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The similarity of the bands in the crystal of a metal to the atomic orbitals can be explained by the band theory of metals. In an atom, when the electrons get excited, the electrons jumps to a higher orbital so as to reach equilibrium. This is analogous to the electrons in the metals which also jumps to another band once excited by an external energy (e.g. electrical energy).
Let us assume that there is a 100g sample present. The respective mass of each element will then be:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Now, we divide each constituent's mass by its Mr to obtain the moles of each
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Dividing by the smallest number:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula is
C10H12NO
Let's assume that change of pressure of PCl₅ is X
According to the ICE table,
PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
Initial Pressure 0.123 - -
Change -X +X +X
Equilibrium 0.123 - X X X
Kp = Ppcl₃(g) x Pcl₂(g) / Ppcl₅(g)
by substitution,
0.0121 = X × X / (0.123 - X)
0.0121 × (0.123 - X) = X²
1.4883 × 10⁻³ - 0.0121X = X²
X² + 0.0121X - 1.4883 × 10⁻³ = 0
X₁ = 0.03735 or X₂ = -0.03865
X cannot be a negative value since X is a pressure.
Hence,
X = 0.03735 atm
Hence, partial pressure of PCl₃ is 0.03735 atm
Answer:
98.4 kPa
Explanation:
Step 1: Given and required data
- Partial pressure of water vapor at 60°C (pH₂O): 19.9 kPa (this info is tabulated)
- Partial pressure of nitrogen (pN₂): 53.0 kPa
- Partial pressure of helium (pHe): 25.5 kPa
Step 2: Calculate the total pressure of the gaseous mixture
The total pressure of the gaseous mixture (P) os equal to the sum of the partial pressures of the gases that form it.
P = pH₂O + pN₂ + pHe
P = 19.9 kPa + 53.0 kPa + 25.5 kPa = 98.4 kPa
