This is to fill in the answer to the question.
The area where the blast originates is referred to as <span>Scene Perimeter/Isolation Zone. This whole area is dangerous for people since it can contain harmful gasses or falling debris depending on the environment of the blast.
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Because it contains vinegar because it does not form layers when mixed with other liquids. Sugar or citric acid because they don't leave sediment.
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%.
The copper wire was sanded before burning in order to make sure that copper metal was exposed on the surface of the wire.
Answer: B
Explanation
The copper wire when placed in atmosphere without coating leads to oxidation of copper metal with respect to the impurities present in the atmosphere.
As copper is electropositive in nature, so electronegative ions present in the universe will try to react with copper and the copper will react easily with other elements.
So generally copper wire is coated with color or polymer coating.
In this case, the copper wire without any coating is sanded, so that the eddy sheets or polishing materials on friction with copper wire will remove the impurities by the electrostatic law of conservation of charges and charge transfer.
As the impurities are removed when copper wire is sanded, the copper atoms will be exposed on the surface of the wire leading to burning of copper in the copper wire.
Mg(OH)₂ ⇄ Mg²⁺ + 2 OH⁻
Ksp = [Mg²⁺] [OH⁻]²
6.0 x 10⁻¹⁰ = 0.10 x [OH⁻]²
[OH⁻] = 7.746 x 10⁻⁵ M
when Mg(OH)₂ 1st precipitates, [OH⁻] = 7.746 * 10⁻⁵ M
Fe(OH)₂ <—> Fe²⁺ + 2OH⁻
Ksp = [Fe²⁺] [OH⁻]²
7.9 x 10⁻¹⁶ = [Fe²⁺] x (7.746 x 10⁻⁵)²
[Fe²⁺] = 1.32 x 10⁻⁷ M
Answer: 1.32 x 10⁻⁷ M
