What is the sum of the numbers of neutrons and electrons in the ion 208 Pb2+?

Chemistry

1 answer:

AysviL [449]2 years ago

3 0

Answer:

For ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = 206

Explanation:

Lead, chemical symbol Pb, is a chemical element which belongs to the group 14 of the periodic table. The atomic number of lead is 82 and it is a member of the p-block.

The isotope of lead with the mass number 208, has 126 neutrons.

Since, atomic number = number of protons = number of electrons for neutral atom

Therefore, for ²⁰⁸Pb: number of electrons= 82

So, for ²⁰⁸Pb²⁺ cation: number of electrons= 82 - 2 = 80

Therefore, for ²⁰⁸Pb²⁺ cation, the sum of the number of neutrons and electrons = number of electrons + number of neutrons = 80 electrons + 126 neutrons = 206.

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How many moles of sodium atoms do you have if you have 5.60 ~

Papessa [141]

Answer:

0.93 mol

Explanation:

Given data:

Number of moles of Na atom = ?

Number of atoms = 5.60× 10²³

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

5.60× 10²³ atoms × 1 mol / 6.022 × 10²³ atoms

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3 years ago

Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th

stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH= $2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}$

Knowing:

ΔH= 75.5 kJ/mol
$H_{NO}$ = 90.25 kJ/mol
$H_{Cl_{2} }$ = 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
$H_{NOCl}$ =?