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Answer:
The equilibrium concentration of NO is 0.02124 M.
Explanation:
Given that,
Initial concentration of NOBr = 0.878 M
Temperature = 24°C
We know that,
The balance equation is
Initial concentration is,
Concentration is,
Equilibrium concentration
We need to calculate the value of x
Using formula of concentration
![k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}](https://tex.z-dn.net/?f=k_%7Bc%7D%3D%5Cdfrac%7B%5BNO%5D%5BBr_%7B2%7D%5D%7D%7B%5BNOBr%5D%5E2%7D)
Put the value into the formula
![3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}](https://tex.z-dn.net/?f=3.07%5Ctimes10%5E%7B-4%7D%3D%5Cdfrac%7B%5B2x%5D%5Bx%5D%7D%7B%5B0.878-2x%5D%5E2%7D)
We need to calculate the equilibrium concentration of NO
Using formula of concentration of NO
Put the value of x
Hence, The equilibrium concentration of NO is 0.02124 M.
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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Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.
