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Scorpion4ik [409]
2 years ago
14

(a) what is the approximate ductility (%el) of a brass that has a yield strength of 230 mpa (33360 psi)?

Chemistry
1 answer:
ss7ja [257]2 years ago
3 0

Based on the calculations, the approximate ductility (%el) of this brass is equal to 2.3%.

<u>Given the following data:</u>

  • Yield strength = 230 mpa (33360 psi).

<h3>What is ductility?</h3>

Ductility can be defined as an important property of a material which determines its ability to become elongated due to the application of stress.

Mathematically, the ductility of a material can be expressed as percentage elongation in length:

\% el = \frac{\Delta L}{L_i} =\frac{L_f - L_i}{L_i}

<u>Where:</u>

  • L_i is the original length.
  • L_f is the final length.
  • \Delta L is the yield strength.

For this exercise, let us assume the original length of this brass is equal to 100 meters.

Substituting the parameters into the formula, we have;

\% el = \frac{230}{100}

Ductility = 2.3%.

Read more on ductility here: brainly.com/question/828860

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The answer is D - near the roof.

Explanation:

Attics don't have a ton of insulation or hard cinder-block to block the goggles form seeing through it. This is why the attic can be a place where the temperature varies with the outside temperature.

5 0
3 years ago
The. bond dissociation enthalpies of the H-H bond and the H-Cl bond are 435 kJ mol^-1 and 431 kJ mol^-1, respectively. The ΔHfO
Novay_Z [31]

The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.

<h3>What is the dissociation enthalpy?</h3>

Given that;

H-H bond energy =  435 kJ mol^-1

H-Cl bond energy = 431 kJ mol^-1

ΔHfO of HCL(g) = -92kJ mol^-1

Bond dissociation enthalpy of the Cl-Cl bond = x

-92 = 435  +  431 + x

x = -92 - (435  +  431)

x = -958 kJ mol^-1

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6 0
1 year ago
A mixture of hydrogen and argon gases is maintained in a 6.47 L flask at a pressure of 3.43 atm and a temperature of 85 °C. If t
11Alexandr11 [23.1K]

Answer:

The mixture contains 8.23 g of Ar

Explanation:

Let's solve this with the Ideal Gases Law

Total pressure of a mixture = (Total moles . R . T) / V

We convert T° from °C to K → 85°C + 273 = 358K

3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L

(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles

0.756= Total moles from the mixture

Moles of Ar + Moles of H₂ = 0.756 moles

Moles of Ar + 1.10 g / 2g/mol = 0.756 moles

Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206

We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g

8 0
3 years ago
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