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3 years ago

5

I AM GIVING BRAINLIEST! PLEASEEEEE HELP I NEED HELPPP PLEASEEEE How many atoms of nitrogen in 2.0 mol of dinitrogen monoxide, N2

O?(WITH STEPS)

1 answer:

sveta [45]3 years ago

8 0

Answer: 2.409 x 10^24 atoms of nitrogen

Explanation:

According to Avogadro

1 mole of any substance = 6.023 x10 ^23 atoms or molecules

1 molecules of N2O has 2 Nitrogen atoms

Number of mol of Nitrogen atom = 2 X number of moles of N2O

= 2 X 2.0 moles

= 4 moles

Number of atomns = number of moles x avogadro number

Number of atomns= 4.0 x 6.023 x 10 ^23= 2.409 x 10^24 atoms of nitrogen

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

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a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

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c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

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Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

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