Answer:
Option A=> -NHCOCH3 and option C = -CH3.
Explanation:
The option A that is -NHCOCH3 is CORRECT because it possesses lone pair of electron with the exception of group 7A elements. It is this lone pair that is used in the Activation of the ring towards substitution. Other groups that falls into this group are; OCH3, alkyls and many others.
Option B that is -COOH is good group for withdrawal of electron through Resonance. Other examples are NO2, -CN and SO3H.
Option C falls to the same category as option A above that is Activation of the ring towards substitution.
Option D falls to the same category as option B above that is group for withdrawal of electron through Resonance.
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
Answer:
Acid(BSA) = CH₃COOH
Base (BSB) = H₂O
Conjugate base (CB) = CH₃COO⁻
Conjugate acid (CA) = H₃O⁺
Explanation:
Equation of reaction;
CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺
Hello,
From my understanding of the question, we are required to identify the
1) Acid
2) Base
3) conjugate acid
4) conjugate base in the reaction
Acid (BSA) = CH₃COOH
Base (BSB) = H₂O
CA = conjugate acid = H₃O⁺
CB = conjugate base = CH₃COO⁻
