Which element is likely to be the best conductor of electricity?
1 answer:
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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.
Answer:
0.1667
Explanation:
Hello,
Dalton's law defines:
A total pressure is:
So, for A (solving for in the previous equation, we get:
Since , we obtain:
}
Best regards.
Answer:
a) The lewis dot structure is shown in the image attached to this answer
b) The formal charge on each of the atoms is zero
c) bromine has an oxidation state of +5 while fluorine has an oxidation state of -1
d) 90 degrees
e) Square Pyramidal
f) polar bonds
g) polar molecule
Explanation:
The molecule BrF5 has a formal charge of zero. It exhibits an sp3d2 hybridization state with a square pyramidal geometry. The bond angle in the molecule is 90 degrees. It is a molecule of the type AX5E. The oxidation state of bromine is +5 while that of fluorine is -1.
The Br-F bonds are polar. The overall molecule is polar due to asymmetric charge distribution concentrating on the central atom since the molecule is square pyramidal.
