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vazorg [7]
3 years ago
15

Compare the models of the superconductor to the CaTiO3 models. What similarities and differences do you notice? How do the coord

ination numbers of the central ions compare?
Chemistry
1 answer:
bulgar [2K]3 years ago
8 0

Answer:

Compare the models of the superconductor to the CaTiO3 models.

What similarities and differences do you notice?

The differences are in the crystal structure unit cell, consisting of five atoms  with calcium atoms at the corners, a titanium atom at the center and oxygen at centers forming an octahedron, and the similarities are in the HTSC cuprates structures.

How do the coordination numbers of the central ions compare?

The Ca+2 cation layers are insulating and donate electrons to the CuO2 planes. The Sro layers are barriers, isolate groups of CuO2 planes from each other, the ca2 and bi2 are charge reservoir layers.

Explanation:

Allow the current to flow without resistence or interruption, through a superconductor material at room temperature, is still a not fullfilled dream in the superconductivity research.

Transition temperatures (Tc) achieving though, have opened the options for many applications, high temperature superconductors are now the main researchs´ focusing, known as perovskites, which are simply ceramics, such as yttrium barium copper oxides (YBCOs) or 1-2-3 compounds and the bismuth strontium calcium copper oxide (BSCCOs) or Pb-BSCO (PBSCCO) are the best insulators known at room temperature, and at liquid nitrogen temperature, the become perfectly superconducting.

The discovery of superconducting transition at 35 K in lanthanum barium copper oxide ceramic system- La2-xBaxCuO4 and the 92 K for 123 systems made a difference among them as these systems contained rare-earth elements.

A great step was gained with the discovery of the first high temperature (Tc) oxide ceramic system, based on Bi-Sr-Cu-O perovskite, which did not have any rare-earth component, followed by several discoveries of these rare-earth free systems, such as the Bi-Sr-Cu-O, which increased Tc to 85 K adding calcium, Br-Sr-Ca-Cu-O system which reached 110 K; Ti-Ba-Ca-Cu-O system which reached a Tc of 125 K, but the Bi systems synthesis and rare-earth systems as YBCO differ in simplicity and getting a monophase superconducting phase is still not fullfilled.

Varying elemental ratios and dopants such as Pb, and so forth has given only partial success, even the Bi compositions which showed to have monophase in Bi systems, reproducibility controlling elemental ratios cause a high percentage of inaccuracy.

In comparisson with the conventional solidstate sintering technique, the glassy precursor route is more efficient and realizable to achieving superconducting monophase with rareearth free BSCCO perovskites or Bi perovskites, with interesting parameters and optimizing factors.

YBCO gave the highest Tc ever in 1987, easy to synthesize and good phase stability, bismuth-based cuprates gave a Tc of 110 Kc, Thalium-based cuprates gave a Tc of 120 to 125 K and mercury-based cuprates that gets a Tc of 135 K, which created a new hope for HTSC based on cuprates.

Most of the known cuprate superconductors belong to a single structural familiy closely realted to each other.

Bistmuth-based cuprates are good HTSC as their grain alignment is along the c-axis, which increases the critical current.

Bi-Sr-Ca-Cu-O

   

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If have a volume of 18 L of a gas at a temperature of 272 K and a pressure of 90 atm, what will be the pressure of the gas if ra
Solnce55 [7]

Answer:

P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)

Explanation:

Given:

P₁ = 90 atm                    P₂ = ?

V₁ = 18 Liters(L)              L₂ = 12 Liters(L)      

=> decrease volume => increase pressure

=> volume ratio that will increase 90 atm is (18L/12L)                                                                  

T₁ = 272 Kelvin(K)          T₂ = 274 Kelvin(K)

=>  increase temperature => increase pressure

=> temperature ratio that will increase 90 atm is (274K/272K)

n₁ = moles = constant    n₂ = n₁ = constant

P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)

By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.

3 0
2 years ago
When considering metal complexation with EDTA, if you are comparing 2 metals, the metal with a higher ____________ will react wi
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If you are comparing 2 metals, the metal with a higher <u>Number of free ions</u> will react with EDTA first

<h3>What is EDTA ?</h3>

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  • Preventing blood clotting of blood samples
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The EDTA will readily react with metals which have a hiogher number of free ions that it can bind with.

Hence we can conclude that If you are comparing 2 metals, the metal with a higher <u>Number of free ions</u> will react with EDTA first.

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7 0
2 years ago
At what height above the earth surface will the gravitational acceleration be 5m
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Answer: -

The acceleration due to gravity at height r = a = GM/r²

Rearranging

r² = GM /a

= (6.674 x 10⁻¹¹ x 5.972 x 10²⁴ ) / 5

r = 8.917 x 10⁶ m

r = 8917 Km

Now Radius of earth = 6371 Km

So height = 8917 - 6371 = 2546 Km

8 0
3 years ago
A mixture of helium and methane gases, at a total pressure of 821 mm Hg, contains 0.723 grams of helium and 3.43 grams of methan
Nookie1986 [14]

<u>Answer:</u>

<u>For 1:</u> The partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

<u>For 2:</u> The mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

<u>Explanation:</u>

<u>For 1:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For helium:</u>

Given mass of helium = 0.723 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

\text{Moles of helium}=\frac{0.723g}{4g/mol}=0.181mol

  • <u>For methane gas:</u>

Given mass of methane gas = 3.43 g

Molar mass of methane gas = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane gas}=\frac{3.43g}{16g/mol}=0.214mol

To calculate the mole fraction , we use the equation:

\chi_A=\frac{n_A}{n_A+n_B}     .......(2)

To calculate the partial pressure of gas, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}       ......(3)

  • <u>For Helium gas:</u>

We are given:

n_{He}=0.181mol\\n_{CH_4}=0.214mol

Putting values in equation 2, we get:

\chi_{He}=\frac{0.181}{0.181+0.214}=0.458

Calculating the partial pressure by using equation 3, we get:

p_T=821mmHg\\\\\chi_{He}=0.458

Putting values in equation 3, we get:

p_{He}=0.458\times 821mmHg=376mmHg

  • <u>For Methane gas:</u>

We are given:

n_{He}=0.181mol\\n_{CH_4}=0.214mol

Putting values in equation 2, we get:

\chi_{CH_4}=\frac{0.214}{0.181+0.214}=0.542

Calculating the partial pressure by using equation 3, we get:

p_T=821mmHg\\\\\chi_{CH_4}=0.542

Putting values in equation 3, we get:

p_{CH_4}=0.542\times 821mmHg=445mmHg

Hence, the partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

  • <u>For 2:</u>

We are given:

Partial pressure of nitrogen gas = 363 mmHg

Partial pressure of carbon dioxide gas = 564 mmHg

Total pressure = (363 + 564) mmHg = 927 mmHg

Calculating the mole fraction of the gases by using equation 3:

<u>For nitrogen gas:</u>

363=\chi_{N_2}\times 927\\\\\chi_{N_2}=\frac{363}{927}=0.392

<u>For carbon dioxide gas:</u>

564=\chi_{CO_2}\times 927\\\\\chi_{CO_2}=\frac{564}{927}=0.608

Hence, the mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

6 0
3 years ago
LiOH+HBr---&gt; LiBr +h20 If you start with 10.0 grams of lithium hydroxide, how many grams of lithium bromide will be produced?
kicyunya [14]
<span>LiOH+HBr---> LiBr +h20. Moles of LiOH = 10/24 = 0.41moles. According to stoichiometry, moles of LiOH = moles of LiBr = 0.41moles. Therefore mass of LiBr =moles of LiBr x molecular weight of LiBr = o.41 x 87 = 35.67g. Hope it helps </span>
7 0
3 years ago
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