Answer: The statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
Explanation:
A strong acid upon dissociation gives a weak conjugate base. This can also be said as stronger is the acid, weaker will be its conjugate base or vice-versa.
Hydrofluoric acid is a strong base as it dissociates completely when dissolved in water.
For example, 
The conjugate base is
which is a weak base.
Acetic acid is a weak acid as it dissociates partially when dissolved in water. So, the conjugate base of acetic acid is a strong base.

Thus, we can conclude that the statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.
Explanation:
Lithium diisopropylamide (LDA) is used in many organic synthesis and is a strong base. It is prepared by the acid base reaction of N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) and butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ).
The equation is show below as:
[(CH₃)₂CH]₂NH + Li⁺⁻CH₂CH₂CH₂CH₃ ⇒ [(CH₃)₂CH]₂N⁻Li⁺ + CH₃CH₂CH₂CH₃
N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) is a weaker acid and hence, LDA ( [(CH₃)₂CH]₂N⁻Li⁺ ) is stronger base. (Weaker acid has stronger conjugate base)
Butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ) is a very strong base and hence, butane ( CH₃CH₂CH₂CH₃ ) is a very weak acid. (Strong base has weaker conjugate acid)
Answer:
the answer is b
Explanation:
convection is caused by density differences whereby warm air with less density rise above a heater and sink near a cold wind.
<u>Answer:</u> The volume of acid should be less than 100 mL for a solution to have acidic pH
<u>Explanation:</u>
To calculate the volume of acid needed to neutralize, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is HCl
are the n-factor, molarity and volume of base which is NaOH
We are given:

Putting values in above equation, we get:

For a solution to be acidic in nature, the pH should be less than the volume of acid needed to neutralize.
Hence, the volume of acid should be less than 100 mL for a solution to have acidic pH