Lol. It is now a solution and not really inseparable, so your only logical option would be to throw it away and try again.
Answer:
Explanation:
![[A]_0](https://tex.z-dn.net/?f=%5BA%5D_0)
= Initial concentration = 1.28 M
![[A]](https://tex.z-dn.net/?f=%5BA%5D)
= Final concentration = ![0.17[A]_0](https://tex.z-dn.net/?f=0.17%5BA%5D_0)
k = Rate constant = 0.0632 s
t = Time taken
For first order reaction we have the relation
![kt=\ln\dfrac{[A]_0}{[A]}\\\Rightarrow t=\dfrac{\ln\dfrac{[A]_0}{[A]}}{k}\\\Rightarrow t=\dfrac{\ln\dfrac{[A]_0}{0.17[A]_0}}{0.0632}\\\Rightarrow t=28.037\ \text{s}](https://tex.z-dn.net/?f=kt%3D%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B%5BA%5D%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B%5BA%5D%7D%7D%7Bk%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B0.17%5BA%5D_0%7D%7D%7B0.0632%7D%5C%5C%5CRightarrow%20t%3D28.037%5C%20%5Ctext%7Bs%7D)
Time taken to reach the required concentration would be .
Missing question:
Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that:
1. The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose.
2. The temperature of the air is constant (body temperature).
3. The air acts as an ideal gas.
4. Salt water has an average density of around 1.03 g/cm^3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm.
T = 37°C = 310 K.
p₁ = 2,5 atm = 253,313 kPa.
p₂ = 1 atm = 101,325 kPa.
Ideal gas law: p·V = n·R·T.
n₁ = 253,313 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₁ = 0,589 mol.
n₂ = 101,325 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₂ = 0,2356 mol.
Δn = 0,589 mol - 0,2356 mol = 0,3534 mol.
The answer is false i just did that question
