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2 years ago

How much energy would be released as an electron moved from the n=4 to the n=3 energy level?

Chemistry

1 answer:

Thepotemich [5.8K]2 years ago

8 0

The energy released when electron move from n=4 to n=3 is 0.66 eV

We know that in an atom energy of nth state is

$E_n = -13.6/n^{2}$ eV

where n is the energy level

Therefore,

$E_4 = -13.6/4^{2} \\E_3 = -13.6/3^{2}$

Thus, $E_4$ = -0.85eV

$E_3$ = -1.51eV

Therefore, total mount of energy released in moving electron from n=4 to n=3 is given by -

$E_4 -E_3$

= -0.85 - ( -1.51)

= 0.66eV

To know more about energy released in electron transition

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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

$2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?$

We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

$\Delta H_{f,CuO}=-157.3 kJ/mol$

$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

The equation for the enthalpy change of the given reaction is:

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

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3 years ago

You are balancing an ionic compound. Let's call it: AB.

rewona [7]

Answer:- $A_3B_2$

Explanations:- It is given that the charge for A is +2 and the charge for B is -3. The over all compound is neutral means the over all charge is zero. For making the over all charge zero, we need 3 positive ions and 2 negative ions. This makes a +6 charge for A and -6 charge for B and the over all charge is zero.

Also, if we think about the criss cross then charge of A becomes the subscript of B and the charge of B becomes the subscript of A.

So, the formula of the ionic compound is $A_3B_2$ . In this compound the ratio of A to B is 3:2.

The balanced equation could be shown as:

$3A+2B\rightarrow A_3B_2$

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