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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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An isotope of cesium-137 has a half-life of 30 years. If 5.0g of cesium-137 decays over 60 years. How many grams will remain.
LenKa [72]

Answer:

1.25 gram of cesium-137 will remain.

Explanation:

Given data:

Half life of cesium-137 = 30 year

Mass of cesium-137 = 5.0 g

Mass remain after 60 years = ?

Solution:

Number of half lives passed = Time elapsed / half life

Number of half lives passed = 60 year / 30 year

Number of half lives passed = 2

At time zero = 5.0 g

At first half life = 5.0 g/2 = 2.5 g

At 2nd half life = 2.5 g/ 2 = 1.25 g

Thus. 1.25 gram of cesium-137 will remain.

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2 years ago
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Inessa05 [86]

Answer:

The answer is <u>applied research</u>

Explanation:

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Hope it help^^

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