Answer:
A. Methanol
B. 2-chloropropan-1-ol
C. 2,2-dichloroethanol
D. 2,2-difluoropropan-1-ol
Explanation:
Primary alcohols are stronger acids than secondary alcohols which are stronger than tertiary alcohols.
This trend is so because of the stability of the alkoxide ion formed(stabilising the base, increases the acidity). A more stabilised alkoxide ion is a weaker conjugate base (dissociation of an acid in water).
By electronic factors, When there are alkyl groups donating electrons, the density of electrons on th O- will increase a d thereby make it less stable.
By stearic factors, More alkyl group bonded to the -OH would mean the bulkier the alkoxide ion which would be harder to stabilise.
Down the group of the periodic table, basicity (metallic character) decreases as we go from F– to Cl– to Br– to I– because that negative charge is being spread out over a larger volume that is electronegativity decreases down the group.
Electronegative atoms give rise to inductive effect and a decrease in indutive effects leads to a decrease in acidity. Therefore an Increasing distance from the -OH group lsads to a decrease in acidity.
From above,
A. Methanol
B. 2-chloropropan-1-ol
C. 2,2-dichloroethanol
D. 2,2-difluoropropan-1-ol
Is a measure of how heavy atoms are. It's the ratio of the average mass per atom of an element from a given sample to 1/12 the mass of a carbon-12 atom."
Answer:
B) 0.025 
Explanation:
Solution of the problem is in picture attached,
Answer:
131 atm
Explanation:
To find the new pressure, you need to use Boyle's Law:
P₁V₁ = P₂V₂
In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new pressure (P₂) by plugging the given values into equation and simplifying.
P₁ = 3.88 atm P₂ = ? atm
V₁ = 7.74 L V₂ = 0.23 L
P₁V₁ = P₂V₂ <----- Boyle's Law
(3.88 atm)(7.74 L) = P₂(0.23 L) <----- Insert values
30.0312 = P₂(0.23 L) <----- Simplify left side
131 = P₂ <----- Divide both sides by 0.23
The answer to the problem is 7/10
