E=hc/λ =6.626×10^-34×3 ×10^8 / 3×10^7 × 10^-9 = 6.626×10 ^-24J.
Answer:
D
Explanation:
This explains how two noble gases molecules can have an attractive force between them.
This force is called as van dar Waals forces.
It plays a fundamental role in fields in as diverse as supramolecular chemistry structural biology .
If no other forces are present, the point at which the force becomes repulsive rather than attractive as two atoms near one another is called the van der Waals contact distance. This results from the electron clouds of two atoms unfavorably coming into contact.[1] It can be shown that van der Waals forces are of the same origin as the Casimir effect, arising from quantum interactions with the zero-point field.[2] The resulting van der Waals forces can be attractive or repulsive.[3] It is also sometimes used loosely as a synonym for the totality of intermolecular forces.[4] The term includes the force between permanent dipoles (Keesom force), the force between a permanent dipole and a corresponding induced dipole (Debye force), and the force between instantaneously induced dipoles
Answer:
The answer to your question is 3% H2SO4 solution
Explanation:
Data
Concentration 2 = C₂ = ?
Concentration 1 = C₁ = 15 %
Volume 1 = V₁ = 50 ml
Volume 2 = V₂ = 250 ml
Formula
C₁V₁ = C₂V₂
Solve for C₂
C₂ = C₁V₁ / V₂
Substitution
C₂ = (15)(50) / 250
Simplification and result
C₂ = 3 %
Answer:
The Structure of "B" is alkene.
Explanation:
The compound "A" having R- configuration and undergoes Hofmann elimination to form an alkene.
The compound "B" on oxidatively cleaving with ozone followed by dimethyl sulfide forms 
and 
The structure of "A" and"B" is as follows.
