5. the surface area of the lungs is about 75 m2, and the thickness of the alveolar diffusion layer is about 0.5 μm. the po2 in alveolar air is 100 mmhg, while the po2 of venous blood is 40 mmhg. the diffusion coefficient of o2 in water is about 1.5 x 10-5 cm2s-1. the solubility of o2 is given by henry’s law as [o2] = 0.024 po2; here [o2] is expressed in ml o2 at stpd (standard temperature and pressure, dry: 0º c and 1 atm pressure) per ml of water and po2 is in atmospheres. only the dissolved o2 diffuses. what is the initial rate of o2 diffusion from the aggregate alveoli to the blood? (initial rate means to pretend that the venous po2 is clamped at 40 mm hg) give the answer in ml o2 per min and in mol o2 per min using the ideal gas law. assume 37°, r = 0.082 l-atm mol-1 °k-1. remember that 1 atm = 760 mm hg.

precipitation

explanation:

when it goes from cloud to rain/back to earth in the form of rain

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

can you attach a picture of the lettuce? : )