Answer:
The Third Law of thermodynamics states that the entropy of a pure substance in a perfect crystalline state at zero temperature is zero.
The atomic mass of the isotope Ni ( 62 over 28 ) = 61.928345 amu.
Mass of the electrons: 28 · 5.4584 · 10^(-4 ) amu = 0.0152838 amu ( g/mol )
Mass of the nuclei:
61.928345 amu - 0.0152838 amu = 61.913062 amu (g/mol)
The mass difference between a nucleus and its constituent nucleons is called the mass defect.
For Ni ( 62 over 28 ): Mass of the protons: 28 · 1.00728 amu = 28.20384 amu
Mass of the neutrons: 34 · 1.00866 amu = 34.299444 amu
In total : 62.49828 amu
The mass defect = 62.49828 - 61.913062 = 0.585218 amu
Nucleus binding energy:
E = Δm · c² ( the Einstein relationship )
E = 0.585218 · ( 2.9979 · 10^8 m/s )² · 1 / (6.022 · 10^23) · 1 kg / 1000 g =
= 0.585218 · 8.9874044 · 10 ^16 : (6.022 · 10^23) · 0.001 =
= ( 5.2595908 : 6.022 ) · 0.001 · 10^(-7 ) =
= 0.0008733 · 10^(-7) J = 8.733 · 10^(-11) J
The nucleus binding energy per nucleon:
8.733 · 10^(-11) J : 62 = 0.14085 · 10 ^(-11) =
= 1.4085 · 10^(-12) J per nucleon.
Answer:
9.4
Explanation:
The equation for the reaction can be represented as:
+ 
⇄ 
The ICE table can be represented as:
+ 
⇄
Initial 0.27 0.49 0.0
Change -x -2x x
Equilibrium 0.27 - x 0.49 -2x x
We can now say that the concentration of at equilibrium is x;
Let's not forget that at equilibrium = 0.11 M
So:
x = [] = 0.11 M
[] = 0.27 - x
[] = 0.27 - 0.11
[] = 0.16 M
[] = (0.49 - 2x)
[] = (0.49 - 2(0.11))
[] = 0.49 - 0.22
[] = 0.27 M
![K_C = \frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_C%20%3D%20%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
= 9.4
∴ The equilibrium constant at that temperature = 9.4
3.
∆E = ∆m x c ² ∆m = E / c ² ∆m = 3,83•10^-12 / 3•10^8 ² ∆m = 4,256•10^-29 kg
Taking this class as well
