The electron has a higher frequency compared to the neutron. It can be explained by the way an electron orbits the nucleus of an atom.

According to Quantum Mechanics, electrons do not really orbit the nucleus of an atom. In fact, the most tightly bound state, the 1s orbital, has no angular momentum at all. This would be the state with the most "kinetic energy" and yet there is no "orbital" motion at all in this state.

<span>However, there are frequencies associated with each orbital.
</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly.

8 0

3 years ago

What is usually released as a product in a combustion reaction?

mojhsa [17]

A for number 1 and C for number 2 im sorry i dont know the rest. :3

4 0

3 years ago

The normal freezing point of a certain liquid

stepladder [879]

Answer : The molal freezing point depression constant of X is $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of X liquid (solvent) = 450.0 g

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X= $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant of X = ?

m = molality

Now put all the given values in this formula, we get

$[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}$

$k_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of X is $4.12^oC/m$

4 0

3 years ago

You need to produce a buffer solution that has a pH of 5.31. You already have a solution that contains 10. mmol (millimoles) of

Karolina [17]

Answer:

37 mmol of acetate need to add to this solution.

Explanation:

Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-

$pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]$

Here pH is 5.31, $pK_{a}$ (acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.

Plug in all the values in the above equation:

$5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]$

or, mmol of $CH_{3}COO^{-}$ = 37

So 37 mmol of acetate need to add to this solution.

3 0

3 years ago