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3 years ago

What are the compounds used in nanotechnology

1 answer:

7 0

The remaining 3 % are nanomaterials made from e.g. aluminium oxide, barium titanate, titanium dioxide, cerium oxide and zinc oxide. Carbon nanotubes, graphene and fullerenes have annual production amounts in the hundred tonnes range. Nanosilver is estimated to be produced in about 20 tonnes per year.

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What is the mass of 531 ml of a liquid that has a density of 0.760 g/ml?

pogonyaev

Mass is density times volume
0.760 x 531 = 403.56g
404g if 3 sig figs

4 0

3 years ago

How many atoms are in each compound ?

noname [10]

Compounds are made of two or more atoms of different elements, such as water (H2O) and methane (CH4). Atoms are not drawn to scale. Molecules of compounds have atoms of two or more different elements.

7 0

3 years ago

What is the energy of a wave with wavelength of 4.2 x 10-7 m. (Hint: Calculate for frequency first.)

erik [133]

Answer:

Option B. 4.74×10¯¹⁹ J.

Explanation:

The following data were obtained from the question:

Wavelength (λ) = 4.2×10¯⁷ m

Energy (E) =.?

Next, we shall determine the frequency of the wave. This can be obtained as follow:

Wavelength (λ) = 4.2×10¯⁷ m

Velocity (v) = constant = 3×10⁸ m/s

Frequency (f) =.?

v = λf

3×10⁸ = 4.2×10¯⁷ × f

Divide both side by 4.2×10¯⁷

f = 3×10⁸ / 4.2×10¯⁷

f = 7.143×10¹⁴ s¯¹

Therefore, the frequency of the wave is 7.143×10¹⁴ s¯¹.

Finally, we shall determine the energy of the wave using the following formula

E = hf

Where

E is the energy.

h is the Planck's constant

f is the frequency

Thus, the enery of the wave can be obtained as follow:

Frequency (f) = 7.143×10¹⁴ s¯¹.

Planck's constant = 6.63×10¯³⁴ Js

Energy (E) =..?

E = hf

E = 6.63×10¯³⁴ × 7.14×10¹⁴

E = 4.74×10¯¹⁹ J

Therefore, the energy of the wave is 4.74×10¯¹⁹ J.

5 0

3 years ago

construct the lewis structure of oo . draw the structure by placing atoms on the grid and connecting them with bonds. include al

ankoles [38]

In Lewis dot structures, you draw the atom in the center and then surround the atom with its valence electrons. The Lewis structure for O is as shown in the attached diagram.

<h3>What is the Lewis structure of O ?</h3>

Lewis Structure of an atom of oxygen contains 6 electrons in the valence shell. Four valence electrons exist in lone pairs. It implies that oxygen atom must participate in two single bonds or one double bond in order to have an octet configuration.

A simplified representation of the valence shell electrons in a molecule is called Lewis Structure. It shows how electrons are arranged around individual atoms in the molecule.

To know more about Lewis structure, refer

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1 year ago

A 50.0 mg sample of iodine-131 was placed in a container 32.4 days ago. if its half-life is 8.1 days, how many milligrams of iod

sertanlavr [38]

3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

What is Half life?

The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

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1 year ago