If 465 cm³ of sulphur(IV)oxide,can be diffuse through porous partition in 30s how long will(a) an equal volume , (b)620cm³ of hydrogen sulphide take to diffuse through the same partition ?(H=1,S=32,O=16)
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how will I substitute the new answer please I am confused please explain more
part a.
mm = molar mass
(rate 1)/(rate 2) = sqrt (mm 2/mm 1)
Substitute the following:
rate 1 = rate SO2 = 465 cc/30 sec
rate 2 = rate H2S = 465 cc/x sec
mm 1 = molar mass SO2 = 64
mm 2 = molar mass H2S = 34
Solve for x sec.
part b. Substitute the new numbers.
Post your work if you get stuck.
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I am confused
th2/to2=√mo2/mo2
To solve this problem, we need to apply 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.
Here are the steps to calculate the diffusion time for hydrogen sulfide (H2S) in comparison to sulfur(IV) oxide (SO2):
Step 1: Determine the molar mass of each gas.
Molar mass of SO2:
Sulfur (S) = 1 atom x 32 g/mol = 32 g/mol
Oxygen (2 O atoms) = 2 atoms x 16 g/mol = 32 g/mol
Total molar mass = 32 g/mol + 32 g/mol = 64 g/mol
Molar mass of H2S:
Hydrogen (2 H atoms) = 2 atoms x 1 g/mol = 2 g/mol
Sulfur (S) = 1 atom x 32 g/mol = 32 g/mol
Total molar mass = 2 g/mol + 32 g/mol = 34 g/mol
Step 2: Calculate the rate of diffusion using Graham's law.
Rate of diffusion of SO2 / Rate of diffusion of H2S = √(Molar mass of H2S / Molar mass of SO2)
Rate of diffusion of SO2 / Rate of diffusion of H2S = √(34 g/mol / 64 g/mol)
Rate of diffusion of SO2 / Rate of diffusion of H2S = √(0.53125)
Rate of diffusion of SO2 / Rate of diffusion of H2S ≈ 0.73
Step 3: Calculate the diffusion time.
Diffusion time of SO2 / Diffusion time of H2S = Rate of diffusion of H2S / Rate of diffusion of SO2
Diffusion time of SO2 / 30s ≈ 0.73
Diffusion time of SO2 ≈ 30s / 0.73 ≈ 41.10s
(a) An equal volume of hydrogen sulfide (620 cm³):
The diffusion time for hydrogen sulfide will be the same as for sulfur(IV) oxide, approximately 41.10 seconds.
(b) 620 cm³ of hydrogen sulfide:
Since the volume is increased by a factor of 1.34 (620 cm³ / 465 cm³), we can use the same ratio to calculate the diffusion time.
Diffusion time of 620 cm³ of H2S = Diffusion time of SO2 x (Volume of H2S / Volume of SO2)
Diffusion time of 620 cm³ of H2S = 41.10s x (620 cm³ / 465 cm³)
Diffusion time of 620 cm³ of H2S ≈ 54.80s
Therefore, 620 cm³ of hydrogen sulfide would take approximately 54.80 seconds to diffuse through the same partition.