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3 years ago

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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Calculate the formula/molecular mass of following

ddd [48]

Answer:

Refer to the period table. Use the atomic mass for calculation. Multiply the atomic mass with the number of atoms. For example, Al2(SO4)3. (27 × 2) + (32 x 3) + (16 x 12) = molecular mass

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3 years ago

What is the mass, in grams, of

DIA [1.3K]

Answer:

The answers to your question are below

Explanation:

a) 6.85×1020 H2O2 molecules

H2O2 MW = 32 + 2 = 34 g

34g -------------------- 6.023 x 10²³ molecules

x ------------------- 6.85 x 10 ²⁰

x = (6.85 x 10 ²⁰)(34)/ 6.023 x 10²³

x = 0.038 g

3.3×1022 SO2 molecules

MW SO2 = 32 + 32 = 64g

64 g -------------------- 6.023 x 10²³ molecules

x -------------------- 3.3×1022 SO2 molecules

x = (3.3×1022 SO2)(64) / 6.023 x 10²³

x = 3.51 g

5.5×1025 O3 molecules

MW = 16 x 3 = 48g

48 g ----------------- 6.023 x 10²³ molecules

x ------------------ 5.5×1025 O3 molecules

x = (5.5×1025 )(48) / 6.023 x 10²³

x = 4383 g

9.30×1019 CH4 molecules

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3 years ago

By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling

vova2212 [387]

It is found that scuba diver’s can safely ascend up to 52.53 ft without Breathing out.

By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling or exhaling then

$p_{1}$ =patm+ $pH_{2} O$

$p_{1}$ = 101325+ρ $gh_{1}$

It is given that height is 125ft. Put the value of h in above formula:

h1 =125ft=38.1m

ρ=1.04g/mL=1040kg/ $m^{3}$

g=9.81

$p_{1}$ =101325Pa+388711.44

$p_{1}$ =490036.44‬Pa

$p_{2}$ =p atm =101325Pa

It is known that volume and pressure can be expressed as:

V*P=const.

where, V is volume and P is pressure.

Now,

$V_{1} *p_{1}$ = $V_{2} *p_{2}$

$V_{2} /V_{1}$ = $p_{2} /p_{1}$

$V_{2} /V_{1}$ =490036.44/101325

$V_{2} /V_{1}$ =4.84

Assume constant temperature

d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL.

now $p_{1}$ =p atm+ $pH_{2} O$ =490036.44Pa

V*p=const

where, V is volume and P is pressure.

Now,

$V_{1} *p_{1}$ = $V_{2} *p_{2}$

$V_{2} /V_{1}$ = $p_{2} /p_{1}$

$V_{2} /V_{1}$ =490036.44/X

$p_{2}$ =490036.44pa/(V2/V1) =326690.96Pa

$p_{2}$ =patm +p $H_{2} O$

$p_{2}$ =101325Pa+ρ $gh_{2}$

326690.96Pa=101325Pa+ρ $gh_{2}$

ρgh1 =151987.5-101325=225365.96‬‬Pa

ρ=1,04g/mL=1040kg/m3

g=9.81

$h_{2}$ =225365.96/‬ρ∗g

$h_{2}$ =225365.96 / ‬1040∗9.81

$h_{2}$ =22.09m= 72.47ft

ΔH= $H_{1} -H_{2}$

=125-72.47

=52.53ft

So she can safely ascend up to 52.53 ft without Breathing out

To know more about Scuba diver here

brainly.com/question/15430942

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Which one of the following would you expect to dissolve in water? limonene sucrose oil atmospheric oxygen none of the above

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

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