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:
ooh sorry, but will this help you now:
Ocean dynamics define and describe the motion of water within the oceans. Ocean temperature and motion fields can be separated into three distinct layers: mixed (surface) layer, upper ocean (above the thermocline), and deep ocean. Ocean currents are measured in sverdrup (sv), where 1 sv is equivalent to a volume flow rate of 1,000,000 m (35,000,000 cu ft) per second.
Surface currents, which make up only 8% of all water in the ocean, are generally restricted to the upper 4…
Explanation:
Hope this helps :)

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its
can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.
Let the three equations with
given be denoted as (1), (2), (3), and the last equation (4). Let
,
, and
be letters such that
. This relationship shall hold for all chemicals involved.
There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance,
shall resemble the number of
left on the product side when the second equation is directly added to the third. Similarly
Thus
and

Verify this conclusion against a fourth species involved-
for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

Atomic mass / mass number / atomic weight
(all of which mean the same thing)