This design enabled them to control for several factors. the total area of the patch plus the corridor, or the patch plus the wings was the same. the total amount of edge (perimeter of each patch) was similar between the two treatments. both patch types were the same distance from the central patch.

Graph 1: This design does not control for distance from the central patch.

Graph 2: This design does not control for area of the patch.

Graph 3: In this design, there is no difference between the treatments.

Graph 4: This design does not control for edge effects.

Explanation:

  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

gravity

explanation: