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ioda
3 years ago
6

PLSSSSS ANSWERRR

Chemistry
1 answer:
Paha777 [63]3 years ago
3 0
The answer should be d
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Using the Bohr model, determine the energy in joules of the photon produced when an electron in a Li2+ ion moves from the orbit
djverab [1.8K]

Answer:

1.64x10⁻¹⁸ J

Explanation:

By the Bohr model, the electrons surround the nucleus of the atom in shells or levels of energy. Each one has it's energy, and the electron doesn't fall to the nucleus because it can reach another level of energy, and then return to its level.

When the electrons go to another level, it absorbs energy, and then, when return, this energy is released, as a photon (generally as luminous energy). The value of the energy can be calculated by:

E = hc/λ

Where h is the Planck constant (6.626x10⁻³⁴ J.s), c is the light speed (3.00x10⁸ m/s), and λ is the wavelength of the photon.

The wavelength can be calculated by:

1/λ = R*(1/nf² - 1/ni²)

Where R is the Rydberg constant (1.097x10⁷ m⁻¹), nf is the final orbit, and ni the initial orbit. So:

1/λ = 1.097x10⁷ *(1/1² - 1/2²)

1/λ = 8.227x10⁶

λ = 1.215x10⁻⁷ m

So, the energy is:

E = (6.626x10⁻³⁴ * 3.00x10⁸)/(1.215x10⁻⁷)

E = 1.64x10⁻¹⁸ J

3 0
3 years ago
If you take a part time job while you are still in school what might your opportunity scost be
yarga [219]

The opportunity costs of working part-time while attending high school could be studying time, grades, and your own free time. It depends on the person, how they do in school, but if it were for me, I would say the benefits outweigh the costs. I have enough time to study and still get good grades in school.

7 0
2 years ago
Read 2 more answers
Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
enyata [817]

Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

5 0
3 years ago
NEED HELP ANSWERING THIS QUESTION
victus00 [196]

Answer:

I think the answer is A.

please give thanks if it helps

and sorry if it doesn't help.

4 0
3 years ago
Is mercury a solution​
Dvinal [7]

Answer:

No.

Explanation:

Mercury is not a solution​. It is a detonator.

4 0
3 years ago
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