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3 years ago

Does anyone have any information about "trititanium pentoxide" or "titanium oxide"?

Chemistry

1 answer:

hjlf3 years ago

7 0

Answer:

Commonly available heat-storage materials cannot usually store the energy for a prolonged period. If a solid material could conserve the accumulated thermal energy, then its heat-storage application potential is considerably widened. Here we report a phase transition material that can conserve the latent heat energy in a wide temperature range, T<530 K and release the heat energy on the application of pressure. This material is stripe-type lambda-trititanium pentoxide, λ-Ti3O5, which exhibits a solid–solid phase transition to beta-trititanium pentoxide, β-Ti3O5. The pressure for conversion is extremely small, only 600 bar (60 MPa) at ambient temperature, and the accumulated heat energy is surprisingly large (230 kJ L−1). Conversely, the pressure-produced beta-trititanium pentoxide transforms to lambda-trititanium pentoxide by heat, light or electric current. That is, the present system exhibits pressure-and-heat, pressure-and-light and pressure-and-current reversible phase transitions. The material may be useful for heat storage, as well as in sensor and switching memory device applications.

Explanation:

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The volume density of atoms for a bcc lattice is 5 x 1026 m-3. Assume that the atoms are hard spheres with each atom touching it

ollegr [7]

Explanation:

It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.

This means that volume occupied by 2 atoms is equal to volume of the unit cell.

So, according to the volume density

$5 \times 10^{26} atoms = 1 [tex]m^{3}$

2 atoms = $\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms$

= $4 \times 10^{-27} m^{3}$

Formula for volume of a cube is $a^{3}$ . Therefore,

Volume of the cube = $4 \times 10^{-27} m^{3}$

As lattice constant (a) = $(4 \times 10^{-27} m^{3})^{\frac{1}{3}}$

= $1.59 \times 10^{-9} m$

Therefore, the value of lattice constant is $1.59 \times 10^{-9} m$ .

And, for bcc unit cell the value of radius is as follows.

r = $\frac{\sqrt{3}}{4}a$

Hence, effective radius of the atom is calculated as follows.

r = $\frac{\sqrt{3}}{4}a$

= $\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m$

= $6.9 \times 10^{-10} m$

Hence, the value of effective radius of the atom is $6.9 \times 10^{-10} m$ .

3 0

3 years ago

What would happen to the results if the plastic wrap was dirty in a water filtration experiment?

weeeeeb [17]

Answer:

The plastic wrap of the covered cup acts like the atmosphere, and traps the water vapor. In a real cloud, the water vapor cools back into liquid water. In the covered cup, the air can only hold so much vapor, and the vapor condenses back to liquid water forming a “rain cloud” on the plastic wrap.

Explanation:

4 0

3 years ago

Why does the velocity of catalyzed reaction reach maximum at certain concentration of enzyme?

OleMash [197]

Answer:

Because the total energy of the enzyme has been used up and actions already complete

Explanation:

Catalyst are substance known to speed up rate of chemical reaction just like taking a short pathway. When velocity of catalyzed reaction reach maximum, at that moment reaction has reach it full potential (there is an equilibrium in reaction). there won't be any changes on further addition, and mostly indicate that Reaction is Complete

4 0

3 years ago

What is the volume of ammonia produced at 243 K at a pressure of 1.38 atm by the unbalanced reaction on the left if 5740 moles o

Neporo4naja [7]

Answer:

49671 L is the produced volume of ammonia

Explanation:

We think the reaction of ammonia 's production:

N₂(g) + 3H₂(g) → 2NH₃ (g)

We have the moles of each reactant so let's determine the limiting reactant:

Ratio is 1:3. 1 mol of nitrogen reacts with 3 moles of H₂

Then, 1720 moles of N₂ will react with (1720 .3) /1 = 5160 moles of H₂

We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂

1 mol of N₂ can produce 2 moles of ammonia

Therefore 1720 moles of N₂ will produce (1720 . 2) /1 = 3440 moles of NH₃

We apply now, the Ideal Gases Law → P . V = n . R .T

V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm

V = 49671 L

We confirm that the nitrogen was the limiting reactant

3 moles of H₂ need 1 mol of nitrogen to react

Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂

It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles

6 0

3 years ago

Which reactions are oxidation-reduction reactions? check all that apply. fes(s) 2hcl(aq) â†’ h2s(g) fecl2(g) agno3(aq) nacl(aq)

Dafna1 [17]

oxidation-reduction reactions are -

2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)
Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

For reaction,

2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)

On reactant side:

Oxidation state of Carbon = +2

Oxidation state of Oxygen = 0

On product side:

Oxidation state of Carbon = +4

Oxidation state of Oxygen = -2

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is oxidizing and the oxygen's oxidation state is decreasing from 0 to -2. As a result, it is decreasing.

For reaction,

Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g)

When reacting:

Iron's oxidation state is +3.

Carbon's oxidation state is +2.

On product side:

Iron's oxidation state is zero.

Carbon's oxidation state is +4.

Here, carbon's oxidation state is rising from +2 to +4. As a result, it is being oxidized and the iron's oxidation state is changing from +3 to 0. As a result, it is decreasing.

To learn more about oxidation-reduction from given link

brainly.com/question/5794822

#SPJ4

7 0

1 year ago

Does anyone have any information about "trititanium pentoxide" or "titanium oxide"? ​

Does anyone have any information about "trititanium pentoxide" or "titanium oxide"?