Answer:
A and C are true , B and D are false
Explanation:
For A)
from the first law of thermodynamics (in differential form)
dU= δQ - δW = δQ - PdV
from the second law
dS ≥ δQ/T
then
dU ≤ T*dS - p*dV
dU - T*dS + p*dV ≤ 0
from the definition of Gibbs free energy
G=H - TS = U+ PV - TS → dG= dU + p*dV + V*dp - T*dS - S*dT
dG - V*dp + S*dT = dU - T*dS + p*dV ≤ 0
dG ≤ V*dp - S*dT
in equilibrium, pressure and temperature remains constant ( dp=0 and dT=0). Thus
dG ≤ 0
ΔG ≤ 0
therefore the gibbs free energy should decrease in an spontaneous process → A reaction with a negative Gibbs standard free energy is thermodynamically spontaneous under standard conditions
For B) Since the standard reduction potential is related with the Gibbs standard free energy through:
ΔG⁰=-n*F*E⁰
then, when ΔG⁰ is negative , E⁰ is positive and therefore a coupled redox reaction with a positive standard reduction potential is thermodynamically spontaneous.
There are 2 different types of atoms there, that's why
This is not possible for the earth has a 13 degrees tilt. this means the earth would need quite an interesting orbit path for it to be possible.
Answer:
In order to obtain eight valence electrons (an octet), the potassium atom will transfer its single valence electron to the fluorine atom.