Answer:
C
Explanation:
This experiment by Rutherford involved the firing of alpha particles at gold foils. It is also. called the gold foil experiment.
He fired these alpha particles at different points. He noticed that at some points, there were deflections, while at some other points, there were no deflections. It is necessary to state that these alpha particles are positively charged. For there to be a deflection, there must have been a kind of repulsion between the gold foil and the alpha particles.
From the basic physics of like repels like, he knew for sure that there must be dense positive core in the atom that is causing the deflection of the alpha particles. This enabled him to come up with the theory that the atom contained a small dense positive core called the nucleus
Answer: The answer is 167
Explanation: This is because that was right on edg. so yea heart this tho plsss
Im going for d i think i took this last year
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
Answer:
Explanation:The pi-molecular orbitals in propene (CH3-CH=CH2) are essentially the ... This central carbon thus provides two p-orbitals – one for each pi bond – and these two different p-orbitals have to be perpendicular, leading to a twisted structure as shown: ... It all comes down to where the location of the electron-deficient carbon
