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3 years ago

If 4 moles of O2 and 8 moles of N2 occupy a 12 L

Chemistry

1 answer:

IRISSAK [1]3 years ago

7 0

Answer:

P(N₂) = 4.02 atm

Explanation:

Number of moles of O₂ = 4 mol

Number of moles of N₂ = 8 mol

Total pressure = 6 atm

Partial pressure of N₂ = ?

Solution:

We will calculate the partial pressure by using mole friction.

P(N₂) = X(N₂)P(total)

X(N₂) = mol friction = moles of N₂ / total number of moles

X(N₂) = 8 mol / 12 mol = 0.67

Partial pressure of N₂:

P(N₂) = X(N₂)P(total)

P(N₂) = 0.67 × 6 atm

P(N₂) = 4.02 atm

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Atrazine is an herbicide widely used for corn and is a common groundwater pollutant in the corn-producing regions of the United

riadik2000 [5.3K]

Answer:

0.94

Explanation:

Let's assume that the total sample of the soil is 1 cm³; which has a porosity of 0.4 cm³; then the original sample of the soil will be 0.6 cm³

The mass of the soil particles in 1 cm³ of soil + water = 0.75 g

The fraction of the total atrazine can be calculated by the expression;

Total atrazine = $C_w(\frac{g}{mL})*V_w(mL)+C_{(soil)}(\frac{g}{soil})*M_{(soil)}(gsoil)$

Also; $C_{(soil)}(\frac{g}{soil})= K_{oc}(\frac{mL}{gOC})*f_{oc} (\frac{gOC}{gsoil})*C_w(\frac{g}{mL})$

NOTE THAT: $K_{oc}( \frac{mL}{gOC})=K_{oc}\frac{L}{kgOC}$

Total atrazine can now be calculated as:

$C_w*0.4+K_{oc}f_{oc}C_w*0.75$ = $\frac{C_{soil}*M_{soil}}{Total atrazine}$

$=\frac{K_{oc}f_{oc}C_w*M_{soil}}{C_w*0.4+K_{oc}f_{oc}C_w*0.75} = \frac{0.75*K_{oc}f_{oc}}{0.4+K_{oc}f_{oc}*0.75}$

$= \frac{10^{2.52}*0.025*0.75}{0.4+10^{2.52}*0.025*0.75}$

= 0.94

Thus,the fraction of total atrazine that will be adsorbed to the soil = 0.94.

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4 years ago