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3 years ago

Explain why a nanowire can be longer than 3000nm but still be considered a metallic nanomaterial

Chemistry

1 answer:

Ksivusya [100]3 years ago

8 0

Answer:

A nanowire is a nanostructure, with the diameter of the order of a nanometre (10−9 meters). It can also be defined as the ratio of the length to width being greater than 1000. Alternatively, nanowires can be defined as structures that have a thickness or diameter constrained to tens of nanometers or less and an unconstrained length. At these scales, quantum mechanical effects are important—which coined the term "quantum wires". Many different types of nanowires exist, including superconducting (e.g. YBCO[1]), metallic (e.g. Ni, Pt, Au, Ag), semiconducting (e.g. silicon nanowires (SiNWs), InP, GaN) and insulating (e.g. SiO2, TiO2). Molecular nanowires are composed of repeating molecular units either organic (e.g. DNA) or inorganic (e.g. Mo6S9−xIx).

Explanation:

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What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

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salantis [7]

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Explain why a nanowire can be longer than 3000nm but still be considered a metallic nanomaterial​

Explain why a nanowire can be longer than 3000nm but still be considered a metallic nanomaterial