Answer:
36290 min = 604.8 hr.
Explanation:
1 lbs = 453.59237 grams.
∴ 2 lbs = 907.18474 grams.
<em><u>Using cross multiplication:</u></em>
500 mg of iron oxide dissolved → 20 minutes.
907184.74 mg of iron oxide dissolved → ??? minutes.
<em>∴ The time needed to dissolve 2 lbs of iron oxide =</em> (907184.74 mg)(20 min)/(500 mg) = <em>36290 min = 604.8 hr.</em>
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:
=1.666 liters
Explanation:
1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.
0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1
=11.2 liters.
Standard pressure= 1 atmosphere (Atm)
Standard temperature = 273.15 Kelvin
According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂
Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.
Therefore P₂ =P₁V₁T₂/T₁V₂
Substituting for the values we get:
P₂= (1 atm× 11.2L ×203K)/ (273K×5L)
=1.666 atm
