Question 1) Nanotechnology, the field of trying to build ultrasmallstructures on

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Question 1) Nanotechnology, the field of trying to build ultrasmallstructures one atom at a time, has progressed in recent years. Onepotential application of nanotechnology is the construction ofartificial cells. The simplest cells would probably mimic red bloodcells, the body's oxygen transporters. For example, nanocontainers,perhaps constructed of carbon, could be pumped full of oxygen andinjected into a person's bloodstream. If the person neededadditional oxygen-due to a heart attack perhaps, or for the purposeof space travel-these containers could slowly release oxygen intothe blood, allowing tissues that would otherwise die to remainalive. Suppose that the nanocontainers were cubic and had an edgelength of 23 nanometers. A)What is the volume of one nanocontainer?(Ignore the thickness of the nanocontainer's wall.)(liters) B)Suppose that each nanocontainer could containpure oxygen pressurized to a density of 86g/L . How many grams of oxygencould be contained by each nanocontainer? (answer ing) C)Normal air contains about 0.28 of oxygen per liter. Anaverage human inhales about 0.50 of air per breath andtakes about 20 breaths per minute. How many grams of oxygen does ahuman inhale per hour? (Assume two significantfigures.) Express your answer using two significant figures. D)What is the minimum number of nanocontainers that a person wouldneed in their bloodstream to provide 1 hour's worth ofoxygen? Express your answer using two significant figures. E)What is the minimum volume occupied by the number ofnanocontainers computed in part D? Express your answer using two significant figures. Vmin=_________L F)Is such a volume feasible, given that total blood volume in anadult is about 5 liters? Yes or no

Explanation / Answer

(a) 1 nm = 10^(-3) micrometer = 10^(-6) mm = 10^(-9) m = 10^(-8) dm Volume of a cube: V = a³ V = 25³*10^(-8)³ dm³ = 15625 x 10^(-24) dm³ = 1.5625 x 1E-020 Liter (b) The density ? = m/V; m = V x ? m = 1.5625 x 1E-020 Liter x 86 g/L = 1.34 x 1E-018 g = 1.34 ag (attogramm) (c) 0.28 g/L x 0.5 L x 20/min x 60 min/h = 168 g/h (d) Nmin. = 168 g/ 1.34 x 1E-018 g =1.25 x E020 (e) V = 1.25 x E020 x 1.5625 x 1E-020 Liter = 1.95 Liter (f) I don't think so.

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