PH+pOH=14
4.5+pOH=14
pOH=14-4.5
pOH = 9.5
We know that
[H⁺][OH⁻]=10⁻¹⁴,so
-log([H⁺][OH⁻])=-log10⁻¹⁴
-log [H⁺]+ (- log[OH⁻])= - log10⁻¹⁴,
- log[H⁺] = pH
- log[OH⁻] = pOH
- log10⁻¹⁴=-(-14)log10=14*1=14
pH + pOH =14
Answer:
The total energy to break all the bonds in 1 mole of 1-propanol, C₃H₈O, is 4411 kJ/mol
Explanation:
We note that propanol, C₃H₈O is also known as 1-propanol is written as follows;
CH₃CH₂CH₂OH which gives
CH₃-CH₂-CH₂-OH
Hence, the total number of bonds are;
C-H Bonds = 3 + 2 + 2 = 7
C-O Bonds = 1
O-H Bond = 1
C-C Bonds = 2
The bond energies are as follows;
C-H Bonds = 413 kJ/mol
C-O Bonds = 358 kJ/mol
O-H Bond = 468 kJ/mol
C-C Bonds = 347 kJ/mol
Energy required to break the bonds in 1-propanol is therefore;
C-H Bonds = 413 kJ/mol × 7 = 2,891 kJ/mol
C-O Bonds = 358 kJ/mol × 1 = 358 kJ/mol
O-H Bond = 468 kJ/mol × 1 = 468 kJ/mol
C-C Bonds = 347 kJ/mol × 2 = 694 kJ/mol
The total energy to break all the bonds in 1 mole of 1-propanol = 4411 kJ/mol.
