Answer:
Black holes are merely the most exotic example of the general principle that ... quantum computer stores bits on protons and uses magnetic fields to flip them. ... Powered by Standard Model software, the universe computes
Explanation:because they are mostly exotic
A chemical element that has an atomic number less than 58 and an atomic mass greater than 135.6m is barium (atomic no. 56 and atomic mass137.13 ) and lanthanum (atomic no. 57 and atomic mass 135.6).
<h3>Give a brief introduction about Barium and Lanthanum.</h3>
Barium is an element with the symbol Ba and atomic number 56. It is an alkaline earth metal that is soft and silvery, and it is the fifth element in group 2. Barium is never found in nature as a free element due to its extreme chemical reactivity. Oil well drilling fluid uses barium sulfate as an insoluble ingredient. It is employed as an X-ray radiocontrast agent in a purer form to image the human gastrointestinal tract. Barium compounds that dissolve in water have been employed as rodenticides despite being hazardous.
Chemical element lanthanum has the atomic number 57 and the symbol La. It is a silvery-white, ductile, soft metal that slowly tarnishes when exposed to air. It serves as the eponym for the group of 15 related elements in the periodic table between lanthanum and lutetium, of which lanthanum is the first and prototype. The rare earth elements traditionally include lanthanum.
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There are 6.022 × 10²³ atoms in 39.948 g of argon and 4.0026 g of helium.
Explanation:
39.945 g/mole is the molar mass of argon so 39.948 g of argon are equal to 1 mole of argon.
4.0026 g/mole is the molar mass of helium so 4.0026 g of helium are equal to 1 mole of helium.
We know that Avogadro's number tell us the number of particles in 1 mole of substance which is 6.022 × 10²³.
So in 39.948 g of argon and 4.0026 g of helium contains the same number of atoms, 6.022 × 10²³.
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Avogadro's number
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To answer this question, you need to know <span>Graham's Law of Effusion/Diffusion formula. In this formula, the rate of diffusion/effusion would be influenced by the mass. As the molecule has bigger mass, the rate should be slower because it will be harder to pass the membrane. The calculation should be:</span>
<span>Rate 1 / Rate 2 = √[M2/M1]
</span>4.11/1= √[M2/2]
M2=33.78 g/mol
