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Nanotechnology, the field of trying to build ultrasmall structures one atom at a time, has progressed in recent years. One potential application of nanotechnology is the construction of artificial cells. The simplest cells would probably mimic red blood cells, the body's oxygen transporters. For example, nanocontainers, perhaps constructed of carbon, could be pumped full of oxygen and injected into a person's bloodstream. If the person needed additional oxygen-due to a heart attack perhaps, or for the purpose of space travel-these containers could slowly release oxygen into the blood, allowing tissues that would otherwise die to remain alive. Suppose that the nanocontainers were cubic and had an edge length of 23 nanometers.

A)What is the volume of one nanocontainer?

B)Suppose that each nanocontainer could contain pure oxygen pressurized to a density of 85 g/Lg/L . How many grams of oxygen could be contained by each nanocontainer?

C)Normal air contains about 0.28 gg of oxygen per liter. An average human inhales about 0.50 LL of air per breath and takes about 20 breaths per minute. How many grams of oxygen does a human inhale per hour?

D)What is the minimum number of nanocontainers that a person would need in their bloodstream to provide 1 hour's worth of oxygen?

E)What is the minimum volume occupied by the number of nanocontainers computed in part D?

Expert Solution

Step 1

Since we answer up to three subparts, we will answer the first three. Please resubmit the question and specify the other subparts you would like to get answered.

Given: The edge length is $23 nm$ or $23 \times 10^{- 9} m$

To determine: (A) The volume of one nanocontainer

The nanocontainers are cubic, so the volume of the cube is

$V = a^{3}$

where a is the edge length

The volume of one nanocontainer is

$V = {(23 \times 10^{- 9} m)}^{3} V = 12167 \times 10^{- 27} m^{3}$

(B) The density of the oxygen is $85 g / L$

Note: $1 m^{3} = 1000 L$ , so $1 L = \frac{1}{1000} m^{3}$

The density is

$85 g / L = 85 \frac{g}{\frac{1}{1000} m^{3}} = 85000 g / m^{3}$

The density is

$d = \frac{m}{V}$ .............................. (1)

where d is the density and m is the mass

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