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.
Explanation:
Let us assume that the given data is as follows.
V = 3.10 L, T =
= (19 + 273)K = 292 K
P = 40 torr (1 atm = 760 torr)
So, P = 
= 0.053 atm
n = ?
According to the ideal gas equation, PV = nRT.
Putting the given values into the above equation to calculate the value of n as follows.
PV = nRT

0.1643 = 
n = 
It is known that molar mass of ethanol is 46 g/mol. Hence, calculate its mass as follows.
No. of moles =
mass =
g
= 0.315 g
Thus, we can conclude that the mass of liquid ethanol is 0.315 g.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of
=
= 0.0458 M
Concentration of
=
= 0.0521 M
GIven that :
Ka = 
Thus; it's pKa = 4.72




pH ≅ 4.80