Answer: The answer is frequency.
Explanation: If the wavelength is long, the frequency will be low. If the wavelength is short, the frequency will be high.
F > N > O > C > Be
Hope this help
Answer:
Explanation:
Hello,
In this case, the Boyle's is mathematically defined via:
Which stands for an inversely proportional relationship between volume and pressure, it means the higher the volume the lower the pressure and vice versa. In such a way, since the volume is decreased to one quarter, we can write:
We can compute the new pressure:
Which means the pressure is increased by a factor of four.
Regards.
The overlapping of two s atomic orbitals produces two molecular orbitals, <span>one bonding molecular orbital and one anti-bonding molecular orbital.
Whenever two atomic orbitals overlap according to molecular orbital theory, it will produce one bonding and one anti bonding molecular orbital. Molecular orbital theory is also a way for determining molecular shape wherein electrons are not assigned to character bonds between atoms, however are dealt with as transferring underneath the effect of the nuclei within the entire molecule.</span>
∆H° of the following reaction H₂(g) + I₂(g) → 2HI(g) is -3kJ/mol.
<h3>What is Bond Enthalpy? </h3>
The minimum amount of energy which is required to braak down or form the bonds in chemical reaction is known as bond enthalpy.
It can be calculated as:
∆Hrxn = sum of ∆H bond broken - sum. of ∆H of bond formed.
In order to Calculate ∆Hrxn for the given equation we have:
Bond energies in kJ/mol
- H—H = 436
- H—I = 295
- I—I = 151
Now, the given reaction is
H₂(g) + I₂(g) → 2HI(g)
Here, 1 mol of H₂ and 1 mole of I₂ breaks to form 2 moles of HI.
Therefore,
We know that,
∆Hrxn = B. E(H—H) + B. E(I—I) - 2B. E(H—I)
= 436 + 151 - 2× 295
= 436+ 151 - 590
∆Hrxn = -3kJ/mol.
Thus, from the above conclusion we can say that ∆Hrxn of the reaction H₂(g) + I₂(g) → 2HI(g) is -3kJ/mol.
learn more about Bond energy:
brainly.com/question/26964179
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