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earnstyle [38]
3 years ago
12

If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature of 1100˚C for 8 hours, wh

at is the depth below the surface at which the concentration is 1016 atoms/cm3 if the surface concentration is maintained at 1018 atoms/cm3? Use D = 2 x 10-12 cm2/s for aluminum diffusing in silicon at 1100˚C.
Chemistry
1 answer:
RideAnS [48]3 years ago
6 0

Answer:

8.354 nanometers

Explanation:

To treat a diffusive process in function of time and distance we need to solve  2nd Ficks Law. This a partial differential equation, with certain condition the solution looks like this:

\frac{C_{s}-C{x}}{C_{s}-C_{o}}=erf(x/2\sqrt{D*t})

Where Cs is the concentration in the surface of the solid

Cx is the concentration at certain deep X

Co is the initial concentration of solute in the solid

and erf is the error function

Then we solve right side,

\frac{C_{s}-C{x}}{C_{s}-C_{o}}=\frac{1018atoms/cm3-1016atoms/cm3}{1018atoms/cm3}=0.001964

And we need to look up the inverse error function of 0.001964 resulting in: 0.00174055

Then we solve for x:

x=0.00174055*2*\sqrt{D*t} =0.00174055*2*\sqrt{2*10^{-12}cm^{2}/s*8h*3600s/h}=8.35464*10^{-7}cm

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3 years ago
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gavmur [86]
<h2>Hello!</h2>

The answer is:

The new temperature will be equal to 4 K.

T_{2}=4K

<h2>Why?</h2>

We are given the volume, the first temperature and the new volume after the gas is compressed. To calculate the new temperature after the gas was compressed, we need to use Charles's Law.

Charles's Law establishes a relationship between the volume and the temperature at a gas while its pressure is constant.

Now, to calculate the new temperature we need to assume that the pressure is kept constant, otherwise, the problem would not have a solution.

From Charle's Law, we have:

\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

So, we are given the following information:

V_{1}=500mL\\T_{1}=20K\\V_{2}=100mL

Then, isolating the new temperature and substituting the given information, we have:

\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

T_{2}=\frac{T_{1}}{V_{1}}*V_{2} \\

T_{2}=\frac{20.00K}{500mL}*100mL\\

T_{2}=4K

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3 years ago
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Crazy boy [7]

Answer:

3.94 L

Explanation:

From the question given above, the following data were obtained:

Mass of O₂ = 5.62 g

Volume of O₂ =?

Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:

Mass of O₂ = 5.62 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mole of O₂ =?

Mole = mass / molar mass

Mole of O₂ = 5.62 / 32

Mole of O₂ = 0.176 mole

Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.

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