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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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The substance called olivine may have any composition between Mg2SiO4 and Fe2SiO4, i.e. the Mg atoms can be replaced by Fe atoms
Ipatiy [6.2K]

Answer:

The answer is "0.147 nm and  99.63 mol %"

Explanation:

In point (a):

\to nk1 = 062

\to \text{Bragg angle} \theta =37.21^{\circ}

\to \text{diffraction angle} 2 \theta = 74.42^{\circ}

\to \lambda = 0.1790 nm

find:

d(062)=?

formula:

\to nx = 2d \sin  \theta

\to  d(062) = \frac{1 \times 0.1790^{\circ}}{2 \times \sin 37.21^{\circ}}\\

               = \frac{0.1790^{\circ}}{2 \times 0.604738126}\\\\= \frac{0.29599589}{2}\\\\= 0.147 \\

In point (b):

\to Mg_2SiO_4\longleftrightarrow  Fe_2SiO_4

d= 0.14774  \ \ \ \ \ olivine = 0.147 \ \ \ \  \ 0.15153

formula:

\to d=\frac{a}{\sqrt{n^2+k^2+i^2}}\\

that's why the composition value equal to 99.63 %

3 0
3 years ago
stimate the force required in extruding 70-30 brass at 700C, if the billet diameter is 125 mm and the extrusion ratio is 20
Jlenok [28]
U have to find the bill it into the internet in order for me to get the answer. Also u have to make it change into the call.
7 0
3 years ago
a steal bar with 1 in diameter has been subjected to a tensile force of 3 tons find tensile stress in the bar
Marysya12 [62]

Answer:

stress = 8,556 Psi.   or  (stress = 59 Mpa)

Explanation:

stress = force / area

force P = 3 tons (convert to lbs. for units consistency)

1 ton = 2240 lbs.

P = 6,720 lbs.

steel bar Diameter D = 1 in. (convert to d

Area of steel bar = (π *  1²) / 4 = 0.785 in²

therefore, stress = 6720 lbs. / 0.785 in²

stress = 8,556 Psi.

in Mpa ----- 8556 Psi * 0.00689476 MPa/Psi = 59 Mpa

6 0
3 years ago
Often an attacker crafts e-mail attacks containing malware designed to take advantage of the curiosity or even greed of the reci
Mademuasel [1]

Answer:

1) The ethical act Amy should display is to open the Email since it came in through her personal inbox. However, further action should not be taken since ordinarily opening the mail will not disclose or corrupt her system either locally of her system on the server and web space.

2)  If such email came in, the first action i will take is to open the mail then take a pause to read through the instruction because most scam or malware comes with an array of further instructions.

After carefully reading through, i will call up Davey to confirm if he sent such mail which answer will obviously be a No and in the case where i am unable to get a call through to Davey, i will immediately delete the mail from my inbox to avoid the mistake of clicking any embedded link within the mail.

5 0
3 years ago
Air initially at 15 psla and 60 F is compressed to 75 psia and 400 F. The power input to air under steady state condition is 5 h
Triss [41]

Answer:\dot{m}=3.46lbm/min

Explanation:

Initial conditions

P_1=15 psia

T_1=60 F^{\circ}

Final conditions

P_2=75 psia

T_2=400F^{\circ}

Steady flow energy equation

\dot{m}\left [ h_1+\frac{v_1^2}{2}+gz_1\right ]+\dot{Q}=\dot{m}\left [ h_2+[tex]\frac{v_2^2}{2}+gz_2\right ]+\dot{W}

\dot{m}\left [ c_pT_1+\frac{0^2}{2}+g0\right ]+\dot{Q}=\dot{m}\left [ c_pT_2+\frac{0^2}{2}+g0\right ]+\dot{W}

\dot{m}c_p\left [ T_1-T_2\right ]+\left [ -5hp\right ]=\dot{W} -5\times 746\times 3.4121

-4\dot{m}-\dot{m}\times 0.24\times \left [ 400-60\right ]

-81.6\dot{m}-4\dot{m}=-4.949 BTU/sec

\dot{m}=0.057821lbm/sec

\dot{m}=3.46lbm/min

3 0
3 years ago
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