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DanielleElmas [232]
3 years ago
6

14. Compare between the plasticity of edge and screw dislocations in BCC metals 15. Why does a screw dislocation in a BCC metal

needs high thermal activation to move?
Chemistry
1 answer:
Alenkasestr [34]3 years ago
4 0

Answer:

14) The edge dislocation is more plastic than the screw dislocation

15) So as to form kinks that are fast moving

Explanation:

14) Edge and screw dislocations are the two main types of mobile dislocations

The three dimensional core of the screw dislocation prevents the slipping of the layers (one over the other) in a BCC metal such that kinks are required to be formed first  by thermal activation (heating) in order. The kinks are edge dislocation that move such that the screw dislocation moves forward

Hence, the edge dislocation is more plastic than the screw dislocation

15) The three dimensional structure of a screw dislocation acts like a wedge which resists the slipping of the layers in the BCC structure such that the screw dislocation needs to be highly thermally activated forming kinks before the surrounding layers can move.

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It is a subscript

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How can criteria be used to help define the problem?
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Answer:

It helps get you to the proof of the matter

Explanation:

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3 years ago
58.0 g of K2SO4 was dissolved in 500 g of water. What is the molality of this solution?
podryga [215]

Answer: The molality of solution is 0.66 mole/kg

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Molarity=\frac{n\times 1000}{W_s}

where,

n = moles of solute

W_s = weight of solvent in g

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Now put all the given values in the formula of molality, we get

Molality=\frac{0.33\times 1000}{500g}=0.66mole/kg

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7 0
3 years ago
A reaction between substances Y and Z is
Anuta_ua [19.1K]

Answer : The value of activation energy for this reaction is 108.318 kJ/mol

Explanation :

The Arrhenius equation is written as:

K=A\times e^{\frac{-Ea}{RT}}

Taking logarithm on both the sides, we get:

\ln k=-\frac{Ea}{RT}+\ln A             ............(1)

where,

k = rate constant  = 2.95\times 10^{-3}L/mol.s

Ea = activation energy  = ?

T = temperature = 435 K

R = gas constant  = 8.314 J/K.mole

A = pre-exponential factor  = 3.00\times 10^{+10}L/mol.s

Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:

\ln (2.95\times 10^{-3}L/mol.s)=-\frac{Ea}{8.314J/K.mol\times 435K}+\ln (3.00\times 10^{10}L/mol.s)

Ea=108318.365J/mol=108.318kJ/mol

Therefore, the value of activation energy for this reaction is 108.318 kJ/mol

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3 years ago
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butalik [34]

Answer:

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2 years ago
Read 2 more answers
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