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
Answer:
the nucleus is the center of the atom, made up of protons and neutrons, without the nucleus you'd just have a bunch of electrons floating around; the nucleus is positively charged
protons are the positively charged particles that sit within the nucleus
neutrons are particles of no charge that sit within the nucleus, and because they have no charge, they do not cancel out the positive charge of the protons, making the nucleus positive
electrons are negatively charged particles that float around the nucleus in an area known as the electron cloud, they orbit around the nucleus because they are attracted to the positive charge of the nucleus (caused by the protons), with charges, opposites attract
Explanation:
Combination
Hope this helps :)
Answer : The rate constant at 785.0 K is, 
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= ?
= activation energy for the reaction = 262 kJ/mole = 262000 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature =
= final temperature =
Now put all the given values in this formula, we get:
![\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B6.1%5Ctimes%2010%5E%7B-8%7Ds%5E%7B-1%7D%7D%29%3D%5Cfrac%7B262000J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B785.0K%7D%5D)
Therefore, the rate constant at 785.0 K is,
