Answer:
4,38%
small molecular volumes
Decrease
Explanation:
The percent difference between the ideal and real gas is:
(47,8atm - 45,7 atm) / 47,8 atm × 100 = 4,39% ≈ <em>4,38%</em>
This difference is considered significant, and is best explained because argon atoms have relatively <em>small molecular volumes. </em>That produce an increasing in intermolecular forces deviating the system of ideal gas behavior.
Therefore, an increasing in volume will produce an ideal gas behavior. Thus:
If the volume of the container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to <em>decrease</em>
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I hope it helps!
Solutions are said to be C. homogeneous mixtures, composed of two or more substances. It is usually liquid, however it may be solid or gas.
Answer:
Coefficient of 
Coefficient of
=8
Explanation:
We are given that a reaction in which
reacts with 
We have to find the coefficient of each reactants in balanced reaction

Coefficient is defined the constant value multiplied with a reactant in a reaction.
Coefficient of
=3
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of 
Coefficient of KOH=2
Hence, Coefficient of
and coefficient of 
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%.
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