The answer is natural gasses
Answer:
Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.
Explanation:
Initially
3.0 atm 0
At equilibrium
(3.0-2p) p
Equilibrium partial pressure of 
p = 0.45 atm
The value of equilibrium constant wil be given by :
After addition of 1.5 atm of nitrogen dioxide gas equilibrium reestablishes it self :
After adding 1.5 atm of 
:
(2.1+1.5) atm 0.45 atm
At second equilibrium:'
(3.6-2P) (0.45+P)
The expression of equilibrium can be written as:
Solving for P:
P = 0.37 atm
Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time:
= (0.45+P) atm = (0.45 + 0.37 )atm = 0.82 atm
Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.
Answer:
will be not soluble in water
Explanation:
LiOH is a strong base. Hence it gets completely dissociated in aqueous solution.
is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.
is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.
is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.
is a sparingly soluble salt. Hence it is not dissociated and hence dissolved in water. This is due to the fact that both 
and 
ions are similar in size. Hence crystal structure of is quite stable. Hence is reluctant to undergo any dissociation in aqueous solution.
Answer:
1.26 × 10¹⁵ s⁻¹
Explanation:
Work function is the minimum energy required to remove an electron from the surface of metal
energy of the electron = hf - Φ
Φ = work function = hf₀ where f₀ = threshold frequency
f₀ = Φ / h where h ( Planck constant = 6.626 × 10⁻³⁴ Js)
Φ = 5.22eV = 5.22 × 1 eV where 1 eV = 1.60217662 × 10⁻¹⁹ J
Φ = 5.22 × 1.60217662 × 10⁻19 J = 8.363362 × 10⁻¹⁹ J
f₀ = (8.363362 ×10⁻¹⁹ J) / (6.626× 10⁻³⁴ Js) = 1.26 × 10¹⁵ s⁻¹
The frequency must be greater than the 1.26 × 10¹⁵ s⁻¹ to observe the emission
