How did the grsnd cayon form

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago

Hich statement describes the process of fusion, but not the process of fission?

lesantik [10]

The answer is A. this process is only in the research phase

The process of fusion involves merging of atomic nuclei to form heavier nuclei resulting in the release of enormous amounts of energy. Fusion takes place when two low mass isotopes, typically isotopes of hydrogen, unite under conditions of extreme pressure and temperature. Scientists continue to work on controlling nuclear fusion in an effort to make a fusion reactor to produce electricity. However, progress is slow due to challenges with understanding how to control the reaction in a contained space.

4 0

3 years ago

*WILL GIVE BRAINLIEST IF SOMEONE ANSWERS THIS CORRECTLY ASAP!!!!*

Lilit [14]

Answer:

D. C > B >A

Hope it helps!

Explanation:

From strongest to weakest, the intermolecular forces rank in the following way:

Strongest: Hydrogen bonding. This occurs when compounds contain #"O"-"H"# , #"N"-"H"# , or #"F"-"H"# bonds. ...

Less strong: Dipole-dipole forces. ...

Weakest: London Dispersion Forces.

5 0

3 years ago

Read 2 more answers

given that the speed of light is 2.998 x 108 m/s, what is the frequency of a wave that has a wavelength of 3.55 x 10-8 meters

Vesna [10]

Frequency, f = v / λ

f = 2.998 * 10⁸ / 3.55*10⁻⁸

f = 8.445 * 10¹⁵ Hz.

8 0

3 years ago

What amount (moles) is represented by each of these samples?

Zinaida [17]

<h3>Answer:</h3>

a) Moles of Caffeine = 1.0 × 10⁻⁴ mol

b) Moles of Ethanol = 4.5 × 10⁻³ mol

<h3>Solution:</h3>

Data Given:

Mass of Caffeine = 20 mg = 0.02 g

M.Mass of Caffeine = 194.19 g.mol⁻¹

Molecules of Ethanol = 2.72 × 10²¹

Calculate Moles of Caffeine as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 0.02 g ÷ 194.19 g.mol⁻¹

Moles = 1.0 × 10⁻⁴ mol

Calculate Moles of Ethanol as,

As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

Number of Moles = Number of Molecules ÷ 6.022 × 10²³

Putting values,

Number of Moles = 2.72 × 10²¹ Molecules ÷ 6.022 × 10²³

Number of Moles = 4.5 × 10⁻³ Moles

5 0

3 years ago