Answer:
They have electrons in their 3d- and 4s-orbital for bond formation.
Explanation:
d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.
The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.
If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.
If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2
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
+
- 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:
1455.6
Explanation: you first convert 2250ml to l by dividing by 1000 so you get 2.25l then you use Boyles law which is p1v1=p2v2 then insert values
35.75*100=p2*2.25 then divide both sides by 2.25 then you get 1455.6