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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6NaC₂H₃O₂ + Fe₂O₃ → 2Fe(C₂H₃O₂)₃ + 3Na₂O
Explanation:
Given equation;
NaC₂H₃O₂ + Fe₂O₃ → Fe(C₂H₃O₂)₃ + Na₂O
To find the coefficient that will balance this we equation, let us set up simple mathematical algebraic expressions that we can readily solve.
Let us have at the back of our mind that, in every chemical reaction, the number of atom is usually conserved.
aNaC₂H₃O₂ + bFe₂O₃ → cFe(C₂H₃O₂)₃ + dNa₂O
a, b, c and d are the coefficients that will balance the equation.
conserving Na; a = 2d
C: 2a = 6c
H: 3a = 9c
O; 2a + 3b = 6c + d
Fe: 2b = c
let a = 1
solving:
2a = 6c
2(1) = 6c
c =
2b = c
b = =
d = 2a + 3b - 6c = 2(1 ) + (3 x ) - (6 x
) =
Now multiply through by 6
a = 6, b = 1, c = 2 and d = 3
6NaC₂H₃O₂ + Fe₂O₃ → 2Fe(C₂H₃O₂)₃ + 3Na₂O
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Balanced equation brainly.com/question/9325293
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Answer:
Explanation:
The hydrocarbon shown has a double bond. Hydrocarbons with double bonds are known as alkenes.
Cyclic alkanes have cyclic structure.
Alkanes only have single bonds.
Alkynes have triple bonds.