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
eV
where n is the energy level
Therefore,
Thus, 
= -0.85eV
= -1.51eV
Therefore, total mount of energy released in moving electron from n=4 to n=3 is given by -
= -0.85 - ( -1.51)
= 0.66eV
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We have to solve this question using the stoichiometry of the reaction:
The equation of the reaction is;
According to the question;
Number of moles of CO2 released = 21.3 g/44 g/mol = 0.48 moles
From the stoichiometry of the reaction:
Since;
24 moles of CO2 released 15,026 KJ
0.48 moles of CO2 will release 0.48 * 15,026/24
= 301 KJ of heat.
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The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.
Answer: A different group of scientists using different methods.
