Nanochemistry Questions and Answers

Answer: C

Correct answer: carry the drug to its target and release it there

The object is to change where the drug goes rather than what it is: a carrier that circulates, accumulates preferentially at the target - often by exploiting the leaky vasculature of a tumour - and releases its cargo there. The benefit is a lower dose and less exposure of healthy tissue, not a larger dose.

Answer: D

Correct answer: electrical conductance measurably

A gas adsorbing on a nanostructured semiconductor donates or withdraws electrons, changing the carrier concentration in a body so thin that the surface effect dominates the whole conductance. The response is large, fast and reversible, which is what makes tin oxide and zinc oxide nanostructures useful as sensors.

Answer: A

Ultraviolet light generates carriers that oxidise adsorbed organic matter to carbon dioxide and water, and the surface also becomes strongly wetting so rain sheets the residue away. Both effects depend on illumination.

Answer: B

Size and surface chemistry govern where a particle can go in the body and how it interacts once there, so a substance harmless in bulk may not be harmless when finely divided. Assessing each material at each size is what makes the field difficult.

Answer: C

Correct answer: 0.6 nm-1

\[ \frac{S}{V} = \frac{6}{d} = \frac{6}{10} = 0.6\ \mathrm{nm}^{-1}. \] For a sphere the ratio is \( 3/r \), equivalently \( 6/d \). Working with the radius rather than the diameter gives 1.2 and is the usual slip here.

Answer: D

Correct answer: 10

Since the ratio varies as \( 1/d \), reducing the diameter by a factor of ten multiplies it by ten. The total surface area for a given mass rises by the same factor, which is the whole basis of nanoscale catalysis - and equally of the tendency to agglomerate.

Answer: A

Correct answer: 10^21

A centimetre is \( 10^7 \) nanometres, so the cube divides into \( 10^7 \) along each edge and \[ (10^7)^3 = 10^{21} \] cubes in all. The total surface area rises from 6 cm² to 6000 m², roughly the area of a football pitch, from a single cubic centimetre of material.

Answer: B

The specific area is 6 divided by the product of density and diameter, that is 6 over (5 \( \times \) 10\( ^{-6} \) m), giving 1.2 \( \times \) 10\( ^6 \) cm\( ^2 \)/g or about 120 square metres per gram. Such figures are why nanoparticles make effective catalysts.

Answer: C

Correct answer: 20

\[ \frac{5\ \mathrm{nm}}{0.25\ \mathrm{nm}} = 20\ \mathrm{atoms}. \] Twenty atoms across is small enough that the outermost shell holds a substantial fraction of the total - roughly a quarter - which is why particles of this size behave so differently from bulk material of the same composition.

Answer: D

Correct answer: 73 per cent

\[ 100 - 27 = 73\ \mathrm{per\ cent}. \] The arithmetic is trivial; the significance is not. More than a quarter of the atoms lie in a single outermost layer, whereas in a particle a micrometre across the figure would be well under a thousandth. That is what makes nanoparticle properties surface-dominated.