Answer:
The partial pressure of SO₃ is measured at 82.0 atm.
Explanation:
The equilibrium constant Kp is defined as the ratio of the equilibrium pressures of the gaseous products, each raised to the power of their respective coefficients in the reaction, divided by the pressures of the gaseous reactants raised to their coefficients.
For the given reaction,
2 SO₂(g) + O₂(g) → 2 SO₃(g)
![Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm](https://tex.z-dn.net/?f=Kp%20%3D%200.345%20%3D%20%5Cfrac%7B%28pSO_%7B3%7D%29%5E%7B2%7D%20%7D%7B%28pSO_%7B2%7D%29%5E%7B2%7D%20%5Ctimes%20pO_%7B2%7D%20%7D%5C%5CpSO_%7B3%7D%20%3D%20%5Csqrt%5B%5D%7B0.345%20%5Ctimes%20%28pSO_%7B2%7D%29%5E%7B2%7D%20%5Ctimes%20pO_%7B2%7D%20%7D%20%5C%5CpSO_%7B3%7D%20%3D%20%5Csqrt%5B%5D%7B0.345%20%5Ctimes%20%2835.0%29%5E%7B2%7D%20%5Ctimes%2015.9%20%7D%20%5C%5CpSO_%7B3%7D%20%3D%2082.0%20atm)
Option d is the correct choice, as both belong to the alkali metals category (group one).
<span>128 g/mol
Applying Graham's law of effusion, we can utilize the formula:
r1/r2 = sqrt(m2/m1)
where
r1 = effusion rate of gas 1
r2 = effusion rate of gas 2
m1 = molar mass of gas 1
m2 = molar mass of gas 2
Given that the atomic weight of oxygen is 15.999, the molar mass of O2 = 2 * 15.999 = 31.998.
We can now insert the known values into Graham's equation to find m2.
r1/r2 = sqrt(m2/m1)
2/1 = sqrt(m2/31.998)
4/1 = m2/31.998
Thus, we find m2 to be 127.992.
Rounding to three significant figures yields 128 g/mol</span>
The slight warm feeling noticed at the valve stem when air is pumped into the tire is likely due to the kinetic energy generated by the friction from the pump and the resultant increase in gas pressure within the tire.
