If he(g) has an average kinetic energy of 5610 j/mol under certain conditions, what is the root mean square speed of o2(g) molecules under the same conditions?

Kinetic energy is the energy possessed by a particle or a body when in motion.
It is calculated by 1/2 mv², where m is the mass and v is the velocity.
But from the Grahams law of diffusion
 V(he)/V(o2) = √(M(o2)/M(he)
                    = √(32/4)
                    = 2.83
This means that Helium diffuses at a rate of 2.83 times faster than oxygen.
Since the conditions are the same then the two gases will have equal kinetic energy,
Hence; ( He= 4g/mol, O2 = 32g/mol
5610 = 0.5 × 4/1000 ×v²
     v² = 2805000
     v = 1674.8 m/s for He
Therefore, since He diffuses at a rate of 2.83 times faster than oxygen
Then, the velocity of Oxygen will be;
           1674.8/2.83
            = 591.8 m/s
Thus, the velocity of oxygen molecules is 591.8 m/s
 

The mole ratio of hydrogen to ammonia is 3: 2.