3 years ago

determine the percent yield for the reaction between 14.9g NH3 and excess oxygen to produce 18g of NO gas

Chemistry

1 answer:

emmainna [20.7K]3 years ago

6 0

Answer:

I don't know really i am sorry

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What happens to the electron cloud at very high atomic numbers, when the innermost electrons would, using a non-relativistic mod

marin [14]

Answer:

tough question

Explanation:

8 0

2 years ago

given that 32.0g sulphur contains 6.02×10^23 sulphur atoms,how many atoms are there in 2.70g of aluminum

dangina [55]

Answer:

6.022 × 10²² atoms

Explanation:

Generally 1 mol of any element contains 6.02×10^23 atoms. The number 6.022 × 10²³ is known as Avogadro's number.

Mass of Aluminium = 2.70g

Molar mass = 27g/mol

Number of moles = Mass / Molar mass = 2.70 / 27 = 0.1 mol

1 mol = 6.022 × 10²³

0.1 mol = x

x = 6.022 × 10²³ * 0.1 = 6.022 × 10²² atoms

5 0

3 years ago

Draw the most stable resonance structure for the intermediate in the electrophilic aromatic bromination of aniline, anisole, and

ASHA 777 [7]

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