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:
Volume = 30cm³
Explanation:
A block is a geometrical figure and its volume, -look at the figure-, follows the equation:
Volume = Width*Length*Height
As the measurements of the block are 5.00cm, 3.00cm and 2.00cm, the volume is:
Volume = 5.00cm*2.00cm*3.00cm
<h3>Volume = 30cm³</h3>
Answer:
88,88 % de O y 11,11 % de H
Explanation:
La composición porcentual se define como la masa que hay de cada mol de átomo en 100g. Las moles de agua en 100g son:
<em>Masa molar agua:</em>
2H = 2*1g/mol = 2g/mol
1O = 1*16g/mol = 16g/mol
Masa molar = 2 + 16 = 18g/mol
100g H2O * (1mol / 18g) = 5.556 moles H2O.
Moles de hidrógeno:
5.556 moles H2O * (2mol H / 1mol H2O) = 11.11 moles H
Moles Oxígeno = Moles H2O = 5.556 moles
La masa de hidrógeno es:
11.11mol * (1g/mol) 11.11g H
La masa de oxígeno es:
5.556 mol * (16g / 1mol) = 88.89g O
Así, el porcentaje de O es 88.89% y el de H es 11.11%. La opción correcta es:
<h3>88,88 % de O y 11,11 % de H</h3>