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GenaCL600 [577]
1 year ago
11

What are the distinguishing features of each extraction process: pyrometallurgy, electrometallurgy, and hydrometallurgy? Explain

briefly how the types of metallurgy are used in the production of(a) Fe;
Chemistry
1 answer:
Tatiana [17]1 year ago
7 0
<h3>1.Pyrometallurgy</h3>
  • This techniques involves the high temperature at which the Chemical reactions took place between the gases and solids.
  • Pyrometallurgy involves calcination, Roasting and smelting process.

 

<h3>  2.Electrometallurgy</h3>
  • This technique involves the metallurgy operations that took place in electrolytic cell.
  • Electro- winning and Electro-refining are some of the electrometallurgy process

<h3>   3. Hydrometallurgy</h3>
  • Reactions involves aqueous solutions to extract metal, use this technique.
  • Hydrometallurgy involves precipitation , distillation , adsorption and solid extraction operations.

Now,

Type of metallurgy are used in the production of Fe is Roasting and Smelting.

<h3>What is Roasting?</h3>
  • Ore is heated strongly with other substances, usually with Oxygen.
  • Process temperature is below the melting point.
  • Some of Impurities is removed as volatile substance.

In this reaction,

 FeO changes to Fe₂O₃ to prevent the loss of iron during smelting.

4FeO +O₂→ Fe₂O₃.

Now,

<h3>   What is Smelting?</h3>
  • The oxides of less electropositive metals like Pb, Zn, Fe, Sn, etc.
  • These are reduced by strongly heating with coal or coke.
  • Reduction of the oxide with Carbon at high temperature is known as Smelting.

Thus from the above conclusion we can say that , metallurgy are used in the production of Fe is Roasting and Smelting.

Learn more about Iron here: brainly.com/question/2535311

#SPJ4

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Answer:

As the temperature of a fixed volume of a gas increase the pressure will increase.

Explanation:

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As the temperature increase, the pressure also increase.

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
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Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

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n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

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