Answer:
3.55atm
Explanation:
We will apply Boyle's law formula in solving this problem.
P1V1 = P2V2
And with values given in the question
P1=initial pressure of gas = 1.75atm
V1=initial volume of gas =7.5L
P2=final pressure of gas inside new piston in atm
V2=final volume of gas = 3.7L
We need to find the final pressure
From the equation, P1V1 = P2V2,
We make P2 subject
P2 = (P1V1) / V2
P2 = (1.75×7.5)/3.7
P2=3.55atm
Therefore, the new pressure inside the piston is 3.55atm
Moles = (6.74*10^23)/(6.02*10^23) =1.119 moles
1.119*44.09=49.36g
Answer:
Here's what I get
Explanation:
(a) Intermediates
The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).
(b) Relative Stabilities
The relative stabilities decrease in the order shown.
N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.
(c) Relative reactivities
The relative reactivities would be
C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃
Unsaturation (IHD) 2 hydrogen Needed
IHD = [(2n+2) -H]/2
(H: X=1, N=-1, O= zero)
Unsaturation:
Double bonds = 1
Rings = 1
Triple Bonds = 2
The degrees of unsaturation in a molecule are additive — a
molecule with one double bond has one degree of unsaturation, a molecule with
two double bonds has two degrees of unsaturation, and so forth.
Hydrogen as it has only 1 proton and therefore only an atomic mass of 1...