We are given with the composition of the air inside a breathing apparatus which is
<span>78% N2
21% O2
The total pressure at a given depth is 500 kPa
The partial pressure of oxygen is
500 kPa (0.21) = 105 kPa</span>
The reaction equation is first order with respect to [H+]
when PH1 = -㏒[H+]1 so, when PH = 6
So by substitution:
∴ 6 = -㏒[H+]1
∴[H+]1 = 1 x 10^-6
and when PH2 = -㏒[H+]2 so, when PH2 = 2
so by substitution:
∴ 2 = -㏒[H]2
∴[H]2 = 1 x 10^-2
So the rate of reaction changes by the factor of:
[H2]2/[H]1 = (1 x 10^-2) / (1 x 10^-6) = 10000
It is 10000 times faster when PH decreases from 6 to 2
PH=-log[H⁺]
pH=-log(1.87×10⁻¹³)
pH=12.72
I hope this helps. Let me know if anything is unclear.
Answer: Formal Charges: Hydrogen = 0 and Oxygen = +1
Unshared Pair of electrons: Hydrogen = 0 and Oxygen = 2
Explanation:
The attachment below shows the Lewis structure and the calculations
