Answer:
a. True
Explanation:
The control objectives seek to support the fulfillment of the critical factors of business success, in order to finally support the achievement of the institutional objectives, on this the CITI is based
It is very important to clearly identify the relationships of the control objectives with other internal control elements such as:
· Critical success factors receive direct support from control objectives to support compliance.
· The control objectives are made up of control goals, which allow them to have a more detailed perspective of the control objectives and at the same time facilitate the evaluation of their compliance.
CITI CONCEPT:
It is the set of administration elements that a company establishes in a coordinated way so that the use of information technology resources effectively supports the institutional objectives of the company
Answer:
5.31x10⁻⁶ C
Explanation:
The cube is located 100 m altitude from the ground, so the superior face is at 100m and has E = 70 N/C, and the inferior face is at the ground with E = 130 N/C.
The electric field is perpendicular to the bottom and the top of the cube, so the total flux is the flux at the superior face plus the flux at the inferior face:
Фtotal = Ф100m + Фground
Where Ф = E*A*cos(α). α is the angle between the area vector and the field (180° at the topo and 0° at the bottom):
Фtotal = E100*A*cos(180°) + Eground*A*cos(0°)
Фtotal = 70A*(-1) + 130*A*1
Фtotal = 60A
By Gauss' Law, the flux is:
Фtotal = q/ε, where q is the charge, and ε is the permittivity constant in vacuum = 8.854x10⁻¹² C²/N.m²
A = 100mx100m = 10000 m²
q = 60*10000*8.854x10⁻¹²
q = 5.31x10⁻⁶ C
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
D. They all contain carbon as an important part of their structure.
