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
Statement 1 is false because heat is the movement of molecules.
Statement 2 is true because matter is composed of elementary particles which are VERY small.
Statement 3 is false because the kinetic energy of a particle determines its heat and it is impossible for a particle to be completely motionless.
Statement 4 is false because neutrons do not have a charge and make up a large portion of all matter.
Answer:
A polar bond is one where the charge distribution between the two atoms in the bond is unequal. A polar molecule is one where the charge distribution around the molecule is not symmetric. It results from having polar bonds and also a molecular structure where the bond polarities do not cancel.
Explanation:
The correct option is (b)
NaNH2 is an effective base. It can be a good nucleophile in the few situations where its strong basicity does not have negative side effects. It is employed in elimination reactions as well as the deprotonation of weak acids.Alkynes, alcohols, and a variety of other functional groups with acidic protons, such as esters and ketones, will all be deprotonated by NaNH2, a powerful base.Alkynes are deprotonated with NaNH2 to produce what are known as "acetylide" ions. These ions are powerful nucleophiles that can react with alkyl halides to create carbon-carbon bonds and add to carbonyls in an addition reaction.Acid/base and nucleophilic substitution are the two types of reactions.Using the right base, terminal alkynes can be deprotonated to produce a carbanion.A good C is the acetylide carbanion.The acetylide carbanion can undergo nucleophilic substitution reactions because it is a potent C nucleophile. (often SN2) with 1 or 2 alkyl halides with electrophilic C to create an internal alkyne (Cl, Br, or I).Elimination is more likely to occur with 3-alkyl halides.It is possible to swap either one or both of the terminal H atoms in ethylene (acetylene) to create monosubstituted (R-C-C-H) and symmetrical (R = R') or unsymmetrical (R not equal to R') disubstituted alkynes (R-C-C-R').
Learn more about NANH2 here :-
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Answer:
900 K
Explanation:
Recall the ideal gas law:
Because only pressure and temperature is changing, we can rearrange the equation as follows:
The right-hand side stays constant. Therefore:
The can explodes at a pressure of 90 atm. The current temperature and pressure is 300 K and 30 atm, respectively.
Substitute and solve for <em>T</em>₂:
Hence, the temperature must be reach 900 K.
