Answer:
E = 5.69x10⁻²⁸m
Explanation:
To solve this question we neeed to convert the wavelength in meters to energy in joules using the equation:
E = hc / λ
<em>Where E is energy in joules, h is Planck's constant = 6.626x10⁻³⁴Js</em>
<em>c is light constant = 3.0x10⁸m/s</em>
<em>And λ is wavelength in meters = 349m</em>
Replacing:
E = 6.626x10⁻³⁴Js*3.0x10⁸m/s / 349m
E = 5.69x10⁻²⁸m
Answer:
has boiling point of 238 K
Explanation:
Boiling point depends on different intermolecular force such as molecular wight, dipole-dipole attraction force, hydrogen bonding, ionic attraction force.
Homonuclear diatomic molecules are covalent non-polar molecules and thereby free from dipole-dipole attraction force, hydrogen bonding and ionic interaction forces.
Hence, boiling point of homonuclear diatomic molecules depends solely on molecular weight.
We know, higher the molecular weight of a molecule, higher will be its boiling point. This phenomenon can be realized in terms of increasing london dispersion force with increase in molecular weight.
Decreasing order of molecular weight of halogen molecules :
>
>
>
So, decresing order of boiling point of halogen molecules:
>
>
>
Hence
has boiling point of 238 K
Answer:
Teaching
Explanation:
Teaching or tutoring is a type of human job that gives strength to other individuals by impacting them with the necessary knowledge or skills they required to survive in life.
Teachers or tutors are the people that carry out this task.
It should be understood that the most important thing that people need to survive in life is information. And when they are given the information they will do exploit.
The new pH is 7.69.
According to Hendersen Hasselbach equation;
The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.
pH = pKa + log10 ([A–]/[HA])
Here, 100 mL of 0.10 m TRIS buffer pH 8.3
pka = 8.3
0.005 mol of TRIS.
∴ ![8.3 = 8.3 + log \frac{[0.005]}{[0.005]}](https://tex.z-dn.net/?f=8.3%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.005%5D%7D%7B%5B0.005%5D%7D)
<em> </em>inverse log 0 = ![\frac{[B]}{[A]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
![\frac{[B]}{[A]} = 1](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D%20%3D%201)
Given; 3.0 ml of 1.0 m hcl.
pka = 8.3
0.003 mol of HCL.
![pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69](https://tex.z-dn.net/?f=pH%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.005-0.003%5D%7D%7B%5B0.005%2B0.003%5D%7D%5C%5CpH%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.002%5D%7D%7B%5B0.008%5D%7D%5C%5C%5C%5CpH%20%3D%208.3%20%2B%20log%20%7B0.25%7D%5C%5C%5C%5CpH%20%3D%208.3%20%2B%20%28-0.62%29%5C%5CpH%20%3D%207.69)
Therefore, the new pH is 7.69.
Learn more about pH here:
brainly.com/question/24595796
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