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
Answer:
CaCO3(aq) → Ca2+(aq ) + CO3 2-(aq)
Explanation:
- Dissolution reactions are reactions that occur when a solute either in gaseous, liquid, or solid form dissolves in a solvent such as water to form a solution.
- In this case we are given Calcium carbonate (CaCO3) which undergoes dissolution according to the equation;
- CaCO3(aq) → Ca2+(aq ) + CO3 2-(aq)
- Then<em><u> the bicarbonate ion combines with two protons from water to form a weak acid H2CO3. The weak acid is then broken down to form CO2 and H2O since its unstable.</u></em>
Richter Scale.
Hope this helps!
Answer:
At temperatures below −78 °C, carbon dioxide changes directly from a gas to a white solid called dry ice through a process called deposition.
It’s Benzene, I do believe :) hope this helped, good luck!
