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.
Answer:
it forms :
1. Gold ( Au )
2. Zinc nitrate ( Zn(NO3)2 )
Explanation:
When a chunk of zinc is added to a solution of gold (III) nitrate to extract the gold. The reaction forms Gold and Zinc nitrate .
it's a single displacement reaction,
here's the balanced equation for above reaction :
3 Zn + 2 Au(NO3)3 =》3 Zn(NO3)2 + 2 Au
The original options for this question were cleavage, luster and hardness. The answer would be cleavage.
Percent strength (% w/w) of a solution is defined as the amount of solute present in 100 g of the solution.
Given data:
Mass of the solute, potassium chloride = 62.5 g
Volume of water (solution) = 187.5 ml
We know that the density of water = 1 g/ml
Therefore, the mass corresponding to the given volume of water
= 187.5 ml * 1 g/1 ml = 187.5 g
We have a solution of 62.5 g of potassium chloride in 187.5 g water
Therefore, amount of solute in 100 g of water= 62.5 * 100/187.5 = 33.33
The percentage strength = 33.33 %
Answer: XF8
Explanation:
Empirical Formular shows the simplest ratio of elements in a compound.
Xe = 46.3% F = 53.7%
Divide the percentage composition of each element by the atomic mass.
Xe = 46.3/ 131.3 F= 53.7/ 19
= 0.353( approx) = 2.826 (approx)
Divide through with the smallest of the answers gotten in previous step.
Xe = 0.353 / 0.353 F = 2.826/ 0.353
= 1 = 8.0
Empirical formular = XF8
