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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Calculate AH for the reaction:<br> 2N2 (g) + 6H20 (g) → 302 (g) + 4NH3(g)

Ratling [72]

Answer:

Explanation:

4NH₃ (g) + 3O₂ (g) ⇒ 2N₂ (g) + 6H₂ O(1)

Δ $H_r={\sum}{ \Delta }H_f(\text{product})-{\sum}{ \Delta }H_f(\text{reactant})$

ΔH r =(2ΔH f(N 2 )+6ΔH f (H 2 O(l)))−(4ΔH f (NH 3 (g))+3ΔH f (O 2 (g)))

ΔH rex =[2×0+6×(−286)]−[4×(−46)+3×0]=−1716+186

ΔH rex =−1532kJ/mol

Thermodynamics is a branch of physical chemistry that studies heat and its effects and interactions. Governed by the four main laws, thermodynamics plays a huge role in physics and chemistry, and is also responsible for the law of conservation of energy, a fundamental rule in science.

7 0

1 year ago

The isomerization reaction, ch3nc → ch3cn, is first order and the rate constant is equal to 0.46 s-1 at 600 k. what is the conce

Harlamova29_29 [7]

According to this formula :
㏑[A] /[Ao] = - Kt
when we have Ao = 0.3 m
and K =0.46 s^-1
t = 20min = 0.2 x 60 =12 s
So by substitution :
㏑[A] / 0.3 = - 0.46 * 12
㏑[A] / 0.3 = - 5.52
by taking e^x for both side of the equation we can get [A]
∴[A] = 0.0012 mol dm^-3

6 0

3 years ago

How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)

zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0

3 years ago

When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C

natka813 [3]

Answer:

$\Delta_{r}U$ of the reaction is -6313 kJ/mol