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podryga [215]
3 years ago
5

Write equations that show the processes that describe the first, second, and third ionization energies for a gaseous gadolinium

atom. Express your answers as chemical equations separated by commas. Identify all of the phases in your answer.
Chemistry
1 answer:
Inessa [10]3 years ago
5 0

Answer:

Gd(g) → Gd^{+}(g) + e^{-} , Gd^{+}(g)→ Gd^{2+}(g) + e^{-}, Gd^{2+}(g) → Gd^{3+}(g) + e^{-}

Explanation:

Ionization energy is the energy required to remove an electron from a gaseous atom or ion.

The first ionization energy is the energy required to remove the valence electron(outermost) from a neutral atom:

Gd(g)→ Gd^{+}(g) + e^{-}

The second ionization energy is the energy required to remove next/second electron from x^{+} ion. The second ionization energy is always higher than the first:

Gd^{+}(g) → Gd^{2+}(g) + e^{-}

The third ionization energy is the energy required to remove third electron from x^{2+} ion:

Gd^{2+}(g) → Gd^{3+}(g) + e^{-}  

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I think you mean miosis because mitosis is cell reproduction, photosynthesis is to do with light, respiration is breathing and meiosis is division of cells
5 0
3 years ago
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Use the half-reaction method to balance the following equation in acidic solution. It is not necessary to include any phases of
yawa3891 [41]

Answer:

2H⁺ + 2MnO4⁻ + 3CN⁻ → 2MnO₂  + H₂O + 3CNO⁻

Explanation:

CN⁻   +   MnO4⁻  →  CNO⁻   +   MnO₂

In right side permangante, with Mn element has +7 as oxidation number.

In the MnO₂, Mn acts with +4.

This is the half reaction of reduction, where the Mn has gained 3 electrons.

in right side, cianide with C element has +2 as oxidation number. In the anion cianate, C acts with +4.

The oxidation number has increased in this half reaction. It's oxidation where the C, has lost 2 elecontrons

( 4H⁺ + MnO4⁻ + 3e⁻  → MnO₂  + 2H₂O ) .2

(H₂O  +  CN⁻  →  CNO⁻  + 2e⁻  + 2H⁺) .3

I have to add water, to ballance the amount of oxygens and protons to ballance H, in the opposite side

To ballance the half reactions, I have to multiply x2 (reduction) and x3 (oxidation)  so I can cancel, the electrons.

2H⁺  +  2MnO4⁻ + 3CN⁻ → 2MnO₂  + H₂O + 3CNO⁻

The electrons are now cancelled, and I can also modify water. In reactant side I have 3H₂O and in product side, I have 4H₂O so, H₂O in right side is gone so I finally obtained 1 H₂O on left. I had 8H⁺ on right, and 6H⁺ on left, so finally I obtained 2H⁺ on right, and the protons of product side are gone.

6 0
3 years ago
Balance the following reaction:
miv72 [106K]

 The balanced reaction is   as below

3A₂B  + 2DC₃→ 6 AC  + D₂B₃


The number that must be  to the left of AC   is 6

 Explanation

  • According to the law  of mass  conservation , the number of atoms in reactant side   must be equal  to number  to the number of atoms  in  product  side.
  • Therefore the equation above is balance  since  it obey the law of mass conservation.
  • For example there is 6 atoms  of A in reactant side and  6 in product side.
5 0
3 years ago
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The percent yield of a chemical reaction is
Veronika [31]

Answer:

the measured amount of product that is made from a given amount of reactant

Explanation:

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4 0
2 years ago
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What is the wavelength if the frequency is 29.2 hz?
ZanzabumX [31]

Answer : The wavelength is 1.027\times 10^7m

Solution : Given,

frequency = 29.2 Hz

Formula used :

\nu=\frac{c}{\lambda}\\\lambda=\frac{c}{\nu}

where,

\nu = frequency

\lambda = wavelength

c = speed of light = 3\times 10^8m/s

Now put all the given values in this formula, we get the wavelength.

\lambda=\frac{3\times 10^8m/s}{29.2Hz}=0.1027\times 10^8m=1.027\times 10^7m                 (1Hz=1s^{-1})

Therefore, the wavelength is 1.027\times 10^7m

8 0
3 years ago
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