Graham's Law of Diffusion Quiz

What is the main focus of lesson nine in the video?

• Review of previous lessons

• Introduction to Graham's Law

• Solving questions 8 and 9

• Introduction to the chemistry of the mole

In the first method of solving question 8, what is the initial step?

• Use the formula for time and molecular mass

• Determine the molecular mass

• Equalize the volumes of the gases

• Calculate the rate of diffusion

What is the molecular mass of Carbon Peroxide used in the first method?

• 46

• 44

• 32

• 30

What is the time taken for 150 cm³ of nitrogen peroxide to diffuse in the first method?

• 60 seconds

• 46 seconds

• 45 seconds

• 30 seconds

In the second method of solving question 8, what is the rate of Carbon Peroxide diffusion?

• 5.55 cm³/s

• 3.33 cm³/s

• 2.22 cm³/s

• 4.44 cm³/s

What is the relationship between rate and molecular mass in the second method?

• Rate is directly proportional to molecular mass

• Rate is independent of molecular mass

• Rate is inversely proportional to the square root of molecular mass

• Rate is equal to molecular mass

What is the molecular mass of sulfur dioxide used in question 9?

• 46

• 64

• 44

• 32

In question 9, what is the rate of oxygen diffusion calculated as?

• 0.8 cm³/s

• 2.4 cm³/s

• 1.2 cm³/s

• 1.5 cm³/s

What is the final volume of sulfur dioxide calculated in question 9?

• 50 cm³

• 70 cm³

• 63.7 cm³

• 60 cm³

What topic will be covered in the next class after this lesson?

• Graham's Law of Diffusion

• The chemistry of the mole

• Extended questions

• Review of form three