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aleksley [76]
3 years ago
8

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:

Hope it helps

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Liquid octane (CH) has a density of 0.7025 g/mL at 20 °C. Find the true mass (murue) of octane when the mass weighed in 18 air i
Goshia [24]

Explanation:

According to Buoyance equation,

          m = [m' \times \frac{1 - \frac{d_{a}}{d_{w}}}{1 - \frac{d_{a}}{d}}]

where,      m = true mass

                 m' = mass read from the balance = 17.320 g

              d_{a} = density of air = 0.0012 g/ml

              d_{w} = density of the balance = 7.5 g/ml

                    d = density of liquid octane = 0.7025 g/ml

Now, putting all the given values into the above formula and calculate the true mass as follows.

      m = [m' \times \frac{1 - \frac{d_{a}}{d_{w}}}{1 - \frac{d_{a}}{d}}]    

          = [17.320 g \times \frac{1 - \frac{0.0012 g/ml}{7.5 g/ml}}{1 - \frac{0.0012 g/ml}{0.7025}}]

          = 17.320 g \times 0.999850                

          = 17.317 g

Thus, we can conclude that the true mass of octane is 17.317 g.

7 0
3 years ago
I need help on #2 a and b. please explain steps and answer
Vaselesa [24]
I think A. o2 is oxygen, h202 is a clear liquid like bleach, And H20 is water lol :P
4 0
3 years ago
If a 2 kg object produces a 16 N force, what is its acceleration?
dolphi86 [110]

Answer:

acceleration (a)=8 m/s square.

7 0
3 years ago
A sample of gas in a sealed container (fixed volume) is heated from room temperature to 80.0°C.
inna [77]
Hello!

A) When the temperature inside the container increases, the pressure inside the container increases as well. 

An example for that is when you have an aerosol can and start heating it. The pressure of the gas inside the aerosol can will start to increase, and that would lead to the exploding of the can if heating is kept for too long. Bombs work on this principle too: The heat from the violent chemical reaction inside the closed compartment increase the pressure of the gases until the fragments are ejected at high velocities. 

B) The effect of temperature on the pressure of a gas illustrates Gay-Lussac's Law.

This law was formulated by the famous French chemist that gives it its name. It relates the expansion of a gas with the increase in temperature when the volume is left constant. The Gay-Lussac's Law can be expressed as follows, for the case of this exercise:

\frac{P1}{T1}= \frac{P2}{T2} \\ \\ P2= \frac{P1}{T1}*T2=P1* \frac{80 degC}{25degC}= P1*3,2

You can see that the factor that is multiplying P1 is higher than 1 for the case of heating from 25 °C to 80 °C, so the pressure will increase.

C) At a molecular level, when the temperature is raised the kinetic energy of the molecules inside the container will increase. This increase in the kinetic energy will cause the molecules to move faster, and to hit the walls of the container more often. This causes an increase in the pressure inside the container because there more hits means more force on the walls of the container, and that is the definition of pressure. 

Have a nice day!
8 0
3 years ago
Which alternative reagents could be used in Place of Benedet's solution<br><br>​
Sliva [168]

Answer:

The Fehling's Test for non-reducing sugar is an alternative to the Benedict's Test. However it is less popular as it less sensitive and requires that the reagents - Fehling's solutions A and B - be kept separate until the experiment is carried out.

Explanation:

7 0
3 years ago
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