A speedy tortoise can run with a speed of and a hare can run with a speed . The hare waits to rest for a time Δt₀ after the race begins and the tortoise starts and then runs as fast as he can, but the tortoise still wins by a shell (length of shell = s). The length of the race track is a distance, d. The race is considered over when the tortoise crosses the finish line.
Using the symbols of the problem, write an equation for the amount of time that the hare ran, .

Explanation:

Thinking process:

Let:

distance = speed × time

for the hare:

for the tortoise:

Let the length of the shell be s. Then the two are related. So:

Let's say the hare ran at x times the speed of the tortoise:

So, the equation still holds :

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂