Explanation:
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<span>The maximum number of electrons in a single d-subshell is:
10</span>
Answer:
The manufacturing processes for liquefied petroleum gas are designed so that the majority, if not all, of the sulfur compounds are removed. The total sulfur level is therefore considerably lower than for other crude oil-based fuels and a maximum limit for sulfur content helps to define the product more completely. The sulfur compounds that are mainly responsible for corrosion are hydrogen sulfide, carbonyl sulfide and, sometimes, elemental sulfur. Hydrogen sulfide and mercaptans have distinctive unpleasant odors. A control of the total sulfur content, hydrogen sulfide and mercaptans ensures that the product is not corrosive or nauseating. Stipulating a satisfactory copper strip test further ensures the control of the corrosion.
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<h3>
Answer:</h3>
8 alpha particles
4 beta particles
<h3>
Explanation:</h3>
<u>We are given;</u>
- Neptunium-237
- Thallium-205
- Neptunium-237 undergoes beta and alpha decay to form Thallium-205.
We are required to determine the number of beta and alpha particles produced to complete the decay series.
- We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
- When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.
In this case;
Neptunium-237 has an atomic number 93, while,
Thallium-205 has an atomic number 81.
Therefore;
²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti
We can get x and y
237 = 4x + y(0) + 205
237-205 = 4x
4x = 32
x = 8
On the other hand;
93 = 2x + (-y) + 81
but x = 8
93 = 16 -y + 81
y = 4
Therefore, the complete decay equation is;
²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti
Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.
