You read a primary source and a secondary source that discuss the same
2 answers:
You might be interested in
<em>The bond between Na + and Cl- ions is an ionic bond</em>
<em>To achieve stability, atoms with low ionization energy such as Na atoms will release electrons and bind to atoms with high-affinity energy such as Cl atoms</em>
Atoms have different stability. Unstable atomic atoms will try to form stable electron configurations like those of noble gases. Where noble gases have the number of outer electrons 2 or 8
The formation of electron configurations such as noble gases can be done by forming shared ions or electron pairs
Atoms with high ionization energy and atoms that are difficult to draw electrons will form bonds with shared electron pairs (electrons can be from one atom or two atoms attached)
In forming ions, atoms will release or attract electrons
Ion bonds occur because atoms that have low ionization energy (easily release electrons) form a + ion, and these electrons are bound by atoms that have large affinity energy (easily pulling electrons)
Generally, ionic bonds occur in metal and nonmetallic elements. Metal elements have low ionization energy and non-metallic elements have a high electron affinity
In Na atoms, to achieve stability, Na will release one electron so that it has an electron configuration like noble gas Ne
Na: 2 8 1
Ne: 2 8
so that it becomes a Na + ion
While the Cl atom will bind one electron, so it has an electron configuration like noble gas Ar
Cl: 2 8 7
Ar: 2 8 8
So that it becomes a Cl- ion
Finally, there will be an attractive attraction between positive ions and negative ions and NaCl is formed
So the bond between Na + and Cl- ions is an ionic bond
the octet rule
a noble gas
ionic bonds and covalent bonds
Keywords: ionic and covalent bonds, the octet rule, noble gases, electron configurations
Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed