This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 27 °C = 300 K
Initial Pressure = P₁ = constant
Initial Volume = V₁ = 8 L
Final Temperature = T₂ = 78 °C = 351 K
Final Pressure = P₂ = constant
Final Volume = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
V₁ / T₁ = V₂ / T₂
Solving for V₂,
V₂ = (V₁ × T₂) ÷ T₁
Putting Values,
V₂ = (8 L × 351 K) ÷ 300 K
V₂ = 9.38 L
Answer:
The final electron acceptor of the electron transport chain is oxygen
Explanation:
Four electrons gotten from cytochrome c are involved in the conversion of a molecule of oxygen (O2) to two molecules of water (H2O). This final electron transfer occurs in complex IV. Complex IV, also known as cytochrome c oxidase, facilitates the the use of four protons from the matrix of the mitochondrion, in the production of water molecules while pumping four protons to the intermembrane space of the mitochondrion.
The ideal gas law:
![pV=nRT \Rightarrow p=\frac{nRT}{V}](https://tex.z-dn.net/?f=pV%3DnRT%20%5CRightarrow%20p%3D%5Cfrac%7BnRT%7D%7BV%7D)
p - pressure, n - number of moles, R - the gas constant, T - temperature, V - volume
The volume and temperature of all three containers are the same, so the pressure depends on the number of moles. The greater the number of moles, the higher the pressure.
The mass of gases is 50 g.
![Ar \\ M \approx 39.948 \ \frac{g}{mol} \\ n=\frac{50 \ g}{39.948 \ \frac{g}{mol}} \approx 1.25 \ mol \\ \\ SF_6 \\ M \approx 146.06 \ \frac{g}{mol} \\ n=\frac{50 \ g}{146.06 \ \frac{g}{mol}} \approx 0.34 \ mol \\ \\ Cl_2 \\ M=70.9 \ \frac{g}{mol} \\ n=\frac{50 \ g}{70.9 \ \frac{g}{mol}} \approx 0.71 \ mol](https://tex.z-dn.net/?f=Ar%20%5C%5C%0AM%20%5Capprox%2039.948%20%5C%20%5Cfrac%7Bg%7D%7Bmol%7D%20%5C%5C%0An%3D%5Cfrac%7B50%20%5C%20g%7D%7B39.948%20%5C%20%5Cfrac%7Bg%7D%7Bmol%7D%7D%20%5Capprox%20%201.25%20%5C%20mol%20%5C%5C%20%5C%5C%0ASF_6%20%5C%5C%0AM%20%5Capprox%20146.06%20%5C%20%5Cfrac%7Bg%7D%7Bmol%7D%20%5C%5C%0An%3D%5Cfrac%7B50%20%5C%20g%7D%7B146.06%20%5C%20%5Cfrac%7Bg%7D%7Bmol%7D%7D%20%5Capprox%200.34%20%5C%20mol%20%5C%5C%20%5C%5C%0ACl_2%20%5C%5C%0AM%3D70.9%20%5C%20%5Cfrac%7Bg%7D%7Bmol%7D%20%5C%5C%0An%3D%5Cfrac%7B50%20%5C%20g%7D%7B70.9%20%5C%20%5Cfrac%7Bg%7D%7Bmol%7D%7D%20%5Capprox%200.71%20%5C%20mol)
The greatest number of moles is in the container with Ar, so there is the highest pressure.
Answer:
The 8-hour TWA exposure for the employee is 101 ppm and it exceeds the OSHA PEL of 100 ppm for ethylbenzene.
Explanation:
The TWA for 8 hours is calculated by the sum of airbone concentrations multiplied by the time it has been exposed to that period. The total is divided by 8 which refers to the 8 hours total the employee has been exposed.
TWA = (125x2+88x2+112x2.5+70x1.5)/8.
The OSHA PEL is a known number for every compound and it can be find in PEL tables. In the case of ethylbenezene, it is 100 ppm.