Part B - Permeability of the lipid bilayer Some solutes are able to pass directly through the lipid bilayer of a plasma membrane, whereas other solutes require a transport protein or other mechanism to cross between the inside and the outside of a cell. The fact that the plasma membrane is permeable to some solutes but not others is what is referred to as selective permeability. Which of the following molecules can cross the lipid bilayer of a membrane directly, without a transport protein or other mechanism

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Complete Question:

Part B - Permeability of the lipid bilayer Some solutes are able to pass directly through the lipid bilayer of a plasma membrane, whereas other solutes require a transport protein or other mechanism to cross between the inside and the outside of a cell. The fact that the plasma membrane is permeable to some solutes but not others is what is referred to as selective permeability. Which of the following molecules can cross the lipid bilayer of a membrane directly, without a transport protein or other mechanism?

a) proteins

b) water

c) ions

d) sucrose

e) lipids

f) carbon dioxide

g) oxygen

b) water

e) lipids

f) carbon dioxide

g) oxygen

Explanation:

Diffusion is the means by which molecules like water, lipids, carbon dioxide and oxygen easily move across the lipid bilayer of a membrane directly without the use of a transport protein or any another mechanism.

Diffusion is the way or means through which molecules( such as gases or liquids) move across a membrane from and region of higher concentration to a region of lower concentration.

Before a molecule can pass through the semi permeable membrane without the use of any mechanism or transport protein, it has to be:

a) a small molecule

b) an uncharged molecule

c) an hydrophobic molecule

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  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