Answer:
a. withdraws electrons inductively
b. donates electrons by hyperconjugation
c. donates electrons by resonance
d. withdraws electrons inductively
Explanation:
a. The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:
F > Cl > Br>I
b. The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons. Because of the delocalization, hyperconjugation is the result effect.
c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.
d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.
Answer: It has the highest number of hydroxide ions
Explanation: I took the review
This is an incomplete question, the table is attached below.
Answer : The correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.
Explanation :
As we know that the intermolecular force of attraction play an important role in the interaction of solute-solute, solute-solvent and solvent solvent solution.
In the solution A, the solute-solute and solvent-solvent interactions are weak. So, their solute-solvent interaction will be strong. That means, the solution will be more exothermic.
In the solution C, the solute-solute and solvent-solvent interactions are strong. So, their solute-solvent interaction will be weak. That means, the solution will be more endothermic.
Thus, the correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.
The concentration of a solution is the number of moles of solute per fixed volume of solution.
Concentration (C) = number of moles of solute (n) / volume of the solution (v)
we have to find the volume of the solution when 36.0 g of Ca(OH)₂ is added to water to make a solution of concentration 0.530 M
mass of Ca(OH)₂ added - 36.0 g
number of moles of Ca(OH)₂ - 36.0 g / 74.1 g/mol = 0.486 mol
we know the concentration of the solution prepared and the number of moles of Ca(OH)₂ added, substituting these values in the above equation, we can find the volume of the solution
C = n/v
0.530 mol/L = 0.486 mol / V
V = 0.917 L
answer is 0.917 L