Answer:
You have to remember this memory aid, diagonals and all.
You have to keep a tight tally of the electrons you’ve used so far so you don’t go over the number of electrons in the element you’re working on.
You have to remember how many electrons fit into each subshell (s, p, d, f).
It takes a lot of time, especially when the element has more than 20 electrons.
Explanation:
Answer:
Mass = 3.84 g
Explanation:
Given data:
Mass of hydrogen sulfide = 2.7 g
Mass of oxygen required = ?
Solution:
Chemical equation:
2H₂S + 3O₂ → 2H₂O + 2SO₂
Number of moles of hydrogen sulfide:
Number of moles = mass/ molar mass
Number of moles = 2.7 g / 34 g/mol
Number of moles = 0.08 mol
Now we will compare the moles of hydrogen sulfide with oxygen.
H₂S : O₂
2 : 3
0.08 : 3/2 ×0.08 = 0.12 mol
Mass of oxygen;
Mass = number of moles × molar mass
Mass = 0.12 mol × 32 g/mol
Mass = 3.84 g
The main class of high-temperature superconductors are in the class of copper oxides (only some particular copper oxides) especially the Rare-earth barium copper oxides (REBCOs) such as Yttrium barium copper oxide (YBCO).
<h3>What superconducting material works with the highest temperature?</h3>
As of 2020, the material with the highest accepted superconducting temperature is an extremely pressurized carbonaceous sulfur hydride with a critical transition temperature of +15°C at 267 GPa.
<h3>How do high-temperature superconductors work?</h3>
High-temperature superconductivity, the ability of certain materials to conduct electricity with zero electrical resistance at temperatures above the boiling point of liquid nitrogen, was unexpectedly discovered in copper oxide (cuprate) materials in 1987.
Learn more about high temperature superconductors here:
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The water molecules are not completely removed so additional heating is required.
Explanation:
We have the copper (II) sulfate pentehydrate with the chemical formula CuSO₄ · 5H₂O.
molar mass of CuSO₄ · 5H₂O = 159.6 + 5 × 18 = 249.6 g/mole
Knowing this, we devise the following reasoning:
if in 249.6 g of CuSO₄ · 5H₂O there are 90 g of H₂O
then in 8 g of CuSO₄ · 5H₂O there are Y g of H₂O
Y = (8 × 90) / 249.6 = 2.88 g of water
mass of dried CuSO₄ = mass of CuSO₄ · 5H₂O - mass of H₂O
mass of dried CuSO₄ = 8 - 2.88 = 5.12 g
5.12 g is less that the weighted mass of 6.50 g. We deduce from this that the sample needs additional heating in order to remove all the water (H₂O) molecules.
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hydrates
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