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GenaCL600 [577]
2 years 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]2 years 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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When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular
Luda [366]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=1.03g

Mass of H_2O=0.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

Thus, the empirical and molecular formula for the given organic compound is C_3H_2O and C_6H_4O_2

6 0
3 years ago
Granite is a ________ that contains the mineral quartz. a. mineral c. rock b. glass d. mica
kow [346]
I think it would be C

7 0
3 years ago
Read 2 more answers
How many H atoms in 12H2O?
Ilya [14]

Answer:

24 atoms of H  or 1.4 x 10²⁵ hydrogen atoms

Explanation:

simple method

1 H₂O has  atoms of hydrogen and 1 atom of oxygen

while 12H₂O will have 24 atoms of hydrogen and 12 atom of oxygen

by Avagadros number

molar mass of water H₂0=18.01528 g/mol

1 mole of H₂0 have 2 moles of Hydrogen  

one mole of water= 6.02⋅10²³water molecules =1.2 x 10²⁴hydrogen atoms

12 mole of H₂O   =    1.2 x 10²⁴ x 12=  1.4 x 10²⁵ hydrogen atoms  

<u />

1.4 x 10²⁵ hydrogen atoms   in 12 moles of H₂O

<u />

3 0
3 years ago
Pls help!<br> Is volume equal to mass over density?
olganol [36]

density = mass/volume

volume= mass/density

Yes, you're correct.

5 0
3 years ago
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A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

5 0
3 years ago
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