Answer:
A jump occurs when a core electron is removed.
Explanation:
A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.
For Beryllium, the electronic configuration of is 1s2 2s2.
There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron
The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron
The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron
The electronic configuration of Lithium is 1s2 2s1
There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
Answer:
Explanation:
A) Formal charges represent an actual separation of charges.(FALSE)
(B) ΔHo rxn can be estimated from the bond enthalpies of reactants and products.(TRUE)
C)All second-period elements obey the octet rule in their compounds(FALSE).
(D)The resonance structures of a molecule can be separated from one another in the laboratory.(FALSE)
Bond enthalpy which is also reffered to as bond energy is the amount of energy that is required to break one mole of a bond.
taking the single bond between Oxygen and Hydrogen into considerationthe bond energy between their single bond is 463 kJ/mol.
formal charge is used for the comparison of the number of electrons present around an atom in a particular molecule with the number of electrons present around a neutral
Answer:
i'm not able to see images but if you type it all i can answer:)
Explanation:
Using electronegativity difference is a good guide to the ionic/ covalent nature. Large differences indicate greater ionic character, small differences more covalent character. The larger the difference in electronegativity the more ionic properties a bond is said to have. The smaller the difference in electronegativity the more covalent properties a bond is said to have.
Ionic bonding is formed through electrostatic attraction between a cation and anion. Foe example, Sodium fluoride has ionic bonding because it is composed by sodium and Fluorine (a non metal). On the other hand, covalent bonding is characterized by atoms sharing pairs of electrons. For example; methane has covalent bonding; carbon has 4 valence electrons and hydrogen has 1; when they bond they have a total of 8 electrons and satisfies the octet rule.
