Delta E = Ef - Ei
E = energy , h = plank constant , v = frequency
h= 6.626 * 10 ^-34 j*s , T = 10 ^ 12 , v = 74 * 10 ^12 Hz , Hz = s^-1
E = ( 6.626 * 10^ -34 j*s) ( 74 * 10 ^ 12 s^ -1 ) = 4.90 * 10 ^ -20 J
Delta E = Ef - Ei
-4.90 * 10 ^ -20 J = -2.18 * 10 ^ -18J ( 1/4 ^2 - 1/x ^2)
0.0225 = 0.0625 - ( 1/x ^ 2)
0.225 - 0.0625 = - 1/ x ^ 2
- 0.0400 = - 1/x ^2 = -1 / - 0.0400 = x^2
25 = x^2
x = 5
Answer: (a) The solubility of CuCl in pure water is
.
(b) The solubility of CuCl in 0.1 M NaCl is
.
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.

Initial: 0 0
Change: +x +x
Equilibm: x x

And, equilibrium expression is as follows.
![K_{sp} = [Cu^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCu%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)

x = 
Hence, the solubility of CuCl in pure water is
.
(b) When NaCl is 0.1 M,
, 
, 
Net equation: 
= 0.1044
So for, 
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' = 
0.1044 = 
x = 
Therefore, the solubility of CuCl in 0.1 M NaCl is
.
Answer:
Incineration is a waste treatment process that involves the combustion of organic substances contained in waste materials. Incineration and other high-temperature waste treatment systems are described as "thermal treatment". Incineration of waste materials converts the waste into ash, flue gas and heat.
NOTE: The other person is drunk.
ΔΗ= -115.6 kJ/mol
H2 has a Mr of 2, so as mol=mass/mr we can work out that there are 4 moles of H reacting
Therefore we multiply -115.6 by 4 = -462.4