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:
11 electrons total and 23amu
Explanation:
This tells us that sodium has 11 protons and because it is neutral it has 11 electrons. The mass number of an element tells us the number of protons AND neutrons in an atom (the two particles that have a measurable mass). Sodium has a mass number of 23amu.
Answer:
The net ionic equation is as follows:
HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)
Explanation:
The reaction between Hydrocyanic acid, HCN, and sodium hydroxide is a neutralization reaction between a weak acid and a strong base.
Hydrocyanic acid being a weak acid ionizes only slightly, while sodium hydroxide being a strong base ionizes completely. The equation for the reaction is given below:
A. HCN(aq) + NaOH-(aq) ----> NaCN(aq) + H2O(l)
Since Hydrocyanic acid is written in the aqueous form as it ionizes only slightly and the ionic equation is given below:
HCN(aq) + Na+(aq)+OH-(aq) ----> Na+(aq)+CN-(aq) + H2O(l)
Na+ being a spectator ion is removed from the net ionic equation given below:
HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)
Answer:
I needed to use good precision in my measurement for my chemistry lab
Explanation:
Answer:
filter the hot mixture.
Explanation:
Solid is stayed undissolved since the arrangement is gotten super saturated. On the off chance that solid molecule is available recrysallization won't happen in this way we need expel the solid molecule by filtarion in hot condition itself . Subsequently, arrangement become totally homogenous and recrysallization item will shaped by moderate cooling
