Answer:
P₂ = 140 KPa
Explanation:
Given data:
Initial volume = 8.0 L
Final volume = 4.0 L
Initial pressure = 70 KPa
Final pressure = ?
Solution:
According to Boyle's law
P₁V₁ = P₂V₂
P₂ = P₁V₁ / V₂
P₂ = 70 KPa ×8.0 L/4.0 L
P₂ = 560 KPa .L / 4.0 L
P₂ = 140 KPa
Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.
Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.
The reaction is as shown in the image.
The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid
is shown in the image.
m-CPBA (meta-perchlorobenzoic acid) is a peracid and forms epoxides on reacting with alkenes.
From a Chemistry perspective it is the electron at 1/1800th the mass of a proton or neutron.
Answer:
yes it can λ =265 nm
Explanation:
Here we will use the relationship
E = h c/λ ∴ λ = E/ hc where
h= Plank's constant
c= Speed of light
λ = Wavelength = ?
Substituting
note need E in J ,
E = 4.7 eV x 1.602 x 10⁻¹⁹ J/eV = 7.5 x 10⁻¹⁹ J)
λ = 7.5 x 10 ⁻¹⁹ J / ( 6.626 x 10⁻³⁴ Js x 3 x 10^8) = 2.65 x 10⁻⁷ m = 2.65
= 2.65 x 10⁻⁷ m x 1 x 10⁹ nm/m = 265 nm