Explanation:
Fusion vs Fission
In fission, energy is gained by splitting apart heavy atoms, for example uranium, into smaller atoms such as iodine, caesium, strontium, xenon and barium, to name just a few. However, fusion is combining light atoms, for example two hydrogen isotopes, deuterium and tritium, to form the heavier helium. Both reactions release energy which, in a power plant, would be used to boil water to drive a steam generator, thus producing electricity.
<span>global wind patterns, rotation of the earth, shape of ocean basins.</span>
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
Many electrophilic aromatic halogenations require the presence of an aluminum trihalide as a catalyst. We generally acetylated the amino group as protection. Now, this acetanilide can be brominated at Ortho or para position. An atom that is attached to an aromatic system usually hydrogen is replaced by an electrophile is an organic reaction which is called Electrophilic aromatic substitution. There are what you called important electrophilic aromatic substitutions they are aromatic nitration, aromatic sulfonation, aromatic halogenation and acylation and alkylating Friedel-Crafts reaction. Aromatic bromination is an electrophilic aromatic substitution (EAS) reaction, which will require benzene to act as a nucleophile to acquire an electrophile. Therefore, any directing groups that activate the ring will make it react more quickly with respect to aromatic bromination. Acetanilide is a moderately-activated ring <span>having a decent EWG.</span>
Answer:
65.21 percent
Explanation:
We are to find the percentage yield
We have this equation,
2no+o2 -->2no2
Such that 1500 kilograms/30grams
= Mno2/46
=1500/30 = mno2/46
We cross multiply from here
1500x46 = 30xMnO2
69000 = 30Mno2
69000/30 = MnO2
2300 = Mno2
The percentage yield would be
1500/2300 *100
= 0.6521 x 100
= 65.21%
This answers the question