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:
Of the following substances, <u>PCl₃</u> has the highest boiling point
Explanation:
The given substances and their boiling point from online resources are;
Krypton, Kr, boiling point = -153.4°C
Chlorine gas, Cl₂, boiling point = -34.6°C
Borane, BH₃, (transient compound), boiling point of the dimer it forms B₂H₆ = -164.85°C
Methane, CH₄, boiling point = -161.6°C
Phosphorus trichloride, PCl₃, boiling point = 76.1°C
Therefore, given that PCl₃ is a volatile liquid at room temperature, while the other substances are gases and that PCl₃ has the highest boiling point of all the substances of 76.1°C, the substance with the highest boiling point is PCl₃.
Answer is: electron in 2pz orbital.
The principal quantum number is one
of four quantum numbers which are assigned to each electron in
an atom to describe that electron's state, n=1,2,3... n=2 - <span>the </span>second energy level.<span>
The azimuthal quantum number is a quantum number for
an atomic orbital that determines its orbital angular
momentum and describes the shape of the orbital. l = 0,1...n-1, when l = 1, that is p </span>subshell.
The magnetic quantum number<span>, </span><span>ml, show</span> orbital<span> in which the electron is located, ml = -l...+l, ml = 0 is pz orbital.</span>
The spin quantum number<span>, </span><span>ms</span><span>, is the spin of the electron; ms = +1/2 or -1/2.</span>
Answer:
frequency = 0.47×10⁴ Hz
Explanation:
Given data:
Wavelength of wave = 6.4× 10⁴ m
Frequency of wave = ?
Solution:
Formula:
Speed of wave = wavelength × frequency
Speed of wave = 3 × 10⁸ m/s
Now we will put the values in formula.
3 × 10⁸ m/s = 6.4× 10⁴ m × frequency
frequency = 3 × 10⁸ m/s / 6.4× 10⁴ m
frequency = 0.47×10⁴ /s
s⁻¹ = Hz
frequency = 0.47×10⁴ Hz
Thus the wave with wavelength of 6.4× 10⁴ m have 0.47×10⁴ Hz frequency.
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