 0.0204 mole

Mole of LiCl = 0.0202 mole

Total mole =?

Total mole = Mole of NaCl + Mole of KCl + Mole of LiCl

Total mole = 0.0097 + 0.0204 + 0.0202

Total mole = 0.0503 mole

Finally, we shall determine the mole fraction of KCl in the mixture. This can be obtained as follow:

Mole of KCl = 0.0204 mole

Total mole = 0.0503 mole

Mole fraction of KCl =?

Mole fraction of KCl = Mole of KCl /Total mole

Mole fraction of KCl = 0.0204 / 0.0503

Mole fraction of KCl = 0.4056

4 0
3 years ago
Read 2 more answers
What is the mass (in mg) of 2.63 moles of nickel?
zaharov [31]
Data:
Molar Mass of Nickel = 58,7 g/mol

Solving:
58,7 g → 1 mol
y -------→ 2.63 mol

Solving: (They are proportional measures, the rule of three is made (directly proportional)

\frac{58.7}{y} = \frac{1}{2.63}
multiply cross
1*y = 58.7*2.63
y = 154.381\:g \stackrel{converting}{\longrightarrow}\:\boxed{y = 154381\:mg}


8 0
3 years ago
If 3.25 mol of Ar occupies 100. L at a particular temp and pressure, what volume does
gtnhenbr [62]

Answer:

435.38 L

Explanation:

From the question,

Applying ideal gas equation,

PV = nRT................... Equation 1

Where P = Pressure, V = Volume, n = number of moles, R = molar gas constant, T = Temperature.

Since Temperature and Pressure were constant,

V ∝ n

V/n = V'/n'.................... Equation 2

Where V and V'  = the initial and final Volume of Ar respectively, n and n'  = the initial and final moles of Ar respectively

make V' the subject of the equation

V' = Vn'/n............... Equation 3

Given: V = 100 L, n = 3.25 mol, n' = 14.15

Substitute into equation 3

V' = (100×14.14)/3.25

V' = 435.38 L

8 0
3 years ago
Write the equilibrium constant expression for this reaction: nh4
pochemuha
Kc= (nh4)
--------
(nh3) + (h2o)
4 0
3 years ago
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