The portion of a nerve cell that conducts signals is called an axon. Many of the electrical properties of axons are governed by ion channels, which are protein molecules that span the axon's cell membrane. When open, each ion channel has a pore that is filled with fluid of low resistivity and connects the interior of the cell electrically to the medium outside the cell. In contrast, the lipid-rich cell membrane in which ion channels reside has very high resistivity. In a simple model of an axon conducting a nerve signal, ions move across the cell membrane through open ion channels, which act as purely resistive elements. Assume that a typical open ion channel spanning an axon's membrane has a resistance of 1×10^ 11Ω. Required:
a. If a typical current density (current per unit cross-sectional area) in the cell membrane is 5 mA/cm2 when the voltage across the membrane (the action potential) is 50 mV, what is the number density of open ion channels in the membrane?
b. If a typical current density (current per unit cross-sectional area) in the cell membrane is 5 when the voltage across the membrane (the action potential) is 50 , what is the number density of open ion channels in the membrane?

n/A = 1*10^10 /cm^2

Explanation:

a) Each open ion can be considered as a resistor in the circuit. Then, you have that the equivalent resistance is:

    (1)

(you assume the resistance are in parallel)

R: resistance of an axon's membrane =  1×10^ 11Ω.

Req: equivalent resistance = ?

n: number density of open ion channels

Next, you take into the Ohm's law:

   (2)

I: current

V: voltage acroos the membrane = 50mV

you replace the equation (1) into the equation (2):

    (3)

But yo know that the current is also

I = JA       (4)

J: current density = 5mA/cm^2

A: cross sectional area of the axon's membrane

You replace equation (4) into the equation (3), an d you solve for n/A

Then, you replace the values of the parameters:

hence, the density of ion channels is 1*10^10 /cm^2

  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

the ima (higher than 1m) will be greater than the ima with the height of 1m.