If excess carbon disulfide reacts with 450 mL of oxygen, 150 mL of carbon dioxide and 300 mL of sulfur dioxide gases would be produced respectively.
<h3>Stoichiometric calculation</h3>
The reaction between liquid carbon disulfide and oxygen is represented by the equations below:

The mole ratio of oxygen to carbon dioxide and sulfur dioxide produced is 3:1:2.
Thus, for 450 mL oxygen, 1/3 x 450 = 150 mL of carbon dioxide will be required.
Also for 450 mL of oxygen, 2/3 x 450 = 300 mL of sulfur dioxide will be required.
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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:
Rank in increasing order of effective nuclear charge:
Explanation:
This explains the meaning of effective nuclear charge, Zeff, how to determine it, and the calculations for a valence electron of each of the five given elements: F, Li, Be, B, and N.
<u>1) Effective nuclear charge definitions</u>
- While the total positive charge of the atom nucleus (Z) is equal to the number of protons, the electrons farther away from the nucleus experience an effective nuclear charge (Zeff) less than the total nuclear charge, due to the fact that electrons in between the nucleus and the outer electrons partially cancel the atraction from the nucleus.
- Such effect on on a valence electron is estimated as the atomic number less the number of electrons closer to the nucleus than the electron whose effective nuclear charge is being determined: Zeff = Z - S.
<u><em>2) Z eff for a F valence electron:</em></u>
- F's atomic number: Z = 9
- Total number of electrons: 9 (same numer of protons)
- Period: 17 (search in the periodic table or do the electron configuration)
- Number of valence electrons: 7 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 9 - 7 = 2
- Zeff = Z - S = 9 - 2 = 7
<u><em>3) Z eff for a Li valence eletron:</em></u>
- Li's atomic number: Z = 3
- Total number of electrons: 3 (same number of protons)
- Period: 1 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 1 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 3 - 1 = 2
- Z eff = Z - S = 3 - 2 = 1.
<em>4) Z eff for a Be valence eletron:</em>
- Be's atomic number: Z = 4
- Total number of electrons: 4 (same number of protons)
- Period: 2 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 2 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 4 - 2 = 2
- Z eff = Z - S = 4 - 2 = 2
<u><em>5) Z eff for a B valence eletron:</em></u>
- B's atomic number: Z = 5
- Total number of electrons: 5 (same number of protons)
- Period: 13 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 3 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 5 - 3 = 2
- Z eff = Z - S = 5 - 2 = 3
<u><em>6) Z eff for a N valence eletron:</em></u>
- N's atomic number: Z = 7
- Total number of electrons: 7 (same number of protons)
- Period: 15 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 5 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 7 - 5 = 2
- Z eff = Z - S = 7 - 2 = 5
<u><em>7) Summary (order):</em></u>
Atom Zeff for a valence electron
- <u>Conclusion</u>: the order is Li < Be < B < N < F