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ra1l [238]
3 years ago
14

What are the principal alloy elements of an AISI 4340 steel? How much carbon does it have? Is it hardenable? By what techniques?

Chemistry
1 answer:
amid [387]3 years ago
3 0

Answer:

The principal elements are Iron, Nickel, Chromium

It has 0.37% to 0.43% Carbon

It is hardenable.

It can be hardened by cold working, annealing, quenching.

Explanation:

A)

The chemical composition of AISI 4340 Steel are as follows:

Iron  ---  95% to 96%

Nickel  ---  1.6% to 2.0%

Chromium  ---  0.7% to 0.9%

Manganese  ---  0.6% to 0.8%

Carbon  ---  0.37% to 0.43%

Molybdenum  ---  0.2% to 0.3%

Silicon  ---  0.15% to 0.30%

Sulfur  ---   0.040%

Phosphorous  --- 0.0350%

So, the principal alloy elements from this composition are <u>Iron, Nickel and Chromium</u>

B)

The carbon content in this alloy is <u>0.37% to 0.43%</u>

C)

<u>Yes, it can be hardened</u>.

D)

<u>It can be Hardened by Cold Working, Annealing or Quenching</u>.

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The concentration of protein in a urine sample is calculated to be 2.77 μg/mL. What is the concentration of this solution in uni
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Answer:

The concentration of this solution in units of pounds per gallon is 2.776*10^{-5} \frac{lb}{gal}

Explanation:

Units of measurement are established models for measuring different quantities. The conversion of units is the transformation of a quantity, expressed in a certain unit of measure, into an equivalent one, which may or may not be of the same system of units.

In this case, the conversion of units is carried out knowing that 1 μg are equal to 2.205*10⁻⁹ Lb and  1 mL equals 0.00022 Gallons. So

2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}

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So, 2.77 μg= 6.10785*10⁻⁹ lb

Then:

2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}=\frac{6.10785*10^{-9}lb }{mL} =\frac{6.10785*10^{-9}lb }{0.00022 gal} =\frac{6.10785*10^{-9}lb }{0.00022 gal}

You get:

2.77 \frac{ug}{mL} = 2.776*10^{-5} \frac{lb}{gal}

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Answer:

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Explanation:

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