Answer:
b.) Br and Br
Explanation:
A covalent bond occurs when electrons are shared between two atoms causing them to form a bond.
A "pure" covalent bond refers to a nonpolar covalent bond. In these bonds, the electrons are shared equally between two atoms as a result of the absence of an (or very small) electronegativity difference. The purest covalent bond would therefore be between two atoms of the same electronegativity. Two bromines (Br) have the same electronegativity, thus making it the purest covalent bond.
Polar covalent bonds occur when electrons are shared unequally between two atoms. There is a larger electronegativity difference between the two atoms, but not large enough to classify the bonds as ionic. In this case, a.) and c.) are polar covalent bonds and d.) is an ionic bond.
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=
![\dfrac{[H^{+}] [A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
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Explanation:
- When an aqueous solution of a certain acid is prepared it is dissociated is as follows-
⇄ 
Here HA is a protonic acid such as acetic acid, 
- The double arrow signifies that it is an equilibrium process, which means the dissociation and recombination of the acid occur simultaneously.
- The acid dissociation constant can be given by -
= ![\dfrac{[H^{+}] [A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
- The reaction is can also be represented by Bronsted and lowry -
⇄ ![[H_3O^+] [A^-]](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%5BA%5E-%5D)
- Then the dissociation constant will be
= ![\dfrac{[H_3O^{+}] [A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5BH_3O%5E%7B%2B%7D%5D%20%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
Here,
is the dissociation constant of an acid.
An intensive property is a property that does not change depending on how much mass of it you are considered. An example of an intensive property is density. No matter how much water you examine, the density of the sample will be 1g/cm³.
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 answer is D, medical diagnosis