Answer:
Above 7
Explanation:
The equivalence point of any titration can be read off from the appropriate titration curve.
A titration curve is a plot of the pH of analyte against the volume of titrant added.
For a strong base and weak acid, the equivalence point lies above 7.
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
My answer will be B and D
Answer:
The answer to your question is: letter D
Explanation:
Data
Isotope mass 1 = 54 amu
Isotope mass 2 = 56 amu
Atomic weight = 54.37 amu
Percent abundance of isotope 2 = ? = X
Percent abundance of isotope 1 = 1 - x
Formula
Atomic weight = (Mass isotope 1)(1 - x) + (mass isotope 2)(x)
Process
Place the data in the formula and solve for x.
54.37 = 54( 1 - x) + 56x
54.37 = 54 - 54x + 56x
54.37 - 54 = -54x + 56x
0.37 = 2x
x = 0.37 / 2
x = 0.185
To find the percent multiply by 100
x = 0.185 x 100
x = 18.5 % = Mass of isotope 2
Mass of isotope 1 = 100 - 18.5 = 81.5 %