Answer:
0.049 mol/L.s
Explanation:
The decomposition of hydrogen peroxide is:

![Rate = -\dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{ 2 \Delta [H_2O_2]}{\Delta t}](https://tex.z-dn.net/?f=Rate%20%3D%20-%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%202%20%20%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D)
The rate of decomposition reaction = the rate of formation of
= 0.098 mol/L.s
∴
Rate of formation of


= 0.049 mol/L.s
<span>Let's assume
that the oxygen gas has ideal gas behavior.
Then we can use ideal gas formula,
PV = nRT</span>
Where, P is the pressure of the gas (Pa), V is the volume of the gas
(m³), n is the number of moles of gas (mol), R is the universal gas
constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.
<span>
P = 2.2 atm = 222915 Pa
V = 21 L = 21 x 10</span>⁻³ m³
n = ?
R = 8.314 J mol⁻¹ K⁻¹
<span>
T = 87 °C = 360 K
By substitution,
</span>222915 Pa x 21 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻<span>¹ x 360 K
n
= 1.56</span><span> mol</span>
<span>
Hence, 1.56 moles of the oxygen gas are </span><span>
left for you to breath.</span><span>
</span>
As the pressure on the on a gas cofined above a liquid increases, the solubility of the gas will increase
this also happen when we lower the temperature