Here, the three different notation of the p-orbital in different sub-level have to generate
The value of azimuthal quantum number (l) for -p orbital is 1. We know that the magnetic quantum number 
depends upon the value of l, which are -l to +l.
Thus for p-orbital the possible magnetic quantum numbers are- -1, 0, +1. So there will be three orbitals for p orbitals, which are designated as 
, 
and 
in space.
The three p-orbital can be distinguish by the quantum numbers as-
For 2p orbitals (principal quantum number is 2)
1) n = 2, l = 1, m = -1
2) n = 2, l = 1, m = 0
3) n = 2, l = 1, m = +1
Thus the notation of different p-orbitals in the sub level are determined.
<span>The answer is 13,902 centiliters.
</span>
Answer 1:
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.
In present case, following reaction is involved:
2NO2 ↔ 2NO + O2
Here, K =
![\frac{[PNO]^2[O2]}{[PNO2]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BPNO%5D%5E2%5BO2%5D%7D%7B%5BPNO2%5D%5E2%7D%20)
Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence, K =
![\frac{[0.0022]^2[0.0011]}{[0.247]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5B0.0022%5D%5E2%5B0.0011%5D%7D%7B%5B0.247%5D%5E2%7D%20)
= 8.727 X 10^-8
Thus, equilibrium constant of reaction = 8.727 X 10^-8
.......................................................................................................................
Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.
Therefore, Reaction quotient = </span>
=
![\frac{[0.021]^2[0.037]}{[0.192]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5B0.021%5D%5E2%5B0.037%5D%7D%7B%5B0.192%5D%5E2%7D%20)
= 4.426 X 10^-4.
Here, Reaction quotient > Equilibrium constant.
Hence, <span>the reaction need to go to
reverse direction to reattain equilibrium </span>
Answer : The pressure in the flask after reaction complete is, 2.4 atm
Explanation :
To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.
where,
P = final pressure in the flask = ?
R = gas constant = 0.0821 L.atm/mol.K
T = temperature = 
V = volume = 4.0 L
= moles of 
= 0.20 mol
= moles of 
= 0.20 mol
Now put all the given values in the above expression, we get:
Thus, the pressure in the flask after reaction complete is, 2.4 atm
Pressure of argon = 546.8 kPa
Conversion factor: 1 atm = 101.325 kPa
Pressure of argon = 546.8 kPa x 1 atm/101.325 kPa = 5.4 atm
Moles of argon = 15.82
Volume of argon = 75.0 L
According to Ideal gas law,
PV = nRT
where P is the pressure, V is the volume , n is the number of moles, R is the universal gas constant, and T is the temperature
T = PV/nR = (5.4 atm x 75.0 L) / (15.82 x 0.0821 L.atm.mol⁻¹K⁻¹)
T = 311.82 K
Hence the temperature of the canister is 311.82 K.
