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%.
Answer:
Explanation:
If the concentration of any substance A in a dynamic equilibrium increases, The equilibrium will be shifted to its opposite side so that Substance A can be created less and the substance opposite to A can be created more so that a "dynamic equilibrium" can again be established.
Argon, it's a noble gas in the last group on the periodic table.
Answer:
32.8g/mole
Explanation:
Given parameters:
Mass of sample of gas = 32.8g
Volume = 22.4L
Unknown:
Molecular weight = ?
Solution:
To solve this problem we must understand that at rtp;
1 mole of gas occupies a volume of 22.4L
Number of mole of the gas = 1 mole
Now;
Mass = number of moles x molecular weight
molecular weight = 
= 
= 32.8g/mole
Answer:
suggested explanation
Explanation:
a hypothesis is like a guess to a question. then you test the guess.
