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Sedaia [141]
2 years ago
15

Based on the relative energy of the absorbed electromagnetic radiation, which absorber, a peptide bond or an aromatic side chain

, exhibits an electronic excited state that is closer in energy to the ground state?.
Chemistry
1 answer:
alina1380 [7]2 years ago
4 0

Aromatic side chain exhibits an electronic excited state that is closer in energy to the ground state.

  • In order to respond to this query, we must decide whether a peptide bond or an aromatic side chain is demonstrating an electronic exited state that is more closely related to the ground state in terms of energy.
  • When our energy is as low as possible, we are in the ground state.
  • What I want to point out is that if we can choose between the two options—peptide bond or aromatic side chain—without knowing the specific reasons, we can immediately rule out two potential answers.
  • Consider what we already know about energy, we have:

                                 E = h x c/λ

  • That indicates that when we have more energy, a wavelength decreases. Lower energy corresponds to higher wavelength.
  • Aromatic side chains absorb between 250 and 290 nm, while peptide bonds do so between 190 and 250 nm.
  • According to our breakdown, we have an electron excited state that is more closely related to the ground state in terms of energy as wavelength increases.

Thus, Aromatic side chain exhibits an electronic excited state that is closer in energy to the ground state.

To view similar questions about energy, refer to:

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The SAE curriculum includes practical farming tasks conducted outside the scheduled classroom and laboratory period by students. SAEs offer a method for students in agricultural education to gain real-world work opportunities that they are most interested in in the field of agriculture. Supervised agricultural experience is an essential component of agricultural education, and all Agriculture, Food and Natural Resources (AFNR) courses are a necessary component.

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A solution has a poh of 7. 1 at 10∘c. what is the ph of the solution given that kw=2. 93×10−15 at this temperature? remember to
Gnom [1K]

A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .

It is given that,

pOH of solution = 7.1

Kw =2.93×10^(-15)

Firstly, we will calculate the value of pKw

The expression which we used to calculate the pKw is,

pKw=-log [Kw]

Now by putting the value of Kw in this expression,

pKw =−log{2.93×10^(-15)}

pKw =15log(2.93)

pKw=14.5

Now we have to calculate the pH of the solution.

As we know that,

pH+pOH=pKw

Now put all the given values in this formula,

pH+7.1=14.5

pH=7.4

Therefore, we find the value of pH of the solution is, 7.4.

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7 0
1 year ago
What is the percent composition of calcium fluoride
vladimir1956 [14]

Answer:

78.07

Explanation:

them seperatly is

Calcium Ca 51.333%

Fluorine F 48.667%

8 0
2 years ago
In the laboratory a student uses a "coffee cup" calorimeter to determine the specific heat of a metal. She heats 19.6 grams of z
storchak [24]

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

<u>Step 1: </u>Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc T1= 98.37 °C

Initial temperature of water T1= 24.16 °C

Final temperature of water (and zinc) T2 = 25.70 °C

Specific heat of water = 4.184 J/g°C

<u>Step 2: </u>Calculate Specific heat of zinc

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

Qzinc = -Qwater

m(zinc)*c(zinc)* ΔT(zinc) = - m(water)*c(water)* ΔT(water)

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

4 0
2 years ago
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