Calculate the mean translational kinetic energy of molecules at 27°C and the root-mean-squared speed of oxygen molecules.

$$E_k = \frac{3}{2} (1.38 \times 10^{-23} \text{ J/K}) (300.15 \text{ K})$$ $$E_k = 1.5 \times 1.38 \times 10^{-23} \times 300.15 \text{ J}$$ $$E_k = 6.21 \times 10^{-21} \text{ J}$$

The mean translational kinetic energy of the molecules is $\boxed{\text{6.21} \times 10^{-21} \text{ J}}$.

The mass of a hydrogen molecule is $3.34 \times 10^{-27} \text{ Kg}$, and an oxygen molecule is 16 times more massive.

$$m_O = 16 \times (3.34 \times 10^{-27} \text{ Kg})$$ $$m_O = 53.44 \times 10^{-27} \text{ Kg}$$

$$v_{rms,O} = \sqrt{\frac{3 \times (1.38 \times 10^{-23} \text{ J/K}) \times (300.15 \text{ K})}{53.44 \times 10^{-27} \text{ Kg}}}$$ $$v_{rms,O} = \sqrt{\frac{1242.621 \times 10^{-23}}{53.44 \times 10^{-27}}} \text{ m/s}$$ $$v_{rms,O} = \sqrt{23.25339 \times 10^{4}} \text{ m/s}$$ $$v_{rms,O} = \sqrt{232533.9} \text{ m/s}$$ $$v_{rms,O} = 482 \text{ m/s}$$

The root-mean-squared speed of the oxygen molecules is $\boxed{\text{482 m/s}}$.