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polet [3.4K]
3 years ago
6

Explain this method (Froth floatation method)..........​

Chemistry
1 answer:
Svetach [21]3 years ago
3 0

Answer:

froth flotation is a technique commonly used in the mining industry. In this technique, particles of interest are physically separated from a liquid phase as a result of differences in the ability of air bubbles to selectively adhere to the surface of the particles, based upon their hydrophobicity.

Explanation:

Froth floatation method is commonly used to concentrate sulphide ore such as galena (PbS), zinc blende (ZnS) etc. (ii) In this method, the metaalic ore particles which are perferentially wetted by oil can be separated from gangue. (iii) In this method, the crushed ore is suspended in water and mixed with frothing agent such as pine oil, eucalyptus oil etc. (iv) A small quantity of sodium ethyl xanthate which act as a collector is also added. (v) A froth is generated by blowing air through this mixture. (vi) The collector molecules attach to the ore particles and make them water repellent. (vii) As a result, ore parrticles, wetted by the oil, rise to the surface along with the froth. (viii) The froth is skimmed off and dried to recover the concentration ore. (ix) The gangue particles that are preferentially wetted by water settle at the bottom.

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<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

  • K: 39.098;
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Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

What's the enthalpy change of this process?

Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

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