Determine the formula mass of KClO3

Decomposition of potassium chlorate yields potassium chloride and oxygen as:

2KClO

3

→2KCl+3O

2

Thus 2 moles of potassium chlorate yields 3 moles of oxygen gas.

2 moles of potassium chlorate =2×122.5=245g of potassium chlorate

At STP, the volume occupied by 1 mol of gas =22.4 dm

3

the volume occupied by three moles of a gas =3×22.4=67.2dm

3

Therefore, 245g of potassium chlorate yields 67.2dm

3

of oxygen gas

To liberate 6.72 dm

3

oxygen amount of potassium chlorate required is

=

67.2

245

×6.72=24.5g

Hence, 24.5 g of potassium chlorate required to liberate 6.72 dm

3

of oxygen at STP

Hence, 24.5 g of potassium chlorate required to liberate 6.72 dm

3

of oxygen at STP

what is the average time it took the car to travel 0.25 meters?

⇒ 2.23 seconds

what is the average time it took the car to travel 0.50 meters?

⇒ 3.13 seconds

the correct answer to this question is 2