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3 years ago

What are nanochemicals and what are there uses?

1 answer:

6 0

Nanochemicals can be defined as chemicals generated by using nanomaterials (materials that possess of size on nanometer dimensions). The nanochemicals are used in multiple different applications including chemical warfare, bicycle making, armor design and military weapons crafting. The most commonly used and observed nanochemicals are carbon nanotubes that are used a ton in industry for applications such as stronger materials (stronger bicycles).

Smart materials are exquisitely designed materials whose property(ies) can be modified with the use of an external stimulus such as temperature, stress, pH, and so on. Some examples of smart materials include shape memory materials, piezoelectric materials, ferrofluids, self-healing materials, and such. Applications involve memory pillows, memory based solar panels (for satellites), light sensitive glasses, and so on.

Specialized materials are made specifically to perform a specified task or function. Applications involve electronic equipment (high purity silicon & germanium), machine tools (high tungsten high carbon steel), dental filling (dental amalgam), and so on.

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Acetic acid, ethanol, acetaldehyde and ethane form a series of 2-carbon molecules which differ in their extent of oxidation. Con

yanalaym [24]

Explanation:

Ethanol can be oxidized to ethanal or acetaldehyde which is further oxidized to acid that is acetic acid.

$CH_3CH_2OH$ → $CH_3CHO$ [oxidation by loss of hydrogen]

An oxidizing agent potassium dichromate(VI) solution is used to remove the hydrogen from the ethanol.

An oxidizing agent used along with dilute sulphuric acid for acidification.

Acetaldehyde can also be reduced back to ethanol again by adding hydrogen to it by using a reducing agent that is sodium tetrahydro borate, NaBH4.

The oxidation of aldehydes to carboxylic acids can be done by the two-step process.

In the first step, one molecule of water is added in the presence of a catalyst that is acidic.

There is a generation of a hydrate. (geminal 1,1-diol).

$CH_3CHO$ → $CH_3CH_2COOH$ [reduction by the gain of electrons]

Here, the oxidizing agent used is $Cr_3O$ in the presence of acetone.

7 0

3 years ago

Problem page in each of the molecules drawn below one chemical bond is colored red. decide whether this bond is likely to be pol

Slav-nsk [51]

The question is incomplete, here is the complete question:

Problem page in each of the molecules drawn below one chemical bond is colored red. Decide whether this bond is likely to be polar or not. if the bond is likely to be polar, write down the chemical symbol for the atom which will have more negative charge.

The image is attached below.

Answer:

For carbon dioxide molecule: The bond is considered as polar and the elecvtronegative atom is oxygen.

For water molecule: The bond is considered as polar and the elecvtronegative atom is oxygen.

Explanation:

There are two types of covalent bonds:

Polar covalent bond: This bond is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond. For Example: $CO_2,NO_2$ etc..
Non-polar covalent bond: This bond is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. For Example: $N_2,O_2$ etc..

In carbon dioxide molecule:

The given bond is present between C and O atom.

Electronegativity value of C = 2.5

Electronegativity value of O = 3.5

Electronegativity difference = (3.5 - 2.5) = 1

As, electronegativity difference is present. So, the bond is considered as polar and the elecvtronegative atom is oxygen.

In water molecule:

The given bond is present between H and O atom.

Electronegativity value of H = 2.1

Electronegativity value of O = 3.5

Electronegativity difference = (3.5 - 2.1) = 1.4

As, electronegativity difference is present. So, the bond is considered as polar and the elecvtronegative atom is oxygen.

8 0

3 years ago

Polar molecules have _____.

Ksju [112]

Answer:

B. Charges ( a slight positive charge on one end, and a slight negative charge on the other).

4 0

2 years ago

Read 2 more answers

5/10

Marianna [84]

Answer: 0.082 atm L k^-1 mole^-1

Explanation:

Given that:

Volume of gas (V) = 62.0 L

Temperature of gas (T) = 100°C

Convert 100°C to Kelvin by adding 273

(100°C + 273 = 373K)

Pressure of gas (P) = 250 kPa

[Convert pressure in kilopascal to atmospheres

101.325 kPa = 1 atm

250 kPa = 250/101.325 = 2.467 atm]

Number of moles (n) = 5.00 moles

Gas constant (R) = ?

To get the gas constant, apply the formula for ideal gas equation

pV = nRT

2.467 atm x 62.0L = 5.00 moles x R x 373K

152.954 atm•L = 1865 K•mole x R

To get the value of R, divide both sides by 1865 K•mole

152.954 atm•L / 1865 K•mole = 1865 K•mole•R / 1865 K•mole

0.082 atm•L•K^-1•mole^-1 = R

Thus, the value of gas constant is 0.082 atm L k^-1 mole^-1

3 0

3 years ago

Read 2 more answers

the concentration of the radio active isotope potassium-40 in a rock sample is found to be 6.25%. what is the age of the rock

julsineya [31]

Answer:

5.0 x 10⁹ years.

Explanation:

It is known that the decay of a radioactive isotope isotope obeys first order kinetics.

Half-life time is the time needed for the reactants to be in its half concentration.

If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).

Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

The half-life of K-40 = 1.251 × 10⁹ years.

For, first order reactions:

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.

Also, we have the integral law of first order reaction:

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (K-40) ([A₀] = 100%).

[A] is the remaining concentration of (K-40) ([A] = 6.25%).

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.

∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.

8 0

3 years ago

Read 2 more answers