Answer:
frequency = 8.22 x 10¹⁴ s⁻¹
Explanation:
An electron's positional potential energy while in a given principle quantum energy level is given by Eₙ = - A/n² and A = constant = 2.18 x 10⁻¹⁸j. So to remove an electron from the valence level of Boron (₅B), energy need be added to promote the electron from n = 2 to n = ∞. That is, ΔE(ionization) = E(n=∞) - E(n=2) = (-A/(∞)²) - (-A/(2)²) = [2.18 x 10⁻¹⁸j/4] joules = 5.45 x 10⁻¹⁹ joules.
The frequency (f) of the wave ionization energy can then be determined from the expression ΔE(izn) = h·f; h = Planck's Constant = 6.63 x 10⁻³⁴j·s. That is:
ΔE(izn) = h·f => f = ΔE(izn)/h = 5.45 x 10⁻¹⁹ j/6.63 x 10⁻³⁴ j·s = 8.22 x 10¹⁴ s⁻¹
<span>Oxidation or reaction with oxygen is a chemical property</span>
Answer:
The hollow ball is more buoyant than the solid ball.
Explanation:
The hollow ball will float while the solid ball sinks because the volume of water displaced by the hollow ball is more than that displaced by the solid ball.
An object floats when it displaces a volume of water equal to its weight. A solid ball is more compact, and hence, does not displace a lot of water when dropped into a bowl full of water. On the other hand, the hollow ball will have a larger curved surface area, since it has no fill. This will make it displace a larger volume of water and hence make it float.
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp =
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104![[P(NO_{2} )]^{2}](https://tex.z-dn.net/?f=%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D)
+ 
- 200 = 0
Resolving the second degree equation:
= 
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are = 98.7 MPa and P(N₂O₄) = 101.3 MPa
