Answer:
it is called waning and waxing It's because the moon is a world in space, just as Earth is. Like Earth, the moon is always half illuminated by the sun; the round globe of the moon has a day side and a night side. And, like Earth, the moon is always moving through space. it is also because of the position of the earth and the moon.
Explanation:
The rate of Formation of Carbocation mainly depends on two factors'
1) Stability of Carbocation: The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.
2) Ease of detaching of Leaving Group: The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,
R-I > R-Cl > R-F
B > C > A
Answer:
A. 
Explanation:
The work function of the Platinum =
For maximum wavelength, the light must have energy equal to the work function. So,
Where,
h is Plank's constant having value
c is the speed of light having value
is the wavelength of the light being bombarded
Thus,



Answer:
<h2>
The equilibrium constant Kc for this reaction is 19.4760</h2>
Explanation:
The volume of vessel used=
ml
Initial moles of NO=
moles
Initial moles of H2=
moles
Concentration of NO at equilibrium=
M

Moles of NO at equilibrium= 
=
moles
2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)
<u>Initial</u> :1.3*10^-2 2.6*10^-2 0 0 moles
<u>Equilibrium</u>:1.3*10^-2 - x 2.6*10^-2-x x x/2 moles
∴
⇒
![Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH2O%5D%5E2%5BN2%5D%7D%7B%5BH2%5D%5E2%5BNO%5D%5E2%7D%20%28volume%20of%20vesselin%20litre%29)
<u>Equilibrium</u>:0.31*10^-2 1.61*10^-2 0.99*10^-2 0.495*10^-2 moles
⇒
⇒