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Leviafan [203]
4 years ago
7

The activation energy for diffusion in BCC iron is 84 kJ/mol. Which of the following explains why the carbide precipitation acti

vation energy is slightly higher than the BCC diffusion activation energy?
a. Steel is FCC and the activation energy for FCC diffusion is higher
b. Precipitation requires the diffusional activation energy plus an additional energy to form the precipitate.
c. Precipitation requires less energy than diffusion
d. The carbide is orthorhombic which is a much higher energy than BCC
Engineering
1 answer:
k0ka [10]4 years ago
3 0

Answer:

  • B. Precipitation require the diffusional activation energy plus an additional energy to form the precipitate.

Explanation:

Precipitation is the creation of a solid from a solution. When the reaction occurs in a liquid solution, the solid formed is called the precipitate.

The formation of a precipitate indicates the occurrence of a chemical reaction.

Precipitation of carbide requires alot of energy which the diffusion activational energy alone cannot achieve and this was calculated to be 225.6 kJ/mol.

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3 years ago
Which equation can be used to find x, the length of the hypotenuse of the right triangle? A triangle has side lengths 63, 16, x.
Kryger [21]

Answer: 16 squared + 63 squared = x squared

Explanation:

Hi, since we have a right triangle we have to apply the Pythagorean Theorem:  

c^2 = a^2 + b^2  

Where c is the hypotenuse of the triangle (the longest side) and a and b are the other sides.  

Replacing with the values given:  

x^2 = 63^2 + 16^2  

So, the correct option is  

16 squared + 63 squared = x squared

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4 years ago
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vitfil [10]

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Explanation:

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3 years ago
Which of these materials are insulators? Select the THREE (3) that apply.
lubasha [3.4K]

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3 years ago
Air at 20 kPa and 5 °C enters a 1.5 cm diameter tube at a uniform velocity of 1.5 m/s. The tube walls are maintained at a unifor
ankoles [38]

Answer:

The distance from the entrance at which the flow becomes fully developed (entrance lenght) is:

L_{E}=1.752 m

Explanation:

First, we need to know if the flow is laminar or turbulent using the equation for the Reynolds number in a circular tube, which is:

Re=\frac{VD}{v}          (Equation 1)

We know that for

Re\leq 2300,   the flow is laminar

2300\leq Re\leq 1x10^{5},    the flow is turbulent

Then, tanking into account that for air at 20 kPa and 5°C, kinematic viscosity (v) is 1.252x10^{-5} \frac{m^{2}}{s} (taken from Table A-9, Cengel's book), we use the equation 1 ,

Re=\frac{(1.5 m/s)(0.015m)}{1.252x10^{-5}m^{2}/s}=1797.12

And, we can conclude that the flow is laminar. Then, we can use the relationship between the entrance length (L_{E}), which is the distance from the entrance at wich the flow becomes fully developed, and diameter for a laminar flow in a circular tube, which is:

\frac{L_{E}}{D}=0.065Re

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L_{E}=0.065ReD\\L_{E}=0.065(1797.12)(0.015 m)\\L_{E}=1.752m

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3 years ago
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