If sulphur (iv) oxide and methane are released simultaneously at the opposite end of a narrow tube the rate of diffusion Rso2 Rch2 will be in the ratio of (s=32 o=16 c=12 h =1)

Answer 1

mm = molar mass
molar mass SO2 = 64
molar mass CH4 = methane = 16
ratio rate SO2/rate CH4 = sqrt(mm CH4/SO2) = sqrt (16/64) = 1/2
The problem didn't ask for it but I drew a tube of length 30 cm, released CH4 at one end and SO2 at the other. The two meet in the tube 20 cm from the CH4 end and 10 cm from the SO2 end so rate of CH4 is twice that of the SO2 or SO2 = 1/2 rate CH4

Answer 2

I need workings

Answer 3

Nice but I need workings just like oluwa femi said but nice work.
I'm just a kid (10 years).