<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.
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Mass in kilograms of liquid air required = 0.78 kg
<u>Given that </u>
1 Litre of liquid air contains 1.3 grams of oxygen ( air )
<u />
<u> Determine the a</u><u>mount of liquid air</u><u> in Kg</u>
volume of air given = 600 L
mass of liquid air required = x
1 litre = 1.3 grams
600 L = x
∴ x ( mass of liquid air ) = 1.3 * 600
= 780 g = 0.78 kg
Hence we can conclude that Mass in kilograms of liquid air required = 0.78 kg
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Im not completely sure but It might be Cycling of Matter
D = m / V
1.025 = m / 100,000 mL
m = 1.025 x 100,000
m = 102,500 g
A mole of any gas occupied 22.4 L at STP. So, the number of moles of nitrogen gas at STP in 846 L would be 846/22.4 = 37.8 moles of nitrogen gas.
Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.
