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Annette [7]
3 years ago
11

Calculate the energy required to ionize a hydrogen atom to an excited state where the electron is initially in the n = 5 energy

level. Report your answer in kilojoules
Chemistry
2 answers:
SVETLANKA909090 [29]3 years ago
7 0

. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:

E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ

Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ


2. The wavelength of a photon with this energy would be:

Energy = hc/wavelength

wavelength = hc/energy

wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m

Converting to nanometers gives: 91.16 nm


3. Repeat the calculation in 1, but using n=5.


4. Repeat the calculation in 2 using the energy calculated in 3.

Furkat [3]3 years ago
3 0

Answer:

The value of ionization energy is 8.716\times 10^{-23} kJ.

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

We are given: Z = 1 , n= 5

E_5=-13.6\times \frac{1^2}{5^2}eV=-0.544 eV

Z = 1 , n = ∞

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

Ionization energy to ionize the hydrogen atom where electron is present in n=5 initially will equal to the difference between energy of electron at infinity from the energy of electron at n = 5 energy level.

I.E=E_{\infty }-E_5

=0 eV-(-0.544 eV)=0.544 eV

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

0.544 eV=0.544\times 1.60218\times 10^{-22} J=8.716\times 10^{-23} kJ

The value of ionization energy is 8.716\times 10^{-23} kJ.

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b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)

Explanation:

The equation is given as;

2HCl(aq) + Zn(s) + H2(g) + ZnCl2(aq)

In writing an ionic equation, only the aqueous compounds dissociates into ions. This means HCl and ZnCl2 would dissociate to form ions.

This is given as;

2H+  +  2Cl-  + Zn(s) --> H2(g) + Zn2+  + 2Cl-

The correct option is;

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2 years ago
What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
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Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

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