Answer:
Lewis structure in attachment.
Explanation:
Atoms of elements in and beyond the third period of the periodic table form some compounds in which more than eight electrons surround the central atom. In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding. These orbitals enable an atom to form an <u>expanded octet</u>.
The central Xe atom in the XeF₄ molecule has <u>two</u> unbonded electron pairs and <u>four</u> bonded electron pairs in its valence shell.
Answer:
(a) Covalent bond. NF₃ (nitrogen trifluoride)
(b) Ionic bond. LiCl (lithium chloride)
Explanation:
<em>(a) N and F</em>
Nitrogen and fluorine are nonmetals, with high and similar electronegativities, so they form covalent bonds, in which they share pairs of electrons to complete the octet in their valence shell. N has 5 valence electrons so it will form 3 covalent bonds while each Cl has 7 valence electrons so it will form 1 covalent bond. As a result, the empirical formula is NF₃ (nitrogen trifluoride).
<em>(b) Li and Cl</em>
Lithium is a metal and Chlorine is a nonmetal. They have different electronegativities so they form an ionic bond, in which Cl gains 1 electron (7 valence e⁻) and Li loses 1 electron (1 valence e⁻). The empirical formula is LiCl (lithium chloride).
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano


1 gramo de metano aporta 50.125 kilojoules.
Octano


1 gramo de metano aporta 48.246 kilojoules.