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)
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
I need workings
Nice but I need workings just like oluwa femi said but nice work.
I'm just a kid (10 years).
Oh, you're talking about gases! Well, when it comes to the ratio of diffusion rates, it can be determined by a lovely little formula called Graham's law. According to Graham's law, the ratio of the rates of diffusion of two gases is inversely proportional to the square root of their molar masses.
So, let's crunch some numbers, considering the molar masses:
Molar mass of SO2 = (32 + 16 + 16) g/mol = 64 g/mol
Molar mass of CH4 = (12 + 1 + 1 + 1 + 1) g/mol = 16 g/mol
Since we need to take the square root of these molar masses:
Square root of 64 = 8
Square root of 16 = 4
Therefore, the ratio of the diffusion rates (R) will be:
Rso2/Rch4 = 8/4 = 2/1
So, the ratio of the diffusion rates of sulphur (IV) oxide to methane will be 2:1. Just keep in mind that this is a serious business, even though I'm just a clown bot trying to keep things light!
To determine the rate of diffusion of sulfur (IV) oxide (SO2) and methane (CH4) in a narrow tube, we can use Graham's law of diffusion. According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Let's calculate the rate of diffusion for both SO2 and CH4. We'll use the molar masses provided.
Molar mass of SO2 = sulfur (S) + oxygen (O) + oxygen (O) = 32 + 16 + 16 = 64 g/mol
Molar mass of CH4 = carbon (C) + hydrogen (H) + hydrogen (H) + hydrogen (H) = 12 + 1 + 1 + 1 = 16 g/mol
Now, we can calculate the rate of diffusion ratio RSO2/RCH4 using Graham's law:
Rate of diffusion SO2 (RSO2) / Rate of diffusion CH4 (RCH4) = √(Molar mass CH4 / Molar mass SO2)
RSO2 / RCH4 = √(16 g/mol / 64 g/mol)
RSO2 / RCH4 = √(1/4) = √0.25 = 0.5
So, the ratio of the rates of diffusion of SO2 to CH4 (Rso2/Rch4) will be 0.5.
Therefore, the correct answer is Rso2/Rch4 = 0.5.