Answer:
Part 1: - 1.091 x 10⁴ J/mol.
Part 2: - 1.137 x 10⁴ J/mol.
Explanation:
Part 1: At standard conditions:
At standard conditions Kp= 81.9.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.
Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.
For the reaction:
I₂(g) + Cl₂(g) ⇌ 2ICl(g).
Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.
Answer:
No, because Flourine can only form 1 bond, thus backbonding is not obtainable
Answer:
pH is calculated by taking negative logarithm of hydrogen ion concentration. Acids have pH ranging from 1 to 6.9, bases have pH ranging from 7.1 to 14 and neutral solutions have pH equal to 7. Thus the sum of pH and pOH is 14
Explanation:
The answer is B) gain 8 electrons
