hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:
En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )
where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:
En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )
where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:
En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )
where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.
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Traditionally they include boron from group 3A, silicon and germanium in group 4A, aresnic and antimony in group 5A and tellurium from group 6A, although sometimes selenium, astatine, polonium and even bismuth have also been considered as metalloids. Typically metalloids are brittle and show a semi-metallic luster.
The six commonly recognised metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. Five elements are less frequently so classified: carbon, aluminium, selenium, polonium, and astatine.
Answer:
IHD = 0
Explanation:
Given that
C₆H₁₅N
Number of carbon atoms(n) = 6
Number of hydrogen atoms(x') = 15
Number of nitrogen atoms = 1
There is nitrogen atoms then x = x' -1
The index of hydrogen deficiency given as

So


IHD = 0
The index of hydrogen deficiency is zero.
The required amount of silver nitrate to produce 16.2g of silver is 25.48 grams.
<h3>What is the relation between mass & moles?</h3>
Relation between the mass and moles of any substance will be represented as:
n = W/M, where
- W = given mass
- M = molar mass
Moles of silver = 16.2g / 107.8g/mol = 0.15mol
From the stoichiometry of the given reaction it is clear that, same moles of silver nitrate is required to produce same moles of silver. So 0.15 moles of silver nitrate is required.
Mass of silver nitrate = (0.15mol)(169.87g/mol) = 25.48g
Hence required mass of silver nitrate is 25.48g.
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