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2 years ago

7

A sample of cast iron with .35 wt%C is slow cooled from 1500C to room temperature. What is the fraction of proeutectoid Cementit

e phase in this material?

1 answer:

Marizza181 [45]2 years ago

6 0

Answer:

So the fraction of proeutectoid cementite is 44.3%

Explanation:

Given that

cast iron with 0.35 % wtC and we have to find out fraction of proeutectoid cementite phase when cooled from 1500 C to room temperature.

We know that

Fraction of proeutectoid cementite phase gievn as

${w_p}'=\dfrac{{C_o}'-0.022}{0.74}$

Now by putting the values

${w_p}'=\dfrac{0.35-0.022}{0.74}$

${w_p}'=0.443$

So the fraction of proeutectoid cementite is 44.3%

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Answer:

work will be positive when it is under polytropic expansion process

Explanation:

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Also work during the process of polytropic is given as

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A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m

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Answer:

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Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

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$= 0.003408\ mol \times 98g/mol$

= 0.333984 g

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We simply applied the above formulas

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