Someone please explain

Use the given relationship to calculate the ionization energy of a thallium atom, given that radiation of 58.4 nm produces elect

Elenna [48]

The ionization energy is 1 × 10^-18 J

<h3>What is ionization energy?</h3>

The ionization energy is the energy required to remove an electron from an atom.

We have the following information;

wavelength of the photon = 58.4 nm

Speed of the electron = 2310 × 10^3 m/s

Since;

hv = I + 1/2mv^2

v = c/λ

hc/λ = I + 1/2mv^2

I = hc/λ - 1/2mv^2

I = (6.6 × 10^-34 × 3 × 10^8/58.4 × 10^-9) - (1/2 × 9.11 × 10^-31 × (2310 × 10^3)^2)

I = 1 × 10^-18 J

Learn more about ionization energy: brainly.com/question/16243729

3 0

2 years ago

Ill upvote

Aliun [14]

The answer to the given question above would be the first option. The one that is an example of <span> the conversion of electrical energy to electromagnetic energy is a NUCLEAR FUSION. Somewhat, electrical energy and electromagnetic energy are similar in a way that their energies both travel in waves. Hope this answers your question.</span>

4 0

3 years ago

Read 2 more answers

The capacity of the air to hold water vapor is basically a function of

Kisachek [45]

It is basically the function of evaporation.
If it is wrong i am sorry.

4 0

3 years ago

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 9.50 MJ 9

tensa zangetsu [6.8K]

Answer:

709415.584 g

Explanation:

This is a heat capacity problem. We use the formula below to calculate.

Q=mcΔT

where Q is the quantity of heat

m is the mass of substance

ΔT is the change in temperature of substance( T2-T1).

Q= 9.5MJ = 9.5*10^(6}J

C.P = 30.8J/(K.mol)

temperature of sodium increases by 10°C which means our final temperature in kelvin T2 will be 510K

initial temperature T1= 500K

m= Q/[c (T2-T1)]

m= 9.5*10^(6}/ {30.8(510-500)]

m= 30844.156 mol of Na

Converting to grams we multiply by the molar mass of Na.

molar mass of Na = 23

mass of Na= 30844.156 *23

=709415.584 g

7 0

3 years ago

Draw the structure of the compound C9H10O2 that might exhibit the 13C-NMR spectrum below. Impurity peaks are omitted from the pe

zhenek [66]

Complete question

Draw the structure of the compound $C_{9}H_{10}O_{2}$ that exhibits the $^{13}C-NMR$ spectrum shown on the first uploaded image(on the second and third uploaded image is closer look at the $^{13}C-NMR$ spectrum ) . Impurity peaks are omitted from the peak list. The triplet at 77 ppm is $CDC_{l3}$ .

Answer:

The structure that might exhibit the $^{13}C-NMR$ spectrum is shown on the fifth uploaded image

Explanation:

In order to get a good understanding of the answer above we need to know that

• Proton NMR spectrum: proton NMR spectroscopy is one of the techniques, which is useful to predict the structure of the compound.

• In $^{\rm{1}}{\rm{H NMR}}$ spectroscopy, peaks are observed at the point where the wavelength of proton nuclei matched to substance nuclei wavelength.

• In same manner there are other spectroscopies are present like $^{{\rm{13}}}{\rm{C NMR}}$

, IR and mass spectroscopy.

• Infrared spectroscopy is used to determine the functional groups present in a compound.

• Infrared bands observed when there is change in dipole moment occurs between the atoms. Infrared bands describe about the bond stretches, which causes due to the dipole moment present in the molecule.

Fundamentals

Double bond equivalence: number of double bonds or number of rings in the structure can be calculated by using double bond equivalence formula.

$DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})$

Where,

$N_{c} =$ number of carbon atoms

$N_{H}=$ number of hydrogen atoms

$N_{Cl} =$ number of chlorine atoms

$N_{N}=$ number of nitrogen atoms

The table for the $^{{\rm{13}}}{\rm{C NMR}}$ is shown on the fourth uploaded image

Molecular formula of the compound is ${{\rm{C}}_9}{{\rm{H}}_{{\rm{10}}}}{{\rm{O}}_{\rm{2}}}$

Double bond equivalence of the compound is calculated below.

$DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})$

Where,

$N_{c} =$ 9

$N_{H}=$ 10

$N_{Cl} =$ 0

$N_{N}=$ 0

$DBE = N_{c} + 1 - (\frac{(10+0) -0}{2}})$

$DBE =5$

Therefore, the compound has five double bonds, which indicating that there is chance of getting aromatic rings too.

Note:

Double bond equivalence is calculated as 5 which indicates that there are 5 double bond (may rings) in the structure of the compound.

Double bond equivalence is calculated by using this formula.

$DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})$

13C NMR data of the compound is explained below.

1.A peak at 166.5 ppm, which indicates the presence of ester group

2.Peaks at 132.7, 130.5, 129.5, 128.2 ppm (aromatic carbons) are indicating a mono substituted aromatic ring

3.A peak at 60.9 ppm means methylene group attached to oxygen atom

4.A peak at 14.3 ppm, which indicates the presence of methyl group

According to this data and the using the double bond equivalence, structure of the compound shown on the fifth uploaded image .

Note:

According to given spectral data, structure of the compound has been predicted. It is clear that; -ester functional group is present in the structure because there is a peak at 166.5ppm. According to given proton $^{13}C NMR$ data, above structure has been drawn. Therefore, the compound is ethyl benzoate.

7 0

4 years ago