Doping Se (group VI elements) with P(group V)elements would produce a P-TYPE semiconductor with HIGHER conductivity compared to pure Se
the reason is P dopant will introduce holes in the Se as P has lesser valence electron
Answer:
The answer is "3.57 and 0.07".
Explanation:
Using the slop formula:
Given:
length path 
from calibration it is found that
Answer:
They are a base and an acid, so they neutralize each other.
Explanation:
Sodium hydroxide is a strong base. Hydrochloric acid is a strong acid.
They react with (neutralize) each other to form a salt (sodium chloride) and water.
NaOH + HCl ⟶ NaCl + H₂O
Answer:
Atomic radius of sodium = 227 pm
Atomic radius of potassium = 280 pm
Explanation:
Atomic radii trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
Consider the example of sodium and potassium.
Sodium is present above the potassium with in same group i.e, group one.
The atomic number of sodium is 11 and potassium 19.
So potassium will have larger atomic radius as compared to sodium.
Atomic radius of sodium = 227 pm
Atomic radius of potassium = 280 pm
Answer:
2,2,4-Trimethyl-pentane
Explanation:
Structural characteristics of the compound is as follows:
- Has five methyl group
- Has one quaternary carbon
- No. double bond
- Gives four monochloro substitution products
Compound must have straight chain of 5 carbons.
Three methyl substituents are attached to 2 and 4 carbons.
Therefore, IUPAC name of the compound will be 2,2,4-Trimethyl-pentane.
