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tekilochka [14]
3 years ago
7

-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63

26 nm and a density of 7.26 g/cm3. The atomic weight of manganese is 54.938 g/mol and the atomic radius is 0.112 nm. Determine the percent volume change that would occur if -Mn transforms to -Mn. (9 poi
Engineering
1 answer:
Fudgin [204]3 years ago
3 0

Answer:

The percentage volume change is -3.0%

Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn

Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

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Answer:

20, 083 L

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then converting this pounds to Kg by multiplying by  0.454 Kg/L one gets

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3 years ago
Block A has a weight of 8 lb. and block B has a weight of 6 lb. They rest on a surface for which the coefficient of kinetic fric
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Answer:

For block A, a = 9.66 ft/s²

For block B, a = 15 ft/s²

Explanation:

A free body diagram for this force system is attached to this solution

Mass of block A = m₁ = 8 lb

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Analysing Block A first

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N = W = 8 lb

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aₓ = 9.66 ft/s²

The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

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Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

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3 years ago
A harmonic oscillator with spring constant, k, and mass, m, loses 3 quanta of energy, leading to the emission of a photon.
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Answer: (a). E = 3.1656×10³⁴ √k/m  

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(c). Infrared region.

Explanation:

From Quantum Theory,

The energy of a proton is proportional to the frequency, from the equation;

E = hf

where E = energy in joules

h = planck's constant i.e. 6.626*10³⁴ Js

f = frequency

(a). from E = hf = 1 quanta

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E = 3hf = 3h/2π × √k/m

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E = 3.1656×10³⁴ √k/m    

(b). given from the question that K = 15 N/m

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f = 3h/2π × √k/m

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Paradox of Organizational Change: Engineering Organizations with Behavioral Systems Analysis. by. Maria E. Malott.

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