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ivann1987 [24]
4 years ago
9

What is a warm front

Chemistry
2 answers:
Olegator [25]4 years ago
5 0
The changeover region where a warm air mass is replacing a cold air mass
Galina-37 [17]4 years ago
3 0

Answer:the Boundary of an advancing mass of warm air, in particular the leading edge of the warm sector of a low-pressure system.

Explanation:

You might be interested in
Mixing vinegar and with baking soda produces carbon dioxide and water. Is this a chemical change or a physical change?
Sergeeva-Olga [200]
This is a simple chemical change due to what it produces and how it is added together. Hope this helps.
3 0
3 years ago
Read 2 more answers
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
What happens when magma cools during the rock cycle
Mumz [18]

Answer:

Si las condiciones para que el magma permanezca líquido no perduran, el magma se enfriará y solidificará en una roca ígnea. Una roca que se enfría en el interior de la Tierra se denomina intrusiva o plutónica y su enfriamiento será muy lento, produciendo una estructura cristalina de granos grueso.

Explanation:

7 0
3 years ago
Question 5 (1 point)
Aleksandr-060686 [28]

Answer : The pressure it exert under these new conditions will be, 87 atm

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 19 atm

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 100 L

V_2 = final volume of gas = 20 L

T_1 = initial temperature of gas = 25^oC=273+25=298K

T_2 = final temperature of gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{19atm\times 100L}{298K}=\frac{P_2\times 20L}{273K}

P_2=87atm

Therefore, the pressure it exert under these new conditions will be, 87 atm

5 0
3 years ago
Calculate the mass percent of ba(no3)2 in this solution, assuming the density of the solution is 1.000 kg/l.
viktelen [127]
You forgot to include the known characteristics of the solution.

I searched them and copy here:

volume: 1.000 liter

M = 0.0190 M

Now, you can start with the definition of mass percent.

mass percent = (grams of solute / grams of solution) * 100

grams of solute are obtained from the molar concentration:

M = (number of moles of solute) / (volume of solution in liters)

where number of moles = (grams) / (molar mass)

=> M = (grams of solute / molar mass of solute) / (volume of solution in liters)

=> grams of solute = M * (volume of solution in liters) * (molar mass of solute)

And density = (kg of solution / volume of solution in liters) =>

kg of solution = density * volume of solution in liters

grams of solution = density * (volume of solution in liters) * 1000 g/kg

=> mass percent = M * (volume of solution in liters) * (molar mass) / (density * volume of solution in liters * 1000 g/ kg) * 100

=> mass percent = M * molar mass * 10 / density

now replace the values known:

M = 0.0190 mol / liter
density = 1,000 kg / liter

molar mass of Ba(NO3)2 = 137.327 g/mol + 2*14.007 g/mol + 2*3*15.999 g/mol = 256.335 g/mol

=> mass percent = 0.0190 mol/liter * 256.335 g/mol * 10  kg/ g / (1.000 kg/liter)

=> mass percent = 48.7%
6 0
4 years ago
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