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 result is presented in proportion which gives a clearer understanding and accurate result.
Explanation: Percentage change in mass is the proportion of the initial mass of a substance changed after sometime. The results is presented as a percentage making it more accurate and can help to give future reference to weight calculations.
Change is Mass is the mass of a substance left after sometime mostly given in grams. It is not as accurate as percentage change in mass. It is generally better to show results in percentage change in mass as it gives a better understanding of what mass of a substance was lost after a given period or after application of energy like Heat or increased temperature.
The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.
The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.
Now, we will consider each reaction individually to decode whether or not a precipitate is possible.
- In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
- In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
- In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.
Learn more: brainly.com/question/11969651
Answer:
0.967mole
Explanation:
Given parameters:
Volume of NH₄Cl = 21.67L
Unknown:
Number of moles = ?
Solution:
If we assume that the volume was taken at standard temperature and pressure,
Then;
Number of moles = 
Number of moles = 
= 0.967mole
