Why are viruses called obligate intracellular parasites?

Igneous rocks with an andesitic composition ________.

slega [8]

Are found along volcanic island arcs

6 0

3 years ago

Is jewelry gold an element, compound or mixture?

suter [353]

Gold is a mixture Becasue it contains more than one element

3 0

3 years ago

In a laboratory setting, chemists strive to be

Kitty [74]

Accurate and precise

3 0

4 years ago

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

bulgar [2K]

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

8 0

4 years ago

Decide which of the following statements are True and which are False about equilibrium systems:A large value of K means the equ

ivanzaharov [21]

Answer:

a. True

b. False

c. True

d. False

e. False

Explanation:

A. (true) The equilibrium constant K is defined as

$\frac{Products}{reagents}$

In any case

aA +Bb ⇌ Cd +dD

where K is:

$K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

A large value on K means that the concentration of products is bigger than the concentrations of reagents, so the forward reaction is favored, and the equilibrium lies to the right.

B. (False) When we work with gases, we use partial pressure to make calculations in the equilibrium, so we estimate Kp as:

$Kp= \frac{(P_{C})^{c}(P_{D})^{d}}{(P_{A})^{a}(P_{B})^{b}}$

Using the ideal gas law, we can get a relationship between K and Kp

Pv=nRT where $P=\frac{n}{v}*RT$ we know that $\frac{n}{v}$ is the molar concentration. When we replace P in the expression for Kp we get:

$Kp= \frac{[C]^{c}*(RT)^{c}[D]^{d}*(RT)^{d}}{[A]^{a}*(RT)^{a}[B]^{b}*(RT)^{b}}$

Reorganizing the equation:

$Kp= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}*\frac{(RT)^{c+d}}{(RT)^{a+b}}$

We can see K in the expression

$Kp= K*(RT)^{c+d-a-b}$

Delta n = c+d-a-b

$Kp= K*(RT)^{delta n}$

For the reaction

$H_{2}(g) + F_{2}(g)-- equilibrium---2HF(g)$

Delta n = 2-1-1=0

$Kp= K*(RT)^{0}$

So Kp=K in this case.

C. (true) The value of K just depends on the temperature that’s why changing the among of products won’t have any effect on its value.

D. (false) as we can see this reaction involve a heterogeneous system with solids and gases. For convention the concentration for solids and liquids can be considered constant during the reaction that’s why they’re not include in the calculation for the equilibrium constant. Taking this into account the expression for the equilibrium for this reaction is:

$CaCO_{3}(s)---equilibrium----CaO(s) + CO_{2}(g)$

$K= [CO_{2}]$

So we can see that $[CaCO_{3}]$ is not include in the expression.

E. (False) The equilibrium is defined as the point where the rate of the forward reaction is the same to the rate of the reverse reaction. The value of K is telling you which reaction is favored but the rate of both reactions is the same in this point. (see picture)

3 0

3 years ago