If sulphur (IV) oxide and methane are released simultaneously at the opposite ends of a narrow tube, the rates of diffusion Rso₂ and Rch₄ will be in the ratio?

(S = 32, O = 16, C = 12, H = 1)

Goodness00000

4 years ago

Rso2/Rch4=√mch4/√mso2
=√16/√64
=4/8
=1/2 or 1:2

Myschool Blessing

10 years ago

Thanks for your contributions, corrections have been made.

Ahmadi2020

3 years ago

r&1/m
r1/r2=√m2/m1
Rso2/Rch4=√16/64=√1/4=1/2
Therefore Rso2 and Rch4 = 1:2

Nomzy39

12 years ago

I dnt agree wif dis... D right ans is c)1:2

Jasmine2008

3 years ago

i feel it's 1:2

excel256divine

2 weeks ago

correct

mofe15

2 years ago

√16 : √64
= 4:8
=1/2 / 1:2 as ans

Prince Nemerem

9 years ago

Pls the answer is 4:1

lemibare

11 years ago

why some are calculating like dis, just change to square root, pls dis is how it goes



So2 : Ch4

64 : 16

8 : 4

2 : 1 - so, how come u dnt know d ratio again.

cobetta1

10 years ago

1:4

Donkaz

12 years ago

Hw is it 2:1? D ryt ans is 4:1...SO2/CH4=Square root of d mass no of CH4/d mass no of SO2...Wic is 16/64....wic is 1/4 bt d questn said d ratio of SO2 to CH4 wic is 4:1