Extreme (1, 2, or 3) in the center of nebulae causes hydrogen atoms to

Extreme (1, 2, or 3) in the center of nebulae causes hydrogen atoms to shed their electrons and collide with one another. The (4, 5, or 6) of the collisions cause the smaller, bare nuclei to fuse into large helium nuclei. 1. expansion 2. heat 3. density 4. size and luminosity 5. force and speed 6. mass and temperature

2 months ago

Solution 1

Guest Guest #2990
2 months ago
Extreme (Heat) in the center of nebulae causes hydrogen atoms to shed their electrons and collide with one another. The (force and speed) of the collisions cause the smaller, bare nuclei to fuse into large helium nuclei.

Solution 2

Guest Guest #2991
2 months ago

Answer:

2. heat

5. force and speed

Explanation:

Nuclear fusion in stars causes the hydrogen atoms to collide with each other and form the heavier helium atom. The fusion reaction is therefore:

₁H²  +  ₁H² →  ₂He⁴

Extreme heat in the center of nebulae causes hydrogen atoms to shed their electrons and collide with one another. The force and speed of the collisions cause the smaller, bare nuclei to fuse into large helium nuclei.

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Balanced nuclear reaction equation for the beta minus decay of iodine-131 is ${}_{{\rm{53}}}^{{\rm{131}}}{\rm{I}} \to {}_{54}^{{\rm{131}}}{\rm{Xe + }}{}_{{\rm{ - 1}}}^0{\rm{e}}$.

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Beta minus decay is one of the example of nuclear reaction, and in this reaction a neutron decays into an electron which is also known as beta particle having no mass in it. And at the same time gamma rays also. In the question, it is given that iodine-131 shows beta minus decay, where the superscript of xenon is 131 and subscript of electron is -1. i.e.

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Solution 2
Answer:


131            131            0
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1) β decay is the radiactive process in which an element (Iodine - 131) in this case undergoes a desintegration by emitting beta radiation.

2) Beta radiation are electrons.


3) The electron has a negative charge 1-, and no mass (almost  not mass), so it is represented with a left supersript 0 (does not change the mass number) and a left underscript 1-  (the atomic number increases in 1).


4)  Then when iodine-131 emits a beta particle (electron) its mass number does not change, but the atomic number is increased from 53 to 54, becoming the isotope xenon - 131.


That is what the superscript to the left of Xe shows (131), while the subscript shows 54.


The subscript of electron is  - 1.


So the balance is:


Iodine              Xenon           electron

 131       =          131       +       0

   53       =            54       -         1
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